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Math Help - Integration: Error Function

  1. #1
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    Integration: Error Function

    Hello,
    I dont have the strongest math foundation, so please excuse me if this is a dumb question.
    I was integrating the following in Maple:

    \int_0^1  e^{-x^2} dx

    which produced the answer:

    \frac{1}{2} \ erf(1) \ \sqrt{\pi}

    I've read that the erf is the error function but I don't really understand what it is. How can I obtain a 'Real' value for this?
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by JavaJunkie View Post
    Hello,
    I dont have the strongest math foundation, so please excuse me if this is a dumb question.
    I was integrating the following in Maple:

    \int_0^1  e^{-x^2} dx

    which produced the answer:

    \frac{1}{2} \ erf(1) \ \sqrt{\pi}

    I've read that the erf is the error function but I don't really understand what it is. How can I obtain a 'Real' value for this?
    By a little snake-oil (i.e. you don't care about the proof of why it's true right? If you do let me know_ we have that \int_0^1e^{-x^2}dx=\sum_{n=0}^{\infty}\int_0^1 \frac{(-1)^nx^{2n}}{n!}dx=\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)n!}
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  3. #3
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    Well, I would like to know why its true even though I don't have to.

    \int_0^1e^{-x^2}dx=\sum_{n=0}^{\infty}\int_0^1 \frac{(-1)^nx^{2n}}{n!}dx=\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)n!}  \ \ \ \Longrightarrow thats a fourier series, isn't it?
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by JavaJunkie View Post
    Well, I would like to know why its true even though I don't have to.

    \int_0^1e^{-x^2}dx=\sum_{n=0}^{\infty}\int_0^1 \frac{(-1)^nx^{2n}}{n!}dx=\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)n!}  \ \ \ \Longrightarrow thats a fourier series, isn't it?
    Not quite! But good guess. It comes from a Maclaurin series. The difficult part comes from why the following step is justified \int_0^1\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n}}{n!}=\sum_{n=0}^{\infty}\int_0^1\frac{(-1)^n x^{2n}}{n!}dx. The idea is that \left|\frac{(-1)^nx^{2n}}{n!}\right|\leqslant\frac{1}{n!} and so by the Weierstrass M-test we may conclude that the sum is uniformly convergent and thus the interchangeability of the series and the integral is justified.
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