- #1
ritwik06
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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??