1. ## Taylor Inequality

Suppose you used the first three nonzero terms of the power series representation to approximate $e^{-(x^2)}$ for $0\le x \le 1/2$.
Obtain an upper bound on the error.
All the terms, are negative, I think? The parentheses confused me. So: power series representation is $1-x^2-x^4/2-x^6/6...$
The approximation is found, I think, just by plugging in x=1/2, for which I got 0.71875.
Then use the Taylor Inequality... $R_3_(x) \le \frac{M}{(n+1)!} /x-a/^{n+1}$
${f^{4}(x) \le M$
$\frac {(1/2)^6}{6} \le M$, so do I set M=1/384?
I stopped here because I wasn't sure if my value for M was right...I have the feeling it's not. Any help would be great.

2. ## Re: Taylor Inequality

Originally Posted by bcahmel
Suppose you used the first three nonzero terms of the power series representation to approximate $e^{-(x^2)}$ for $0\le x \le 1/2$.
Obtain an upper bound on the error.
All the terms, are negative, I think? The parentheses confused me. So: power series representation is $1-x^2-x^4/2-x^6/6...$
The approximation is found, I think, just by plugging in x=1/2, for which I got 0.71875.
Then use the Taylor Inequality... $R_3_(x) \le \frac{M}{(n+1)!} /x-a/^{n+1}$
${f^{4}(x) \le M$
$\frac {(1/2)^6}{6} \le M$, so do I set M=1/384?
I stopped here because I wasn't sure if my value for M was right...I have the feeling it's not. Any help would be great.
$e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + ...$

$e^{-x^2} = 1 + (-x^2) + \frac{(-x^2)^2}{2!} + \frac{(-x^2)^3}{3!} + ...$

so, the first three terms of the series for $e^{-x^2}$ are ...

$e^{-x^2} \approx 1 - x^2 + \frac{x^4}{2!}$

... and the series is alternating.

3. ## Re: Taylor Inequality

ok thanks, the parentheses dont make a difference then. I still don't understand M, however...