# Simplify equation using binomial theorem

#### Perplexed

##### New member
I'm sure this is easy but it has got me baffled.

I'm told that the binomial theorem can be used to simplify the following formula

$$\displaystyle x = \dfrac{1 - ay/2}{\sqrt{1-ay}}$$

to (approximately)

$$\displaystyle x = 1 + a^2 y^2 / 8$$

if a << 1.

Thanks for any help or pointers on this one in particular, and/or general help on how the binomial theorem is used in cases like this.

Perplexed

#### Ackbach

##### Indicium Physicus
Staff member
I'm sure this is easy but it has got me baffled.

I'm told that the binomial theorem can be used to simplify the following formula

$$\displaystyle x = \dfrac{1 - ay/2}{\sqrt{1-ay}}$$

to (approximately)

$$\displaystyle x = 1 + a^2 y^2 / 8$$

if a << 1.

Thanks for any help or pointers on this one in particular, and/or general help on how the binomial theorem is used in cases like this.

Perplexed
Hmm. I don't know if the binomial theorem would help here or not. It looks to me more like an application of a Maclaurin series:
$$f(x) \approx f(0) + f'(0) x + \frac{ f''(0) x^{2}}{2!}+ \frac{f'''(0)x^{3}}{3!}+\dots.$$
So we'll need $f'(0)$. To find the derivative, we use the Quotient Rule:
$$f'(y)= \frac{ \left( \sqrt{1-ay} \right)(-a/2)-(1 - ay/2)(1/2)(1-ay)^{-1/2}(-a)}{1-ay}.$$
Evaluating this at $y=0$ yields
$$f'(0)=(-a/2)-(1)(1/2)(-a)=0.$$
Hence, the linear term disappears. We'll need the quadratic term, then. Simplifying the first derivative yields
\begin{align*}
f'(y)&= \frac{ -(a/2) \sqrt{1-ay}+ \frac{(1 - ay/2)(a/2)}{ \sqrt{1-ay}}}{1-ay} \\
&= \frac{-(a/2)(1-ay)+(a/2)(1-ay/2)}{(1-ay)^{3/2}} \\
&=\frac{a^{2}y}{4(1-ay)^{3/2}}.
\end{align*}
The second derivative yields
\begin{align*}
f''(y)&= \frac{4(1-ay)^{3/2}(a^{2})-6(a^{2}y)(1-ay)^{1/2}(-a)}{16(1-ay)^{3}} \\
&= \frac{2(1-ay)(a^{2})+3a^{3}y}{8(1-ay)^{5/2}} \\
&= \frac{a^{2}(2+ay)}{8(1-ay)^{5/2}}.
\end{align*}
Hence, $f''(0)=2a^{2}/8=a^{2}/4.$ It follows, then, that the Maclaurin expansion yields
$$f(y) \approx 1 + \frac{a^{2}y^{2}/4}{2!}=1+ \frac{a^{2}y^{2}}{8},$$
as desired.

#### Perplexed

##### New member
Hmm. I don't know if the binomial theorem would help here or not. It looks to me more like an application of a Maclaurin series:
Wow! Thanks for that, I will need some time to absorb and understand it, but thanks for working through it for me.

#### Perplexed

##### New member
Hmm. I don't know if the binomial theorem would help here or not. It looks to me more like an application of a Maclaurin series:
Would you mind explaining something for me please? When I first saw

$$\displaystyle x = \dfrac{1 - ay/2}{\sqrt{1-ay}}$$

and given that a << 1, I thought I might be able to simplify it using the rule
$$\displaystyle \sqrt{1-x} \rightarrow (1-x/2)$$ and $$\displaystyle \dfrac{1}{\sqrt{1-x}} \rightarrow (1+x/2)$$
so that it became

$$\displaystyle x = \dfrac{1 - ay/2}{1-ay/2} \rightarrow (1 - ay/2)(1+ay/2)$$

but multiplying this out gave me $$\displaystyle 1 - \dfrac{a^2y^2}{4}$$
so there is something wrong with my simple simplification, but I can't see what.

Still Perplexed

#### Ackbach

##### Indicium Physicus
Staff member
Would you mind explaining something for me please? When I first saw

$$\displaystyle x = \dfrac{1 - ay/2}{\sqrt{1-ay}}$$

and given that a << 1, I thought I might be able to simplify it using the rule
$$\displaystyle \sqrt{1-x} \rightarrow (1-x/2)$$ and $$\displaystyle \dfrac{1}{\sqrt{1-x}} \rightarrow (1+x/2)$$
so that it became

$$\displaystyle x = \dfrac{1 - ay/2}{1-ay/2} \rightarrow (1 - ay/2)(1+ay/2)$$

but multiplying this out gave me $$\displaystyle 1 - \dfrac{a^2y^2}{4}$$
so there is something wrong with my simple simplification, but I can't see what.

Still Perplexed
Yes, I thought of doing that as well, and got the same result. It comes down to a different approximation - probably not as good, since you're only treating part of the function as a series.

#### Guest

##### Active member
$\displaystyle (1+z)^{s} = 1+sz+\frac{s(s-1)}{2}z^2+\cdots$ (Binomial Theorem)

Let $\displaystyle s= -1/2$ and $\displaystyle z= -ay$ and multiply both sides by $(1-\frac{1}{2}ay)$.

#### Perplexed

##### New member
$\displaystyle (1+z)^{s} = 1+sz+\frac{s(s-1)}{2}z^2+\cdots$ (Binomial Theorem)

Let $\displaystyle s= -1/2$ and $\displaystyle z= -ay$ and multiply both sides by $(1-\frac{1}{2}ay)$.
That does it! Thanks.

Not quite so Perplexed.