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Math Help - Ok I am drawing a blank

  1. #1
    MHF Contributor Mathstud28's Avatar
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    Ok I am drawing a blank

    integral(e^(-x))...from there I went to e^(-x)≡ sigma[(-x)^n/n!]= sigma[(-1)^n*x^(2n)/n!] so integral(e^(-x))= integral[!]= sigma[(-1)^n*x^(2n)/n!] dx]...but I am stuck on where to go from here if there is route
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Sorry

    It should read integral(e^(-x))= sigma[(-1)^n*x^(2n)/n!] dx]...
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  3. #3
    Math Engineering Student
    Krizalid's Avatar
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    Well it's just a direct application of Maclaurin series for e^x.
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    This is embarrassing

    I've learned it but forgotten could you elaborate
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  5. #5
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
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    Quote Originally Posted by Mathstud28 View Post
    integral(e^(-x))...from there I went to e^(-x)≡ sigma[(-x)^n/n!]= sigma[(-1)^n*x^(2n)/n!] so integral(e^(-x))= integral[!]= sigma[(-1)^n*x^(2n)/n!] dx]...but I am stuck on where to go from here if there is route
    \int \sum_{n=0}^{\infty}\frac{(-1)^nx^{2n}}{n!}dx=\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n+1}}{n!(2n+1)}
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  6. #6
    MHF Contributor Mathstud28's Avatar
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    Just checking

    but that holds true because n! and (-1)^n are constants and you just yanked them out evaluated your integral and then reintroduced them?
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