Use Taylor Series to approximate f(x) = e^(-x^2): Find a Taylor series represent

Question

Use Taylor Series to approximate f(x) = e^(-x^2):

Find a Taylor series representation for f(x) = e-x2 centered at x = 0. Write out the first six terms of this series [don't bother to expand any factorials that show up]. Using your answer from (a), find a Taylor series for F(x) = e-t2 dt centered at x = 0. [Hint: be careful with constants of integration that pop up, and make sure your Taylor series agrees with F(0) = 0.] Use your answer from (b) to approximate F( 1/10) = e t2 dt. Argue that your approximation differs from the actual answer by no more than 0.0000000001 = 10-10.

Explanation / Answer

The Taylor series for e^x is the sum of x^n/n! for n=0 to infinity (note that the n'th derivative of e^x is e^x, which evaluates to 1 at x = 0, and the denominator is n!)

Thus, for the first six terms of the Taylors series for e^-x^2, I would simply substitute x^2 for x in this sum.

a)1 - x^2/1! + x^4/2! - x^6/3! + x^8/4! - x^10/5! =

1 - x^2 + x^4/2 - x^6/6 + x^8/24- x^10/120

b) The integral from 0 to x of 1 - x^2 + x^4/2 - x^6/6 + x^8/24- x^10/120 =

x - x^3/3 + x^5/2*5 - x^7/6*7 + x^9/9*24 - x^11/11*120 =

x - x^3/3 + x^5/10 - x^7/42 + x^9/216 - x^11/1320

As there are no constant terms, F(0) = 0

Our approximation for F(.1) =

.1 - .1^3/3 + .1^5/10 - .1^7/42 + .1^9/216 - .1^11/1320 = 0.0996676642903363

As x = .1, this equals 10^-13/13*720= 1.06837606837607E-17

This is much less that 10^-10

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