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How to transform $$\left({1+\sqrt{1-4ab}\over2a}\right)^n-\left({1-\sqrt{1-4ab}\over2a}\right)^n\over \left({1+\sqrt{1-4ab}\over2a}\right)^m-\left({1-\sqrt{1-4ab}\over2a}\right)^m$$ into $$\left({b\over a}\right)^n -1\over \left({b\over a}\right)^m -1$$?


Added: As per comments, there is an additional assumption that $a,b\gt 0$, $b=1-a$, and $b\gt a$.

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@smanoos: I am not quite sure what "divide each expression by one of the expressions" means...? – dan Oct 13 '11 at 13:33
What makes you think you can? If $b=0$, then the first expression simplifies to $a^{m-n}$ and the second expression simplifies to $1$. – Thomas Andrews Oct 13 '11 at 13:37
@ThomasAndrews: Because I am so told... What if I add the constraint that $b>a$ and $b=1-a$ and $a,b>0$? – dan Oct 13 '11 at 13:40
If $b=1-a$, then $1-4ab = 1-4(1-a)a = 1-4a+4a^2 = (1-2a)^2$; since $1-a\gt a$, then $2a\lt 1$, so $1\pm\sqrt{1-4ab} = 1\pm|1-2a| = 1 \pm (1-2a)$. So you have $$\frac{\left(\frac{2-2a}{2a}\right)^n - 1}{\left(\frac{2-2a}{2a}\right)^m - 1}$$and since $2-2a=2(1-a)=2b$, the simplification follows immediately. – Arturo Magidin Oct 13 '11 at 13:56
@ArturoMagidin: Thanks sooo much! :D – dan Oct 13 '11 at 13:57

With the added information, which is key (otherwise, the two expressions are just not equal in general, as noted by Ross):

If $b=1-a$ and $b\gt a$, then $1-a\gt a$ so $1\gt 2a$. Then $$1 -4ab = 1-4a(1-a) = 1-4a+4a^2 = (1-2a)^2,$$ and since $1-2a\gt 0$, we have $$\sqrt{1-4ab} = \sqrt{(1-2a)^2} = |1-2a| = 1-2a.$$ So $$\begin{align*} \frac{1+\sqrt{1-4ab}}{2a} &= \frac{1+(1-2a)}{2a} = \frac{2(1-a)}{2a} = \frac{b}{a},\\ \frac{1-\sqrt{1-4ab}}{2a} &= \frac{1-(1-2a)}{2a} = \frac{2a}{2a} = 1. \end{align*}$$ That means that $$\begin{align*} \left({1+\sqrt{1-4ab}\over2a}\right)^n-\left({1-\sqrt{1-4ab}\over2a}\right)^n\over \left({1+\sqrt{1-4ab}\over2a}\right)^m-\left({1-\sqrt{1-4ab}\over2a}\right)^m & = \frac{\left(\frac{b}{a}\right)^n - 1^n}{\left(\frac{b}{a}\right)^m - 1^m}\\ &=\frac{\left(\frac{b}{a}\right)^n - 1}{\left(\frac{b}{a}\right)^m - 1}. \end{align*}$$

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If $a=0.1, b=0.2, n=5, m=2$ I find the top fraction to be $940.4$ and the second $10.333$ so such a transformation won't work.

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