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Sequences and Series

3.6 Taylor Series

53 problems · hints, answers and solutions shown beside each one

1Stage 1Conceptual

Do you understand the idea? Usually little or no calculation.

Q1Stage 1

Below is a graph of y=f(x)y=f(x), along with the constant approximation, linear approximation, and quadratic approximation centred at a=2a=2. Which is which?

Figure from prob_s3.6, line 2

Figure from prob_s3.6, line 2

Hint

Which of the functions are constant, linear, and quadratic?

Answer

A: linear B: constant C: quadratic

Full solution

All functions AA, BB, and CC intersect the function y=f(x)y=f(x) when x=2x=2. BB is a constant function, so this is the constant approximation. AA is the tangent line, so AA is the linear approximation. CC is a tangent parabola, so CC is the quadratic approximation.

Q2Stage 1

Suppose T(x)T(x) is the Taylor series for f(x)=arctan⁡3(ex+7)f(x)=\arctan^3\left(e^x+7\right) centred at a=5a=5. What is T(5)T(5)?

Hint

You don't have to actually calculate the entire series T(x)T(x) to answer the question.

Answer

T(5)=arctan⁡3(e5+7)T(5)=\arctan^3\left(e^5+7\right)

Full solution

Following how a Taylor series is constructed, the Taylor series and the function agree at the point chosen as the centre. So, T(5)=arctan⁡3(e5+7)T(5)=\arctan^3\left(e^5+7\right).

If we were evaluating a Taylor series at a point other than its centre, we would generally need to check that (a) the series converges, and (b) it converges to the same value as the function we used to create it.

Q3Stage 1

Below are a list of common functions, and their Taylor series representations. Match the function to the Taylor series and give the radius of convergence of the series.

functionseries
A. 11−x\dfrac{1}{1-x}I. ∑n=0∞(−1)nxn+1n+1\displaystyle\sum_{n=0}^\infty(-1)^n\dfrac{x^{n+1}}{n+1}
[20pt] B. log⁡(1+x)\log(1+x)II. ∑n=0∞(−1)nx2n+1(2n+1)!\displaystyle\sum_{n=0}^\infty(-1)^n\dfrac{x^{2n+1}}{(2n+1)!}
[20pt] C. arctan⁡x\arctan xIII. ∑n=0∞(−1)nx2n(2n)!\displaystyle\sum_{n=0}^\infty(-1)^n\dfrac{x^{2n}}{(2n)!}
[20pt] D. exe^xIV. ∑n=0∞(−1)nx2n+12n+1\displaystyle\sum_{n=0}^\infty(-1)^n\dfrac{x^{2n+1}}{2n+1}
[20pt] E. sin⁡x\sin xV. ∑n=0∞xn\displaystyle\sum_{n=0}^\infty x^n
[20pt] F. cos⁡x\cos xVI. ∑n=0∞xnn!\displaystyle\sum_{n=0}^\infty \frac{x^n}{n!}
Hint

If you don't have these memorized, it's good to be able to derive them. For instance, log⁡(1+x)\log(1+x) is the antiderivative of 11+x\dfrac{1}{1+x}, whose Taylor series can be found by modifying the geometric series ∑xn\sum x^n.

Answer
A - V, radius=1=1B - I, radius=1=1C - IV, radius=1=1
D - VI, radius=+∞=+\inftyE - II, radius=+∞=+\inftyF - III, radius=+∞=+\infty
Full solution

These are listed in Theorem 3.6.5 in the CLP–2 text. However, it's possible to figure out many of them without a lot of memorization. For example, e0=cos⁡(0)=11−0=1e^0=\cos(0)=\frac{1}{1-0}=1, while sin⁡(0)=log⁡(1+0)=arctan⁡(0)=0\sin(0)=\log(1+0)=\arctan(0)=0. So by plugging in x=0x=0 to the series listed, we can divide them into these two categories.

The derivative of sine is cosine, so we can also look for one series that is the derivative of another. The derivative of exe^x is exe^x, so we can look for a series that is its own derivative.

Furthermore, sine and arctangent are odd functions and only II and IV are odd. Cosine is an even function and only III is even.

Alternately, we can find the first few terms of each series using the definition of a Taylor series, and match them up.

In any event, here is what Theorem 3.6.5 in the CLP–2 text gives.

  1. The Taylor series representation of 11−x\frac{1}{1-x} is given in V. The series converges for −1<x<1-1<x<1. So by Definition 3.5.3 in the CLP–2 text, the series has radius of convergence 11.

  2. The Taylor series representation of log⁡(1+x)\log(1+x) is given in I. The series converges for −1<x≤1-1<x\le 1. In particular, it converges for all ∣x∣<1|x|<1 and diverges for all ∣x∣>1|x|>1. So by Definition 3.5.3 in the CLP–2 text, the series has radius of convergence 11.

  3. The Taylor series representation of arctan⁡x\arctan x is given in IV. The series converges for −1≤x≤1-1\le x\le 1. In particular, it converges for all ∣x∣<1|x|<1 and diverges for all ∣x∣>1|x|>1. So it has radius of convergence 11.

  4. The Taylor series representation of exe^x is given in VI. The series converges for all xx. So it has infinite radius of convergence.

  5. The Taylor series representation of sin⁡x\sin x is given in II. The series converges for all xx. So it has infinite radius of convergence.

  6. The Taylor series representation of cos⁡x\cos x is given in III. The series converges for all xx. So it has infinite radius of convergence.

Q4Stage 1

  1. Suppose f(x)=∑n=0∞n2(n!+1)(x−3)nf(x)=\displaystyle\sum_{n=0}^\infty \frac{n^2}{(n!+1)}(x-3)^n for all real xx. What is f(20)(3)f^{(20)}(3) (the twentieth derivative of f(x)f(x) at x=3x=3)?

  2. Suppose g(x)=∑n=0∞n2(n!+1)(x−3)2ng(x)=\displaystyle\sum_{n=0}^\infty \frac{n^2}{(n!+1)}(x-3)^{2n} for all real xx. What is g(20)(3)g^{(20)}(3)?

  3. If h(x)=arctan⁡(5x2)x2h(x)=\dfrac{\arctan(5x^2)}{x^2}, what is h(20)(0)h^{(20)}(0)? What is h(22)(0)h^{(22)}(0)?

Hint

See Example 3.6.16 in the CLP-2 text.

Answer

(a) f(20)(3)=202(20!20!+1)f^{(20)}(3)=20^2\left(\dfrac{20!}{20!+1}\right) (b) g(20)(3)=102(20!10!+1)g^{(20)}(3)=10^2\left(\dfrac{20!}{10!+1}\right)
(c) h(20)(0)=−20!⋅51111h^{(20)}(0)=-\dfrac{20!\cdot 5^{11}}{11}; h(22)(0)=0h^{(22)}(0)=0

Full solution

  1. Using the definition of a Taylor series, we know

    ∑n=0∞n2(n!+1)(x−3)n=∑n=0∞f(n)(3)n!(x−3)n\displaystyle\sum_{n=0}^\infty \frac{n^2}{(n!+1)}(x-3)^n=\displaystyle\sum_{n=0}^\infty \frac{f^{(n)}(3)}{n!}(x-3)^n

    So, the coefficient of (x−3)20(x-3)^{20} is f(20)(3)20!\frac{f^{(20)}(3)}{20!} (using the definition). Using the given series, the coefficient of (x−3)20(x-3)^{20} is 20220!+1\frac{20^2}{20!+1}. So,

    f(20)(3)20!=20220!+1⇒f(20)(3)=202(20!20!+1)\begin{align*} \frac{f^{(20)}(3)}{20!}&=\frac{20^2}{20!+1}\\ \Rightarrow \qquad f^{(20)}(3)&=20^2\left(\frac{20!}{20!+1}\right)\\ \end{align*}

    (which is extremely close to 20220^2).

  2. Using the definition of a Taylor series, we know

    ∑n=0∞n2(n!+1)(x−3)2n=∑k=0∞g(k)(3)k!(x−3)k\displaystyle\sum_{n=0}^\infty \frac{n^2}{(n!+1)}(x-3)^{2n}=\displaystyle\sum_{k=0}^\infty \frac{g^{(k)}(3)}{k!}(x-3)^{k}

    So, the coefficient of (x−3)20(x-3)^{20} is g(20)(3)20!\frac{g^{(20)}(3)}{20!} (using the definition). Looking at the given series, the coefficient of (x−3)20(x-3)^{20} occurs when n=10n=10, so it is 10210!+1\frac{10^2}{10!+1}. So,

    g(20)(3)20!=10210!+1⇒g(20)(3)=102(20!10!+1)\begin{align*} \frac{g^{(20)}(3)}{20!}&=\frac{10^2}{10!+1}\\ \Rightarrow \qquad g^{(20)}(3)&=10^2\left(\frac{20!}{10!+1}\right)\\ \end{align*}

  3. With the previous two examples in mind, we find the Maclaurin series for h(x)h(x). (Using the series representation will be much easier than differentiating h(x)h(x) directly twenty times.) Recall from the text that we know the Maclaurin series for arctan⁡x\arctan x.

    arctan⁡(x)=∑n=0∞(−1)nx2n+12n+1arctan⁡(5x2)=∑n=0∞(−1)n(5x2)2n+12n+1=∑n=0∞(−1)n52n+12n+1x4n+2arctan⁡(5x2)x2=∑n=0∞(−1)n52n+12n+1x4n∑k=0∞h(k)(0)k!xk=∑n=0∞(−1)n52n+12n+1x4n\begin{align*} \arctan(x)&=\sum_{n=0}^\infty (-1)^n\frac{x^{2n+1}}{2n+1}\\ \arctan(5x^2)&=\sum_{n=0}^\infty (-1)^n\frac{{(5x^2)}^{2n+1}}{2n+1} =\sum_{n=0}^\infty (-1)^n\frac{5^{2n+1}}{2n+1}x^{4n+2}\\ \frac{\arctan(5x^2)}{x^2}&=\sum_{n=0}^\infty (-1)^n\frac{5^{2n+1}}{2n+1}x^{4n}\\ \sum_{k=0}^\infty \frac{h^{(k)}(0)}{k!}x^k&=\sum_{n=0}^\infty (-1)^n\frac{5^{2n+1}}{2n+1}x^{4n} \end{align*}

    Using the definition of a Maclaurin series, the coefficient of x20x^{20} is h(20)(0)20!\dfrac{h^{(20)}(0)}{20!}. This occurs in the given series when n=5n=5, so

    h(20)(0)20!=(−1)552×5+12×5+1=−51111⇒h(20)(0)=−20!⋅51111\begin{align*} \dfrac{h^{(20)}(0)}{20!}&=(-1)^5\frac{5^{2\times5+1}}{2\times5+1}=-\frac{5^{11}}{11}\\ \Rightarrow\qquad h^{(20)}(0)&=-\frac{20!\cdot 5^{11}}{11} \end{align*}

    Similarly, the coefficient of x22x^{22} in the Maclaurin series is h(22)(0)22!\dfrac{h^{(22)}(0)}{22!}. Since no term x22x^{22} occurs in our series, that coefficient is 0, so h(22)(0)=0h^{(22)}(0)=0.

2Stage 2Procedural

Practising the skill itself, until applying it is automatic.

In Questions 5 through 8, you will create Taylor series from scratch. In practice, it is often preferable to modify an existing series, rather than creating a new one, but you should understand both ways.

Q5Stage 2

Using the definition of a Taylor series, find the Taylor series for f(x)=log⁡(x)f(x)=\log(x) centred at x=1x=1.

Hint

The series will bear some resemblance to the Maclaurin series for log⁡(1+x)\log(1+x).

Answer

∑n=1∞(−1)n−1n(x−1)n\displaystyle\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n}(x-1)^n

Full solution

The definition of a Taylor series tells us we will be computing the coefficients in the series

∑n=0∞f(n)(1)n!(x−1)n\sum_{n=0}^\infty \frac{f^{(n)}(1)}{n!}(x-1)^n

That is, we need a general description of f(n)(1)f^{(n)}(1). To find this, we take a few derivatives, and look for a pattern.

f(x)=log⁡(x)f(1)=0f′(x)=x−1f′(1)=1f′′(x)=(−1)x−2f′′(1)=−1f(3)(x)=(−2)(−1)x−3f(3)(1)=2!f(4)(x)=(−3)(−2)(−1)x−4f(4)(1)=−3!f(5)(x)=(−4)(−3)(−2)(−1)x−5f(5)(1)=4!f(6)(x)=(−5)(−4)(−3)(−2)(−1)x−6f(6)(1)=−5!⋮⋮f(n)(x)=(−1)n−1(n−1)! x−nf(n)(1)=(−1)n−1(n−1)!\begin{align*} f(x)&=\log(x) & f(1)&=0\\ f'(x)&=x^{-1} & f'(1)&=1\\ f''(x)&=(-1)x^{-2} & f''(1)&=-1\\ f^{(3)}(x)&=(-2)(-1)x^{-3}& f^{(3)}(1)&=2!\\ f^{(4)}(x)&=(-3)(-2)(-1)x^{-4}& f^{(4)}(1)&=-3!\\ f^{(5)}(x)&=(-4)(-3)(-2)(-1)x^{-5}& f^{(5)}(1)&=4!\\ f^{(6)}(x)&=(-5)(-4)(-3)(-2)(-1)x^{-6}& f^{(6)}(1)&=-5!\\ &\vdots&&\vdots\\ f^{(n)}(x)&=(-1)^{n-1}(n-1)!\,x^{-n} & f^{(n)}(1)&=(-1)^{n-1}(n-1)! \end{align*}

Using the convention 0!=10!=1, our pattern for f(n)(1)f^{(n)}(1) begins when n=1n=1.

∑n=0∞f(n)(1)n!(x−1)n=0+∑n=1∞(−1)n−1(n−1)!n!(x−1)n=∑n=1∞(−1)n−1n(x−1)n\sum_{n=0}^\infty \frac{f^{(n)}(1)}{n!}(x-1)^n =0+\sum_{n=1}^\infty \frac{(-1)^{n-1}(n-1)! }{n!}(x-1)^n =\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n}(x-1)^n
Q6Stage 2

Find the Taylor series for f(x)=sin⁡xf(x)=\sin x centred at a=πa=\pi.

Hint

The terms f(n)(π)f^{(n)}(\pi) are going to be similar to the terms f(n)(0)f^{(n)}(0) that we used in the Maclaurin series for sine.

Answer

∑n=0∞(−1)n+1(2n+1)!(x−π)2n+1\displaystyle \sum_{n=0}^\infty \frac{(-1)^{n+1}}{(2n+1)!}(x-\pi)^{2n+1}

Full solution

To find the Taylor series for sine, centred at a=πa=\pi, we'll need to know the various derivatives of sine at π\pi.

f(x)=sin⁡xf(π)=0f′(x)=cos⁡xf′(π)=−1f′′(x)=−sin⁡xf′′(π)=0f′′′(x)=−cos⁡xf′′′(π)=1f(4)(x)=sin⁡x=f(x)f(4)(π)=0\begin{align*} f(x)&=\sin x & f(\pi)&=0\\ f'(x)&=\cos x & f'(\pi)&=-1\\ f''(x)&=-\sin x & f''(\pi)&=0\\ f'''(x)&=-\cos x & f'''(\pi)&=1\\ f^{(4)}(x)&=\sin x = f(x) & f^{(4)}(\pi)&=0 \end{align*}

Even derivatives are 0; odd derivatives alternate between −1-1 and +1+1. (If you're following along with the derivation of the Maclaurin series for sine in the text, note f(n)(π)=−f(n)(0)f^{(n)}(\pi)=-f^{(n)}(0).)

In our Taylor series, every even-indexed term will be zero, and we will be left with only odd-indexed terms. If we let nn be our index, then the term 2n+12n+1 will capture all the odd numbers. Since the signs alternate, f(2n+1)(π)=(−1)n+1f^{(2n+1)}(\pi)=(-1)^{n+1}. So, our Taylor series is:

∑k=0∞f(k)(π)k!(x−π)k=∑n=0∞f(2n+1)(π)(2n+1)!(x−π)2n+1(since the even terms are all zero)=∑n=0∞(−1)n+1(2n+1)!(x−π)2n+1\begin{align*} \sum_{k=0}^\infty \frac{f^{(k)}(\pi)}{k!}(x-\pi)^k &= \sum_{n=0}^\infty \frac{f^{(2n+1)}(\pi)}{(2n+1)!}(x-\pi)^{2n+1} \qquad\text{(since the even terms are all zero)} \\ &=\sum_{n=0}^\infty \frac{(-1)^{n+1}}{(2n+1)!}(x-\pi)^{2n+1} \end{align*}
Q7Stage 2

Using the definition of a Taylor series, find the Taylor series for g(x)=1xg(x)=\dfrac{1}{x} centred at x=10x=10. What is the interval of convergence of the resulting series?

Hint

The Taylor series will look similar to a geometric series.

Answer

110∑n=0∞(10−x10)n\displaystyle\frac{1}{10}\sum_{n=0}^\infty\left(\frac{10-x}{10}\right)^n with interval of convergence (0,20)(0,20).

Full solution

The definition of a Taylor series tells us we will be computing the coefficients in the series

∑n=0∞g(n)(10)n!(x−10)n\sum_{n=0}^\infty \frac{g^{(n)}(10)}{n!}(x-10)^n

That is, we need a general description of g(n)(10)g^{(n)}(10). To find this, we take a few derivatives, and look for a pattern.

g(x)=x−1g(10)=110g′(x)=(−1)x−2g′(10)=−1102g′′(x)=(−2)(−1)x−3g′′(10)=(−1)22!103g(3)(x)=(−3)(−2)(−1)x−4g(3)(10)=(−1)33!104g(4)(x)=(−4)(−3)(−2)(−1)x−5g(4)(10)=(−1)44!105g(5)(x)=(−5)(−4)(−3)(−2)(−1)x−6g(5)(10)=(−1)55!106⋮⋮g(n)(x)=(−1)nn!x−(n+1)g(n)(10)=(−1)nn!10n+1\begin{align*} g(x)&=x^{-1} & g(10)&=\frac{1}{10}\\ g'(x)&=(-1)x^{-2} & g'(10)&=\frac{-1}{10^2}\\ g''(x)&=(-2)(-1)x^{-3}& g''(10)&=\frac{(-1)^22!}{10^3}\\ g^{(3)}(x)&=(-3)(-2)(-1)x^{-4}& g^{(3)}(10)&=\frac{(-1)^33!}{10^4}\\ g^{(4)}(x)&=(-4)(-3)(-2)(-1)x^{-5}&g^{(4)}(10)&=\frac{(-1)^44!}{10^5}\\ g^{(5)}(x)&=(-5)(-4)(-3)(-2)(-1)x^{-6}& g^{(5)}(10)&=\frac{(-1)^55!}{10^6}\\ &\vdots&&\vdots\\ g^{(n)}(x)&=(-1)^{n}n!x^{-(n+1)} & g^{(n)}(10)&=\frac{(-1)^{n}n!}{10^{n+1}} \end{align*}

Using the convention 0!=10!=1, our pattern for g(n)(10)g^{(n)}(10) begins when n=0n=0.

∑n=0∞g(n)(1)n!(x−10)n=∑n=0∞(−1)nn!n!10n+1(x−10)n=−∑n=0∞(x−10)n(−10)n+1=110∑n=0∞(10−x10)n\begin{align*} \sum_{n=0}^\infty \frac{g^{(n)}(1)}{n!}(x-10)^n &=\sum_{n=0}^\infty \frac{(-1)^{n}n!}{n! 10^{n+1}}(x-10)^n\\ &=-\sum_{n=0}^\infty \frac{(x-10)^n}{(-10)^{n+1}}\\ &=\frac{1}{10}\sum_{n=0}^\infty\left(\frac{10-x}{10}\right)^n \end{align*}

For fixed xx, we recognize this as a geometric series with r=10−x10r=\frac{10-x}{10}. So it converges precisely when ∣r∣<1|r|<1, i.e.

∣10−x10∣<1∣10−x∣<10−10<x−10<100<x<20\begin{align*} \left| \frac{10-x}{10} \right|&<1\\ \left|10-x \right|&<10\\ -10<x-10&<10\\ 0<x&<20 \end{align*}

So, its interval of convergence is (0,20)(0,20).

