# Thread: Simple issue using Taylor's Thm

1. ## Simple issue using Taylor's Thm

I'm trying to figure out how to show
$\displaystyle 1-x^2\leq e^{-x^2}$

I am using $\displaystyle f(x) = e^{-x^2}$ and $\displaystyle x_0 = 0$.

I'm getting the 1 for the first term but then I keep getting zeroes, am I approaching this wrong?

2. Originally Posted by CrazyCat87
I'm trying to figure out how to show
$\displaystyle 1-x^2\leq e^{-x^2}$

I am using $\displaystyle f(x) = e^{-x^2}$ and $\displaystyle x_0 = 0$.

I'm getting the 1 for the first term but then I keep getting zeroes, am I approaching this wrong?
$\displaystyle e^{-x^2}=1-x^2+R_2$. What's the remainder term and why is it greater than zero?

3. I don't know why I'm not getting that $\displaystyle -x^2$ term, but then R(x) should then be $\displaystyle \frac{2c^2 e^{-c^2}}{2!} (x^2)$, which is obviously a positive term

4. Originally Posted by CrazyCat87
I don't know why I'm not getting that $\displaystyle -x^2$ term, but then R(x) should then be $\displaystyle \frac{2c^2 e^{-c^2}}{2!} (x^2)$, which is obviously a positive term
Taylor series for $\displaystyle e^x$ is $\displaystyle 1 + \frac{x^1}{1!} + \frac{x^2}{2!} + \dots$.

Just sub $\displaystyle -x^2$ into that. Second term is therefore $\displaystyle \frac{-x^2}{1!} = -x^2$.

5. I was given a hint to split this into 2 cases, $\displaystyle |x| \leq 1$ and $\displaystyle |x|>1$ , but I can't figure out why to do this?