Why Does the Taylor Series of exp(-x^2) at x = 0 Start with 1 - x^2?

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The Taylor Series of f(x) = exp(-x^2) at x = 0 begins with 1 - x^2 due to the evaluation of the function and its derivatives at zero. Although the first derivative f'(0) is zero, the second derivative f''(0) is -2, which contributes to the x^2 term in the expansion. The series can also be derived by substituting t = -x^2 into the Taylor expansion of exp(t), leading to the series for e^-t^2. The absence of an x^4 term in the second derivative is explained by the definition of the Taylor series, where the second derivative contributes only to the x^2 term. Thus, the first two nontrivial terms of the expansion are correctly represented as 1 - x^2.
cytochrome
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The Taylor Series of f(x) = exp(-x^2) at x = 0 is 1-x^2...

Why is this?

The formula for Taylor Series is f(x) = f(0) + (x/1!)(f'(0)) + (x^2/2!)(f''(0)) + ...

and f'(x) = -2x(exp(-x^2)) therefore f'(0) = 0?

Can someone please explain why it is 1-x^2?
 
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Although f'(0) = 0, yet f''(0) will NOT be zero.
 
The formula is (x^n)/n! (with the sigma crap in front)
0! is 1 and x^0 is 1

Just substitute x=-t^2 and you get the series for e^-t^2
 
HomogenousCow said:
The formula is (x^n)/n! (with the sigma crap in front)
0! is 1 and x^0 is 1

Just substitute x=-t^2 and you get the series for e^-t^2

Then why doesn't the second derivative term have an x^4?
 
Here's what HomogenousCow is trying to say. Consider the expansion of ##\exp(t)## where ##t=-x^2##. The Taylor expansion of ##\exp(t)## is $$\sum_{n=0}^\infty \frac{t^n}{n!},$$ so substituting gives you $$e^{-x^2}=\sum_{n=0}^\infty \frac{(-x^2)^n}{n!}=\sum_{n=0}^\infty \frac{(-1)^n x^{2n}}{n!}.$$

The second derivative term does not have an ##x^4## by its definition: Each term is given by ##\frac{f^{(n)}(0)x^n}{n!}## (for MacLaurin; for Taylor about ##c##, it's ##\frac{f^{(n)}(c)(x-c)^n}{n!}##). When you plug in 2, you get $$\frac{f''(0)x^2}{2!}.$$ In this case, ##f''(0)=-2##. This cancels with the ##2!## in the denominator, so you end up with ##1-x^2## as the first two nontrivial terms of the expansion.
 
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cytochrome said:
Then why doesn't the second derivative term have an x^4?
It looks like you are trying to combine two different ways of finding the Taylor series of e^{-x^2}

1) Using the basic definition: the "x^2" term has coefficient f''(0)/2 so that the second derivative term, by definition, involves x^2, not x^4.

2) Replacing x in the Taylor's series for e^x with -x^2. In that case, the term that you get from the second derivative of e^x has x^4 but that has nothing to do with the second derivative of e^{-x^2}. That term comes from replacing x with -x^2 in the f'(0) x term of the Taylor series for e^x.
 
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