Q8Stage 2

Using the definition of a Taylor series, find the Taylor series for h(x)=e3xh(x)=e^{3x} centred at x=ax=a, where aa is some constant. What is the radius of convergence of the resulting series?

Hint

Your answer will depend on aa.

Answer

∑n=0∞3ne3an!(x−a)n\displaystyle \sum_{n=0}^\infty \frac{3^ne^{3a} }{n!}(x-a)^n, with infinite radius of convergence

Full solution

The definition of a Taylor series tells us we will be computing the coefficients in the series

∑n=0∞h(n)(a)n!(x−a)n\sum_{n=0}^\infty \frac{h^{(n)}(a)}{n!}(x-a)^n

That is, we need a general description of h(n)(a)h^{(n)}(a). To find this, we take a few derivatives, and look for a pattern.

h(x)=e3xh(a)=e3ah′(x)=3e3xh′(a)=3e3ah′′(x)=32e3xh′′(a)=32e3ah′′′(x)=33e3xh′′′(a)=33e3a⋮⋮h(n)(x)=3ne3xh(n)(a)=3ne3a\begin{align*} h(x)&=e^{3x} & h(a)&=e^{3a}\\ h'(x)&=3e^{3x} & h'(a)&=3e^{3a}\\ h''(x)&=3^2e^{3x} & h''(a)&=3^2e^{3a}\\ h'''(x)&=3^3e^{3x} & h'''(a)&=3^3e^{3a}\\ &\vdots&&\vdots\\ h^{(n)}(x)&=3^{n}e^{3x} & h^{(n)}(a)&=3^{n}e^{3a} \end{align*}

The pattern for h(n)(a)h^{(n)}(a) holds for all (whole numbers) n≥0n \ge 0. So, our Taylor series for h(x)h(x) is

∑n=0∞3ne3an!(x−a)n\sum_{n=0}^\infty \frac{3^ne^{3a} }{n!}(x-a)^n

To find its radius of convergence, we use the ratio test.

∣an+1an∣=∣3n+1e3a(x−a)n+1(n+1)!⋅n!3ne3a(x−a)n∣=∣3n+13n⋅e3ae3a⋅n!(n+1)!⋅(x−a)n+1(x−a)n∣=3⋅1n+1⋅∣x−a∣lim⁡n→∞∣an+1an∣=lim⁡n→∞[3n+1⋅∣x−a∣]=0\begin{align*} \left|\frac{a_{n+1}}{a_n}\right|&=\left|\frac{3^{n+1}e^{3a} (x-a)^{n+1}}{(n+1)!}\cdot {\frac{n!}{3^ne^{3a}(x-a)^n }}\right|\\ &=\left|\frac{3^{n+1}}{3^n}\cdot\frac{e^{3a}}{e^{3a}}\cdot \frac{n!}{(n+1)!}\cdot\frac{(x-a)^{n+1}}{(x-a)^n}\right|\\ &=3\cdot\frac{1}{n+1}\cdot |x-a|\\ \lim_{n \to \infty }\left|\frac{a_{n+1}}{a_n}\right|&=\lim_{n \to \infty}\left[\frac{3}{n+1}\cdot |x-a|\right]=0 \end{align*}

Our series converges for every value of xx, so its radius of convergence is ∞\infty.

In Questions 9 through 16, practice creating new Taylor series by modifying known Taylor series, rather than creating your series from scratch.

Q9Stage 2Past exam · M105 2013A

Find the Maclaurin series for f(x)=12x−1f(x) = \dfrac{1}{2x-1}.

Hint

You should know the Maclaurin series for 11−x\dfrac{1}{1-x}. Use it.

Answer

−∑n=0∞2nxn\displaystyle-\sum\limits_{n=0}^\infty 2^nx^n

Full solution

Substituting y=2xy=2x into 11−y=∑n=0∞yn\displaystyle \frac{1}{1-y} = \sum\limits_{n=0}^\infty y^n (which is valid for all −1<y<1-1<y<1) gives

f(x)=12x−1=−11−2x=−∑n=0∞(2x)n=−∑n=0∞2nxnfor all −12<x<12\begin{align*} f(x) = \frac{1}{2x-1} = -\frac{1}{1-2x} =-\sum_{n=0}^\infty (2x)^n =-\sum_{n=0}^\infty 2^nx^n \qquad\text{for all }-\tfrac{1}{2}<x<\tfrac{1}{2} \end{align*}
Q10Stage 2Past exam · M105 2014A

Let ∑n=0∞bnxn\displaystyle\sum\limits_{n=0}^\infty b_nx^n be the Maclaurin series for f(x)=3x+1−12x−1\displaystyle f(x) = \frac{3}{x+1} - \frac{1}{2x-1},
i.e. ∑n=0∞bnxn=3x+1−12x−1\displaystyle\sum\limits_{n=0}^\infty b_nx^n = \frac{3}{x+1} - \frac{1}{2x-1}. Find bnb_n.

Hint

You should know the Maclaurin series for 11−x\dfrac{1}{1-x}. Use it.

Answer

bn=3(−1)n+2nb_n = 3(-1)^n + 2^n

Full solution

Substituting first y=−xy=-x and then y=2xy=2x into 11−y=∑n=0∞yn\dfrac{1}{1-y} = \displaystyle\sum\limits_{n=0}^\infty y^n (which is valid for all −1<y<1-1<y<1) gives

11−(−x)=∑n=0∞(−x)n=∑n=0∞(−1)nxnfor all −1<x<111−(2x)=∑n=0∞(2x)n=∑n=0∞2nxnfor all −12<x<12\begin{alignat*}{3} \frac{1}{1-(-x)} &= \sum_{n=0}^\infty (-x)^n&&=\sum_{n=0}^\infty (-1)^nx^n \qquad\text{for all }-1<x<1 \\ \frac{1}{1-(2x)} &= \sum_{n=0}^\infty (2x)^n&&=\sum_{n=0}^\infty 2^nx^n \qquad\text{for all }-\tfrac{1}{2}<x<\tfrac{1}{2} \end{alignat*}

Hence, for all −12<x<12-\tfrac{1}{2}<x<\tfrac{1}{2},

f(x)=3x+1−12x−1=31−(−x)+11−2x=3∑n=0∞(−1)nxn+∑n=0∞2nxn=∑n=0∞(3(−1)n+2n)xn\begin{align*} f(x) &=\frac{3}{x+1} - \frac{1}{2x-1} = \frac{3}{1-(-x)} + \frac{1}{1-2x} =3\sum_{n=0}^\infty (-1)^nx^n + \sum_{n=0}^\infty 2^nx^n \\ &=\sum_{n=0}^\infty \big(3(-1)^n + 2^n\big)x^n \end{align*}

So bn=3(−1)n+2nb_n = 3(-1)^n + 2^n.

Q11Stage 2Past exam · 2014A

Find the coefficient c5c_5 of the fifth degree term in the Maclaurin series ∑n=0∞cnxn\displaystyle\sum_{n=0}^\infty c_nx^n for e3xe^{3x}.

Hint

You should know the Maclaurin series for exe^x. Use it.

Answer

c5=355!c_5=\dfrac{3^5}{5!}

Full solution

We found the Taylor series for e3xe^{3x} from scratch in Question 8. If we hadn't just done that, we could easily find it by modifying the series for exe^x.

Substituting y=3xy=3x into the exponential series

ey=∑n=0∞ynn!\begin{equation*} e^y=\sum_{n=0}^\infty\frac{y^n}{n!} \end{equation*}

gives

e3x=∑n=0∞(3x)nn!=∑n=0∞3n xnn!\begin{equation*} e^{3x}=\sum_{n=0}^\infty\frac{(3x)^n}{n!} =\sum_{n=0}^\infty\frac{3^n\,x^n}{n!} \end{equation*}

so that c5c_5, the coefficient of x5x^5, which appears only in the n=5n=5 term, is c5=355!c_5=\dfrac{3^5}{5!}

Q12Stage 2Past exam · M105 2012A

Express the Taylor series of the function

f(x)=log⁡(1+2x)\begin{equation*} f(x) = \log(1 + 2x) \end{equation*}

about x=0x = 0 in summation notation.

Hint

Review Example 3.5.20 in the CLP–2 text.

Answer

∑n=0∞(−1)n2n+1xn+1n+1\displaystyle\sum_{n=0}^\infty (-1)^n \frac{2^{n+1} x^{n+1}}{n+1} for all ∣x∣<12|x|<\frac{1}{2}

Full solution

Since

f′(t)=ddtlog⁡(1+2t)=21+2t=2∑n=0∞(−2t)nif ∣2t∣<1 i.e. ∣t∣<12\begin{equation*} f'(t) =\diff{}{t}\log(1+2t) = \frac{2}{1+2t} = 2\sum_{n=0}^\infty(-2t)^n \qquad\text{if }|2t|<1\text{ i.e. }|t|<\frac{1}{2} \end{equation*}

and f(0)=0f(0)=0, we have

f(x)=∫0xf′(t) dt=2∑n=0∞∫0x(−1)n2ntn dt=∑n=0∞(−1)n2n+1xn+1n+1 for all ∣x∣<12\begin{align*} f(x) &=\int_0^x f'(t) \,\dee{t} =2 \sum_{n=0}^\infty \int_0^x (-1)^n 2^n t^n\,\dee{t} =\sum_{n=0}^\infty (-1)^n 2^{n+1} \frac{x^{n+1}}{n+1} \qquad\text{ for all }|x|<\frac{1}{2} \end{align*}
Q13Stage 2Past exam · 2013A

The first two terms in the Maclaurin series for x2sin⁡(x3)x^2 \sin(x^3) are ax5+bx11ax^5 + bx^{11} , where aa and bb are constants. Find the values of aa and bb.

Hint

You should know the Maclaurin series for sin⁡x\sin x. Use it.

Answer

a=1a=1, b=−13!=−16b=-\dfrac{1}{3!}=-\dfrac{1}{6}.

Full solution

We just need to substitute y=x3y=x^3 into the known Maclaurin series for sin⁡y\sin y, to get the Maclaurin series for sin⁡(x3)\sin(x^3), and then multiply the result by x2x^2.

sin⁡y=y−y33!+⋯sin⁡(x3)=x3−x93!+⋯x2sin⁡(x3)=x5−x113!+⋯\begin{align*} \sin y&= y-\frac{y^3}{3!}+\cdots \cr \sin(x^3)&= x^3-\frac{x^9}{3!}+\cdots \cr x^2\sin(x^3)&= x^5-\frac{x^{11}}{3!}+\cdots \cr \end{align*}

so a=1a=1 and b=−13!=−16b=-\frac{1}{3!}=-\frac{1}{6}.

Q14Stage 2Past exam · 2014D

Give the first two nonzero terms in the Maclaurin series for ∫e−x2−1x dx\displaystyle{\int \frac{e^{-x^2}-1}{x} \,\dee{x}}.

Hint

You should know the Maclaurin series for exe^x. Use it.

Answer

∫e−x2−1x dx=C−x22+x48+⋯\displaystyle\int \frac{e^{-x^2}-1}{x} \,\dee{x} = C -\frac{x^2}{2} + \frac{x^4}{8} + \cdots.

It is not clear from the wording of the question whether or not the arbitrary constant CC is to be counted as one of the “first two nonzero terms”.

Full solution

Recall that

ey=∑n=0∞ynn!=1+y+y22+y33!+⋯\begin{equation*} e^y = \sum_{n=0}^\infty\frac{y^n}{n!} =1 + y + \frac{y^2}{2} +\frac{y^3}{3!} + \cdots \end{equation*}

Setting y=−x2y=-x^2, we have

e−x2=1−x2+x42−x63!+⋯e−x2−1=−x2+x42−x66+⋯e−x2−1x=−x+x32−x56+⋯∫e−x2−1x dx=C−x22+x48−x636+⋯\begin{align*} e^{-x^2} & = 1 -x^2 + \frac{x^4}{2} - \frac{x^6}{3!} + \cdots \\ e^{-x^2} - 1& = -x^2 + \frac{x^4}{2} - \frac{x^6}{6} + \cdots \\ \frac{e^{-x^2} - 1}{x}& = -x + \frac{x^3}{2} - \frac{x^5}{6} + \cdots \\ \int \frac{e^{-x^2}-1}{x} \,\dee{x}& = C -\frac{x^2}{2} + \frac{x^4}{8} - \frac{x^6}{36} + \cdots \\ \end{align*}
Q15Stage 2Past exam · 2015A

Find the Maclaurin series for ∫x4arctan⁡(2x) dx\displaystyle{\int x^4\arctan(2x) \,\dee{x}}.

Hint

You should know the Maclaurin series for arctan⁡(x)\arctan(x). Use it.

Answer

∑n=0∞(−1)n22n+1x2n+6(2n+1)(2n+6)+C=∑n=0∞(−1)n22nx2n+6(2n+1)(n+3)+C\displaystyle\sum\limits_{n=0}^\infty (-1)^n\frac{2^{2n+1} x^{2n+6}}{(2n+1)(2n+6)} +C =\sum\limits_{n=0}^\infty (-1)^n\frac{2^{2n} x^{2n+6}}{(2n+1)(n+3)} +C

Full solution

Recall that

arctan⁡(y)=∑n=0∞(−1)ny2n+12n+1\begin{equation*} \arctan(y) = \sum_{n=0}^\infty(-1)^n\frac{y^{2n+1}}{2n+1} \end{equation*}

Setting y=2xy=2x, we have

∫x4arctan⁡(2x) dx=∫(x4∑n=0∞(−1)n(2x)2n+12n+1)dx=∫(∑n=0∞(−1)n22n+1x2n+52n+1)dx=∑n=0∞(−1)n22n+1x2n+6(2n+1)(2n+6)+C=∑n=0∞(−1)n22nx2n+6(2n+1)(n+3)+C\begin{align*} \int x^4\arctan(2x) \,\dee{x}&= \int\left( x^4\sum_{n=0}^\infty (-1)^n\frac{(2x)^{2n+1}}{2n+1} \right)\dee{x} \\&= \int\left( \sum_{n=0}^\infty (-1)^n\frac{2^{2n+1}x^{2n+5}}{2n+1} \right)\dee{x} \\ &= \sum_{n=0}^\infty (-1)^n\frac{2^{2n+1} x^{2n+6}}{(2n+1)(2n+6)} +C\\ &=\sum_{n=0}^\infty (-1)^n\frac{2^{2n} x^{2n+6}}{(2n+1)(n+3)} +C \end{align*}
Q16Stage 2Past exam · M105 2015A

Suppose that dfdx=x1+3x3\displaystyle\diff{f}{x}=\frac{x}{1+3x^3} and f(0)=1f(0)=1. Find the Maclaurin series for f(x)f(x).

Hint

You should know the Maclaurin series for 11−x\dfrac{1}{1-x}. Use it.

Answer

f(x)=1+∑n=0∞(−1)n3n3n+2x3n+2f(x) = 1+ \displaystyle\sum\limits_{n=0}^\infty (-1)^n \frac{3^n }{3n+2} x^{3n+2}

Full solution

Substituting y=−3x3y=-3x^3 into 11−y=∑n=0∞yn\dfrac{1}{1-y} = \sum\limits_{n=0}^\infty y^n gives

dfdx=x⋅11+3x3=x∑n=0∞(−3x3)n=∑n=0∞(−1)n3nx3n+1\begin{align*} \diff{f}{x} = x\cdot\frac{1}{1+3x^3} =x\sum_{n=0}^\infty {\big(-3x^3\big)}^n =\sum_{n=0}^\infty (-1)^n 3^n x^{3n+1} \end{align*}

Now integrating,

f(x)=∑n=0∞(−1)n3nx3n+23n+2+C\begin{align*} f(x) = \sum_{n=0}^\infty (-1)^n 3^n \frac{x^{3n+2}}{3n+2} +C \end{align*}

To have f(0)=1f(0)=1, we need C=1C=1. So, finally

f(x)=1+∑n=0∞(−1)n3n3n+2x3n+2\begin{align*} f(x) = 1+ \sum_{n=0}^\infty (-1)^n \frac{3^n }{3n+2} x^{3n+2} \end{align*}

In past chapters, we were only able to exactly evaluate very specific types of series: geometric and telescoping. In Questions 17 through 25, we expand our range by relating given series to Taylor series.

Q17Stage 2Past exam · M105 2012A

The Maclaurin series for arctan⁡x\arctan x is given by

arctan⁡x=∑n=0∞(−1)nx2n+12n+1\begin{equation*} \arctan x = \sum_{n=0}^\infty (-1)^n\frac{x^{2n+1}}{2n+1} \end{equation*}

which has radius of convergence equal to 11. Use this fact to compute the exact value of the series below:

∑n=0∞(−1)n(2n+1)3n\begin{equation*} \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1) 3^n} \end{equation*}
Hint

Set (−1)nx2n+12n+1=C(−1)n(2n+1)3n(-1)^n\dfrac{x^{2n+1}}{2n+1}=C\dfrac{(-1)^n}{(2n+1)3^n}, for some constant CC. What are xx and CC?

Answer

π23\dfrac{\pi}{2\sqrt{3}}

Full solution

We're given a big hint: that our series resembles the Taylor series for arctangent.

The terms of arctangent are (−1)nx2n+12n+1(-1)^n\dfrac{x^{2n+1}}{2n+1}. Our terms resemble those terms, with x2n+1x^{2n+1} replaced by 13n\dfrac{1}{3^n}.

Since 3n=(3)2n=13(3)2n+13^n=\big(\sqrt{3}\big)^{2n}=\frac{1}{\sqrt{3}}\big(\sqrt{3}\big)^{2n+1}:

∑n=0∞(−1)n(2n+1)3n=3∑n=0∞(−1)n(2n+1)(3)2n+1=3∑n=0∞(−1)nx2n+12n+1∣x=13=3 arctan⁡13=3 π6=π23\begin{align*} \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1) 3^n} &=\sqrt{3} \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)\big(\sqrt{3}\big)^{2n+1} } =\sqrt{3} \sum_{n=0}^\infty (-1)^n\frac{x^{2n+1}}{2n+1} \bigg|_{x=\frac{1}{\sqrt{3}}} =\sqrt{3}\,\arctan\frac{1}{\sqrt{3}} \\ &=\sqrt{3}\,\frac{\pi}{6} =\frac{\pi}{2\sqrt{3}} \end{align*}
Q18Stage 2Past exam · 2014D

Evaluate ∑n=0∞(−1)nn!{\displaystyle\sum_{n=0}^\infty\frac{(-1)^n}{n!}}.

Hint

There is an important Taylor series, one of the series in Theorem 3.6.5 of the

CLP-2 text, that looks a lot like the given series.

Answer

1e\dfrac{1}{e}

Full solution

Recall that ex=∑n=0∞xnn!e^x =\displaystyle \sum\limits_{n=0}^\infty\frac{x^n}{n!}. So

∑n=0∞(−1)nn!=[∑n=0∞xnn!]x=−1=[ex]x=−1=e−1\begin{align*} \sum_{n=0}^\infty\frac{(-1)^n}{n!} =\Big[\sum_{n=0}^\infty\frac{x^n}{n!}\Big]_{x=-1} =\Big[e^x\Big]_{x=-1} =e^{-1} \end{align*}
Q19Stage 2Past exam · M105 2013A

Evaluate ∑k=0∞1ekk!{\displaystyle\sum_{k=0}^\infty\frac{1}{e^k k!}}.

Hint

There is an important Taylor series, one of the series in Theorem 3.6.5 of the

CLP-2 text, that looks a lot like the given series.

Answer

e1/ee^{1/e}

Full solution

Recall that ex=∑k=0∞xkk!e^x = \displaystyle\sum\limits_{k=0}^\infty\frac{x^k}{k!}. So

∑k=0∞1ekk!=[∑k=0∞xkk!]x=1/e=[ex]x=1/e=e1/e\begin{align*} \sum_{k=0}^\infty\frac{1}{e^k k!} =\Big[\sum_{k=0}^\infty\frac{x^k}{k!}\Big]_{x=1/e} =\Big[e^x\Big]_{x=1/e} =e^{1/e} \end{align*}
Q20Stage 2Past exam · M105 2013A

Evaluate the sum of the convergent series ∑k=1∞1πkk!{\displaystyle\sum_{k=1}^\infty\frac{1}{\pi^k k!}}.

