# Math Help - Leibnitz Formula.

1. ## 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?

2. 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.

3. In that case, see what happens if you treat it as .
This should do the trick. Thanks. I wonder why I didn't think of this myself! :-/
Maybe I'm too sleepy at the moment.

4. $\frac{d^{n}}{dx^{n}} x^{k} = \frac{k!}{(k-n)!} x^{k-n}$

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

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

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

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