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Is there a way of approximating e^-x for large x

  1. Dec 22, 2012 #1
    I'm trying to evaluate an integral with e^-x where x is huge in the domain of the integral so I can't evaluate it numerically without making an approximation.
     
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  3. Dec 22, 2012 #2

    mfb

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    Can you add more context? There might be a clever way to approximate e^(-x), but if that value is not added to something, e^(-x)=e^(-x-x0)*e^(x0) where the second factor is independent of x and the first factor can be chosen to be about e^0.
     
  4. Dec 22, 2012 #3
    The integral is 4*(r^2)*exp(-2*r/a)/a^3 integrated between 1 and ∞ (The probability an electron in the ground state of hydrogen is more than 1 metre away from the nucleus) a=0.529*10^-10
     
  5. Dec 22, 2012 #4

    Hurkyl

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    Note that your integrand is one you can anti-differentiate without much trouble, so you can get the exact answer. (e.g. integration by parts. Or computer algebra package)
     
  6. Dec 22, 2012 #5

    mfb

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    If you just want an estimate: The exponential will drop very quick, so regions with r>1m+eps are irrelevant and r^2 is nearly constant and =1m^2. Therefore, the integral is simply 4m^2/a^3 * exp(-2r/a) which can be evaluated as 2m^2/a^2 * exp(-2m/a) ≈ 8*10^20 * exp(-4*10^10) ≈ 10^(-10^10) where the last approximation is very rough.
     
  7. Dec 22, 2012 #6

    mathman

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    e-x = 10-0.43429448190325x. The constant is log10e.

    Using the above for large x, you can separate the integer and fractional parts of the exponent. I assume you know how to proceed from here.
     
  8. Dec 22, 2012 #7

    lurflurf

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    Are you talking about Gauss–Laguerre quadrature?

    [tex]\int_0^\infty f(x) e^{-x} \mathop{\text{dx}}\sim \sum_{i=1}^n w_i f(x_i)[/tex]

    where xi are zeros of a Laguerre polynomial and

    [tex]w_i=\frac{x_i}{(n+1)^2[L_{n+1}(x_i)]^2}[/tex]
     
  9. Dec 23, 2012 #8
    Fair enough, yeah I realised after posting this the integral could be solved analytically >__< and I got a valule of something like 10^(-10^10) but thanks for the responses :)
     
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