Hint

There is an important Taylor series, one of the series in Theorem 3.6.5 of the

CLP-2 text, that looks a lot like the given series. Be careful about the limits of summation.

Answer

e1/π−1e^{1/\pi}-1

Full solution

Recall that ex=∑k=0∞xkk!e^x = \displaystyle\sum\limits_{k=0}^\infty\frac{x^k}{k!}. So

∑k=0∞1πkk!=[∑k=0∞xkk!]x=1/π=[ex]x=1/π=e1/π\begin{align*} \sum_{k=0}^\infty\frac{1}{\pi^k k!} =\Big[\sum_{k=0}^\infty\frac{x^k}{k!}\Big]_{x=1/\pi} =\Big[e^x\Big]_{x=1/\pi} =e^{1/\pi} \end{align*}

This series differs from the given one only in that it starts with k=0k=0 while the given series starts with k=1k=1. So

∑k=1∞1πkk!=∑k=0∞1πkk!−1⏟k=0=e1/π−1\begin{align*} \sum_{k=1}^\infty\frac{1}{\pi^k k!} =\sum_{k=0}^\infty\frac{1}{\pi^k k!} -\underbrace{1}_{k=0} =e^{1/\pi}-1 \end{align*}
Q21Stage 2Past exam · 2012A

Evaluate ∑n=1∞(−1)n−1n 2n{\displaystyle\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n\, 2^n}}.

Hint

There is an important Taylor series, one of the series in Theorem 3.6.5 of the

CLP-2 text, that looks a lot like the given series.

Answer

log⁡(3/2)\log(3/2)

Full solution

Recall, from Theorem 3.6.5 in the CLP-2 text, that, for all −1<x≤1-1<x\le 1,

log⁡(1+x)=∑k=0∞(−1)kxk+1k+1=∑n=1∞(−1)n−1xnn\begin{equation*} \log(1+x) = \sum_{k=0}^\infty(-1)^k\frac{x^{k+1}}{k+1} = \sum_{n=1}^\infty(-1)^{n-1}\frac{x^n}{n} \end{equation*}

(To get from the first sum to the second sum we substituted n=k+1n=k+1. If you don't see why the two sums are equal, write out the first few terms of each.) So

∑n=1∞(−1)n−1n 2n=[∑n=1∞(−1)n−1xnn]x=1/2=[log⁡(1+x)]x=1/2=log⁡(3/2)\begin{align*} \sum_{n=1}^\infty\frac{(-1)^{n-1}}{n\, 2^n} =\Big[\sum_{n=1}^\infty(-1)^{n-1}\frac{x^n}{n}\Big]_{x=1/2} =\Big[\log(1+x)\Big]_{x=1/2} =\log(3/2) \end{align*}
Q22Stage 2Past exam · M121 2012A

Evaluate ∑n=1∞n+2n!en{\displaystyle\sum_{n=1}^\infty\frac{n+2}{n!}e^n}.

Hint

Split the series into a sum of two series. There is an important Taylor series, one of the series in Theorem 3.6.5 of the

CLP-2 text, that looks a lot like each of the two series.

Answer

(e+2)ee−2(e+2)e^e-2

Full solution

Write

∑n=1∞n+2n!en=∑n=1∞nn!en+∑n=1∞2n!en=∑n=1∞en(n−1)!+2∑n=1∞enn!=e∑n=1∞en−1(n−1)!+2∑n=1∞enn!=e∑n=0∞enn!+2∑n=1∞enn!\begin{align*} \sum_{n=1}^\infty\frac{n+2}{n!}e^n &=\sum_{n=1}^\infty\frac{n}{n!}e^n + \sum_{n=1}^\infty\frac{2}{n!}e^n \\ &=\sum_{n=1}^\infty\frac{e^n}{(n-1)!} + 2\sum_{n=1}^\infty\frac{e^n}{n!}\\ &=e\sum_{n=1}^\infty\frac{e^{n-1}}{(n-1)!} + 2\sum_{n=1}^\infty\frac{e^n}{n!} \\ &=e\sum_{n=0}^\infty\frac{e^n}{n!} + 2\sum_{n=1}^\infty\frac{e^n}{n!} \end{align*}

Recall that ex=∑n=0∞xnn!e^x = \displaystyle\sum\limits_{n=0}^\infty\frac{x^n}{n!}. So

∑n=1∞n+2n!en=e[∑n=0∞xnn!]x=e+2[∑n=1∞xnn!]x=e=e[ex]x=e+2[ex−1]x=e=ee+1+2(ee−1)=(e+2)ee−2\begin{align*} \sum_{n=1}^\infty\frac{n+2}{n!}e^n &=e\Big[\sum_{n=0}^\infty\frac{x^n}{n!}\Big]_{x=e} +2\Big[\sum_{n=1}^\infty\frac{x^n}{n!}\Big]_{x=e} =e\Big[e^x\Big]_{x=e} + 2\Big[e^x-1\Big]_{x=e} = e^{e+1} + 2(e^e-1) \\ &=(e+2)e^e-2 \end{align*}
Q23Stage 2

Evaluate ∑n=1∞2nn\displaystyle\sum_{n=1}^\infty \frac{2^n}{n}, or show that it diverges.

Hint

Try the ratio test.

Answer

The sum diverges–see the solution.

Full solution

Let's use the ratio test:

lim⁡n→∞∣an+1an∣=lim⁡n→∞∣2n+1n+12nn∣=lim⁡n→∞2nn+1=2>1\begin{align*} \lim_{n \to \infty}\left|\frac{a_{n+1}}{a_n} \right|&= \lim_{n \to \infty}\left|\frac{\frac{2^{n+1}}{n+1}}{\frac{2^n}{n}} \right|\\ &= \lim_{n \to \infty}2\frac{n}{n+1}=2>1 \end{align*}

So, the series diverges.

Remark: it's tempting to note that log⁡(1+y)=∑n=0∞(−1)nyn+1n+1=−∑n=1∞(−y)nn\log(1+y)=\displaystyle\sum_{n=0}^\infty (-1)^n\frac{y^{n+1}}{n+1}=-\sum_{n=1}^\infty \frac{(-y)^n}{n}, and try to substitute in y=−2y=-2. But, the Maclaurin series for log⁡(1+y)\log(1+y) has radius of convergence R=1R=1, so it doesn't converge at y=−2y=-2. Furthermore, log⁡(1+(−2))=log⁡(−1)\log(1+(-2))=\log(-1), but this is undefined.

Q24Stage 2

Evaluate

∑n=0∞(−1)n(2n+1)!(π4)2n+1(1+22n+1)\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!}\left(\frac{\pi}{4} \right)^{2n+1}\left(1+2^{2n+1} \right)

or show that it diverges.

Hint

Write it as the sum of two Taylor series.

Answer

1+22\dfrac{1+\sqrt{2}}{\sqrt{2}}

Full solution

Our series looks something like the Taylor series for sine, sin⁡x=∑n=0∞(−1)n(2n+1)!x2n+1\sin x = \displaystyle\sum_{n=0}^\infty \frac{(-1)^{n}}{(2n+1)!}x^{2n+1}.

∑n=0∞(−1)n(2n+1)!(π4)2n+1(1+22n+1)=∑n=0∞(−1)n(2n+1)![(π4)2n+1+(π2)2n+1]=∑n=0∞(−1)n(2n+1)!(π4)2n+1+∑n=0∞(−1)n(2n+1)!(π2)2n+1=sin⁡(π4)+sin⁡(π2)=12+1=1+22\begin{align*} \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!}\left(\frac{\pi}{4} \right)^{2n+1}\left(1+2^{2n+1} \right)&= \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!}\left[\left(\frac{\pi}{4}\right)^{2n+1}+\left(\frac{\pi}{2}\right)^{2n+1} \right]\\ &=\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!}\left(\frac{\pi}{4}\right)^{2n+1} +\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!}\left(\frac{\pi}{2}\right)^{2n+1}\\ &=\sin\left(\frac{\pi}{4} \right)+\sin\left(\frac{\pi}{2} \right)\\ &=\frac{1}{\sqrt{2}}+1=\frac{1+\sqrt{2}}{\sqrt{2}} \end{align*}
Q25Stage 2Past exam · M121 2000A

(a) Show that the power series ∑n=0∞x2n(2n)!\displaystyle\sum_{n=0}^\infty \frac{x^{2n}}{(2n)!} converges absolutely for all real numbers xx.

(b) Evaluate ∑n=0∞1(2n)!\displaystyle\sum_{n=0}^\infty \frac{1}{(2n)!}.

Hint

Can you think of a way to eliminate the odd terms from ex=∑n=0∞xnn!e^x=\displaystyle \sum_{n=0}^\infty \frac{x^n}{n!} ?

Answer

(a) See the solution. (b) 12(e+1e)\displaystyle\frac12\left(e+\frac{1}{e}\right)

Full solution

(a)

  • The naive strategy is to set an=x2n(2n)!a_n=\dfrac{x^{2n}}{(2n)!} and apply the ratio test.

    lim⁡n→∞∣an+1an∣=lim⁡n→∞∣x2n+2(2n+2)!x2n(2n)!∣=∣x2n+2x2n⋅(2n)!(2n+2)(2n+1)(2n)!∣=lim⁡n→∞x2(2n+2)(2n+1)=0\begin{align*} \lim_{n\rightarrow\infty}\Big|\frac{a_{n+1}}{a_n}\Big| &=\lim_{n\rightarrow\infty} \left|\frac{ \frac{x^{2n+2}}{(2n+2)!} } {\frac{x^{2n}}{(2n)!}}\right| =\left| \frac{x^{2n+2}}{x^{2n}}\cdot \frac{(2n)!}{(2n+2)(2n+1)(2n)!} \right|\\ &=\lim_{n\rightarrow\infty} \frac{x^2}{(2n+2)(2n+1)} \\ &=0 \end{align*}

    This is smaller than 11 no matter what xx is. So the series converges for all xx.

  • Alternatively, the sneaky way is to observe that both ex=∑n=0∞xnn!e^x=\displaystyle \sum_{n=0}^\infty \frac{x^n}{n!} and e−x=∑n=0∞(−x)nn!e^{-x}=\displaystyle \sum_{n=0}^\infty \frac{(-x)^n}{n!} are known to converge for all xx. So

    12(ex+e−x)=∑n evenxnn!=∑n=0∞x2n(2n)!\begin{equation*} \frac{1}{2}\big(e^x+e^{-x}\big) = \sum_{{n{\rm\ even}}}\frac{x^n}{n!} = \sum_{n=0}^\infty \frac{x^{2n}}{(2n)!} \end{equation*}

    also converges for all xx.

(b) Recall that ex=∑n=0∞xnn!e^x=\displaystyle\sum\limits_{n=0}^\infty \frac{x^n}{n!}. Then:

e=∑n=0∞1n!e−1=∑n=0∞(−1)nn!e+e−1=∑n=0∞1+(−1)nn!=2∑n even∞1n!=2∑n=0∞1(2n)!\begin{align*} e&=\sum\limits_{n=0}^\infty \frac{1}{n!}\cr e^{-1}&=\sum_{n=0}^\infty \frac{(-1)^n}{n!}\\ e+e^{-1}&=\sum_{n=0}^\infty \frac{1+(-1)^n}{n!} =2\sum_{n{\rm\ even}}^\infty \frac{1}{n!} =2\sum_{n=0}^\infty \frac{1}{(2n)!} \end{align*}

Hence ∑n=0∞1(2n)!=12(e+1e)\displaystyle\sum\limits_{n=0}^\infty \frac{1}{(2n)!}=\frac12\left(e+\frac{1}{e}\right).

Q26Stage 2

  1. Using the fact that arctan⁡(1)=π4\arctan(1)=\dfrac{\pi}{4}, how many terms of the Taylor series for arctangent would you have to add up to approximate π\pi with an error of at most 4×10−54\times 10^{-5}?

  2. Example 3.6.13 in the CLP–2 text mentions the formula

    π=16arctan⁡15−4arctan⁡1239\pi=16\arctan\frac15-4\arctan\frac{1}{239}

    Using the Taylor series for arctangent, how many terms would you have to add up to approximate π\pi with an error of at most 4×10−54\times 10^{-5}?

  3. Assume without proof the following:

    arctan⁡12+arctan⁡13=arctan⁡(3+22⋅3−1)\arctan\frac12+\arctan\frac13=\arctan\left(\frac{3+2}{2\cdot3-1}\right)

    Using the Taylor series for arctangent, how many terms would you have to add up to approximate π\pi with an error of at most 4×10−54\times 10^{-5}?

Hint

The series you're adding up are alternating, so it's simple to bound the error using a partial sum.

Answer

(a) 50,000 (b) three terms (n=0n=0 to n=2n=2) (c) six terms (n=0n=0 to n=5n=5)

Full solution

All three series we're adding up are alternating, so we can bound the absolute error in the approximation SNS_N (the NN-th partial sum) by ∣aN+1∣|a_{N+1}|.

The Taylor series for arctangent is

arctan⁡(x)=∑n=0∞(−1)nx2n+12n+1\arctan(x)=\sum_{n=0}^\infty (-1)^n\frac{x^{2n+1}}{2n+1}

for every real xx.

  1. Using the Taylor series for arctangent when x=1x=1, we see

    π4=arctan⁡(1)=∑n=0∞(−1)n12n+1π=∑n=0∞(−1)n42n+1\begin{align*} \frac{\pi}{4}=\arctan(1)&=\sum_{n=0}^\infty(-1)^n \frac{1}{2n+1}\\ \pi &=\sum_{n=0}^\infty (-1)^n\frac{4}{2n+1} \end{align*}

    The error involved in approximating π\pi with the partial sum SNS_N is at most ∣aN+1∣=42N+3|a_{N+1}|=\frac{4}{2N+3}. In order for this to be at most 4×10−54\times 10^{-5}, we need:

    42N+3≤4×10−52N+3≥105N≥105−32=5×104−32=50,000−1.5\begin{align*} \frac{4}{2N+3}&\le 4\times 10^{-5}\\ 2N+3 &\ge 10^5\\ N&\ge \frac{10^5-3}{2}=5\times 10^4-\frac{3}{2}=50,000-1.5 \end{align*}

    Since nn must be an integer, we need to add up the terms from n=0n=0 to n=49,999n=49,999. That is, we add up the first 50,000 terms.

  2. Using the Taylor series for arctangent:

    π=16arctan⁡15−4arctan⁡1239=16∑n=0∞(−1)n1(2n+1)52n+1−4∑n=0∞(−1)n1(2n+1)⋅2392n+1=∑n=0∞(−1)n2n+1(1652n+1−42392n+1)\begin{align*} \pi=16\arctan\frac15-4\arctan\frac{1}{239}&=16\sum_{n=0}^\infty(-1)^n\frac{1}{(2n+1)5^{2n+1}}- 4\sum_{n=0}^\infty(-1)^n\frac{1}{(2n+1)\cdot 239^{2n+1}}\\ &=\sum_{n=0}^\infty\frac{(-1)^n}{2n+1}\left(\frac{16}{5^{2n+1}}-\frac{4}{239^{2n+1}} \right) \end{align*}

    This is an alternating sum, so the absolute error in using the partial sum SNS_N is at most:

    ∣aN+1∣=12N+3(1652N+3−42392N+3)|a_{N+1}|=\frac{1}{2N+3}\left(\frac{16}{5^{2N+3}} -\frac{4}{239^{2N+3}}\right)

    So, we want to find a value of NN that makes this at most 4×10−54\times 10^{-5}. Several values of NN are given below.

    NN∣aN+1∣|a_{N+1}|
    115(1655−42395)≈0.001\dfrac{1}{5}\left(\dfrac{16}{5^5}-\dfrac{4}{239^5} \right)\approx 0.001
    217(1657−42397)≈0.000029<4×10−5\dfrac{1}{7}\left(\dfrac{16}{5^7}-\dfrac{4}{239^7} \right)\approx 0.000029<4\times 10^{-5}

    So, it suffices to add up the first three terms (n=0n=0, n=1n=1, and n=2n=2) of the series.

  3. Again, we use the Taylor series for arctangent.

    arctan⁡12+arctan⁡13=arctan⁡(3+22⋅3−1)=arctan⁡(1)=π4π=4(arctan⁡12+arctan⁡13)=4∑n=0∞(−1)n1(2n+1)22n+1+4∑n=0∞(−1)n1(2n+1)32n+1=∑n=0∞(−1)n42n+1(122n+1+132n+1)\begin{align*} \arctan\frac12+\arctan\frac13&=\arctan\left(\frac{3+2}{2\cdot3-1}\right)=\arctan(1)=\frac{\pi}{4}\\ \pi&=4\left(\arctan\frac12+\arctan\frac13 \right)\\ &=4\sum_{n=0}^\infty (-1)^n\frac{1}{(2n+1)2^{2n+1}}+4\sum_{n=0}^\infty (-1)^n\frac{1}{(2n+1)3^{2n+1}}\\ &=\sum_{n=0}^\infty (-1)^n\frac{4}{2n+1}\left(\frac{1}{2^{2n+1}}+\frac{1}{3^{2n+1}} \right) \end{align*}

    If we use the partial sum SNS_N, our absolute error is at most

    ∣aN+1∣=42N+3(122N+3+132N+3).|a_{N+1}|=\dfrac{4}{2N+3}\left(\dfrac{1}{2^{2N+3}}+\dfrac{1}{3^{2N+3}} \right).

    Several of these values are given below.

    NN∣aN+1∣|a_{N+1}|
    145(125+135)≈0.028\displaystyle \frac{4}{5}\left(\frac{1}{2^5}+\frac{1}{3^5} \right)\approx 0.028
    247(127+137)≈0.0047\displaystyle\frac{4}{7}\left(\frac{1}{2^7}+\frac{1}{3^7} \right)\approx 0.0047
    349(129+139)≈0.00089\displaystyle\frac{4}{9}\left(\frac{1}{2^9}+\frac{1}{3^9} \right)\approx 0.00089
    4411(1211+1311)≈0.00018\displaystyle\frac{4}{11}\left(\frac{1}{2^{11}}+\frac{1}{3^{11}} \right)\approx 0.00018
    5413(1213+1313)≈0.000038<4×10−5\displaystyle\frac{4}{13}\left(\frac{1}{2^{13}}+\frac{1}{3^{13}} \right)\approx 0.000038<4\times10^{-5}

    So, it suffices to add the first six terms (n=0n=0 to n=5n=5) of the series.

Remark: if we actually wanted to approximate π\pi this way, the series from part (a) is probably not ideal–adding 50,000 terms sounds rough. The series from (b) and (c) seem much more practical.

Q27Stage 2

Suppose you wanted to approximate the number log⁡(1.5)\log(1.5) as a rational number using the Taylor expansion of log⁡(1+x)\log(1+x). How many terms would you need to add to get 10 decimal places of accuracy? (That is, an absolute error less than 5×10−115\times10^{-11}.)

Hint

The Taylor Series is alternating, so bounding the error in a partial-sum approximation is straightforward.

Answer

29

Full solution

Using the Taylor series for log⁡(1+x)\log(1+x):

log⁡(1+x)=∑n=1∞(−1)n+1xnnlog⁡(1.5)=log⁡(1+12)=∑n=1∞(−1)n+11n2n\begin{align*} \log(1+x)&=\sum_{n=1}^\infty (-1)^{n+1}\frac{x^n}{n}\\ \log(1.5)=\log\left(1+\frac12\right)&=\sum_{n=1}^\infty (-1)^{n+1}\frac{1}{n2^n} \end{align*}

Since this is an alternating series, the error involved in using the partial sum SNS_N is at most

∣aN+1∣=1(N+1)2N+1.|a_{N+1}|=\frac{1}{(N+1)2^{N+1}}.

We want this to be at most 5×10−115 \times 10^{-11}.

