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Taylor series question 
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#1
Jan1313, 01:57 AM

P: 162

The Taylor Series of f(x) = exp(x^2) at x = 0 is 1x^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 1x^2??? 


#2
Jan1313, 02:14 AM

P: 950

Although f'(0) = 0, yet f''(0) will NOT be zero.



#3
Jan1313, 05:30 AM

P: 362

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 


#4
Jan1313, 12:09 PM

P: 162

Taylor series question



#5
Jan1713, 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)(xc)^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 ##1x^2## as the first two nontrivial terms of the expansion. 


#6
Jan1813, 09:34 AM

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P: 39,510

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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