# Estimate Remainder of Taylor Series

#### JG89

1. The problem \statement, all variables and given/known data

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

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:

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

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.

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#### nickjer

Taylor expansion of an exponential is...

$$e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!}$$

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

$$e^{-t^2} = \sum_{n=0}^{\infty} \frac{(-1)^n t^{2n}}{n!}$$

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

$$\int_0^x \frac{(-1)^{(n+1)} t^{2(n+1)}}{(n+1)!} dt$$

#### JG89

Ok, I just saw your edit, let me go work that out...

#### nickjer

Oops, misread the question. Hah, so I can't help much, sorry.

#### JG89

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?

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#### nickjer

Well... See if this works...

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

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

$$F^m (x) = \sum_{n=m-1}^{\infty} \frac{(-1)^n}{n!} \frac{(2n)!x^{2n-(m-1)}}{(2n-(m-1))!}$$

Again, this could exactly what you are not asking for, hah. Just thought I'd throw this out there, just in case.

#### JG89

My expression for the vth derivative of e^(-t^2) is:

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

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?

Bump...

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