NN∣aN+1∣|a_{N+1}|
10111⋅211≈4×10−5\displaystyle \frac{1}{11\cdot 2^{11}}\approx 4\times 10^{-5}
15116⋅216≈9.5×10−7\displaystyle\frac{1}{16\cdot 2^{16}}\approx 9.5\times 10^{-7}
20121⋅221≈2×10−8\displaystyle\frac{1}{21\cdot 2^{21}}\approx 2 \times 10^{-8}
25126⋅226≈6×10−10\displaystyle\frac{1}{26\cdot 2^{26}}\approx 6 \times 10^{-10}
26127⋅227≈3×10−10\displaystyle\frac{1}{27\cdot 2^{27}}\approx 3 \times 10^{-10}
27128⋅228≈1×10−10\displaystyle\frac{1}{28\cdot 2^{28}}\approx 1 \times 10^{-10}
28129⋅229≈6×10−11\displaystyle\frac{1}{29\cdot 2^{29}}\approx 6 \times 10^{-11}
29130⋅230≈3×10−11\displaystyle\frac{1}{30\cdot 2^{30}}\approx 3 \times 10^{-11}

So, it suffices to add up the first 29 terms.

Q28Stage 2

Suppose you wanted to approximate the number ee as a rational number using the Maclaurin expansion of exe^x. How many terms would you need to add to get 10 decimal places of accuracy? (That is, an absolute error less than 5×10−115\times10^{-11}.)

You may assume without proof that 2<e<32<e<3.

Hint

The Taylor Series is not alternating, so use Theorem 3.6.1-b in the CLP-2 text to bound the error in a partial-sum approximation.

Answer

S13S_{13} or higher

Full solution

The Taylor Series for exe^x is not alternating, so we'll use Theorem 3.6.1-b in the CLP-2 text to bound the error in a partial-sum approximation. The error in the partial-sum approximation SNS_N is

EN=f(N+1)(c)(N+1)!(x−a)N+1E_N=\frac{f^{(N+1)}(c)}{(N+1)!}(x-a)^{N+1}

for some cc strictly between aa and xx. In our case, a=0a=0 and x=1x=1. So, we want to find a value of NN such that

∣f(N+1)(c)(N+1)!(1−0)N+1∣=ec(N+1)!<5×10−11\left|\frac{f^{(N+1)}(c)}{(N+1)!}\left(1-0\right)^{N+1}\right|=\frac{e^c}{(N+1)!}<5\times 10^{-11}

for all cc in (0,1)(0,1).

If cc is between 0 and 1, then ece^c is between 1 and ee. However, since the purpose of this problem is to approximate ee precisely, it doesn't make much sense to use ee in our bound. Since ee is less than 3, then ec<3e^c<3 for all cc in (0,1)(0,1). Now we can search for an appropriate value of NN.

NN3(N+1)!\dfrac{3}{(N+1)!}
10311!=1910≈8×10−8\displaystyle \frac{3}{11!}=\frac{1}{9^{10}}\approx 8\times 10^{-8}
11312!≈6×10−9\displaystyle \frac{3}{12!}\approx 6\times 10^{-9}
12313!≈5×10−10\displaystyle \frac{3}{13!}\approx 5\times 10^{-10}
13314!≈3×10−11\displaystyle \frac{3}{14!}\approx 3\times 10^{-11}

So, it suffices to use the partial sum S13S_{13}.

Q29Stage 2

Suppose you wanted to approximate the number log⁡(0.9)\log(0.9) as a rational number using the Taylor expansion of log⁡(1−x)\log(1-x). Which partial sum should you use to get 10 decimal places of accuracy? (That is, an absolute error less than 5×10−115\times10^{-11}.)

Hint

The Taylor Series is not alternating, so use Theorem 3.6.1-b in the CLP-2 text to bound the error in a partial-sum approximation.

Answer

S9S_9 or higher

Full solution

The Taylor Series for log⁡(1−x)\log(1-x) is not alternating, so we'll use Theorem 3.6.1-b in the CLP-2 text to bound the error in a partial-sum approximation. The error in the partial-sum approximation SNS_N is

EN=f(N+1)(c)(N+1)!(x−a)N+1E_N=\frac{f^{(N+1)}(c)}{(N+1)!}(x-a)^{N+1}

for some cc strictly between aa and xx. In our case, a=0a=0 and x=110x=\dfrac1{10}. So, we want to find a value of NN such that

∣f(N+1)(c)(N+1)!(110)N+1∣<5×10−11\left|\frac{f^{(N+1)}(c)}{(N+1)!}\left(\frac{1}{10}\right)^{N+1}\right|<5\times 10^{-11}

for all cc in (0,110)(0,\frac1{10}).

To find this NN, we to know f(N+1)(x)f^{(N+1)}(x). Just like when we create a Taylor polynomial from scratch, we'll differentiate f(x)f(x) several times, and look for a pattern.

f(x)=log⁡(1−x)f(6)(x)=−2(3)(4)(5)(1−x)6f′(x)=−11−xf(7)(x)=−2(3)(4)(5)(6)(1−x)7f′′(x)=−1(1−x)2⋮f′′′(x)=−2(1−x)3f(N+1)(x)=−N!(1−x)N+1f(4)(x)=−2(3)(1−x)4f(5)(x)=−2(3)(4)(1−x)5\begin{align*} f(x)&=\log(1-x)&f^{(6)}(x)&=\frac{-2(3)(4)(5)}{(1-x)^6}\\ f'(x)&=\frac{-1}{1-x}&f^{(7)}(x)&=\frac{-2(3)(4)(5)(6)}{(1-x)^7}\\ f''(x)&=\frac{-1}{(1-x)^2}&&\vdots\\ f'''(x)&=\frac{-2}{(1-x)^3}&f^{(N+1)}(x)&=\frac{-N!}{(1-x)^{N+1}}\\ f^{(4)}(x)&=\frac{-2(3)}{(1-x)^4}\\ f^{(5)}(x)&=\frac{-2(3)(4)}{(1-x)^5} \end{align*}

Now we want a reasonable bound on f(N+1)(c)f^{(N+1)}(c), when cc is in (0,110)(0,\frac{1}{10}). Note that in this range, 1−c>01-c>0.

0<c<110⇒910<1−c<1⇒(910)N+1<(1−c)N+1<1⇒1<1(1−c)N+1<(109)N+1⇒N!<N!(1−c)N+1<N!(109)N+1\begin{alignat*}{3} &&0 &<c<\frac{1}{10}\\ &\Rightarrow&\qquad \frac{9}{10} &<1-c<1\\ &\Rightarrow&\qquad \left(\frac{9}{10}\right)^{N+1} &<(1-c)^{N+1}<1\\ &\Rightarrow&\qquad 1&<\frac{1}{(1-c)^{N+1}}<\left(\frac{10}{9}\right)^{N+1}\\ &\Rightarrow&\qquad N!&<\frac{N!}{(1-c)^{N+1}}<N!\left(\frac{10}{9}\right)^{N+1} \end{alignat*}

This bound provides us with a “worst-case scenario" error. We don't know exactly what cc is, but we don't need to–the bound above holds for all cc between 0 and 110\frac{1}{10}.

Now we're ready to choose an NN that results in a sufficiently small error bound.

∣f(N+1)(c)(N+1)!(110)N+1∣<N!(109)N+1(N+1)!(110)N+1=19N+1⋅(N+1)So, we want:19N+1⋅(N+1)<5×10−11\begin{align*} \left|\frac{\textcolor{blue}{f^{(N+1)}(c)}}{(N+1)!}\left(\frac{1}{10}\right)^{N+1}\right|&< \frac{\textcolor{blue}{N!\left(\tfrac{10}{9}\right)^{N+1}}}{(N+1)!}\left(\frac{1}{10}\right)^{N+1}=\frac{1}{9^{N+1}\cdot (N+1)}\\ \text{So, we want:}\qquad \frac{1}{9^{N+1}\cdot (N+1)}&<5\times 10^{-11} \end{align*}

To find an appropriate NN, we test several values.

NN19N+1⋅(N+1)\dfrac{1}{9^{N+1}\cdot (N+1)}
819⋅99=1910≈3×10−10\displaystyle \frac{1}{9\cdot 9^{9}}=\frac{1}{9^{10}}\approx 3\times 10^{-10}
9110⋅910≈3×10−11\displaystyle \frac{1}{10\cdot 9^{10}}\approx 3\times 10^{-11}

So, it suffices to use the partial sum S9S_9.

Q30Stage 2

Define the hyperbolic sine function as

sinh⁡x=ex−e−x2.\sinh x = \frac{e^{x}-e^{-x}}{2}.

Suppose you wanted to approximate the number sinh⁡(b)\sinh(b) using the Maclaurin series of sinh⁡x\sinh x, where bb is some number in (−2,1)(-2,1). Which partial sum should you use to guarantee 10 decimal places of accuracy? (That is, an absolute error less than 5×10−115\times10^{-11}.)

You may assume without proof that 2<e<32<e<3.

Hint

Use Theorem 3.6.1-b in the CLP-2 text to bound the error in a partial-sum approximation. This theorem requires you to consider values of cc between xx and x=0x=0; since xx could be anything from −2-2 to 11, you should think about values of cc between −2-2 and 11.

Answer

S18S_{18} or higher

Full solution

We'll use Theorem 3.6.1-b in the CLP-2 text to bound the error of a partial-sum approximation. The error in the partial-sum approximation SNS_N is

EN=f(N+1)(c)(N+1)!(x−a)N+1E_N=\frac{f^{(N+1)}(c)}{(N+1)!}(x-a)^{N+1}

for some cc strictly between aa and xx. In our case, a=0a=0 and xx is in (−2,1)(-2,1). So, we want to find a value of NN such that

∣f(N+1)(c)(N+1)!(x)N+1∣<5×10−11\left|\frac{f^{(N+1)}(c)}{(N+1)!}\left(x\right)^{N+1}\right|<5\times 10^{-11}

for all xx in (−2,1)(-2,1), and all cc in (−2,1)(-2,1).

To find this NN, we to know f(N+1)(x)f^{(N+1)}(x). Just like when we create a Taylor polynomial from scratch, we'll differentiate f(x)f(x) several times, and look for a pattern.

f(x)=sinh⁡(x)=ex−e−x2f′(x)=ex+e−x2f′′(x)=ex−e−x2f′′′(x)=ex+e−x2\begin{align*} f(x)=\sinh(x)&=\frac{e^x-e^{-x}}{2}\\ f'(x)&=\frac{e^x+e^{-x}}{2}\\ f''(x)&=\frac{e^x-e^{-x}}{2}\\ f'''(x)&=\frac{e^x+e^{-x}}{2} \end{align*}

That is, even derivatives of f(x)f(x) are f(x)f(x), and odd derivatives of f(x)f(x) are ex+e−x2\frac{{e^x}+e^{-x}}{2} (which, incidentally, is the function called cosh⁡x\cosh x).

Now we want a reasonable bound on f(N+1)(c)f^{(N+1)}(c), when cc is in (−2,1)(-2,1). Since powers of ee are always positive, we begin by noting that 0<ex−e−x2<ex+e−x20<\frac{e^x-e^{-x}}{2}<\frac{e^x+e^{-x}}{2}. So, all derivatives of f(x)f(x) are bounded above by ex+e−x2\frac{e^x+e^{-x}}{2}.

−2<c<1⇒e−2<ec<eand e−1<e−c<e2⇒f(N+1)(c)<ec+e−c2<e2+e22=e2<9\begin{alignat*}{3} &&-2 &<c<1\\ &\Rightarrow&\qquad e^{-2} &<e^c<e &\text{and } e^{-1}&<e^{-c}<e^2\\ &\Rightarrow&\qquad f^{(N+1)}(c)&<\frac{e^c+e^{-c}}{2} & &<\frac{e^2+e^2}{2}=e^2<9 \end{alignat*}

This bound provides us with a “worst-case scenario" error. We don't know exactly what cc is, but we don't need to–the bound above holds for all cc between −2-2 and 11.

We also don't know exactly what xx will be, only that it's between −2-2 and 11. So, we note ∣x∣N+1<2N+1|x|^{N+1}<2^{N+1}.

Now we're ready to choose an NN that results in a sufficiently small error bound.

∣f(N+1)(c)(N+1)!(x)N+1∣<9⋅2N+1(N+1)!So, we want:9⋅2N+1(N+1)!<5×10−11\begin{align*} \left|\frac{\textcolor{blue}{f^{(N+1)}(c)}}{(N+1)!}\textcolor{red}{\left(x\right)^{N+1}}\right|&< \frac{\textcolor{blue}{9}\cdot \textcolor{red}{2^{N+1}}}{(N+1)!}\\ \text{So, we want:}\qquad \frac{9\cdot2^{N+1}}{(N+1)!}&<5\times 10^{-11} \end{align*}

To find an appropriate NN, we test several values.

NN9⋅2N+1(N+1)!\dfrac{9\cdot 2^{N+1}}{(N+1)!}
109⋅211(11)!≈5×10−4\displaystyle \frac{9\cdot 2^{11}}{(11)!}\approx 5\times 10^{-4}
159⋅216(16)!≈3×10−8\displaystyle \frac{9\cdot 2^{16}}{(16)!}\approx 3\times 10^{-8}
179⋅218(18)!≈4×10−10\displaystyle \frac{9\cdot 2^{18}}{(18)!}\approx 4\times 10^{-10}
189⋅219(19)!≈4×10−11<5×10−11\displaystyle \frac{9\cdot 2^{19}}{(19)!}\approx 4\times 10^{-11}<5\times10^{-11}

So, it suffices to use the partial sum S18S_{18}.

Q31Stage 2

Let f(x)f(x) be a function with

f(n)(x)=(n−1)!2[(1−x)−n+(−1)n−1(1+x)−n]f^{(n)}(x)=\frac{(n-1)!}{2}\left[(1-x)^{-n}+(-1)^{n-1}(1+x)^{-n} \right]

for all n≥1n \ge 1.

Give reasonable bounds (both upper and lower) on the error involved in approximating f(−13)f\left(-\frac13 \right) using the partial sum S6S_6 of the Taylor series for f(x)f(x) centred at a=12a=\frac12.

Remark: One function with this quality is the inverse hyperbolic tangent function. (Of course it is! Actually, hyperbolic tangent is tanh(x)=ex−e−xex+e−x\mathrm{tanh}(x) = \dfrac{e^x-e^{-x}}{e^x+e^{-x}}, and inverse hyperbolic tangent is its functional inverse.)

Hint

Use Theorem 3.6.1-b in the CLP-2 text to bound the error in a partial-sum approximation.

To bound the derivative over the appropriate range, remember how to find absolute extrema.

Answer

The error is in the interval (−5714⋅37[1+137],−577⋅67)≈(−0.199,−0.040)\displaystyle\left(\dfrac{-5^7}{14\cdot 3^7}\left[1+\frac{1}{3^7} \right]\quad,\quad \dfrac{-5^7}{7\cdot 6^7} \right)\approx\left(-0.199, -0.040\right)

Full solution

We'll use Theorem 3.6.1-b in the CLP-2 text to bound the error in a partial-sum approximation. The error in the partial-sum approximation SNS_N is

EN=f(N+1)(c)(N+1)!(x−a)N+1E_N=\frac{f^{(N+1)}(c)}{(N+1)!}(x-a)^{N+1}

for some cc strictly between aa and xx. In our case, a=12a=\dfrac12, x=−13x=-\dfrac1{3}, and we are given the nnth derivative of f(x)f(x):

E6=f(7)(c)7!(−13−12)7=17!⋅6!2[(1−c)−7+(−1)6(1+c)−7](−56)7=−5714⋅67⋅[(1−c)−7+(1+c)−7]=\begin{align*}E_6&=\frac{\textcolor{blue}{f^{(7)}(c)}}{7!}\left(-\frac13-\frac12 \right)^{7}\\ &=\frac{1}{7!}\cdot\textcolor{blue}{\frac{6!}{2}\left[\left(1-c\right)^{-7}+(-1)^{6}\left(1+c\right)^{-7} \right]}\left(-\frac56 \right)^7\\ &=\frac{-5^7}{14\cdot 6^7}\cdot\left[\left(1-c\right)^{-7}+\left(1+c\right)^{-7} \right] &=\end{align*}

for some cc in (−13,12)(-\frac13,\frac1{2}).

We want to provide actual numeric bounds for this expression. That is, we want to find the absolute max and min of

E(c)=−5714⋅67⋅[(1−c)−7+(1+c)−7]E(c)=\frac{-5^7}{14\cdot 6^7}\cdot\left[\left(1-c\right)^{-7}+\left(1+c\right)^{-7} \right]

over the interval (−13,12)\left( -\frac{1}{3},\frac{1}{2}\right). Absolute extrema occur at endpoints and critical points. So, we'll start by differentiating E(c)E(c), and finding its critical points (if any) in the interval (−13,12)\left( -\frac{1}{3},\frac{1}{2}\right).

E(c)=−5714⋅67⋅[(1−c)−7+(1+c)−7]E′(c)=−5714⋅67⋅[7(1−c)−8−7(1+c)−8]=0(1−c)−8=(1+c)−81−c=1+cc=0\begin{align*} E(c)&=\frac{-5^7}{14\cdot 6^7}\cdot\left[\left(1-c\right)^{-7}+\left(1+c\right)^{-7} \right]\\ E'(c)&=\frac{-5^7}{14\cdot 6^7}\cdot\left[7\left(1-c\right)^{-8}-7\left(1+c\right)^{-8} \right]=0\\ \left(1-c\right)^{-8}&=\left(1+c\right)^{-8}\\ 1-c&=1+c\\ c&=0 \end{align*}

Since E′(c)E'(c) is defined over our entire interval, its only critical point is c=0c=0.

  • E(0)=−5714⋅67[2]E(0)=\dfrac{-5^7}{14\cdot 6^7}[2]

  • E(−13)=−5714⋅67[(43)−7+(23)−7]=−5714⋅67[(34)7+(32)7]E\left(-\frac{1}{3}\right)=\dfrac{-5^7}{14\cdot 6^7}\left[\left(\frac43 \right)^{-7}+\left(\frac23 \right)^{-7} \right]=\dfrac{-5^7}{14\cdot 6^7}\left[\left(\frac34 \right)^{7}+\left(\frac32 \right)^{7} \right]

  • E(12)=−5714⋅67[(12)−7+(32)−7]=−5714⋅67[27+(23)7]E\left(\frac{1}{2}\right)=\dfrac{-5^7}{14\cdot 6^7}\left[\left(\frac12 \right)^{-7}+\left(\frac32 \right)^{-7} \right]=\dfrac{-5^7}{14\cdot 6^7}\left[2^7+\left(\frac23 \right)^{7} \right]

We want to decide which of these numbers is biggest, and which smallest. Note that 272^7 is much, much bigger than (3/2)7(3/2)^7, and both (3/4)7(3/4)^7 and (2/3)7(2/3)^7 are less than one. Furthermore, (3/2)7(3/2)^7 is much larger than 2. So: [27+(2/3)7]>[(3/2)7+(3/4)7]>2.\left[2^7+(2/3)^7\right]>\left[(3/2)^7+(3/4)^7\right]>2. Therefore,

−5714⋅67[27+(23)7]<−5714⋅67[(34)7+(32)7]<−5714⋅67[2]\dfrac{-5^7}{14\cdot 6^7}\left[2^7+\left(\frac23 \right)^{7} \right]<\dfrac{-5^7}{14\cdot 6^7}\left[\left(\frac34 \right)^{7}+\left(\frac32 \right)^{7} \right]<\dfrac{-5^7}{14\cdot 6^7}[2]

We conclude that the error E6E_6 is in the interval

(−5714⋅67[27+(23)7],−5714⋅67[2])\left(\dfrac{-5^7}{14\cdot 6^7}\left[2^7+\left(\frac23 \right)^{7} \right]\quad,\quad \dfrac{-5^7}{14\cdot 6^7} [2]\right)

or, equivalently,

(−5714⋅37[1+137],−577⋅67)\left(\dfrac{-5^7}{14\cdot 3^7}\left[1+\frac{1}{3^7} \right]\quad,\quad \dfrac{-5^7}{7\cdot 6^7} \right)

which is approximately (−0.199,−0.040)\left(-0.199, -0.040\right).

