MHB Simplify equation using binomial theorem

AI Thread Summary
The discussion revolves around simplifying the equation x = (1 - ay/2) / √(1 - ay) using the binomial theorem. A participant suggests that a Maclaurin series is more appropriate for this simplification, leading to the result x ≈ 1 + a²y² / 8 when a is much smaller than 1. Another participant initially attempts to use basic approximations for square roots but ends up with an incorrect result, highlighting the importance of proper series expansion. The conversation concludes with a clarification that the binomial theorem can indeed be applied effectively to achieve the desired simplification. Understanding the correct application of these mathematical tools is essential for accurate simplification.
Perplexed
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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

[math]x = \dfrac{1 - ay/2}{\sqrt{1-ay}}[/math]

to (approximately)

[math]x = 1 + a^2 y^2 / 8[/math]

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
 
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Perplexed said:
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

[math]x = \dfrac{1 - ay/2}{\sqrt{1-ay}}[/math]

to (approximately)

[math]x = 1 + a^2 y^2 / 8[/math]

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.
 
Ackbach said:
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.
 
Ackbach said:
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

[math]x = \dfrac{1 - ay/2}{\sqrt{1-ay}}[/math]

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

[math]x = \dfrac{1 - ay/2}{1-ay/2} \rightarrow (1 - ay/2)(1+ay/2)[/math]

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

Thanks for your help.

Still Perplexed
 
Perplexed said:
Would you mind explaining something for me please? When I first saw

[math]x = \dfrac{1 - ay/2}{\sqrt{1-ay}}[/math]

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

[math]x = \dfrac{1 - ay/2}{1-ay/2} \rightarrow (1 - ay/2)(1+ay/2)[/math]

but multiplying this out gave me [math]1 - \dfrac{a^2y^2}{4}[/math]
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.
 
$ \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 $(1-\frac{1}{2}ay)$.
 
Guest said:
$ \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.
 
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