Taylor series example

Question

Task : Write down the McLaurin series till the 4th power ( Hope the translation from german is good :) ).

$f(x) = (e^{-x} - 1)^2$

What I did , because I thought that expanding $(e^{-x} - 1)^2$ till the 2nd power and having 2nd power above the paranthesis will give me the 4th power:

$(e^{-x} - 1)^2 = (1 - x + x^2\frac1{2!} + ... - 1)^2$ $$(x(-1 + x\frac12)^2= x^2(x^2\frac14 - x + 1)$$

What the author did :

$(e^{-x} - 1)^2 = (- x + x^2\frac1{2!} - x^3\frac16 + x^4\frac1{24}-+...) * (- x + x^2\frac1{2!} - x^3\frac16 + x^4\frac1{24}-+...) $

$f(x) = (e^{-x} - 1)^2 => x^2 - x^3 + x^4(\frac16 + \frac14 + \frac16)$

That's clearly different from mine , why is my thought wrong ?

Answer 1

The author is correct.

You could have done $$\left(x\left(-1 + x\frac12 - x^2\frac16+\cdots\right)\right)^2 = x^2\left(1-x+\frac7{12}x^2-\cdots\right).$$

For the fourth power you have taken $x^2\left(x\frac12\right)^2$ but missed $x^2\left(2 \times -1 \times -x^2 \frac16\right)$.

Answer 2

This is correct : you must not stop so early because terms would appear when squaring.

For your example, start with $$e^{-x}=1-x+\frac{x^2}{2}-\frac{x^3}{6}+\frac{x^4}{24}+O\left(x^5\right)$$ $$e^{-x}-1=x+\frac{x^2}{2}-\frac{x^3}{6}+\frac{x^4}{24}+O\left(x^5\right)$$ $$e^{-x}-1\approx x\Big(1+\frac{x}{2}-\frac{x^2}{6}+\frac{x^3}{24} \Big)$$ $$(e^{-x}-1)^2\approx x^2\times\Big(1+\frac{x}{2}-\frac{x^2}{6}+\frac{x^3}{24} \Big)\times\Big(1+\frac{x}{2}-\frac{x^2}{6}+\frac{x^3}{24} \Big)$$and now use multiplication of polynomials; this will show how are composed the different powers of $x$.

I am sure that you can take from here.