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Math Help - Reduction Formula

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    Reduction Formula

    f(x) = \displaystyle\int^\infty_0 t^{x-1}e^{-t} \, \mathrm{d}t. Show that: f(x+1) = xf(x)

    I used integration by parts. I've come across a limit part, what's that value? \left[-e^{-t}t^x\right]^\infty_0. <-- Is that just zero, if so why?
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    Quote Originally Posted by Air View Post
    f(x) = \displaystyle\int^\infty_0 t^{x-1}e^{-t} \, \mathrm{d}t. Show that: f(x+1) = xf(x)

    I used integration by parts. I've come across a limit part, what's that value? \left[-e^{-t}t^x\right]^\infty_0. <-- Is that just zero, if so why?
    Yes, it's zero. As for a proof, consider

    \lim_{t \rightarrow \infty} t^x e^{-t} = \lim_{t \rightarrow \infty}\frac{t^x}{e^t}

    So handwaving - exponentials grow much facter than powers. If you look at a power series for the exponetial

    e^t = 1 + t + \frac{t^2}{2!}<br />
+ \frac{t^3}{3!} \cdots
    so

    \lim_{t \rightarrow \infty} t^x e^{-t} = \lim_{t \rightarrow \infty}\frac{t^x}{1 + t + \frac{t^2}{2!}<br />
+ \frac{t^3}{3!} \cdots}

    and you can always go out far enough in the power series so that the powers in the denominator are larger than that in the numerator.

    Do you know about L'Hopitals rule?
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    Quote Originally Posted by danny arrigo View Post
    Yes, it's zero. As for a proof, consider

    \lim_{t \rightarrow \infty} t^x e^{-t} = \lim_{t \rightarrow \infty}\frac{t^x}{e^t}

    So handwaving - exponentials grow much facter than powers. If you look at a power series for the exponetial

    e^t = 1 + t + \frac{t^2}{2!}<br />
+ \frac{t^3}{3!} \cdots
    so

    \lim_{t \rightarrow \infty} t^x e^{-t} = \lim_{t \rightarrow \infty}\frac{t^x}{1 + t + \frac{t^2}{2!}<br />
+ \frac{t^3}{3!} \cdots}

    and you can always go out far enough in the power series so that the powers in the denominator are larger than that in the numerator.

    Do you know about L'Hopitals rule?
    L'Hopitals rule is valid when limits put into the numerator and denominator give zero. We keep differentiating till numberator and denominator have a different value.

    But if we apply L'Hopitals rule then wouldn't we get a number other than zero unless numerator gives zero. However, isn't the numerator, \lim_{t \rightarrow \infty} t^x = \infty Because greater the value, then greater it's power will increase it by.
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    Quote Originally Posted by Air View Post
    L'Hopitals rule is valid when limits put into the numerator and denominator give zero. We keep differentiating till numberator and denominator have a different value.

    But if we apply L'Hopitals rule then wouldn't we get a number other than zero unless numerator gives zero. However, isn't the numerator, \lim_{t \rightarrow \infty} t^x = \infty Because greater the value, then greater it's power will increase it by.
    First, L'Hoptial's rule applies if

    \lim_{x \rightarrow a} \frac{f(x)}{g(x)} =``\frac{0}{0}" \;\text{or}\; ``\frac{\infty}{\infty}"

    For your problem

    \lim_{t \rightarrow \infty} \frac{t^x}{e^t} has this form \frac{\infty}{\infty}. Let's assume for the sake of argument that x = 5

    \lim_{t \rightarrow \infty} \frac{t^5}{e^t} = \lim_{t \rightarrow \infty} \frac{5t^4}{e^t} = \lim_{t \rightarrow \infty} \frac{5\cdot 4 t^3}{e^t}= \cdots = \lim_{t \rightarrow \infty} \frac{5!}{e^t} = 0 (using L'Hopital's rule repeatedly)

    If x is not an integer but real, then repeated use would eventually get a form of

    \lim_{t \rightarrow \infty} \frac{1}{t^n e^t}

    which is zero.
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  5. #5
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    Hello,

    \left[-e^{-t}t^x\right]^\infty_0=\left[\lim_{t \to 0} \frac{t^x}{e^t}\right]-\left[\lim_{t \to \infty} \frac{t^x}{e^t}\right]

    The first limit is easy since e^t goes to 1 and since x>0, t^x goes to 0.

    For the second one, you can ideed use l'Hospital's rule if you haven't learnt an other way, implying the comparison of increasing "speeds"..
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