# Leibnitz Formula.

• Aug 21st 2009, 10:14 AM
bandedkrait
Leibnitz Formula.
Okay I'm not sure if this is University level or below, but anyway

Apply the Leibnitz theorem to the function $y=x^{2n}$

and hence prove that,

$1 + (n^2)/1 + ((n^2)(n-1)^2)/(1^2)(2^2) + ........ = (2n!)/(n!)^2$

Now I can easily prove this using the Binomial theorem as the terms on the LHS are merely squares of the binomial coefficients in the expansion $(1+x)^n$
And the RHS is the coefficient of $x^n$ in the expansion of
$(1+x)^{2n}$

Using this data the above relation can be proved easily.

But I am not able to prove it using the method the question demands, that is, by using the Lebnitz formula for derivative of a function.

Can anyone help?
• Aug 21st 2009, 10:33 AM
Matt Westwood
Is the Leibniz Formula that thing which is used to determine the n'th derivative of a product?

In that case, see what happens if you treat it as $x^2n = x^n x^n$.

From my limited knowledge:

$D^n fg = \sum_{j=0}^n \binom n j D^j f D^{n-j} g$

You find that $D^j f D^{n-j} g$ gives you the appropriate factorial stuff in it and it looks as though everything works. Haven't actually tried it as I'm too lazy.
• Aug 21st 2009, 10:39 AM
bandedkrait
Quote:

In that case, see what happens if you treat it as http://www.mathhelpforum.com/math-he...9a2d4fb7-1.gif.
This should do the trick. Thanks. I wonder why I didn't think of this myself! :-/
Maybe I'm too sleepy at the moment.
• Aug 21st 2009, 10:50 AM
Random Variable
$\frac{d^{n}}{dx^{n}} x^{k} = \frac{k!}{(k-n)!} x^{k-n}$

$\frac{d^{n}}{dx^{n}} x^{2n} = \frac{(2n)!}{n!} x^{n}$

$\frac{d^{n}}{dx^{n}} x^{2n} = \frac{d^{n}}{dx^{n}} (x^{n}\cdot x^{n}) = \sum_{r=0}^{n} \binom{n}{r} \frac{d^{r}}{dx^{r}} x^{n} \frac{d^{n-r}}{dx^{n-r}} x^{n}$

$= x^{n}n!x^{0} + nnx^{n-1}n!x^{1} + \frac{n(n-1)}{2!}n(n-1)x^{n-2}\frac{n!}{2!} x^{2} +$ $\frac{n(n-1)(n-2)}{3!}n(n-1)(n-2)x^{n-3}\frac{n!}{3!}x^{3} + ... n!x^{0} x^{n}= \frac{(2n)!}{n!} x^{n}$

$1 + \frac{n^{2}}{(1!)^{2}} + \frac{n^{2}(n-1)^{2}}{(2!)^{2}} + \frac{n^{2}(n-1)^{2}(n-2)^{2}}{(3!)^{2}} + ... + 1 = \frac{(2n)!}{(n!)^{2}}$