9.6 Error in Taylor polynomials

$R(10)=f(10)-F(10)=-3-5$

(a) False (b) True (c) True (d) True

Equation 9.6.5 tells us

for some $c$ strictly between 0 and 2.

Equation 9.6.5 gives us the bound $|f(2)-T_3(2)|<6$. A calculator tells us actually $|f(2)-T_3(2)|\approx 1.056$.

You are approximating a third-degree polynomial with a fifth-degree Taylor polynomial. You should be able to tell how good your approximation will be without a long calculation.

$|f(37)-T(37)|=0$

Draw a picture–it should be clear how the two approximations behave.

You do, you clever goose!

In this case, Equation 9.6.5 tells us that

for some $c$ strictly between 11 and 11.5.

$|f(11.5)-T_5(11.5)|<\dfrac{9}{7\cdot 2^6}<0.02$

In this case, Equation 9.6.5 tells us that $\left|f(0.1)-T_2(0.1)\right| = \left|\dfrac{f^{(3)}(c)}{3!}(0.1-0)^{3}\right|$ for some $c$ strictly between 0 and 0.1.

$\left|f(0.1)-T_2(0.1)\right| < \dfrac{1}{1125}$

In our case, Equation 9.6.5 tells us

for some $c$ between $-\dfrac{1}{4}$ and 0.

$\left|f\left(-\dfrac{1}{4}\right)-T_5\left(-\dfrac{1}{4}\right)\right| < \dfrac{1}{6\cdot 4^6}<0.00004$

In this case, Equation 9.6.5 tells us that $\left|f(30)-T_3(30)\right| = \left|\dfrac{f^{(4)}(c)}{4!}(30-32)^{4}\right|$ for some $c$ strictly between 30 and 32.

Your answer may vary. One reasonable answer is
$\left|f(30)-T_3(30)\right|<\dfrac{14}{5^7\cdot9\cdot15}<0.000002$. Another reasonable answer is
$\left|f(30)-T_3(30)\right|<\dfrac{14}{5^7\cdot9}<0.00002$.

In our case, Equation 9.6.5 tells us
$\left|f\left(0.01\right)-T_1\left(0.01\right)\right| = \left|\dfrac{f^{(2)}(c)}{2!}\left(0.01-\frac{1}{\pi}\right)^2\right|$ for some $c$ between 0.01 and $\dfrac{1}{\pi}$.

Equation 9.6.5 gives the bound $|f(0.01)-T_n(0.01)| \leq 100^2\left(\frac{100}{\pi}-1\right)^2$.

A more reasonable bound on the error is that it is less than 5.

Using Equation 9.6.5, $\left|f\left(\dfrac{1}{2}\right)-T_2\left(\dfrac{1}{2}\right)\right| = \left|\dfrac{f^{(3)}(c)}{3!}\left(\dfrac{1}{2}-0\right)^{3}\right|$ for some $c$ in $\left(0,\dfrac{1}{2}\right)$.

Using Equation 9.6.5,

The actual error is

which is about 0.02.

It helps to have a formula for $f^{(n)}(x)$. You can figure it out by taking several derivatives and noticing the pattern, but also this has been given previously in the text.

Any $n$ greater than or equal to 3.

You can approximate the function $f(x)=x^{\tfrac{1}{7}}$.
It's a good bit of trivia to know $3^7=2187$.
A low-degree Taylor approximation will give you a good enough estimation. If you guess a degree, and take that Taylor polynomial, the error will probably be less than 0.001 (but you still need to check).

$\sqrt[7]{2200}\approx3+\dfrac{13}{7\cdot 3^6} \approx 3.00255$

Use the 6th-degree Maclaurin approximation for $f(x)=\sin x$.

If we're going to use Equation 9.6.5, then we'll probably be taking a Taylor polynomial. Using Example 9.5.5, the 6th-degree Maclaurin polynomial for $\sin x$ is

so let's play with this a bit. Equation 9.6.5 tells us that the error will depend on the seventh derivative of $f(x)$, which is $-\cos x$:

for some $c$ between 0 and 1. Since $-1 \leq \cos c \leq 1$,

Remark: there are lots of ways to play with this idea to get better estimates. One way is to take a higher-degree Maclaurin polynomial. Another is to note that, since $0<c<1<\dfrac{\pi}{3}$, then $\dfrac{1}{2}<\cos c < 1$, so

If you got tighter bounds than asked for in the problem, congratulations!

For part (c), after you plug in the appropriate values to Equation 9.6.5, simplify the upper and lower bounds for $e$ separately. In particular, for the upper bound, you'll have to solve for $e$.

(a) $T_4(x)=\sum_{n=0}^4\frac{x^n}{n!}$ (b) $T_4(1)=\frac{65}{24}$ (c) See the solution.