# Reduction Formula

• January 10th 2009, 09:12 AM
Simplicity
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?
• January 10th 2009, 09:32 AM
Jester
Quote:

Originally Posted by Air
$f(x) = \displaystyle\int^\infty_0 t^{x-1}e^{-t} \, \mathrm{d}t$. 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

$e^t = 1 + t + \frac{t^2}{2!}
+ \frac{t^3}{3!} \cdots$

so

$\lim_{t \rightarrow \infty} t^x e^{-t} = \lim_{t \rightarrow \infty}\frac{t^x}{1 + t + \frac{t^2}{2!}
+ \frac{t^3}{3!} \cdots}$

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?
• January 10th 2009, 09:56 AM
Simplicity
Quote:

Originally Posted by danny arrigo
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

$e^t = 1 + t + \frac{t^2}{2!}
+ \frac{t^3}{3!} \cdots$

so

$\lim_{t \rightarrow \infty} t^x e^{-t} = \lim_{t \rightarrow \infty}\frac{t^x}{1 + t + \frac{t^2}{2!}
+ \frac{t^3}{3!} \cdots}$

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.
• January 10th 2009, 10:23 AM
Jester
Quote:

Originally Posted by Air
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

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

$\lim_{t \rightarrow \infty} \frac{t^x}{e^t}$ 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.
• January 11th 2009, 01:17 AM
Moo
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"..