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Simplify equation using binomial theorem

Perplexed

New member
Feb 7, 2014
6
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
Jan 26, 2012
4,193
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
Feb 7, 2014
6
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
Feb 7, 2014
6
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.

Thanks for your help.

Still Perplexed
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,193
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.

Thanks for your help.

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
Jan 4, 2014
199
$ \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
Feb 7, 2014
6
$ \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.