# Math Help - Tn(x) Error Bound

1. ## Tn(x) Error Bound

Good afternoon,

I think I can finally wrap my head around Taylor (& Maclaurin) polynomials. However, I seem to really be struggling with the error bounds. I am looking for an error bound of $<10^{-4}$ for the $n^{th}$ Maclaurin polynomial of $f(x)=e^x$ for $x=2$.

I'm not even sure I know how to set this up. Should it be simply:

$
\frac{|2|^{n+1}}{(n+1)!}<10^{-4}
$

If this is what it should be, can someone explain why (or point me to some "easy to digest" reading)?

2. The McLaurin expasion for $e^{x}$ is...

$\displaystyle e^{x} = 1 + x + \frac{x^{2}}{2} + \frac{x^{3}}{6} + \dots + \frac{x^{n}}{n!} + R_{n} (x)$ (1)

... where $R_{n} (x)$ is an 'error term' that has different interpretations. One of the most simple expression of $R_{n} (x)$ has been found by the french mathematician Augustin Cauchy and is in that case...

$\displaystyle R_{n}(x) = \frac{x^{n+1}\ (1-\theta)^{n}}{n!}\ e^{\theta x}$ , $0 < \theta < 1$ (2)

At this point You have to do the following steps...

a) setting in (2) $x=2$ find the value of $\theta_{0}$ that maximise $R_{n}$ ...

b) stting in (2) $\theta=\theta_{0}$ find the value of n for which is $R_{n} < 10^{-4}$ ...

Kind regards

$\chi$ $\sigma$

3. I am provided with the formula:

$|T_n(x)-f(x)|\leq K\frac{|x-a|^{n+1}}{(n+1)!}$

let $K\leq|f^{(n+1)}(u)}|$ for all $u$ between $x$ and $a$

I am not really sure how set that up to find the correct value of n. I was able to construct a matrix that I believe provides me with the correct answer, but I would like to be able to utilize the formula directly.

4. Recalling that is...

$\displaystyle R_{n} (x) = \frac{x^{n+1}\ (1-\theta)^{n}}{n!}\ e^{\theta x}$ , $0 < \theta < 1$ (1)

... setting $x=2$ we find that $R_{n} (2)$ has the maximum possible value for $\theta=0$, so that is...

$\displaystyle R_{n} (2) \le \frac{2^{n+1}}{n!}}$ (2)

... and in particular...

$\displaystyle n=2 \rightarrow R_{n} (2) \le 4$

$\displaystyle n=3 \rightarrow R_{n} (2) \le 2.666666...$

$\displaystyle n=4 \rightarrow R_{n} (2) \le 1,333333...$

$\displaystyle n=5 \rightarrow R_{n} (2) \le .533333...$

$\displaystyle n=6 \rightarrow R_{n} (2) \le .177777...$

$\displaystyle n=7 \rightarrow R_{n} (2) \le 5.079365... \ 10^{-2}$

$\displaystyle n=8 \rightarrow R_{n} (2) \le 1.269841... \ 10^{-2}$

$\displaystyle n=9 \rightarrow R_{n} (2) \le 2.821869... \ 10^{-3}$

$\displaystyle n=10 \rightarrow R_{n} (2) \le 5.643738... \ 10^{-4}$

$\displaystyle n=11 \rightarrow R_{n} (2) \le 1.026134... \ 10^{-4}$

$\displaystyle n=12 \rightarrow R_{n} (2) \le 1.710223... \ 10^{-5}$ (3)

... so that the required value for n seems to be 12...

Kind regards

$\chi$ $\sigma$

5. Originally Posted by chisigma
Recalling that is...

$\displaystyle R_{n} (x) = \frac{x^{n+1}\ (1-\theta)^{n}}{n!}\ e^{\theta x}$ , $0 < \theta < 1$ (1)

... setting $x=2$ we find that $R_{n} (2)$ has the maximum possible value for $\theta=0$, so that is...

$\displaystyle R_{n} (2) \le \frac{2^{n+1}}{n!}}$ (2)
I find this to be equal to:

$\displaystyle R_{n} (2) \le \frac{2^{n+1}}{n!}}$

As $e^{\theta 2} = 1$ for $\theta = 0$

Producing:

$\displaystyle n=12 \rightarrow R_{n} (2) \le .000103$

$\displaystyle n=13 \rightarrow R_{n} (2) \le .000017$

So...

$n=13$

Am I confused by something?

6. Is my assumption on the last post incorrect? I definitely don't mean to be insulting, just confused.