3Stage 3Application

Further than practice: several ideas at once, or an unfamiliar situation.

Q32Stage 3Past exam · 2014A

Use series to evaluate lim⁡x→01−cos⁡x1+x−ex\displaystyle \lim\limits_{x\rightarrow 0}\frac{1-\cos x}{1+x-e^x}.

Hint

See Example 3.6.21 in the

CLP-2 text

Answer

−1-1

Full solution

Using the Maclaurin series expansions of cos⁡x\cos x and exe^x,

cos⁡x=1−x22!+x44!+⋯1−cos⁡x=x22!−x44!+⋯ex=1+x+x22!+x33!+⋯1+x−ex=−x22!−x33!+⋯1−cos⁡x1+x−ex=x22!−x44!+⋯−x22!−x33!+⋯=12!−x24!+⋯−12!−x3!+⋯\begin{align*} \cos x& = 1-\frac{x^2}{2!} +\frac{x^4}{4!} +\cdots \\ 1-\cos x& = \frac{x^2}{2!} -\frac{x^4}{4!} +\cdots \\ e^x &= 1+x+ \frac{x^2}{2!} +\frac{x^3}{3!} +\cdots \\ 1+x-e^x&=-\frac{x^2}{2!} -\frac{x^3}{3!} +\cdots \\ \frac{1-\cos x}{1+x-e^x} &=\frac{\frac{x^2}{2!} -\frac{x^4}{4!} +\cdots } {-\frac{x^2}{2!} -\frac{x^3}{3!} +\cdots} =\frac{\frac{1}{2!} -\frac{x^2}{4!} +\cdots } {-\frac{1}{2!} -\frac{x}{3!} +\cdots} \end{align*}

we have

lim⁡x→01−cos⁡x1+x−ex=lim⁡x→012!−x24!+⋯−12!−x3!+⋯=12!−12!=−1\begin{equation*} \lim_{x\rightarrow 0}\frac{1-\cos x}{1+x-e^x} =\lim_{x\rightarrow 0}\frac{\frac{1}{2!} -\frac{x^2}{4!} +\cdots } {-\frac{1}{2!} -\frac{x}{3!} +\cdots} =\frac{\frac{1}{2!}} {-\frac{1}{2!}} =-1 \end{equation*}
Q33Stage 3Past exam · 2012A

Evaluate lim⁡x→0sin⁡x−x+x36x5\displaystyle \lim\limits_{x\rightarrow 0}\frac{\sin x -x +\frac{x^3}{6}}{x^5}.

Hint

See Example 3.6.21 in the

CLP-2 text

Answer

15!=1120\dfrac{1}{5!}=\dfrac{1}{120}

Full solution

Using the Maclaurin series expansion of sin⁡x\sin x,

sin⁡x=x−x33!+x55!−x77!+⋯sin⁡x−x+x36=x55!−x77!+⋯sin⁡x−x+x36x5=15!−x27!+⋯\begin{align*} \sin x& = x-\frac{x^3}{3!} +\frac{x^5}{5!} -\frac{x^7}{7!} +\cdots \\ \sin x-x +\frac{x^3}{6} & = \frac{x^5}{5!}-\frac{x^7}{7!} +\cdots \\ \frac{\sin x -x +\frac{x^3}{6}}{x^5} &=\frac{1}{5!}-\frac{x^2}{7!} +\cdots \end{align*}

we have

lim⁡x→0sin⁡x−x+x36x5=lim⁡x→0(15!−x27!+⋯)=15!=1120\begin{equation*} \lim_{x\rightarrow 0}\frac{\sin x -x +\frac{x^3}{6}}{x^5} =\lim_{x\rightarrow 0}\Big(\frac{1}{5!}-\frac{x^2}{7!} +\cdots\Big) =\frac{1}{5!} =\frac{1}{120} \end{equation*}

Remark: to solve this using l'H^opital's rule we would differentiate five times, making series a practical alternative.

Q34Stage 3

Evaluate lim⁡x→0(1+x+x2)2/x\displaystyle \lim\limits_{x\rightarrow 0}\left(1+x+x^2\right)^{2/x} using a Taylor series for the natural logarithm.

Hint

Set f(x)=(1+x+x2)2/xf(x)=\left(1+x+x^2\right)^{2/x}, and find lim⁡x→0log⁡(f(x))\lim\limits_{x\rightarrow 0}\log\left(f(x)\right).

Answer

e2e^2

Full solution

Our limit has the indeterminate form 1∞1^{\infty}; as with l'H^opital's rule, we can change it to a friendlier form using the natural logarithm.

f(x)=(1+x+x2)2/xlog⁡(f(x))=log⁡[(1+x+x2)2/x]=2xlog⁡(1+x+x2)\begin{align*} f(x)&=\left(1+x+x^2\right)^{2/x}\\ \log(f(x))&=\log\left[\left(1+x+x^2\right)^{2/x}\right]=\frac{2}{x}\log\left(1+x+x^2\right) \end{align*}

Recall log⁡(1+y)=∑n=1∞(−1)n+1ynn\log(1+y)=\displaystyle\sum_{n=1}^\infty \frac{(-1)^{n+1}y^n}{n}, and set y=x+x2y=x+x^2. The series converges when ∣y∣<1|y|<1, and since we only consider values of xx that are very close to 0, we can assume ∣x+x2∣<1|x+x^2|<1.

log⁡(f(x))=2xlog⁡(1+(x+x2))=2x∑n=1∞(−1)n+1(x+x2)nn=2x[(x+x2)−(x+x2)22+(x+x2)33−⋯ ]=2+2x−(x+x2)22x+(x+x2)33x−⋯=2+2x−(x2+x)(1+x)2+(x2+x)2(1+x)3−⋯lim⁡x→0log⁡(f(x))=lim⁡x→0[2+2x−(x2+x)(1+x)2+(x2+x)2(1+x)3−⋯ ]=2+0+0⋯=2lim⁡x→0f(x)=lim⁡x→0elog⁡f(x)=e2\begin{align*} \log(f(x))&=\frac{2}{x}\log\left(1+(x+x^2)\right)=\frac{2}{x}\sum_{n=1}^\infty\frac{(-1)^{n+1}(x+x^2)^n}{n}\\ &=\frac{2}{x}\left[(x+x^2)-\frac{(x+x^2)^2}{2}+\frac{(x+x^2)^3}{3}-\cdots\right]\\ &=2+2x-\frac{(x+x^2)^2}{2x}+\frac{(x+x^2)^3}{3x}-\cdots\\ &=2+2x-\frac{(x^2+x)(1+x)}{2}+\frac{(x^2+x)^2(1+x)}{3}-\cdots\\ \lim_{x \to 0}\log(f(x))&=\lim_{x \to 0}\left[2+2x-\frac{(x^2+x)(1+x)}{2}+\frac{(x^2+x)^2(1+x)}{3}-\cdots\right] \\ &=2+0+0\cdots=2\\ \lim_{x \to 0} f(x)&=\lim_{x \to 0}e^{\log f(x)}=e^2 \end{align*}
Q35Stage 3

Use series to evaluate

lim⁡x→∞(1+12x)x\lim_{x \to \infty} \left(1+\frac{1}{2x}\right)^{x}
Hint

Use the substitution y=1xy=\dfrac{1}{x}, and compare to Question 34.

Answer

e\sqrt{e}

Full solution

We have an indeterminate form 1∞1^\infty. We can use a natural logarithm to change this to a friendlier form. Furthermore, to avoid negative powers, we substitute y=12xy=\dfrac{1}{2x}. As xx grows larger and larger, yy gets closer and closer to zero, while staying positive.

log⁡[(1+12x)x]=xlog⁡(1+12x)=12ylog⁡(1+y)=12y∑n=1∞(−1)n+1nyn=12y[y−y22+y33−y44+⋯ ]=[12−y4+y26−y38+⋯ ]lim⁡x→∞log⁡[(1+12x)x]=lim⁡y→0+[12−y4+y26−y38+⋯ ]=12lim⁡x→∞[(1+12x)x]=e1/2=e\begin{align*} \log\left[\left(1+\frac1{2x}\right)^{x}\right]&=x\log\left(1+\frac1{2x}\right)=\frac{1}{2y}\log(1+y)\\ &=\frac{1}{2y}\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}y^{n}\\ &=\frac{1}{2y}\left[y-\frac{y^2}{2}+\frac{y^3}{3}-\frac{y^4}{4}+\cdots\right]\\ &=\left[\frac{1}{2}-\frac{y}{4}+\frac{y^2}{6}-\frac{y^3}{8}+\cdots\right]\\ \lim_{x \to \infty}\log\left[\left(1+\frac1{2x}\right)^{x}\right]&=\lim_{y \to 0^+}\left[\frac{1}{2}-\frac{y}{4}+\frac{y^2}{6}-\frac{y^3}{8}+\cdots\right]=\frac{1}{2}\\ \lim_{x \to \infty}\left[\left(1+\frac1{2x}\right)^{x}\right]&=e^{1/2}=\sqrt{e} \end{align*}
Q36Stage 3

Evaluate the series ∑n=0∞(n+1)(n+2)7n\displaystyle\sum_{n=0}^\infty\frac{ (n+1)(n+2)}{7^n} or show that it diverges.

Hint

Start by differentiating ∑n=0∞xn\displaystyle\sum_{n=0}^\infty x^n.

Answer

2(6/7)3=343108\displaystyle\frac{2}{(6/7)^3}=\frac{343}{108}

Full solution

The factor (n+1)(n+2)(n+1)(n+2) reminds us of a derivative. Start with the geometric series.

11−x=∑n=0∞xnddx{11−x}=ddx{∑n=0∞xn}1(1−x)2=∑n=0∞nxn−1=∑n=1∞nxn−1ddx{1(1−x)2}=ddx{∑n=1∞nxn−1}2(1−x)3=∑n=1∞n(n−1)xn−2=∑n=2∞n(n−1)xn−2=∑n=0∞(n+2)(n+1)xn\begin{align*} \frac{1}{1-x}&=\sum_{n=0}^\infty x^n\\ \diff{}{x}\left\{\frac{1}{1-x} \right\}&=\diff{}{x}\left\{ \sum_{n=0}^\infty x^n \right\}\\ \frac{1}{(1-x)^2}&=\sum_{n=0}^\infty nx^{n-1}=\sum_{n=1}^\infty nx^{n-1}\\ \diff{}{x}\left\{\frac{1}{(1-x)^2} \right\}&=\diff{}{x}\left\{ \sum_{n=1}^\infty nx^{n-1} \right\}\\ \frac{2}{(1-x)^3}&=\sum_{n=1}^\infty n(n-1)x^{n-2}=\sum_{n=2}^\infty n(n-1)x^{n-2}\\ &=\sum_{n=0}^\infty (n+2)(n+1)x^n \end{align*}

Let x=17x=\frac{1}{7}. Then ∣x∣<1|x|<1, so our series converges.

2(1−1/7)3=∑n=0∞(n+2)(n+1)(17)n2(6/7)3=∑n=0∞(n+2)(n+1)7n\begin{align*} \frac{2}{(1-1/7)^3} &=\sum_{n=0}^\infty (n+2)(n+1)\left(\frac{1}{7}\right)^n\\ \frac{2}{(6/7)^3} &=\sum_{n=0}^\infty \frac{(n+2)(n+1)}{7^n} \end{align*}
Q37Stage 3

Write the series f(x)=∑n=0∞(−1)nx2n+4(2n+1)(2n+2)f(x)=\displaystyle\sum_{n=0}^\infty\frac{(-1)^nx^{2n+4}}{(2n+1)(2n+2)} as a combination of familiar functions.

Hint

The series bears a resemblance to the Taylor series for arctangent.

Answer

∑n=0∞(−1)nx2n+4(2n+1)(2n+2)=x3arctan⁡x−x22log⁡(1+x2)\displaystyle \sum_{n=0}^\infty (-1)^n\frac{x^{2n+4}}{(2n+1)(2n+2)}=x^3\arctan x - \frac{x^2}{2}\log(1+x^2)

Full solution

Recall the Taylor series for arctangent is:

arctan⁡x=∑n=0∞(−1)nx2n+12n+1\arctan x = \sum_{n=0}^\infty (-1)^n\frac{x^{2n+1}}{2n+1}

There are similarities between this and our given series: skipping powers of xx, and a denominator that's not factorial. We'll try to manipulate it to look like our series. First, we antidifferentiate, to get a factor of (2n+2)(2n+2) on the bottom.

∫arctan⁡x dx=∑n=0∞(−1)nx2n+2(2n+1)(2n+2)+C\begin{align*}\int \arctan x\,\dee{x} &= \sum_{n=0}^\infty (-1)^n\frac{x^{2n+2}}{(2n+1)(2n+2)}+C\end{align*}

We can find the antiderivative of arctangent using integration by parts. Let u=arctan⁡xu=\arctan x and dv=dx\dee{v}=\dee{x}; then du=11+x2dx\dee{u}=\frac{1}{1+x^2}\dee{x} and v=xv=x.

∫arctan⁡x dx=xarctan⁡x−∫x1+x2 dx+C\begin{align*}\int \arctan x \,\dee{x}&=x\arctan x - \int \frac{x}{1+x^2}\,\dee{x}+C\end{align*}

Now, we use the substitution w=1+x2w=1+x^2, dw=2xdx\dee{w}=2x\dee{x}.

=xarctan⁡x−12log⁡(1+x2)+CSo, ∑n=0∞(−1)nx2n+2(2n+1)(2n+2)=xarctan⁡x−12log⁡(1+x2)+C\begin{align*}&=x\arctan x - \frac{1}{2}\log(1+x^2)+C\\ \text{So, }\qquad \sum_{n=0}^\infty (-1)^n\frac{x^{2n+2}}{(2n+1)(2n+2)}&=x\arctan x - \frac{1}{2}\log(1+x^2)+C\end{align*}

To find CC, we evaluate both sides of the equation at x=0x=0.

0=0arctan⁡0−12log⁡(1)+C=CTherefore, ∑n=0∞(−1)nx2n+2(2n+1)(2n+2)=xarctan⁡x−12log⁡(1+x2)\begin{align*}0&=0\arctan 0 -\frac{1}{2}\log(1)+C=C\\ \text{Therefore, }\qquad \sum_{n=0}^\infty (-1)^n\frac{x^{2n+2}}{(2n+1)(2n+2)}&=x\arctan x - \frac{1}{2}\log(1+x^2)\end{align*}

Multiplying both sides by x2x^2,

∑n=0∞(−1)nx2n+4(2n+1)(2n+2)=x3arctan⁡x−x22log⁡(1+x2)\begin{align*}\sum_{n=0}^\infty (-1)^n\frac{x^{2n+4}}{(2n+1)(2n+2)}&=x^3\arctan x - \frac{x^2}{2}\log(1+x^2)\end{align*}
Q38Stage 3

  1. Find the Maclaurin series for f(x)=(1−x)−1/2f(x) = (1-x)^{-1/2}. What is its radius of convergence?

  2. Manipulate the series you just found to find the Maclaurin series for g(x)=arcsin⁡xg(x)=\arcsin x. What is its radius of convergence?

Hint

For simplification purposes, note (1)(3)(5)(7)⋯(2n−1)=(2n)!2n n!\displaystyle(1)(3)(5)(7)\cdots(2n-1)=\dfrac{(2n)!}{2^n \,n!} .

Answer

(a) the Maclaurin series for f(x)f(x) is ∑n=0∞(2n)!22n (n!)2xn\displaystyle \sum_{n=0}^\infty\frac{(2n)!}{2^{2n}\,(n!)^2}x^n, and its radius of convergence is R=1R=1.

(b) the Maclaurin series for arcsin⁡x\arcsin x is ∑n=0∞(2n)!22n (n!)2(2n+1)x2n+1\displaystyle\sum_{n=0}^\infty\frac{(2n)!}{2^{2n}\,(n!)^2(2n+1)}x^{2n+1}, and its radius of convergence is R=1R=1.

Full solution

  1. We'll start, as we usually do, by finding a pattern for f(n)(0)f^{(n)}(0).

    f(x)=(1−x)−1/2f′(x)=12(1−x)−3/2f′′(x)=1⋅322(1−x)−5/2f′′′(x)=1⋅3⋅523(1−x)−7/2f(4)(x)=1⋅3⋅5⋅724(1−x)−9/2⋮f(n)(x)=1⋅3⋅5⋅…⋅(2n−1)2n(1−x)−(2n+1)/2f(n)(0)=1⋅3⋅5⋅…⋅(2n−1)2n\begin{align*}f(x)&=(1-x)^{-1/2}\\ f'(x)&=\frac{1}{2}(1-x)^{-3/2}\\ f''(x)&=\frac{1\cdot3}{2^2}(1-x)^{-5/2}\\ f'''(x)&=\frac{1\cdot3\cdot5}{2^3}(1-x)^{-7/2}\\ f^{(4)}(x)&=\frac{1\cdot3\cdot5\cdot7}{2^4}(1-x)^{-9/2}\\ \vdots&\\ f^{(n)}(x)&=\frac{1\cdot3\cdot5\cdot \ldots \cdot (2n-1)}{2^n}(1-x)^{-(2n+1)/2}\\ f^{(n)}(0)&=\frac{1\cdot3\cdot5\cdot \ldots \cdot (2n-1)}{2^n}\end{align*}

    We could leave it like this, but we simplify, to make our work cleaner later on.

    =12n⋅(2n)!2⋅4⋅6⋅…⋅(2n)=12n⋅(2n)!2n⋅n!=(2n)!22n n!\begin{align*}&=\frac{1}{2^n}\cdot\frac{(2n)!}{2\cdot4\cdot6\cdot\ldots\cdot(2n)}\\ &=\frac{1}{2^n}\cdot\frac{(2n)!}{2^n\cdot n!}\\ &=\frac{(2n)!}{2^{2n}\,n!}\end{align*}

    This pattern holds for n≥0n\ge0. Now, we can write our Maclaurin series for f(x)f(x).

    (1−x)−1/2=∑n=0∞f(n)(0)n!xn=∑n=0∞(2n)!22n (n!)2xn(1-x)^{-1/2}=\sum_{n=0}^\infty\frac{f^{(n)}(0)}{n!}x^n=\sum_{n=0}^\infty\frac{(2n)!}{2^{2n}\,(n!)^2}x^n

    To find the radius of convergence, we use the ratio test.

    ∣an+1an∣=(2n+2)!22n+2((n+1)!)2⋅22n (n!)2(2n)!⋅∣x∣=(2n+2)!(2n)!(n!(n+1)!)2⋅22n22n+2∣x∣=(2n+2)(2n+1)(1n+1)2⋅14∣x∣=4n2+4n+24n2+8n+4∣x∣lim⁡n→∞∣an+1an∣=lim⁡n→∞[4n2+4n+24n2+8n+4∣x∣]=∣x∣\begin{align*} \left|\frac{a_{n+1}}{a_n}\right| &=\frac{(2n+2)!}{2^{2n+2}((n+1)!)^2}\cdot\frac{2^{2n}\,(n!)^2}{(2n)!}\cdot|x|\\ &=\frac{(2n+2)!}{(2n)!}\left(\frac{n!}{(n+1)!} \right)^2\cdot\frac{2^{2n}}{2^{2n+2}}|x|\\ &=(2n+2)(2n+1)\left(\frac{1}{n+1} \right)^2\cdot\frac{1}{4}|x|\\ &=\frac{4n^2+4n+2}{4n^2+8n+4}|x|\\ \lim_{n \to \infty}\left|\frac{a_{n+1}}{a_n}\right|&=\lim_{n \to \infty}\left[\frac{4n^2+4n+2}{4n^2+8n+4}|x| \right]=|x| \end{align*}

    So, the radius of convergence is R=1R=1.

  2. We note the derivative of the arcsine function is 11−x2=f(x2)\dfrac{1}{\sqrt{1-x^2}}=f(x^2). With this insight, we can manipulate our Taylor series for f(x)f(x) into a Taylor series for arcsine.

