For small $z, (1 + z)^{−2} \sim 1 − 2z$...

Question

I came across the following statement while reading Holmes book on Perturbation Methods -

To reduce the differential equation, recall that, for small $z, (1 + z)^{−2} \sim 1 − 2z$

I don't know why he says 'recall' as I can't see where he mentioned this fact earlier in the book...

For small $z$, the $z$ terms on both sides will disappear and we are left with $1 \sim 1$ which is obviously true. But we could just as easily have $6z, 2000z$ etc.. and we could still say that for small $z$ the LHS and RHS are approximately the same. So what is the significance of saying $2z$ in particular? I think I am missing something here?

Answer 1

You can also use the Taylor's approximation: $f(z)\approx f(0)+f'(0)z$. In your case, $f(x)=\frac{1}{(1+x)^2}$ so $f(0)=1$, $f'(x)=\frac{-2}{(1+x)^3}$ and $f'(0)=-2$.

Answer 2

For $\;|z|<1\;$ we have that

$$\frac1{1+z}=1-z+z^2-z^3+\ldots\implies -\left(\frac1{1+z}\right)'=\frac1{(1+z)^2}=1-2z+3z^2-\ldots$$

since we can differentiate elementwise within the convergence interval.

Answer 3

As people have pointed out, this is pretty much standard. But if this wasn't obvious to you, I'd recommend trying some explicit calculations to see that $(1+z)^{-2}$ and $1 - 2z$ are close to each other. For example, with $z = 0.01$, you have $(1+0.01)^{-2} \approx 0.9803$ and $1-2 \times 0.01 = 0.98$.