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Homework Help: Taylor Series Question

  1. Feb 15, 2008 #1
    Problem Statement
    Compute the Taylor Series expansion of f(x) = exp(-x^2) around 0 and use it to find an approximate value of the integral (from 0 to 0.1) of exp(-t^2) dt

    Solution

    Part1:
    First to compute the Taylor Series - I am pretty sure about this step so I will not give details. But if I am wrong, please correct.

    Taylor Series = 1 - (x^2) + (x^4)/2! - (x^6)/3! + ... --- EQUATION 1

    And a closed form solution is from Sum (i from 0 to inf) of (-1^i)* (x^2*i)/i!

    Part2:
    This is the part I am not doing right - maybe I am not approaching the problem in the right way.
    To solve the integral evaluate the Taylor Series in equation 1 above at 0.1 and 0 and subtract. Also, I took just the first 4 terms of equation 1. Is there a way I can determine how many terms I should take?

    At x = 0.1: Equation 1 is evaluated to 0.99004983375
    At x = 0.0: Equation 1 is evaluated to 1.000000000000
    Subtracting above gives me an integral value of -0.00995016625.

    So my two questions are:

    1- Now this I know is clearly wrong as the value should be positive but I cannot figure out what I am doing wrong. The absolute value above is right but why am I getting a negative. I tried above method for positive exponentials and it worked but any negative exponential, I am always getting the negative answer.

    2- How do I determine how many terms I should use in my Taylor Series expression.


    Thanks


    Asif
     
  2. jcsd
  3. Feb 15, 2008 #2

    CompuChip

    User Avatar
    Science Advisor
    Homework Helper

    Actually, don't you think you should rather calculate
    [tex]\int_0^{0.1} T_n(x) \, \mathrm{d}x[/tex]
    instead of
    [tex]T_n(0.1) - T_n(0)[/tex]
    where [itex]T_n(x)[/itex] is the n-th order Taylor series for the function
    (so for example, [itex]T_4(x) = T_5(x) = 1 - x^2 + x^4/2[/itex] and [itex]T_6(x) = 1 - x^2 + x^4/2 - x^6/6[/itex]).

    Then you can just take it up to order 6 and already get an error of less than [itex]10^{-10}[/itex]. You could do look at the terms you have neglected to estimate the error (e.g. neglecting x^8 and higher, the error you make will be at most of order 1/9 (0,1)^9).
     
  4. Feb 16, 2008 #3
    i dont know abput you. but i only got notes on these questions. cheak then for refernbce. my blog can be found at the top blogs section. kaixuan
     
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