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Taylor series question

by cytochrome
Tags: series, taylor
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cytochrome
#1
Jan13-13, 01:57 AM
P: 162
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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grzz
#2
Jan13-13, 02:14 AM
P: 950
Although f'(0) = 0, yet f''(0) will NOT be zero.
HomogenousCow
#3
Jan13-13, 05:30 AM
P: 356
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

cytochrome
#4
Jan13-13, 12:09 PM
P: 162
Taylor series question

Quote Quote by HomogenousCow View Post
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?
THSMathWhiz
#5
Jan17-13, 06:31 PM
P: 11
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.
HallsofIvy
#6
Jan18-13, 09:34 AM
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Quote Quote by cytochrome View Post
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 [itex]e^{-x^2}[/itex]

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

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


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