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$$A_f=\frac{A_o}{1+\beta A_o}$$ rearranging for $\beta $

I get $A_f(1+\beta A_o)=A_o$

$A_f+A_f\beta A_o=A_o$

$A_f\beta A_o=A_o-A_f$

$$\beta = \frac{A_o-A_f}{A_oA_f}$$

The book shows, I don't see how they get this?

$$\beta=\frac{(A_o/A_f)-1}{A_o}$$

bob a
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1 Answers1

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Has to be

$\beta=\frac{(A_o/A_f)-1}{A_o}$

which you get by diving the numerator and denominator of the expression you get by $A_f$

Therefore a typo

Abhimanyu Arora
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