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Simplify: $$\frac{1}{P(1-\frac{P}{N})(1-\frac{m}{P})}$$ To show that it is equal to:

$$\frac{N}{N-m} \left[\frac{1}{N-P}+\frac{1}{P-m}\right].$$

I honestly have no idea how to even start the question. I am wondering if the question is asking to go backward, i.e. to go from the second part to the first, but it doesn't seem that way. If someone could show me how to do this or even get me started would be very helpful!

Harry Alli
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dsjkd
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1 Answers1

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Hint:

$$\frac{1}{P(1-\frac{P}{N})(1-\frac{m}{P})}=\frac{1}{(1-\frac{P}{N})(P-m)}=\frac{N}{(N-P)(P-m)}.$$

Then use partial fractions.

$$\frac{N}{(N-P)(P-m)}=\frac{\alpha}{N-P}+\frac{\beta}{P-m}$$

Expand the expression on the right and compare it with the expression on the left-hand side to determine $\alpha$ and $\beta$.

Euler's trick: Instead of expanding the right-hand side you can also use Euler's trick. For example to obtain $\alpha$ multiply by the associated denominator $N-P$ and then set $N-P=0\implies P=N$ to obtain:

$$\frac{N}{P-m}\biggl|_{P=N}=\alpha+\frac{N-P}{P-m} \biggl|_{P=N}\beta \implies \alpha = \frac{N}{N-m}$$

MrYouMath
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    Thankyou so much! this was exactly the hint I needed! – dsjkd Sep 28 '17 at 10:30
  • BTW I added Euler's trick which always can be used if the zeros of the denominator only have a multiplicity of one. – MrYouMath Sep 28 '17 at 10:34
  • Would you be able to show me how to do it using partial fractions? Ive tried to do it but can't seem to get an answer @MrYouMath – dsjkd Sep 28 '17 at 11:32