9.5 Some examples

The derivatives of $f(x)$ repeat themselves.

If we were to find the 16th-order Maclaurin polynomial for a generic function, we might expect to have to differentiate 16 times (ugh). But, we know that the derivatives of sines and cosines repeat themselves. So, it's enough to figure out the pattern:

Since $f^{(4)}(x)=f(x)$, our derivatives repeat from here. They follow the pattern:
$\textcolor{blue}{+1}$, $\textcolor{blue}{+1}$, $\textcolor{red}{-1}$, $\textcolor{red}{-1}$.

You are approximating a polynomial with a polynomial.

$T_{100}(t)=127.5+48(t-5)+4.9(t-5)^2=4.9t^2-t+10$

A Taylor polynomial gives a polynomial approximation for a function $s(t)$. Since $s(t)$ is itself a polynomial, any $n$th-order Taylor polynomial, with $n$ greater than or equal to the degree of $s(t)$, will simply give $s(t)$. So, in our case, $T_{100}(t)=s(t)=4.9t^2-t+10$. If that isn't satisfying, you can go through the normal method of calculating $T_{100}(t)$.

The rest of the derivatives of $s(t)$ are identically zero, so they are (in particular) zero when $t=5$. Therefore,

We can now check that $T_{100}(t)$ really is the same as $s(t)$.

as expected.

Recall $\ds\diff{}{x}\left\{2^x\right\}=2^x\log 2$, where $\log 2$ is the constant $\log_e2$.

$T_n(x)=\ds\sum_{k=0}^n \frac{2(\log 2)^k}{k!}(x-1)^k$

Let's start by differentiating $f(x)$ and looking for a pattern. Remember that $\log 2 = \log_e 2$ is a constant number.

So, in general,

We notice that this formula works even when $k=0$ and $k=1$. When $x=1$,

The $n$th-order Taylor polynomial of $f(x)$ about $x=1$ is

Just keep differentiating–it gets easier!

$T_6(x)=7+5(x-1)+\frac{7}{2}(x-1)^2+\frac{1}{3}(x-1)^3-\frac{1}{12}(x-1)^4+\frac{1}{30}(x-1)^5-\frac{1}{60}(x-1)^6$

We need to know the first six derivatives of $f(x)$ at $x=1$. Let's get started.

Now, we can plug in.

Start by differentiating, and finding the pattern for $f^{(k)}(0)$. Remember the chain rule!

$T_n(x)=\ds\sum_{k=0}^n x^k$

We'll start by differentiating and looking for a pattern.

Using the chain rule,

Recognizing the pattern,

The $n$th-order Maclaurin polynomial for $f(x)$ is

Manipulate a known polynomial, rather than working from the definition.

$x^4-\frac{x^{12}}{3!}+\frac{x^{20}}{5!}$

The 5th-order Maclaurin polynomial for $\sin(x)$ is $T_5(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}$, so the 20th-order Maclaurin polynomial for $\sin(x^4)$ is $T_5(x^4)=x^4-\frac{x^{12}}{3!}+\frac{x^{20}}{5!}$. (We use here Theorem 9.5.8 from the textbook.)

Manipulate a known polynomial, rather than working from the definition.

$1-x^2+x^4$

Let $f(x)=\frac{1}{1+x^2}$ and $g(x)=\frac{1}{1-x}$. To use Theorem 9.5.8 from the textbook, we note $\frac{1}{1+x^2}=\frac{1}{1-[-x^2]}$. That is, $f(x)=g(-x^2)$.

(We note here that the coefficient of $x^5$ is 0.)

The lowest-order terms in the numerator cancel out.

$\frac13$

For $x$ close to 0, lower powers dominate higher powers.

Then,

You'll need to differentiate $x^x$. This is accomplished using logarithmic differentiation, covered in Section 4.4.

