1. ## Taylor's Theorem Estimation

Hello again,

This question is more toward the definition of taylor polynomials, why does it guarantee that we can use a taylor polynomial to estimate "e" to any degree of accuracy? I know that the longer the polynomial, the closer the approximation and "e" is one of the easier functions for taylor polynomails. Is there a better way to explain why? other than restating the definition of a taylor polynomial?

- oh, I forgot to say in my last post, Thanks

2. Originally Posted by icedragontt
Hello again,

This question is more toward the definition of taylor polynomials, why does it guarantee that we can use a taylor polynomial to estimate "e" to any degree of accuracy? I know that the longer the polynomial, the closer the approximation and "e" is one of the easier functions for taylor polynomails. Is there a better way to explain why? other than restating the definition of a taylor polynomial?

- oh, I forgot to say in my last post, Thanks
You could look at the Taylor remainder. For the nth degree polynomial

$\displaystyle P_n(x) = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots + \frac{x^n}{n!}$

$\displaystyle R_n = \frac{e^c x^{n+1}}{(n+1)!}$ where $\displaystyle c$ is between $\displaystyle x$ and $\displaystyle 0$

and
$\displaystyle P_n(1) = 1 + 1+ \frac{1}{2!} + \frac{1}{3!} + \cdots + \frac{1}{n!}$

$\displaystyle R_n(1) = \frac{e^c}{(n+1)!} < \frac{3}{(n+1)!}$

Clearly as we increase $\displaystyle n, R_n$ decreases give a better and better answer.

3. Read this section in the wiki Taylor's theorem - Wikipedia, the free encyclopedia

Short story is

Since the derivative of $\displaystyle \frac{d}{dx}e^x=e^x$ for all x.

for any fixed number m we can find an N large enough so that

$\displaystyle \frac{m^{N+1}e^m}{(N+1)!}$ is as small as we want.

This tells us the error can't be any larger than the above term so we would need to take n terms in the taylor expation to get at least this accurate.

Note this is an upper bound on the error it may be much smaller than this.

If we take the limit at $\displaystyle n \to \infty$ and the remainder goes to 0 it tells us that the taylor polynomial does indeed converge to the function.

I hope this helps.