Compute Taylor Series & Approximate Integral of Exponential Function

[asif zaidi]

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

 

[CompuChip]

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

Then you can just take it up to order 6 and already get an error of less than 10^{-10}. 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).

 

[kaisxuans]

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