Assuming x is small, expand $\frac{\sqrt{1-x}}{\sqrt{1+2x}}$ up to and including the term in $x^{2}$

Question

I have tried this many times but can't quite land on the correct answer.

The correct answer: $1-\frac{3x}{2}+\frac{15x^{2}}{8}$

These are the steps I took:

1. Re wrote it as: $\left ( 1-x \right )^{\frac{1}{2}}\left ( \left ( 1+2x \right )^{\frac{1}{2}} \right )^{-1}$

2. Using bionmal expansion I expanded $\left ( 1-x \right )^{\frac{1}{2}}$ up to $x^{2}$

• For this I get : $\left (1-\frac{x}{2}-\frac{x^{2}}{8} \right)$

3. Then to solve the second bracket written in step 1 I did $\left (1+2x \right )^{\frac{1}{2}}$

• For this I got $\left (1-x+\frac{x^{2}}{4} \right)$

• Then finally I raised this to the power one so $\left (1-x+\frac{x^{2}}{4} \right)^{-1}$

• For this I got $\left (1+x+\frac{3x^{2}}{4} \right)$

4. Then I just expaned those two brakcets so: $\left (1-\frac{x}{2}-\frac{x^{2}}{8} \right)$$\left (1+x+\frac{3x^{2}}{4} \right)$

Once expanded, I then neglected powers bigger than $x^{2}$ (mentioned in question). Then collected like terms. However either my method or the algebra is going wrong, and I just need some help with this.

Answer 1

The first terms of the Taylor expansion of $\sqrt{1+2x}$ near $0$ are $1+x-\frac{x^2}{\color{red}2}$; therefore, the first terms of the Taylor expansion of $\frac1{\sqrt{1+2x}}$ near $0$ are $1-x+\frac{3x^2}2$. And if you multiply $1-\frac x2-\frac{x^2}8$ with $1-x+\frac{3x^2}2$ and then you eliminate those terms whose degree is greater than $2$, you do get indeed $1-\frac{3 x}2+\frac{15 x^2}8$.

Answer 2

$$\frac{\sqrt{1-x}}{\sqrt{1+2x}}=(1-x)^{1/2}(1+2x)^{-1/2}$$ Consider that $(1+t)^{\alpha}\sim 1+\alpha t+\frac{1}{2}(\alpha -1)\alpha t^2$, hence you obtain: $$ (1-x)^{1/2}\sim 1-\frac{x}{2}-\frac{x^2}{8} \quad \text{and}\quad (1+2x)^{-1/2}\sim 1-x+\frac{3x^2}{2}$$ It follows that: $$\left(1-\frac{x}{2}-\frac{x^2}{8}\right)\left(1-x+\frac{3x^2}{2}\right)\sim 1-x+\frac{3x^2}{2}-\frac{x}{2}+\frac{x^2}{2}-\frac{x^2}{8}=1-\frac{3x}{2}+\frac{15x^2}{8}$$

Answer 3

$\frac{1}{1+2x}=1-2x+4x^2-8x^3+O(x^4)$

so

$\frac{1-x}{1+2x}=1-2x+4x^2-8x^3-x+2x^2-4x^3+8x^4+O(x^5)=$

$1-3x+6x^2-12x^3+8x^4+O(x^5).$

Therefore, when $x$ is small,

$\sqrt{\frac{1-x}{1+2x}}\sim 1+\frac{1}{2}\left(-3x+6x^2-12x^3+8x^4\right)-\frac{1}{8}\left(-3x+6x^2-12x^3+8x^4\right)^2\sim$

$1-\frac{3}{2}x+3x^2-\frac{9}{8}x^2=1-\frac{3}{2}x+\frac{15}{8}x^2$