Binomial Theorem(Approximation)

[ritwik06]

Homework Statement

If p is nearly equal to q and n>1, show that $$\frac{(n+1)p+(n-1)q}{(n-1)p+(n+1)q}=(\frac{p}{q})^{\frac{1}{n}}$$
Note: the index 1/n is on the whole fraction (p/q)

I think it might be helpful if I specify th chapter from which I got this question. Its the binomial thorem for indexes other than the one which are positive integral.

Homework Equations

I wonder if this needs to be used:
$$1+nx\approx(1+x)^{n}$$

when |x|<<1

The Attempt at a Solution

Solving L.H.S.
$$\frac{np+p+nq-q}{np-p+nq+q}$$


$$\frac{n(p+q)+p-q}{n(p+q)-p+q}$$



$$\frac{n(p+q)-p+q+2p-2q}{n(p+q)-p+q}$$


$$1+\frac{2(p-q)}{n(p+q)-(p-q)}$$


as p tends to q, p-q should be a very small number (which might help me in the approximation)
Can the fraction aded to x be converted to the format mentioned in the relevant equations?
Or is there any other way out??

 

[ritwik06]

Another way to think about the question which I thought makes things simpler:
$$\frac{n(p+q)+p-q}{n(p+q)-p+q}$$



can be written as:


$$\frac{n(p+q)+(p-q)}{n(p+q)-(p-q)}$$


this generated another format:
$$\frac{A+B}{A-B}$$

Multiplying numerator and denominator by A+B

and neglecting the square terms of (p-q), I get

$$1+\frac{2(p-q)}{n(p+q)}$$

Can $$\frac{2(p-q)}{(p+q)}$$ be proved approximately equal to (p/q)?

 

[tiny-tim]

If p is nearly equal to q and n>1, show that $$\frac{(n+1)p+(n-1)q}{(n-1)p+(n+1)q}\ =\ \left(\frac{p}{q}\right)^{\frac{1}{n}}$$

Hi ritwik06!

I haven't worked this out, so I don't know that it works, but I would think that the clue " p is nearly equal to q" means that you should start by saying

"Let q = p(1 + k) where k << 1"

Does that help?