Question

if n is a positive integer, generalize the binomial coefficient ( p ) for arbitrary n by setting:
n

( p ) = ( p(p-1)....(p-n+1) ) / k!
n

put ( p ) = 1
0

a)show that sum ( p ) x^n converges for abs(x) <1
n
b) show that (1+x)f'(x) -pf(x) =0

c) Deduce from b) that f(x) = 1+x)^p for abs x <1

Answer 1

\[\binom pn=\frac{p(p-1)\cdots(p-n+1)}{n!}~~?\]

Answer 2

The first question is to show that
\[\large\sum_{n=0}^p\binom pnx^n~~\text{converges for }|x|<1\]
which looks like a perfect candidate for the ratio test.

The second question is a differential equation - a separable one.

The third question could involve an application of the binomial theorem.

Answer 3

@ssoleman got it? Or would you like some more in-depth help?

Answer 4

@SithsAndGiggles sorry the site was down... its /k! not n! though...

Answer 5

and do you just use the equation (p(p-1)...p-n+1 ) /k! to do the ratio test?

Answer 6

I think the proff made a mistake and its n!

Answer 7

Right, it has to be \(n\). There's no info about \(k\).

First part: Ratio test:
\[\begin{align*}\lim_{n\to\infty}\left|\frac{\dbinom p{n+1}x^{n+1}}{\dbinom pnx^n}\right|&=\lim_{n\to\infty}\left|\frac{\dfrac{p!}{(n+1)!(p-n-1)!}x^{n+1}}{\dfrac{p!}{n!(p-n)!}x^n}\right|\\\\
&=\lim_{n\to\infty}\left|\frac{1}{(n+1)(p-n-1)}x\right|\\\\
&=|x|\lim_{n\to\infty}\left|\frac{1}{(n+1)(p-n-1)}\right|\\\\
&=0\end{align*}\]
Hmm, looks like the series converges for all \(x\)... Do I have the indices on the series right?

Answer 8

yes thats right! I did that too for the first part

Answer 9

now for b) we have to differentiate...but what do you mean by a seperable one?

Answer 10

and i also dont understand how to show (1+x)f'(x) - pf(x) = 0 ... what is the -p(fx)??

Answer 11

A separable differential equation is of the form
\[A(x,y)\frac{dy}{dx}=B(x,y)\]
which can be "separated" so that you can express the equation as a product of functions of different independent variables \(x\) and \(y\):
\[a(x)b(y)\frac{dy}{dx}=\alpha(x)\beta(y)\]
so that you can rewrite it as separate differentials:
\[\frac{b(y)}{\beta(y)}~dy=\frac{\alpha(x)}{a(x)}~dx\]
This allows you to integrate with respect to a single variable on both sides.

You are given the differential equation
\[(1+x)f'(x)-pf(x)=0\]
Rewriting, you have
\[\begin{align*}
(1+x)f'(x)-pf(x)&=0\\
(1+x)f'(x)&=pf(x)\\
\frac{f'(x)}{f(x)}&=\frac{p}{1+x}\\
\frac{1}{f(x)}\frac{df(x)}{dx}&=\frac{p}{1+x}&\text{(another notation for derivative)}\\
\frac{1}{f(x)}~df(x)&=\frac{p}{1+x}~dx\\
\int\frac{1}{f(x)}~df(x)&=\int\frac{p}{1+x}~dx\\
\ln f(x)&=p\ln(1+x)+C\\
f(x)&=e^{p\ln(1+x)+C}\\
f(x)&=(e^{\ln(1+x)})^pe^C\\
f(x)&=C(1+x)^p\end{align*}\]
Let me know if any of this doesn't make sense. Differential equations can be a lot to handle if this is your first time dealing with them.

In solving the equation, you've established what you need to show for part (c).

Answer 12

So what will part (c) look like?

Answer 13

Part (c) seems to have at least two effective approaches. One is solving the differential equation above, and the other is an application of the binomial theorem.
\[\large\sum_{n=0}^p\binom pna^nb^{p-n}=(a+b)^n\]\
Setting \(a=x\) and \(b=1\), you have
\[\large\sum_{n=0}^p\binom pna^n1^{p-n}=\sum_{n=0}^p\binom pna^n=(x+1)^n\]\
Part (b) is a matter of deriving the differential equation, which isn't too difficult. It's a matter of differentiating the the given power series and algebraic rearrangements.