Maclaurin Series of $e^{-x^2}$
The question is:
Find the first 3 non-zero terms in the MacLaurin series for the function: $$y =e^{-x^2}$$ I have been told to simply substitute the $-x^2$ into the standard MacLaurin series for $e^x$ like so: $$e^{(-x^2)} = \sum_{n=0}^\infty \frac {x^n}{n!}= 1 + (-x^2) + \frac {{(-x^2)}^2}2 + \frac {{(-x^2)}^3}{3!}+...$$ Giving: $$e^{-x^2} = \sum_{n=0}^\infty \frac {{(-x^2)}^n}{n!} = 1 - x^2 + \frac {x^4}2 - \frac {x^6}{3!}+...$$ I realise that this is supposed to be the correct answer, but I can't seem to make it work alongside the definition of MacLaurin series given to me in lectures. We were told that the terms of a MacLaurin series were based on the formula: $$\sum_{n=0}^\infty a_nx^n$$ where $$a_n = \frac {f^n(0)}{n!}$$ assuming $0!=1$.
However, when I use this formula for $e^{-x^2}$, I get the following: $$a_0x^0=\frac {f^0(0)}{0!}x^0=\frac {e^{-(0)^2}}{1}\times1=\frac 11=1$$ $$a_1x^1=\frac {f^1(0)}{1!}x=\frac {-2(0)e^{-(0)^2}}{1}x=\frac 01x=0$$ $$a_2x^2=\frac {f^2(0)}{1!}x^2=\frac {4(0)^2e^{-(0)^2}}{1}x^2=\frac 01x^2=0$$ and so on. Giving:$$e^{-x^2}=1+0+0+0+...$$ Can somebody explain this to me?
$\endgroup$ 32 Answers
$\begingroup$The substitution method is probably best for what you are trying to do.
On the other hand, if you want to compute the derivatives of $e^{-x^2}$, then you will have to use a combination of the chain rule and the product rule.
$$f(x) = e^{-x^2}$$
$$f'(x) = -2x e^{-x^2}$$
$$f''(x) = - 2 e^{x^2}+ (-2x)^2 e^{-x^2} = (4x^2 -2) e^{-x^2}$$
$$f'''(x) = 8x e^{-x^2} - (4x^2 - 2) (2x) e^{-x^2} = (-8x^3 + 12x) e^{-x^2}$$
This yields:
$f(0)=1$, $f'(0) = 0$, $f''(0) = -2$, $f'''(0) = 0$, etc.
$\endgroup$ $\begingroup$The second derivative is $$[(-2x)e^{-x^2}]'=(-2+4x^2)e^{-x^2}, $$ which evaluated at $x=0$ gives $-2$.
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