    11−x=∑n=0∞(2n)!22n (n!)2xn11−x2=∑n=0∞(2n)!22n (n!)2x2n∫11−x2 dx=∫(∑n=0∞(2n)!22n (n!)2x2n) dxarcsin⁡x=∑n=0∞(2n)!22n (n!)2(2n+1)x2n+1+Carcsin⁡x=∑n=0∞(2n)!22n (n!)2(2n+1)x2n+1\begin{align*} \frac{1}{\sqrt{1-x}}&=\sum_{n=0}^\infty\frac{(2n)!}{2^{2n}\,(n!)^2}x^n\\ \frac{1}{\sqrt{1-x^2}}&=\sum_{n=0}^\infty\frac{(2n)!}{2^{2n}\,(n!)^2}x^{2n}\\ \int\frac{1}{\sqrt{1-x^2}}\,\dee{x}&=\int\left(\sum_{n=0}^\infty\frac{(2n)!}{2^{2n}\,(n!)^2}x^{2n}\right)\,\dee{x}\\ \arcsin x &=\sum_{n=0}^\infty\frac{(2n)!}{2^{2n}\,(n!)^2(2n+1)}x^{2n+1}+C\\ \arcsin x &=\sum_{n=0}^\infty\frac{(2n)!}{2^{2n}\,(n!)^2(2n+1)}x^{2n+1} \end{align*}

    where we found the value of CC by setting x=0x=0. Its radius of convergence is also 1, by Theorem 3.5.13.

Q39Stage 3Past exam · 2014A

Find the Taylor series for f(x)=log⁡(x)f(x) = \log(x) centred at a=2a = 2. Find the interval of convergence for this series.

Hint

You know the Maclaurin series for log⁡(1+y)\log(1+y). Use it! Remember that you are asked for a series expansion in powers of x−2x-2. So you want yy to be some constant times x−2x-2.

Answer

log⁡(x)=log⁡2+∑n=1∞(−1)n−1n 2n(x−2)n\log(x) =\displaystyle \log 2 +\sum\limits_{n=1}^\infty \frac{(-1)^{n-1}}{n\,2^n}(x-2)^n. It converges when 0<x≤40<x\le 4.

Full solution

We use that

log⁡(1+y)=∑n=1∞(−1)n−1ynnfor all −1<y≤1\begin{equation*} \log(1+y)= \sum_{n=1}^\infty (-1)^{n-1} \frac{y^n}{n} \qquad \text{for all } -1 < y \le 1 \end{equation*}

with y=x−22y=\dfrac{x-2}{2} to give

log⁡(x)=log⁡(2+x−2)=log⁡[2(1+x−22)]=log⁡2+log⁡(1+x−22)=log⁡2+∑n=1∞(−1)n−1n 2n(x−2)n\begin{align*} \log(x) & = \log(2 +x -2) =\log\left[2\left(1+\frac{x-2}{2}\right)\right]\\ &= \log 2 +\log\Big(1+\frac{x-2}{2}\Big) =\log 2 +\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n\,2^n}(x-2)^n \end{align*}

It converges when −1<y≤1-1<y\le 1, or equivalently, 0<x≤40<x\le 4.

Q40Stage 3Past exam · 1996A

Let I(x)=∫0x11+t4 dt\displaystyle I(x)=\int_0^x\frac{1}{1+t^4}\ \dee{t}.

  1. Find the Maclaurin series for I(x)I(x).

  2. Approximate I(1/2)I(1/2) to within ±0.0001\pm0.0001.

  3. Is your approximation in (b) larger or smaller than the true value of I(1/2)I(1/2)? Explain.

Hint

See Example 3.5.21 in the

CLP-2 text. For parts (b) and (c), review § 3.3.4 in the

CLP-2 text.

Answer

(a) ∑n=0∞(−1)n x4n+14n+1\displaystyle\sum\limits_{n=0}^\infty (-1)^n\ \frac{x^{4n+1}}{4n+1} (b) 0.4939670.493967

(c) The approximate value of part (b) is larger than the true value of I(1/2)I(1/2)

Full solution

(a) Using the geometric series expansion with r=−t4r=-t^4,

11−r=∑n=0∞rn  ⟹  11+t4=∑n=0∞(−t4)n=∑n=0∞(−1)nt4n\begin{align*} \frac{1}{1-r} =\sum_{n=0}^\infty r^n &\implies\frac{1}{1+t^4} =\sum_{n=0}^\infty {(-t^4)}^n =\sum_{n=0}^\infty (-1)^nt^{4n} \end{align*}

Substituting this into our integral,

I(x)=∫0x11+t4 dt=∫0x(∑n=0∞(−1)nt4n)dt=[∑n=0∞(−1)n t4n+14n+1]t=0t=x=∑n=0∞(−1)n x4n+14n+1\begin{align*} I(x) &=\int_0^x\frac{1}{1+t^4}\ \dee{t} \\ &=\int_0^x \left( \sum_{n=0}^\infty (-1)^n t^{4n}\right)\dee{t}\\ &=\left[\sum\limits_{n=0}^\infty (-1)^n\ \frac{t^{4n+1}}{4n+1}\right]_{t=0}^{t=x}\\ &=\sum\limits_{n=0}^\infty (-1)^n\ \frac{x^{4n+1}}{4n+1} \end{align*}

(b) Substituting in x=12x=\frac{1}{2},

I(1/2)=∑n=0∞(−1)n 1(4n+1)24n+1=12−15×25+19×29−113×213+⋯=0.5−0.00625+0.000217−0.0000094+⋯=0.493967−0.0000094+⋯\begin{align*} I(1/2)&=\sum_{n=0}^\infty (-1)^n\ \frac{1}{(4n+1)2^{4n+1}}\\ &=\frac12-\frac{1}{5\times2^5}+\frac{1}{9\times 2^9} -\frac{1}{13\times 2^{13}}+\cdots\\ &=0.5-0.00625+0.000217-0.0000094+\cdots =\boxed{0.493967}-0.0000094+\cdots \end{align*}

See part (c) for the error analysis.

(c) The series for I(x)I(x) is an alternating series (that is, the sign alternates) with successively smaller terms that converge to zero. So the error introduced by truncating the series is between zero and the first omitted term. In this case, the first omitted term was negative (−0.0000094-0.0000094). So the exact value of I(1/2)I(1/2) is the approximate value found in part (b) plus a negative number whose magnitude is smaller than 0.00001=10−50.00001=10^{-5}. So the approximate value of part (b) is larger than the true value of I(1/2)I(1/2).

Q41Stage 3Past exam · 2012A

Using a Maclaurin series, the number a=1/5−1/7+1/18a = 1/5-1/7+1/18 is found to be an approximation for I=∫01x4e−x2 dx\displaystyle I = \int_0^1 x^4 e^{-x^2}\,\dee{x}. Give the best upper bound you can for ∣I−a∣|I - a|.

Hint

Look at the signs of successive terms in the series.

Answer

166\dfrac{1}{66}

Full solution

Expanding the exponential using its Maclaurin series,

I=∫01x4e−x2 dx=∑n=0∞∫01x4(−x2)nn! dx=∑n=0∞(−1)nn!∫01x2n+4 dx=∑n=0∞(−1)nn!(2n+5)=15⏟n=0−17⏟n=1+118⏟n=2−13!(11)⏟n=3+⋯\begin{align*} I&=\int_0^1 x^4 e^{-x^2}\ \dee{x} =\sum_{n=0}^\infty\int_0^1 x^4 \frac{{(-x^2)}^n}{n!}\ \dee{x} =\sum_{n=0}^\infty \frac{(-1)^n}{n!}\int_0^1 x^{2n+4} \ \dee{x} \\ &=\sum_{n=0}^\infty \frac{(-1)^n}{n!(2n+5)} = \underbrace{\frac{1}{5}}_{n=0} -\underbrace{\frac{1}{7}}_{n=1} +\underbrace{\frac{1}{18}}_{n=2} -\underbrace{\frac{1}{3!(11)}}_{n=3} +\cdots \end{align*}

The signs of successive terms in this series alternate. Futhermore the magnitude of the nthn^{\rm th} term decreases with nn. Hence, by the alternating series test, II lies between 15−17+118\frac{1}{5}-\frac{1}{7}+\frac{1}{18} and 15−17+118−13!(11)\frac{1}{5}-\frac{1}{7}+\frac{1}{18}-\frac{1}{3!(11)}. So

∣I−a∣≤13!(11)=166\begin{equation*} |I - a| \le \frac{1}{3!(11)} =\frac{1}{66} \end{equation*}
Q42Stage 3Past exam · M121 1999A

Find an interval of length 0.00020.0002 or less that contains the number

I=∫012x2e−x2 dx\begin{equation*} I=\int_0^{\frac{1}{2}} x^2 e^{-x^2}\ \dee{x} \end{equation*}
Hint

The magic word is “series”.

Answer

Any interval of length 0.0002 that contains 0.035920.03592 and 0.036000.03600 is fine.

Full solution

Expanding the exponential using its Taylor series,

I=∫012x2e−x2 dx=∑n=0∞∫012x2(−x2)nn! dx=∑n=0∞(−1)nn!∫012x2n+2 dx=∑n=0∞(−1)nn!(2n+3)122n+3\begin{align*} I&=\int_0^{\frac{1}{2}} x^2 e^{-x^2}\ \dee{x} =\sum_{n=0}^\infty\int_0^{\frac{1}{2}} x^2 \frac{{(-x^2)}^n}{n!}\ \dee{x} =\sum_{n=0}^\infty \frac{(-1)^n}{n!}\int_0^{\frac{1}{2}} x^{2n+2} \ \dee{x} \\ &=\sum_{n=0}^\infty \frac{(-1)^n}{n!(2n+3)}\frac{1}{2^{2n+3}} \end{align*}

The signs of successive terms in this series alternate. Futhermore the magnitude of the nthn^{\rm th} term decreases with nn. Hence, by the alternating series test, II lies between ∑n=0N(−1)nn!(2n+3)122n+3\sum\limits_{n=0}^N \frac{(-1)^n}{n!(2n+3)}\frac{1}{2^{2n+3}} and ∑n=0N+1(−1)nn!(2n+3)122n+3\sum\limits_{n=0}^{N+1} \frac{(-1)^n}{n!(2n+3)}\frac{1}{2^{2n+3}}, for every NN. The first few terms are, to five decimal places,

nn0123
(−1)nn!(2n+3)122n+3\dfrac{(-1)^n}{n!(2n+3)}\frac{1}{2^{2n+3}}0.04167-0.006250.00056-0.00004

Allowing for a roundoff error of 0.0000050.000005 in each of these, II must be between

0.04167−0.00625+0.00056+0.000005×3=0.035995\begin{equation*} 0.04167-0.00625+0.00056+0.000005\times 3=0.035995 \end{equation*}

and

0.04167−0.00625+0.00056−0.00004−0.000005×4=0.035920\begin{equation*} 0.04167-0.00625+0.00056-0.00004-0.000005\times 4=0.035920 \end{equation*}

where the multiples of 0.000005 are the maximum possible accumulated roundoff errors in the added terms.

Q43Stage 3Past exam · 1996D

Let I(x)=∫0xe−t−1t dt\displaystyle I(x)=\int_0^x\frac{e^{-t}-1}{t}\,\dee{t}.

  1. Find the Maclaurin series for I(x)I(x).

  2. Approximate I(1)I(1) to within ±0.01\pm0.01.

  3. Explain why your answer to part (b) has the desired accuracy.

Hint

See Example 3.6.14 in the

CLP-2 text. For parts (b) and (c), review § 3.3.4 in the

CLP-2 text.

Answer

(a) ∑n=1∞(−1)n xnn n!\displaystyle\sum\limits_{n=1}^\infty (-1)^n\ \frac{x^n}{n\ n!} (b) −0.80-0.80 (c) See the solution.

Full solution

(a) Using the Taylor series expansion of exe^x with x=−tx=-t,

e−t=∑n=0∞(−t)nn!  ⟹  e−t−1=∑n=1∞(−1)ntnn!  ⟹  e−t−1t=∑n=1∞(−1)ntn−1n!\begin{align*} e^{-t} =\sum_{n=0}^\infty \frac{(-t)^n}{n!} &\implies e^{-t}-1 =\sum_{n=1}^\infty (-1)^n\frac{t^n}{n!} \implies \frac{e^{-t}-1}{t} =\sum_{n=1}^\infty (-1)^n\frac{t^{n-1}}{n!} \end{align*}

Substituting this into our integral,

I(x)=∫0xe−t−1t dt=∑n=1∞(−1)n∫0xtn−1n! dt=∑n=1∞(−1)n xnn⋅n!\begin{align*} I(x) =\int_0^x\frac{e^{-t}-1}{t}\,\dee{t} =\sum_{n=1}^\infty (-1)^n\int_0^x \frac{t^{n-1}}{n!}\,\dee{t} =\sum_{n=1}^\infty (-1)^n\ \frac{x^n}{n\cdot n!} \end{align*}

(b) Substituting in x=1x=1,

I(1)=∑n=1∞(−1)n 1n⋅n!=−1+12⋅2!−13⋅3!+14⋅4!−15⋅5!+⋯=−1+0.25−0.0556+0.0104−0.0017+⋯=−0.80\begin{align*} I(1)&=\sum_{n=1}^\infty (-1)^n\ \frac{1}{n\cdot n!}\\ &=-1+\frac{1}{2\cdot 2!}-\frac{1}{3\cdot 3!}+\frac{1}{4\cdot 4!}-\frac{1}{5\cdot 5!} +\cdots\\ &=-1+0.25-0.0556+0.0104-0.0017+\cdots =-0.80 \end{align*}

See part (c) for the error analysis.

(c) The series for I(x)I(x) is an alternating series (that is, the sign alternates) with successively smaller terms that converge to zero. So the error introduced by truncating the series is no larger than the first omitted term. So the magnitude of −15 5!+⋯-\frac{1}{5\,5!}+\cdots is no larger than 0.00170.0017. Allowing for a roundoff error of at most 0.00010.0001 in each of the two terms −0.0556+0.0104-0.0556+0.0104

I(1)=−1+0.25−0.0556+0.0104±0.0019=−0.7952±0.0019\begin{align*} I(1)=-1+0.25-0.0556+0.0104\pm 0.0019=-0.7952\pm 0.0019 \end{align*}
Q44Stage 3Past exam · 1997A

The function Σ(x)\Si(x) is defined by Σ(x)=∫0xsin⁡tt dt\Si(x)=\displaystyle\int_0^x\frac{\sin t}{t}\,\dee{t}.

  1. Find the Maclaurin series for Σ(x)\Si(x).

  2. It can be shown that Σ(x)\Si(x) has an absolute maximum which occurs at its smallest positive critical point (see the graph of Σ(x)\Si(x) below). Find this critical point.

  3. Use the previous information to find the maximum value of Σ(x)\Si(x) to within ±0.01\pm 0.01.

    Figure from prob_s3.6, line 1833

    Figure from prob_s3.6, line 1833

Hint

See Example 3.6.14 in the

CLP-2 text. For part (b), review the fundamental theorem of calculus in § 1.3 of the

CLP-2 text. For part (c), review § 3.3.4 in the

CLP-2 text.

Answer

(a) Σ(x)=∑n=0∞(−1)n x2n+1(2n+1)(2n+1)!\displaystyle\Si(x) =\sum\limits_{n=0}^\infty (-1)^n\ \frac{x^{2n+1}}{(2n+1)(2n+1)!} (b) x=πx=\pi (c) 1.85251.8525

Full solution

(a) Using the Taylor series expansion of sin⁡x\sin x with x=tx=t,

sin⁡t=∑n=0∞(−1)nt2n+1(2n+1)!  ⟹  sin⁡tt=∑n=0∞(−1)nt2n(2n+1)!\begin{align*} \sin t =\sum_{n=0}^\infty (-1)^n\frac{t^{2n+1}}{(2n+1)!} &\implies \frac{\sin t}{t} =\sum_{n=0}^\infty (-1)^n\frac{t^{2n}}{(2n+1)!} \end{align*}

So

Σ(x)=∫0xsin⁡tt dt=∑n=0∞(−1)n∫0xt2n(2n+1)! dt=∑n=0∞(−1)n x2n+1(2n+1)(2n+1)!\begin{align*} \Si(x) &=\int_0^x\frac{\sin t}{t}\,\dee{t} =\sum_{n=0}^\infty (-1)^n\int_0^x \frac{t^{2n}}{(2n+1)!}\,\dee{t} =\sum_{n=0}^\infty (-1)^n\ \frac{x^{2n+1}}{(2n+1)(2n+1)!} \end{align*}

(b) The critical points of Σ(x)\Si(x) are the solutions of Σ′(x)=0\Si'(x)=0. By the fundamental theorem of calculus Σ′(x)=sin⁡xx\Si'(x)=\frac{\sin x}{x}, so the critical points of Σ(x)\Si(x) are x=±π,±2π, ⋯x=\pm \pi, \pm 2\pi,\ \cdots. The absolute maximum occurs at x=πx=\pi.

(c) Substituting in x=πx=\pi,

Σ(π)=∑n=0∞(−1)n π2n+1(2n+1)(2n+1)!=π−π33⋅3!+π55⋅5!−π77⋅7!+⋯=3.1416−1.7226+0.5100−0.0856+0.0091−0.0007+⋯\begin{align*} \Si(\pi)&=\sum_{n=0}^\infty (-1)^n\ \frac{\pi^{2n+1}}{(2n+1)(2n+1)!}\\ &=\pi-\frac{\pi^3}{3\cdot 3!}+\frac{\pi^5}{5\cdot 5!}-\frac{\pi^7}{7\cdot 7!}+\cdots\\ &=3.1416-1.7226+0.5100-0.0856+0.0091-0.0007+\cdots \end{align*}

The series for Σ(π)\Si(\pi) is an alternating series (that is, the sign alternates) with successively smaller terms that converge to zero. So the error introduced by truncating the series is no larger than the first omitted term. So

Σ(π)=3.1416−1.7226+0.5100−0.0856+0.0091=1.8525\begin{align*} \Si(\pi)=3.1416-1.7226+0.5100-0.0856+0.0091=1.8525 \end{align*}

with an error of magnitude at most 0.0007+0.00050.0007+0.0005 (the 0.0005 is the maximum possible accumulated roundoff error in all five retained terms).

Q45Stage 3Past exam · 1997D

Let I(x)=∫0xcos⁡t−1t2 dt\displaystyle I(x)=\int_0^x\frac{\cos t-1}{t^2}\,\dee{t}.

  1. Find the Maclaurin series for I(x)I(x).

  2. Use this series to approximate I(1)I(1) to within ±0.01\pm0.01

  3. Is your estimate in (b) greater than I(1)I(1)? Explain.

Hint

See Example 3.6.14 in the

CLP-2 text. For parts (b) and (c), review § 3.3.4 in the

CLP-2 text.

Answer

(a) I(x)=∑n=1∞(−1)nx2n−1(2n)!(2n−1)\displaystyle I(x)=\sum\limits_{n=1}^\infty(-1)^n\frac{x^{2n-1}}{(2n)!(2n-1)} (b) I(1)=−12+14!3±16!5=−0.486±0.001\displaystyle I(1)= -\frac{1}{2}+\frac{1}{4!3} \pm \frac{1}{6!5} =-0.486\pm0.001

(c) I(1)<−12+14!3I(1)<-\dfrac{1}{2}+\dfrac{1}{4!3}

Full solution

(a) Using the Taylor series expansion of cos⁡t\cos t,

cos⁡t=1−t22!+t44!−t66!+⋯=∑n=0∞(−1)nt2n(2n)!cos⁡t−1t2=−12!+t24!−t46!+⋯=∑n=1∞(−1)nt2n−2(2n)!I(x)=∫0xcos⁡t−1t2 dt=−x2!+x34!3−x56!5+⋯=∑n=1∞(−1)nx2n−1(2n)!(2n−1)\begin{alignat*}{3} \cos t&=1-\frac{t^2}{2!}+\frac{t^4}{4!}-\frac{t^6}{6!}+\cdots &&=\sum_{n=0}^\infty(-1)^n\frac{t^{2n}}{(2n)!}\\ \frac{\cos t-1}{t^2}&= -\frac{1}{2!}+\frac{t^2}{4!}-\frac{t^4}{6!}+\cdots &&=\sum_{n=1}^\infty(-1)^n\frac{t^{2n-2}}{(2n)!}\\ I(x)=\int_0^x \frac{\cos t-1}{t^2}\,\dee{t} &=-\frac{x}{2!}+\frac{x^3}{4!3}-\frac{x^5}{6!5}+\cdots &&=\sum_{n=1}^\infty(-1)^n\frac{x^{2n-1}}{(2n)!(2n-1)} \end{alignat*}

(b), (c) Substituting in x=1x=1,

I(1)=−12+14!3−16!5+⋯=−0.5+0.0139−0.0003−⋯=−0.486±0.001\begin{align*} I(1)&=-\frac{1}{2}+\frac{1}{4!3}-\frac{1}{6!5} +\cdots\\ &=-0.5+0.0139-0.0003-\cdots\cr &=-0.486\pm0.001 \end{align*}

The series for I(1)I(1) is an alternating series with decreasing successive terms that converge to zero. So approximating I(1)I(1) by −12+14!3-\frac{1}{2}+\frac{1}{4!3} introduces an error between 00 and −16!5-\frac{1}{6!5}. Hence I(1)<−12+14!3I(1)<-\frac{1}{2}+\frac{1}{4!3}.

