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Estimate Remainder of Taylor Series

  • Thread starter JG89
  • Start date
726
1
1. The problem \statement, all variables and given/known data

Estimate the error involved in using the first n terms for the function [tex] F(x) = \int_0^x e^{-t^2} dt [/tex]


2. Relevant equations



3. The attempt at a solution

I am using the Lagrange form of the remainder. I need to know the n+1 derivative of e^(-t^2) but I found that the derivatives get more complicated as n increases, so I dropped that idea.

I rewrote the function like this:

[tex] \int_0^x e^{-t^2} dt = \int_0^x 1 - t^2 + \frac{t^4}{2!} - \frac{t^6}{3!} + ... + \frac{(-1)^n t^{2(n-1)}}{n!} + R_n [/tex]

I can easily integrate each term, and then the integral from 0 to x of R_n would be the remainder. Note that R_n is the remainder term for e^(-t^2), so right now I am trying what R_n is.

Rewriting e^(-t^2) as its Taylor series, I differentiated term by term for the first, second, third, etc derivative to see if I can find a general expression for the nth derivative, but I can't find one.

Any suggestions?


EDIT: I was also thinking that if there is a constant value M such that all derivatives of F are bounded above by M, and I was able to find this value M, then there would be no problem obtaining an expression to estimate the remainder.
 
Last edited:
674
2
Taylor expansion of an exponential is...

[tex]e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!}[/tex]

So you would replace x by -t^2...

[tex]e^{-t^2} = \sum_{n=0}^{\infty} \frac{(-1)^n t^{2n}}{n!}[/tex]

So you just need to integrate the term in the sum for n+1 to get the value for the next term...

[tex]\int_0^x \frac{(-1)^{(n+1)} t^{2(n+1)}}{(n+1)!} dt[/tex]
 
726
1
Ok, I just saw your edit, let me go work that out...
 
674
2
Oops, misread the question. Hah, so I can't help much, sorry.
 
726
1
EDIT: Just saw your new reply :(

I have an upperbound for the n+1 term though, does this help in finding an estimate for the remainder?
 
Last edited:
674
2
Well... See if this works...

[tex]F'(x) = e^{(-x^2)} = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{n!}[/tex]

So the taylor expansion of the m'th derivative is...

[tex]F^m (x) = \sum_{n=m-1}^{\infty} \frac{(-1)^n}{n!} \frac{(2n)!x^{2n-(m-1)}}{(2n-(m-1))!}[/tex]

Again, this could exactly what you are not asking for, hah. Just thought I'd throw this out there, just in case.
 
726
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My expression for the vth derivative of e^(-t^2) is:

[tex] \sum_{n=0}^{\infty} \frac{(-1)^n 2n(2n-1)(2n-2) \hdots (2n - v + 1) t^{2n - v}}{n!} [/tex]

It's pretty easy to find the Lagrange remainder term now, but it's written as an infinite series. How can I write it in a closed-form expression?
 
726
1
Bump...
 

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