$T_3(x)=1+(x-1)+(x-1)^2+\frac{1}{2}(x-1)^3$

We'll need to know the first three derivatives of $x^x$ at $x=1$. This is a good review of logarithmic differentiation, covered in Section 4.4.

Now, we can plug in:

What is $6\arctan \left(\dfrac{1}{\sqrt{3}}\right)$?

$\pi=6\arctan\left(\dfrac{1}{\sqrt3}\right)\approx\dfrac{82}{45}\sqrt{3}\approx3.156$

We note that $6\arctan\left(\dfrac{1}{\sqrt3}\right)=6\left(\dfrac{\pi}{6}\right)=\pi$. We will find the 5th-order Maclaurin polynomial $T_5(x)$ for $f(x)=6\arctan x$. Then $\pi=f\left(\dfrac{1}{\sqrt{3}}\right) \approx T_5\left(\dfrac{1}{\sqrt{3}}\right)$. Let's begin by finding the first five derivatives of $f(x)=6\arctan x$.

We now use these values to compute the 5th-order Maclaurin polynomial for $f(x)$.

Now, if we want to approximate $f\left(\dfrac{1}{\sqrt{3}}\right)=6\arctan\left(\dfrac{1}{\sqrt{3}}\right)=\pi$:

Remark: There are numerous methods for computing $\pi$ to any desired degree of accuracy. Many of them use the Maclaurin expansion of $\arctan x$. In 1706 John Machin computed $\pi$ to 100 decimal digits by using the Maclaurin expansion together with
$\pi = 16 \arctan\frac{1}{5} -4\arctan\frac{1}{239}$.

After a few derivatives, this will be very similar to Example 9.5.2.

$T_{100}(x)=-1+\ds\sum_{k=2}^{100}\frac{(-1)^k}{k(k-1)}(x-1)^k$

Let's start by differentiating, and looking for a pattern.

When $k \geq 2$, making use of the fact that $0!=1$ and $(-1)^{k-2}=(-1)^k$:

Now we use the standard form of a Taylor polynomial. Since the first two terms don't fit the pattern, we add those outside of the sigma.

Treat the even and odd powers separately.

$T_{2n}(x)=\sum_{\ell=0}^{n}\frac{(-1)^\ell}{(2\ell)!\sqrt{2}}\left(x-\frac{\pi}{4}\right)^{2\ell} +\sum_{\ell=0}^{n-1}\frac{(-1)^\ell}{(2\ell+1)!\sqrt{2}}\left(x-\frac{\pi}{4}\right)^{2\ell+1}$

Recall that

Let's start by taking some derivatives. Of course, since we're differentiating sine, the derivatives will repeat every four iterations.

So, the pattern of derivatives is $\dfrac{1}{\sqrt{2}}$, $\dfrac{1}{\sqrt{2}}$, $-\dfrac{1}{\sqrt{2}}$, $-\dfrac{1}{\sqrt{2}}$, $\dfrac{1}{\sqrt{2}}$, $\dfrac{1}{\sqrt{2}}$, $-\dfrac{1}{\sqrt{2}}$, $-\dfrac{1}{\sqrt{2}}$, etc. This is a little tricky to write in sigma notation. We can deal with the “doubles" by separating the even and odd powers. The first few terms of $T_{2n}$ that contain even powers of $\left(x-\frac{\pi}{4}\right)$ are

Observe that the signs alternate between successive terms. So if we rename $k$ to $2\ell$ these terms are

and the $\ell^{\rm th}$ term here is $\frac{(-1)^\ell}{(2\ell)!\sqrt{2}}\left(x-\frac{\pi}{4}\right)^{2\ell}$. To verify that this really is the $\ell^{\rm th}$ term, evaluate this for $\ell=0,1,2$ explicitly. When $k=2n$, $\ell=n$ so that

Now for the odd powers. The first few terms of $T_{2n}$ that contain odd powers of $\left(x-\frac{\pi}{4}\right)$ are