Q46Stage 3Past exam · 1998A

Let I(x)=∫0xcos⁡t+tsin⁡t−1t2 dt\displaystyle I(x)=\int_0^x\frac{\cos t+t\sin t-1}{t^2}\,\dee{t}

  1. Find the Maclaurin series for I(x)I(x).

  2. Use this series to approximate I(1)I(1) to within ±0.001\pm0.001

  3. Is your estimate in (b) greater than or less than I(1)I(1)?

Hint

See Example 3.6.14 in the

CLP-2 text. For parts (b) and (c), review § 3.3.4 in the

CLP-2 text.

Answer

(a) I(x)=∑n=1∞(−1)n+1x2n−1(2n)!=12!x−14!x3+16!x5−18!x8+⋯\displaystyle I(x)=\sum_{n=1}^\infty (-1)^{n+1}\frac{x^{2n-1}}{(2n)!}= \frac{1}{2!}x-\frac{1}{4!}x^3+\frac{1}{6!}x^5-\frac{1}{8!}x^8+\cdots

(b) 0.460 (c) I(1)<12!−14!+16!<0.460\displaystyle I(1)<\frac{1}{2!}-\frac{1}{4!}+\frac{1}{6!}<0.460

Full solution

(a) Using the Taylor series expansions of sin⁡x\sin x and cos⁡x\cos x with x=tx=t,

sin⁡t=∑n=0∞(−1)nt2n+1(2n+1)!=t−t33!+t55!−t77!+⋯tsin⁡t=∑n=0∞(−1)nt2n+2(2n+1)!=1− t2−t43!+t65!−t87!+⋯=−∑n=1∞(−1)nt2n(2n−1)!cos⁡t=∑n=0∞(−1)nt2n(2n)!=1−t22!+t44!−t66!+t88!+⋯cos⁡t−1=∑n=1∞(−1)nt2n(2n)!=−t22!+t44!−t66!+t88!+⋯cos⁡t+tsin⁡t−1=∑n=1∞(−1)nt2n(2n)!−∑n=1∞(−1)nt2n(2n−1)!=∑n=1∞(−1)nt2n(1(2n)!−1(2n−1)!)=(1−12!)t2−(13!−14!)t4+⋯=∑n=1∞(−1)nt2n(1(2n)!−2n(2n)!)=∑n=1∞(−1)nt2n(1−2n(2n)!)=(22!−12!)t2−(44!−14!)t4+⋯=∑n=1∞(−1)n+1t2n(2n−1(2n)!)=12!t2−34!t4+56!t6−78!t8+⋯cos⁡t+tsin⁡t−1t2=∑n=1∞(−1)n+1t2n−2(2n−1(2n)!)=12!t−34!t2+56!t4−78!t6+⋯\begin{align*} \sin t &= \sum_{n=0}^\infty (-1)^n\frac{t^{2n+1}}{(2n+1)!}&=& t-\frac{t^3}{3!}+\frac{t^5}{5!}-\frac{t^7}{7!}+\cdots\\ \color{red}t\sin t &= \sum_{n=0}^\infty (-1)^n\frac{t^{2n+2}}{(2n+1)!}&=&\color{red}\phantom{1-\ }t^2-\frac{t^4}{3!}+\frac{t^6}{5!}-\frac{t^8}{7!}+\cdots\\ &=\color{red} -\sum_{n=1}^\infty (-1)^n\frac{t^{2n}}{(2n-1)!}\\ \cos t &=\sum_{n=0}^\infty (-1)^n\frac{t^{2n}}{(2n)!}&=& 1-\frac{t^2}{2!}+\frac{t^4}{4!}-\frac{t^6}{6!}+\frac{t^8}{8!}+\cdots\\ \color{blue}\cos t-1 &=\color{blue}\sum_{n=1}^\infty (-1)^n\frac{t^{2n}}{(2n)!}&=&\color{blue} -\frac{t^2}{2!}+\frac{t^4}{4!}-\frac{t^6}{6!}+\frac{t^8}{8!}+\cdots\\ \textcolor{blue}{\cos t} + \textcolor{red}{t\sin t}\textcolor{blue}{ -1}&=\color{blue}\sum_{n=1}^\infty (-1)^n\frac{t^{2n}}{(2n)!}\color{red}-\sum_{n=1}^\infty (-1)^n\frac{t^{2n}}{(2n-1)!} \\&=\sum_{n=1}^\infty (-1)^nt^{2n}\left(\frac{1}{(2n)!}-\frac{1}{(2n-1)!}\right)&=&\Big(1-\frac{1}{2!}\Big)t^2 -\Big(\frac{1}{3!}-\frac{1}{4!}\Big)t^4 +\cdots\\ &=\sum_{n=1}^\infty (-1)^nt^{2n}\left(\frac{1}{(2n)!}-\frac{2n}{(2n)!}\right)\\ &=\sum_{n=1}^\infty (-1)^nt^{2n}\left(\frac{1-2n}{(2n)!}\right) &=&\Big(\frac{2}{2!}-\frac{1}{2!}\Big)t^2 -\Big(\frac{4}{4!}-\frac{1}{4!}\Big)t^4 +\cdots\\ &=\sum_{n=1}^\infty (-1)^{n+1}t^{2n}\left(\frac{2n-1}{(2n)!}\right) &=&\frac{1}{2!}t^2-\frac{3}{4!}t^4 +\frac{5}{6!}t^6 -\frac{7}{8!}t^8+\cdots\\ \frac{\cos t + t\sin t -1}{t^2}&=\sum_{n=1}^\infty (-1)^{n+1}t^{2n-2}\left(\frac{2n-1}{(2n)!}\right) &=&\frac{1}{2!}t-\frac{3}{4!}t^2 +\frac{5}{6!}t^4 -\frac{7}{8!}t^6+\cdots\\ \end{align*}

Now, we're ready to integrate.

I(x)=∫0x(cos⁡t+tsin⁡t−1t2)=∫0x(∑n=1∞(−1)n+1t2n−2(2n−1(2n)!))dt=[∑n=1∞(−1)n+1t2n−1(2n)!]0x=∑n=1∞(−1)n+1x2n−1(2n)!\begin{align*} I(x)=\int_0^x\left( \frac{\cos t + t\sin t -1}{t^2}\right)&=\int_0^x \left(\sum_{n=1}^\infty (-1)^{n+1}t^{2n-2}\left(\frac{2n-1}{(2n)!} \right)\right) \dee{t}\\ &= \left[\sum_{n=1}^\infty (-1)^{n+1}\frac{t^{2n-1}}{(2n)!}\right]_0^x\\ &=\sum_{n=1}^\infty (-1)^{n+1}\frac{x^{2n-1}}{(2n)!} \end{align*}

(b) I(1)=12!−14!+16!−18!+⋯=0.5−0.0416˙+0.00139−0.000024+⋯=0.460I(1)=\frac{1}{2!}-\frac{1}{4!}+\frac{1}{6!}-\frac{1}{8!}+\cdots =0.5-0.041\dot6+0.00139-0.000024+\cdots=\boxed{0.460}. The error analysis is in part (c).

(c) The series for I(1)I(1) is an alternating series with decreasing successive terms that convege to zero. So approximating I(1)I(1) by 12!−14!+16!\frac{1}{2!}-\frac{1}{4!}+\frac{1}{6!} introduces an error between 00 and −18!-\frac{1}{8!}. So I(1)<12!−14!+16!<0.460I(1)<\frac{1}{2!}-\frac{1}{4!}+\frac{1}{6!}<0.460.

Q47Stage 3Past exam · 2016A

Define f(x)=∫0x1−e−tt dt{\displaystyle f(x) = \int_0^x\frac{1-e^{-t}}{t}\ \dee{t}}.

  1. Show that the Maclaurin series for f(x)f(x) is ∑n=1∞(−1)n−1n⋅n!xn\displaystyle\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n\cdot n!} x^n.

  2. Use the ratio test to determine the values of xx for which the Maclaurin series ∑n=1∞(−1)n−1n⋅n!xn\displaystyle\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n\cdot n!} x^n converges.

Answer

(a) See the solution. (b) The series converges for all xx.

Full solution

(a) Substituting x=−tx=-t into the known power series ex=1+x+x22!+x33!+x44!+⋯e^x = 1 + x +\frac{x^2}{2!}+\frac{x^3}{3!} + \frac{x^4}{4!} + \cdots, we see that:

e−t=1−t+t22!−t33!+t44!−⋯1−e−t=t−t22!+t33!−t44!+⋯1−e−tt=1−t2!+t23!−t34!+⋯∫1−e−tt dt=C+x−x22⋅2!+x33⋅3!−x44⋅4!+⋯\begin{align*} e^{-t}&= 1 - t +\frac{t^2}{2!}-\frac{t^3}{3!} + \frac{t^4}{4!} - \cdots \\ 1-e^{-t}&= t - \frac{t^2}{2!}+ \frac{t^3}{3!} - \frac{t^4}{4!} + \cdots \\ \frac{1-e^{-t}}{t}&= 1 - \frac{t}{2!}+ \frac{t^2}{3!} - \frac{t^3}{4!} + \cdots \\ \int \frac{1-e^{-t}}{t}\ \dee{t}&= C + x - \frac{x^2}{2\cdot 2!} + \frac{x^3}{3\cdot 3!} - \frac{x^4}{4\cdot 4!} + \cdots \end{align*}

Finally, f(0)=0f(0) = 0 (since f(0)f(0) is an integral from 00 to 00) and so C=0C=0. Therefore

f(x)=∫0x1−e−tt dt=x−x22⋅2!+x33⋅3!−x44⋅4!+⋯ .\begin{align*} f(x) = \int_0^x\frac{1-e^{-t}}{t}\ \dee{t} = x - \frac{x^2}{2\cdot 2!} + \frac{x^3}{3\cdot 3!} - \frac{x^4}{4\cdot 4!} + \cdots. \end{align*}

We can also do this calculation entirely in summation notation: ex=∑n=0∞xnn!e^{x}= \displaystyle\sum\limits_{n=0}^\infty \frac{x^n}{n!}, and so

e−t=∑n=0∞(−t)nn!=1+∑n=1∞(−1)ntnn!1−e−t=−∑n=1∞(−1)ntnn!=∑n=1∞(−1)n−1tnn!1−e−tt=∑n=1∞(−1)n−1tn−1n!f(x)=∫0x1−e−tt dt=∑n=1∞(−1)n−1xnn⋅n!\begin{align*} e^{-t}&= \sum_{n=0}^\infty \frac{(-t)^n}{n!} =1+ \sum_{n=1}^\infty \frac{(-1)^nt^n}{n!} \\ 1-e^{-t}&= - \sum_{n=1}^\infty \frac{(-1)^nt^n}{n!} = \sum_{n=1}^\infty \frac{(-1)^{n-1} t^n}{n!} \\ \frac{1-e^{-t}}{t}&= \sum_{n=1}^\infty \frac{(-1)^{n-1} t^{n-1}}{n!} \\ \hskip-90pt f(x) = \int_0^x\frac{1-e^{-t}}{t}\ \dee{t} &= \sum_{n=1}^\infty \frac{(-1)^{n-1} x^n}{n\cdot n!} \end{align*}

(b) We set an=Anxn=(−1)n−1n⋅n!xna_n=A_nx^n = \dfrac{(-1)^{n-1}}{n\cdot n!} x^n and apply the ratio test.

lim⁡n→∞∣an+1an∣=lim⁡n→∞∣(−1)nxn+1/((n+1)⋅(n+1)!)(−1)n−1xn/(n⋅n!)∣=lim⁡n→∞(∣x∣n+1∣x∣nn⋅n!(n+1)⋅(n+1)!)=lim⁡n→∞(∣x∣n(n+1)2)since (n+1)!=(n+1) n!=0\begin{align*} \lim_{n\rightarrow\infty}\Big|\frac{a_{n+1}}{a_n}\Big| &=\lim_{n\rightarrow\infty} \bigg|\frac{{(-1)^{n}}x^{n+1}/{((n+1)\cdot (n+1)!)}} {{(-1)^{n-1}}x^n/{(n\cdot n!)}}\bigg| \\ &=\lim_{n\rightarrow\infty} \bigg( \frac{|x|^{n+1}}{|x|^n} \frac{n\cdot n!}{(n+1)\cdot (n+1)!} \bigg) \\ &=\lim_{n\rightarrow\infty} \bigg( |x|\frac{n}{(n+1)^2} \bigg) \qquad\text{since }(n+1)!=(n+1)\,n! \\ &=0 \end{align*}

This is smaller than 11 no matter what xx is. So the series converges for all xx.

Q48Stage 3Past exam · 1998A

Show that ∫01x3ex−1 dx≤13\displaystyle \int_0^1\frac{x^3}{e^x-1}\,\dee{x}\le\frac{1}{3}.

Hint

Use the Maclaurin series for exe^x.

Answer

See the solution.

Full solution
ex=1+x+x22!+x33!+⋯≥1+xfor all x≥0  ⟹  ex−1≥x  ⟹  x3ex−1≤x3x=x2  ⟹  ∫01x3ex−1 dx≤∫01x2 dx=13\begin{align*} &e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots \ge 1+x\qquad\hbox{for all $x\ge 0$}\\ &\implies e^x-1\ge x \\ &\implies \frac{x^3}{e^x-1}\le \frac{x^3}{x} = x^2 \\ &\implies \int_0^1\frac{x^3}{e^x-1}\,\dee{x} \le \int_0^1 x^2\,\dee{x}=\frac{1}{3} \end{align*}
Q49Stage 3Past exam · M121 2012A

Let cosh⁡(x)=ex+e−x2\displaystyle \cosh(x) =\frac{e^x+e^{-x}}{2}.

  1. Find the power series expansion of cosh⁡(x)\cosh(x) about x0=0x_0 = 0 and determine its interval of convergence.

  2. Show that 323≤cosh⁡(2)≤323+0.13\frac{2}{3}\le \cosh(2) \le 3\frac{2}{3} + 0.1.

  3. Show that cosh⁡(t)≤e12t2\cosh(t) \le e^{\frac{1}{2}t^2} for all tt.

Hint

For part (c), compare two power series term-by-term.

Answer

(a) cosh⁡(x)=∑n=0n even∞xnn!=∑n=0∞x2n(2n)!\cosh(x)=\displaystyle \sum\limits_{\genfrac{}{}{0pt}{}{n=0}{n\text{ even}}}^\infty \frac{x^n}{n!} = \sum\limits_{n=0}^\infty\frac{x^{2n}}{(2n)!} for all xx.

Full solution

(a) We know that ex=∑n=0∞xnn!e^x = \sum\limits_{n=0}^\infty \frac{x^n}{n!} for all xx. Replacing xx by −x-x, we also have e−x=∑n=0∞(−x)nn!e^{-x} = \sum\limits_{n=0}^\infty \frac{(-x)^n}{n!} for all xx and hence

cosh⁡(x)=12[ex+e−x]=12[∑n=0∞xnn!+∑n=0∞(−x)nn!]=∑n=0n even∞xnn!=∑n=0∞x2n(2n)!\begin{align*} \cosh(x) =\frac{1}{2}\big[e^x+e^{-x}\big] =\frac{1}{2}\Big[\sum_{n=0}^\infty \frac{x^n}{n!} + \sum_{n=0}^\infty \frac{(-x)^n}{n!}\Big] = \sum_{\genfrac{}{}{0pt}{}{n=0}{n\text{ even}}}^\infty\frac{x^n}{n!} = \sum_{n=0}^\infty\frac{x^{2n}}{(2n)!} \end{align*}

for all xx. In particular, the interval of convergence is all real numbers.

(b) Using the power series expansion of part (a),

cosh⁡(2)=1+222!+244!+∑n=3∞22n(2n)!=323+∑n=3∞22n(2n)!\begin{align*} \cosh(2) &= 1 +\frac{2^2}{2!}+\frac{2^4}{4!} +\sum_{n=3}^\infty\frac{2^{2n}}{(2n)!} = 3\frac{2}{3} +\sum_{n=3}^\infty\frac{2^{2n}}{(2n)!} \end{align*}

So it suffices to show that ∑n=3∞22n(2n)!≤0.1\sum_{n=3}^\infty\frac{2^{2n}}{(2n)!}\le 0.1. Let's write bn=22n(2n)!b_n = \frac{2^{2n}}{(2n)!}. The first term in ∑n=3∞22n(2n)!\sum_{n=3}^\infty\frac{2^{2n}}{(2n)!} is

b3=266!=266×5×4×3×2=445\begin{equation*} b_3 = \frac{2^6}{6!} = \frac{2^6}{6 \times 5 \times 4 \times 3\times 2} = \frac{4}{45} \end{equation*}

The ratio between successive terms in ∑n=3∞22n(2n)!\sum_{n=3}^\infty\frac{2^{2n}}{(2n)!} is

bn+1bn=22n+2/22n(2n+2)!/(2n)!=4(2n+2)(2n+1)≤48×7=114for all n≥3\begin{align*} \frac{b_{n+1}}{b_n} = \frac{2^{2n+2}/2^{2n}}{(2n+2)!/(2n)!} =\frac{4}{(2n+2)(2n+1)} \le \frac{4}{8\times 7} =\frac{1}{14}\qquad \text{for all }n\ge 3 \end{align*}

Hence

∑n=3∞22n(2n)!≤445⏞b3+445×114⏞b4≤+445×1142⏞b5≤+445×1143⏞b6≤+⋯=445 11−114=445 1413=56585<110\begin{align*} \sum_{n=3}^\infty\frac{2^{2n}}{(2n)!} &\le \overbrace{\frac{4}{45}}^{b_3} + \overbrace{\frac{4}{45}\times \frac{1}{14}}^{b_4\le} + \overbrace{\frac{4}{45}\times \frac{1}{14^2}}^{b_5\le} + \overbrace{\frac{4}{45}\times \frac{1}{14^3}}^{b_6\le} + \cdots\\ &=\frac{4}{45}\,\frac{1}{1-\frac{1}{14}} =\frac{4}{45}\,\frac{14}{13} =\frac{56}{585}<\frac{1}{10} \end{align*}

(c) Comparing

cosh⁡(t)=∑n=0∞t2n(2n)!=∑n=0∞(t2)n(2n)!ande12t2=∑n=0∞(12t2)nn!=∑n=0∞(t2)n2nn!\begin{equation*} \cosh(t) = \sum_{n=0}^\infty\frac{t^{2n}}{(2n)!} = \sum_{n=0}^\infty\frac{{(t^2)}^n}{(2n)!} \qquad\text{and}\qquad e^{\frac{1}{2}t^2} = \sum_{n=0}^\infty\frac{{(\frac{1}{2}t^2)}^n}{n!} = \sum_{n=0}^\infty\frac{{(t^2)}^n}{2^n n!} \end{equation*}

we see that it suffices to show that (2n)!≥2nn!(2n)! \ge 2^n n!. Now. for all n≥1n\ge 1,

(2n)!=1×2×⋯×n⏞n factors(n+1)×(n+2)×⋯×2n⏞n factors≥1×2×⋯×n⏞n factors2×2×⋯×2⏞n factors=2n n!\begin{align*} (2n)! &=\overbrace{1\times 2\times\cdots\times n}^{n\text{ factors}} \overbrace{(n+1)\times (n+2)\times\cdots\times 2n}^{n\text{ factors}} \\ &\ge\overbrace{1\times 2\times\cdots\times n}^{n\text{ factors}} \overbrace{2\times 2\times\cdots\times 2}^{n\text{ factors}} \\ &=2^n\, n! \end{align*}
Q50Stage 3

The law of the instrument says “If you have a hammer then everything looks like a nail” — it is really a description of the “tendency of jobs to be adapted to tools rather than adapting tools to jobs.” (Quote from Silvan Tomkins's Computer Simulation of Personality: Frontier of Psychological Theory. See also Birmingham screwdrivers.) Anyway, this is a long way of saying that just because we know how to compute things using Taylor series doesn't mean we should neglect other techniques.

