show equality - binomial formula, taylor?

Question

I am trying to show that this is true using the binomial formula or some taylor expansion:

$\frac{1}{(1+\epsilon \sum\limits_{n=0}^\infty Z_n(t) \epsilon^n)^2} = 1 - 2Z_0\epsilon + (3Z_0^2-2Z_1)\epsilon^2+O(\epsilon^3)$

So far i tried doing it using the binomial formula but I can't get it to work:

$\frac{1}{(1+\epsilon \sum\limits_{n=0}^\infty Z_n(t) \epsilon^n)^2} = \frac{1}{1+(\epsilon \sum\limits_{n=0}^\infty Z_n(t) \epsilon^n)^2+2\epsilon \sum\limits_{n=0}^\infty Z_n(t) \epsilon^n} = \frac{1}{1+(\epsilon(Z_0\epsilon^0+Z_1\epsilon^1+Z_2\epsilon^2))^2+2\epsilon(Z_0\epsilon^0+Z_1\epsilon^1+Z_2\epsilon^2) + O(\epsilon^3)} = \frac{1}{\epsilon^2 Z_0^2+2\epsilon^2 Z_1+2\epsilon Z_0+1+O(\epsilon^3)} $

Answer 1

Consider first the denominator as $$A=\left(1+\epsilon \sum _{n=0}^p Z_n \epsilon ^n\right)^2$$ Expand it using the binomial theorem to get $$A=1+2 Z_0 \epsilon +\left(Z_0^2+2 Z_1\right) \epsilon ^2+2 (Z_0 Z_1+Z_2) \epsilon ^3+\left(Z_1^2+2 Z_0 Z_2+2 Z_3\right) \epsilon ^4+\cdots $$ Now, use the long division to get $$\frac 1 A=1-2 Z_0 \epsilon +\left(3 Z_0^2-2 Z_1\right) \epsilon ^2+\left(-4 Z_0^3+6 Z_1 Z_0-2 Z_2\right) \epsilon ^3+\cdots$$

Another way to do it (since you are not concerned by high order expansions) could be to write $$1=A \sum_{i=0}^k a_i \epsilon^i$$ Expand the product for a few terms and cancel the coefficient for each power of $\epsilon^i$. You will get the same result.

Answer 2

By Taylor, $$\frac1{(1+t)^2}=1-2t+3t^2+O(t^3).$$

The relevant terms from the sum (up to $\epsilon^2$) yield

$$1-2(Z_0\epsilon+Z_1\epsilon^2)+3Z_0^2\epsilon^2+O(\epsilon^3).$$

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For the next approximation,

$$\frac1{(1+t)^2}=1-2t+3t^2-4t^3+O(t^4)$$

and up to $\epsilon^3$, $$1-2(Z_0\epsilon+Z_1\epsilon^2+Z_2\epsilon^3)+3(Z_0^2\epsilon^2+2Z_0Z_1\epsilon^3)-4Z_0^3\epsilon^3+O(\epsilon^4).$$

Answer 3

$$ \begin{align} &\frac1{\left(1+\epsilon\sum\limits_{n=0}^\infty Z_n\epsilon^n\right)^2}\\ &=\frac1{\left(1+\epsilon Z_0+\epsilon^2Z_1+O\!\left(\epsilon^3\right)\right)^2}\\[2pt] &=\frac1{1+2\left(\epsilon Z_0+\epsilon^2Z_1\right)+\epsilon^2Z_0^2+O\!\left(\epsilon^3\right)}\\[6pt] &=1-\left(2\epsilon Z_0+2\epsilon^2Z_1+\epsilon^2Z_0^2\right)+\left(2\epsilon Z_0+2\epsilon^2Z_1+\epsilon^2Z_0^2\right)^2+O\!\left(\epsilon^3\right)\\[12pt] &=1-2\epsilon Z_0-2\epsilon^2Z_1-\epsilon^2Z_0^2+4\epsilon^2 Z_0^2+O\!\left(\epsilon^3\right)\\[12pt] &=1-2Z_0\epsilon+\left(3Z_0^2-2Z_1\right)\epsilon^2+O\!\left(\epsilon^3\right) \end{align} $$