Binomial Theorem(Approximation)

[ritwik06]

Homework Statement

If p is nearly equal to q and n>1, show that [tex]\frac{(n+1)p+(n-1)q}{(n-1)p+(n+1)q}=(\frac{p}{q})^{\frac{1}{n}}[/tex]
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 postive integral.

Homework Equations

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

when |x|<<1

The Attempt at a Solution

Solving L.H.S.
[tex]\frac{np+p+nq-q}{np-p+nq+q}[/tex]


[tex]\frac{n(p+q)+p-q}{n(p+q)-p+q}[/tex]



[tex]\frac{n(p+q)-p+q+2p-2q}{n(p+q)-p+q}[/tex]


[tex]1+\frac{2(p-q)}{n(p+q)-(p-q)}[/tex]


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:
[tex]\frac{n(p+q)+p-q}{n(p+q)-p+q}[/tex]



can be written as:


[tex]\frac{n(p+q)+(p-q)}{n(p+q)-(p-q)}[/tex]


this generated another format:
[tex]\frac{A+B}{A-B}[/tex]

Multiplying numerator and denominator by A+B

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

[tex]1+\frac{2(p-q)}{n(p+q)}[/tex]

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

 

[tiny-tim]

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

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?