  1. Using Newton's method, approximate the constant 23\sqrt[3]{2} as a root of the function g(x)=x3−2g(x)=x^3-2. Using a calculator, make your estimation accurate to within 0.01.

  2. You may assume without proof that

    x3=1+16(x−1)+∑n=2∞(−1)n−1(2)(5)(8)⋯(3n−4)3n n!(x−1)n.\sqrt[3]{x}=1+\frac{1}{6}(x-1)+\sum_{n=2}^\infty(-1)^{n-1}\frac{(2)(5)(8)\cdots(3n-4)}{3^n\, n!}(x-1)^n.

    for all real numbers xx. Using the fact that this is an alternating series, how many terms would you have to add for the partial sum to estimate 23\sqrt[3]{2} with an error less than 0.01?

Hint

For Newton's method, recall we approximate a root of the function g(x)g(x) in iterations: given an approximation xnx_n, our next approximation is xn+1=xn−g(xn)g′(xn)x_{n+1}=x_n-\dfrac{g(x_n)}{g'(x_n)}.

To gauge your error, note that from approximation to approximation, the first digits stabilize. Keep refining your approximation until the first two digits stop changing.

Answer

(a) 33≈1.26\sqrt[3]{3}\approx 1.26 (b) 12 terms (S11S_{11})

Full solution

  1. For Newton's method, recall we approximate a root of the function g(x)g(x) in iterations: given an approximation xnx_n, our next approximation is xn+1=xn−g(xn)g′(xn)x_{n+1}=x_n-\dfrac{g(x_n)}{g'(x_n)}. In our case,

    xn+1=xn−xn3−23xn2=23(xn+1xn2).x_{n+1}=x_n - \frac{x_n^3-2}{3x_n^2}=\frac23\left(x_n+\frac{1}{x_n^2}\right).

    We want to start somewhere reasonably close to the actual root we want, so let's set x0=1x_0=1. (Your starting point may vary.)

    x0=1  ⟹  x1=23(1+11)=43≈1.3333x1=43  ⟹  x2=23(43+916)=9172≈1.2639x2=9172  ⟹  x3=23(9172+722912)=1126819894348≈1.2599x3=1126819894348  ⟹  x4=23(1126819894348+894348211268192)≈1.2599\begin{align*} x_0&=1 &&\implies x_1=\dfrac23\left(1+\frac11\right)=\dfrac{4}{3}&\approx&1.3333\\ x_1&=\dfrac43 &&\implies x_2=\dfrac23\left(\frac43+\frac{9}{16}\right)=\dfrac{91}{72}&\approx&1.2639\\ x_2&=\dfrac{91}{72} &&\implies x_3=\dfrac23\left(\dfrac{91}{72}+\dfrac{72^2}{91^2}\right)=\dfrac{1126819 }{894348 }&\approx&1.2599\\ x_3&=\dfrac{1126819 }{894348 } &&\implies x_4=\dfrac23\left(\dfrac{1126819 }{894348 }+\dfrac{894348^2}{1126819^2}\right)&\approx&1.2599 \end{align*}

    So, 23≈1.26\sqrt[3]{2}\approx 1.26.

  2. We'll evaluate the given series at x=2x=2. This yields the series

    23=1+16+∑n=2∞(−1)n−1(2)(5)(8)⋯(3n−4)3n n!.\sqrt[3]{2}=1+\frac{1}{6}+\sum_{n=2}^\infty(-1)^{n-1}\frac{(2)(5)(8)\cdots(3n-4)}{3^n\, n!}.

    This series is alternating, so if we use the partial sum SNS_N, our absolute error is at most

    ∣aN+1∣=(2)(5)(8)⋯(3N−1)3N+1 (N+1)!|a_{N+1}|=\frac{(2)(5)(8)\cdots(3N-1)}{3^{N+1}\, (N+1)!}

    (if N≥2N \ge 2). We want to know which value of NN makes this at most 0.01. We test several values.

    | m1cm | m10cm | NN∣aN+1∣|a_{N+1}|
    [1em] 3(2)(5)(8)34⋅4!≈0.04\dfrac{(2)(5)(8)}{3^4\cdot 4!}\approx 0.04
    [1em] 4(2)(5)(8)(11)35 5!≈0.03\dfrac{(2)(5)(8)(11)}{3^5\, 5!}\approx 0.03
    [1em] 5(2)(5)(8)(11)(14)36 6!≈0.023\dfrac{(2)(5)(8)(11)(14)}{3^6\, 6!}\approx 0.023
    [1em] 6(2)(5)(8)(11)(14)(17)37 7!≈0.019\dfrac{(2)(5)(8)(11)(14)(17)}{3^7\, 7!}\approx 0.019
    [1em] 7(2)(5)(8)(11)(14)(17)(20)38 8!≈0.016\dfrac{(2)(5)(8)(11)(14)(17)(20)}{3^8\, 8!}\approx 0.016
    [1em] 8(2)(5)(8)(11)(14)(17)(20)(23)39 9!≈0.013\dfrac{(2)(5)(8)(11)(14)(17)(20)(23)}{3^9\, 9!}\approx 0.013
    [1em] 9(2)(5)(8)(11)(14)(17)(20)(23)(26)310 10!≈0.012\dfrac{(2)(5)(8)(11)(14)(17)(20)(23)(26)}{3^{10}\, 10!}\approx 0.012
    [1em] 10(2)(5)(8)(11)(14)(17)(20)(23)(26)(29)311 11!≈0.0103\dfrac{(2)(5)(8)(11)(14)(17)(20)(23)(26)(29)}{3^{11}\, 11!}\approx 0.0103
    [1em] 11(2)(5)(8)(11)(14)(17)(20)(23)(26)(29)(32)312 12!≈0.009\dfrac{(2)(5)(8)(11)(14)(17)(20)(23)(26)(29)(32)}{3^{12}\, 12!}\approx 0.009
    [1em]

    So, the approximation S11S_{11} has a sufficiently small error. That is, we would add up the first twelve terms.

Q51Stage 3

Let f(x)=arctan⁡(x3)f(x)=\arctan(x^3). Write f(10)(15)f^{(10)}\left(\frac{1}{5} \right) as a sum of rational numbers with an error less than 10−610^{-6} using the Maclaurin series for arctangent.

Hint

First, modify your known Maclaurin series for arctangent into a Maclaurin series for f(x)f(x). This series is not hard to repeatedly differentiate, so use it to find a power series for f(10)(x)f^{(10)}(x).

Answer

15!5!⋅56−21!7!⋅11!⋅511+27!9!⋅17!⋅517−33!11!⋅23!⋅523\displaystyle \frac{15!}{5!\cdot5^6}-\frac{21!}{7!\cdot 11!\cdot 5^{11}}+\frac{27!}{9!\cdot 17!\cdot 5^{17}}-\frac{33!}{11!\cdot 23!\cdot 5^{23}}

Full solution

Our plan is as follows:

  • Make a Taylor series for f(x)f(x)

  • Calculate the tenth derivative of the Taylor series of f(x)f(x).

  • Decide how many terms we need to add to achieve the desired accuracy.

  • Approximate f(10)(15)f^{(10)}\left(\frac15\right) with a partial sum.

We know that the Taylor series for arctan⁡x\arctan x is ∑n=0∞(−1)nx2n+12n+1\displaystyle\sum_{n=0}^\infty (-1)^n\frac{x^{2n+1}}{2n+1}, which converges for −1≤x≤1-1\le x\le1. So, the Taylor series for arctan⁡(x3)\arctan(x^3) is

f(x)=arctan⁡(x3)=∑n=0∞(−1)n(x3)2n+12n+1=∑n=0∞(−1)nx6n+32n+1f(x)=\arctan(x^3)=\sum_{n=0}^\infty (-1)^n\frac{(x^3)^{2n+1}}{2n+1} =\sum_{n=0}^\infty (-1)^n\frac{x^{6n+3}}{2n+1}

It is much easier to differentiate this series many times than it is to differentiate arctan⁡(x3)\arctan(x^3) directly many times.

f′(x)=∑n=0∞(−1)n(6n+3)x6n+22n+1f′′(x)=∑n=0∞(−1)n(6n+3)(6n+2)x6n+12n+1f′′′(x)=∑n=0∞(−1)n(6n+3)(6n+2)(6n+1)x6n2n+1⋮f(10)(x)=∑n=0∞(−1)n(6n+3)(6n+2)(6n+1)⋯(6n−6)x6n−72n+1=∑n=2∞(−1)n(6n+3)(6n+2)(6n+1)⋯(6n−6)x6n−72n+1=∑n=2∞(−1)n(6n+3)!(2n+1)(6n−7)!x6n−7f(10)(15)=∑n=2∞(−1)n(6n+3)!(2n+1)(6n−7)!⋅56n−7\begin{align*} f'(x)&=\sum_{n=0}^\infty (-1)^n\frac{(6n+3)x^{6n+2}}{2n+1}\\ f''(x)&=\sum_{n=0}^\infty (-1)^n\frac{(6n+3)(6n+2)x^{6n+1}}{2n+1}\\ f'''(x)&=\sum_{n=0}^\infty (-1)^n\frac{(6n+3)(6n+2)(6n+1)x^{6n}}{2n+1}\\ \vdots&\\ f^{(10)}(x)&=\sum_{n=0}^\infty (-1)^n\frac{(6n+3)(6n+2)(6n+1)\cdots(6n-6)x^{6n-7}}{2n+1}\\ &=\sum_{n=2}^\infty (-1)^n\frac{(6n+3)(6n+2)(6n+1)\cdots(6n-6)x^{6n-7}}{2n+1}\\ &=\sum_{n=2}^\infty (-1)^n\frac{(6n+3)!}{(2n+1)(6n-7)!}x^{6n-7}\\ f^{(10)}\left(\frac{1}{5}\right)&=\sum_{n=2}^\infty (-1)^n\frac{(6n+3)!}{(2n+1)(6n-7)!\cdot 5^{6n-7}} \end{align*}

(Notice, after ten differentiations, the terms a0a_0 and a1a_1 are both zero.)

Since this is an alternating series, the absolute error involved in using the approximation SNS_N is at most

∣aN+1∣=(6N+9)!(2N+3)(6N−1)!⋅56N−1|a_{N+1}|=\frac{(6N+9)!}{(2N+3)(6N-1)!\cdot 5^{6N-1}}

By testing a few values of NN, we find

∣a6∣=∣a5+1∣=39!(13)(29!)⋅529≈0.00000095<10−6|a_6|=|a_{5+1}|=\frac{39!}{(13)(29!)\cdot 5^{29}}\approx 0.00000095<10^{-6}

So, S5S_5 is a sufficient approximation. That is,

f(10)(15)≈∑n=25(−1)n(6n+3)!(2n+1)(6n−7)!⋅56n−7=(−1)215!5⋅5!⋅55+(−1)321!7!⋅11!⋅511+(−1)427!9!⋅17!⋅517+(−1)533!11!⋅23!⋅523=15!5!⋅56−21!7!⋅11!⋅511+27!9!⋅17!⋅517−33!11!⋅23!⋅523\begin{align*} f^{(10)}\left(\frac15\right)&\approx \sum_{n=2}^5 (-1)^n\frac{(6n+3)!}{(2n+1)(6n-7)!\cdot 5^{6n-7}}\\ &=(-1)^2\frac{15!}{5\cdot 5!\cdot 5^{5}}+(-1)^3\frac{21!}{7!\cdot 11!\cdot 5^{11}}+ (-1)^4\frac{27!}{9!\cdot 17!\cdot 5^{17}}+ (-1)^5\frac{33!}{11!\cdot 23!\cdot 5^{23}}\\ &=\frac{15!}{5!\cdot5^6}-\frac{21!}{7!\cdot 11!\cdot 5^{11}}+\frac{27!}{9!\cdot 17!\cdot 5^{17}}-\frac{33!}{11!\cdot 23!\cdot 5^{23}} \end{align*}

Remark: if we had calculated f(10)(1/5)f^{(10)}(1/5) directly, using derivative rules instead of series, we would have found an exact value; however, our value here is easier to find, and is highly accurate (if not exact).

Q52Stage 3

Consider the following function:

f(x)={e−1/x2x≠00x=0f(x)=\begin{cases} e^{-1/x^2} & x \neq 0\\ 0 & x=0 \end{cases}

  1. Sketch y=f(x)y=f(x).

  2. Assume (without proof) that f(n)(0)=0f^{(n)}(0)=0 for all whole numbers nn. Find the Maclaurin series for f(x)f(x).

  3. Where does the Maclaurin series for f(x)f(x) converge?

  4. For which values of xx is f(x)f(x) equal to its Maclaurin series?

Hint

Remember exe^x is never negative for any real number xx.

Answer

(a)

Figure from prob_s3.6, line 2

Figure from prob_s3.6, line 2

(b) the constant function 0 (c) everywhere (d) only at x=0x=0

Full solution

  1. To sketch y=f(x)y=f(x), we note the following:

    • f(x)f(x) is never negative.

    • lim⁡x→±∞f(x)=e0=1\lim\limits_{x \to \pm \infty} f(x)=e^0=1, so the curve has horizontal asymptotes in both directions at y=1y=1.

    • lim⁡x→±0f(x)=lim⁡x→±01e1/x2=lim⁡u→+∞1eu=0=f(0)\lim\limits_{x \to \pm 0} f(x)=\lim\limits_{x \to \pm 0} \frac{1}{e^{1/x^2}}=\lim\limits_{u \to +\infty}\frac{1}{e^u}=0=f(0), so the curve is continuous at x=0x=0.

    • For x≠0x\neq 0, f′(x)=2x3e−1/x2f'(x)=\frac{2}{x^3}e^{-1/x^2}, so our curve is decreasing on (−∞,0)(-\infty,0) and increasing on (0,∞)(0,\infty)

    • For x≠0x\neq 0, f′′(x)=2x−6(2−3x2)e−1/x2f''(x)=2x^{-6}(2-3x^2)e^{-1/x^2}, so our curve is concave up on (−2/3,2/3)(-\sqrt{2/3},\sqrt{2/3}), and concave down elsewhere.

    Figure from prob_s3.6, line 2

    Figure from prob_s3.6, line 2

  2. Since f(n)(0)=0f^{(n)}(0)=0 for all whole nn (that is, the graph is really quite flat at the origin), and since f(0)=0f(0)=0, the Maclaurin series for f(x)f(x) is ∑n=0∞0n!xn=0\displaystyle\sum_{n=0}^\infty \frac{0}{n!}x^n=0.

  3. The Maclaurin series converges for all real values of xx (to the constant 0).

  4. Since ey>0e^y>0 for any real yy, we see f(x)=0f(x)=0 only when x=0x=0. So, f(x)f(x) is only equal to its Maclaurin series at the single point x=0x=0.

Remark: the function f(x)f(x) is an example of a function whose Maclaurin series converges, but not to f(x)f(x)! To describe this behaviour, we say f(x)f(x) is non-analytic.

Q53Stage 3

Suppose f(x)f(x) is an odd function, and f(x)=∑n=0∞f(n)(0)n!xnf(x)=\displaystyle\sum_{n=0}^\infty\frac{f^{(n)}(0)}{n!}x^n. Simplify ∑n=0∞f(2n)(0)(2n)!x2n\displaystyle\sum_{n=0}^\infty \dfrac{f^{(2n)}(0)}{(2n)!}x^{2n}.

Hint

Since f(x)f(x) is odd, f(−x)=−f(x)f(-x)=-f(x) for all xx in its domain. Consider the even-indexed terms and odd-indexed terms of the Taylor series.

Answer

00

Full solution

  • Since f(x)f(x) is odd, f(−x)=−f(x)f(-x)=-f(x) for all xx in its domain. We plug this into our power series, then consider the even-indexed terms and the odd-indexed terms separately.

    f(−x)=−f(x)∑n=0∞f(n)(0)n!(−x)n=−∑n=0∞f(n)(0)n!xn∑n=0∞f(2n+1)(0)(2n+1)!(−x)2n+1+∑n=0∞f(2n)(0)(2n)!(−x)2n=−∑n=0∞f(2n+1)(0)(2n+1)!x2n+1−∑n=0∞f(2n)(0)(2n)!x2n−∑n=0∞f(2n+1)(0)(2n+1)!x2n+1+∑n=0∞f(2n)(0)(2n)!x2n=−∑n=0∞f(2n+1)(0)(2n+1)!x2n+1−∑n=0∞f(2n)(0)(2n)!x2n∑n=0∞f(2n)(0)(2n)!x2n=−∑n=0∞f(2n)(0)(2n)!x2n2∑n=0∞f(2n)(0)(2n)!x2n=0∑n=0∞f(2n)(0)(2n)!x2n=0\begin{align*} f(-x)&=-f(x)\\ \sum_{n=0}^\infty\frac{f^{(n)}(0)}{n!}(-x)^n&=-\sum_{n=0}^\infty\frac{f^{(n)}(0)}{n!}x^n\\ \textcolor{red}{\sum_{n=0}^\infty\frac{f^{(2n+1)}(0)}{(2n+1)!}(-x)^{2n+1}}+ \textcolor{blue}{\sum_{n=0}^\infty\frac{f^{(2n)}(0)}{(2n)!}(-x)^{2n}}&=\textcolor{red}{-\sum_{n=0}^\infty\frac{f^{(2n+1)}(0)}{(2n+1)!}x^{2n+1}}\textcolor{blue}{-\sum_{n=0}^\infty\frac{f^{(2n)}(0)}{(2n)!}x^{2n}}\\ \textcolor{red}{-\sum_{n=0}^\infty\frac{f^{(2n+1)}(0)}{(2n+1)!}x^{2n+1}}+ \textcolor{blue}{\sum_{n=0}^\infty\frac{f^{(2n)}(0)}{(2n)!}x^{2n}}&=\textcolor{red}{-\sum_{n=0}^\infty\frac{f^{(2n+1)}(0)}{(2n+1)!}x^{2n+1}}\textcolor{blue}{-\sum_{n=0}^\infty\frac{f^{(2n)}(0)}{(2n)!}x^{2n}}\\ \textcolor{blue}{\sum_{n=0}^\infty\frac{f^{(2n)}(0)}{(2n)!}x^{2n}}&=\textcolor{blue}{-\sum_{n=0}^\infty\frac{f^{(2n)}(0)}{(2n)!}x^{2n}}\\ 2\textcolor{blue}{\sum_{n=0}^\infty\frac{f^{(2n)}(0)}{(2n)!}x^{2n}}&=0\\ \textcolor{blue}{\sum_{n=0}^\infty\frac{f^{(2n)}(0)}{(2n)!}x^{2n}}&=0 \end{align*}

  • Alternately, we could note the following:

    • Since all derivative of f(x)f(x) exist, all its derivatives are continuous.

    • The derivative of an odd function is even, and the derivative of an even function is odd.

    • So, the even-indexed derivatives of f(x)f(x) are continuous, odd functions.

    • Every continuous, odd function passes through the origin. That is, f(2n)(0)=0f^{(2n)}(0)=0.

    • So, every term in the series is 00.

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From the UBC Math 101 open textbook project, © Joel Feldman, Andrew Rechnitzer, Elyse Yeager and others. Licensed CC BY-NC-SA 4.0; this HTML adaptation is released under the same licence. Full attribution.