Find nth derivative of $\frac{x^{n}}{(1-x)^{2}}$, please?

Question

I need to find the nth derivative of $\frac{x^{n}}{(1-x)^{2}}$ for $0<x<1$

So far, I tried the same method used for $\frac{x^{n}}{1-x}$ and here's what I got: \begin{equation} \frac{x^{n}}{(1-x)^{2}}=x^{n}(1+2x+3x^{2}+4x^{3}+....)=(x^{n}+2x^{n+1}+3x^{n+2}+...) \end{equation} Take nth derivative to get: \begin{align} \frac{\partial^n }{\partial x^n} \frac{x^{n}}{(1-x)^{2}} \notag\\ = & (n!+2(n+1)!x+3\frac{(n+2)!}{2!}x^{2}+...) \notag\\ = & n!(1+2(n+1)x+3\frac{(n+2)(n+1)}{2!}x^{2}+4\frac{(n+3)(n+2)(n+1)}{3!}x^{3}+..) \end{align} Next would be finding a a function whose taylor expansion is the series in the parenthesis but I couldn't think of one. Any ideas for a function or for another method to do this? Thanks!

Now you have two almost complete answers, do you know how to "accept" an answer? Some think it is rude not to accept answers - which is done by a click in the check mark to the left of the answer post. — AD - Stop Putin -, Jun 24 '13 at 10:42

score 8 · Answer 1 · answered Jun 24 '13 at 08:51

8

Write $x = 1-t$, so you're looking at $$(-1)^n \dfrac{d^n}{dt^n} \dfrac{(1-t)^n}{t^2} = \dfrac{d^n}{dt^n} \sum_{j=0}^n {n \choose j} (-1)^{n+j} t^{j-2}$$ The terms with $j \ge 2$ all vanish, and you're left with $$ \dfrac{d^n}{dt^n} \left((-1)^n (t^{-2} - n t^{-1})\right)$$

From there it's easy. You can write the answer as $$\dfrac{n!(nx+1)}{(1-x)^{n+2}}$$

answered Jun 24 '13 at 08:51

Robert Israel

470,583

What a great derivation, thank you so much! – johnny israeli Jun 24 '13 at 10:31

Christian Blatter · Answer 2 · 2013-06-24T09:06:57.273

3

Consider the function $$g(u):={(1-u)^n\over u^2}={1\over u^2}\sum_{k=0}^n{n\choose k}(-u)^k ={1\over u^2}-{n\over u}+\sum_{k=2}^n{n\choose k}(-u)^{k-2}\ .$$ It follows that $$g^{(n)}(u)={(-1)^n (n+1)!\over u^{n+2}}-{n\>(-1)^n\> n!\over u^{n+1}}=(-1)^n n!{n(1-u)+1\over u^{n+2}}\ .$$ Now $f(x)=g(1-x)$, and therefore $$f^{(n)}(x)=(-1)^n g^{(n)}(1-x)=n!{n x+1\over (1-x)^{n+2}}\ .$$

edited Jun 24 '13 at 09:06

answered Jun 24 '13 at 08:50

Christian Blatter

232,144

score 0 · Answer 3 · edited Apr 13 '17 at 12:20

0

A related problem. Here is a closed form

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

Added: Here is the $n$th derivative for the wanted function

$$ \left(\frac{x^n}{(1-x)^2}\right)^{(n)}=\frac{n!(nx+1)}{(1-x)^{n+2}}. $$

You can apply the technique I used to solve this problem.

edited Apr 13 '17 at 12:20

Community

1

answered Jun 24 '13 at 07:32

Mhenni Benghorbal

48,099

1

wow, very interesting result! can you show a step by step derivation please? – johnny israeli Jun 24 '13 at 07:34
@johnnyisraeli: I'll try to do it later. – Mhenni Benghorbal Jun 24 '13 at 07:39
@johnnyisraeli: I think you can prove it by induction. – Mhenni Benghorbal Jun 24 '13 at 07:46
Do we really need the $\Gamma$-function? – AD - Stop Putin - Jun 24 '13 at 07:50
I just saw the answer on yahoo answers - I don't think my question is as simple as that.. – johnny israeli Jun 24 '13 at 07:50
@AD.: we don't need it, all you need is a simple factorial – johnny israeli Jun 24 '13 at 07:52
1

Whoever voted this answer to be good should undo it, it makes others think this actually answers my question but it doesn't really get us closer to the answer as far as I can tell.. – johnny israeli Jun 24 '13 at 07:54
@johnnyisraeli: What's the title of your post and what is the first line in your post says? – Mhenni Benghorbal Jun 24 '13 at 08:10
3

Note that the denominator in the question is $(1-x)^2$, and not $1-x^2$. – Christian Blatter Jun 24 '13 at 08:27
@ChristianBlatter: Thanks for comment. I'll be looking for it. I was considering the wrong function $\frac{x^n}{1-x^2}$. – Mhenni Benghorbal Jun 24 '13 at 09:21
@Downvoter: It was the first answer to be posted. The only error was that I considered the wrong function. Yet I referred the OP to two different techniques for solving his problem. By the it is ny subject and I have been working on finding the $n$th derivative and the $n$th anti-derivative for long time. I am just what is the down vote for? – Mhenni Benghorbal Jun 26 '13 at 10:50

score -1 · Answer 4 · answered Jun 24 '13 at 07:10

-1

You could write $$\frac{\partial^n}{\partial x^n}x^n(1-x)^{-2}=\sum_{k=0}^n\binom{n}{k}\left(\frac{\partial^k}{\partial x^k}x^n\right)\left(\frac{\partial^{n-k}}{\partial x^{n-k}}(1-x)^{-2}\right)$$ and proceed from there. You'd have it as a finite sum.

answered Jun 24 '13 at 07:10

2'5 9'2

56,991

I need a closed form expression like I'd get with that same method for $\frac{x^{n}}{1-x}$, any ideas in that direction? – johnny israeli Jun 24 '13 at 07:14
You say you are looking for a closed form expression for this derivative. Just curious - why the specification that $0<x<1$? Surely a closed form expression or even a finite sum will be valid on almost all $x$. It's the kind of specification one would have if the expected answer were an infinite series. – 2'5 9'2 Jun 24 '13 at 07:27
In my specific application with protein concentrations that just happens to be the case so I mentioned it in case infinite series can help in the same way that yields $\frac{\partial^n }{\partial x^n} \frac{x^{n}}{1-x}=n!(1-x)^{-(n+1)}$ – johnny israeli Jun 24 '13 at 07:33

Find nth derivative of $\frac{x^{n}}{(1-x)^{2}}$, please?

4 Answers4