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I need to show the following:

$$ \frac{\sin^2 x}{x^2 \sqrt{1-\frac{\sin^2 x}{3}}} \approx 1-\frac{x^2}{6} $$ when $ x $ is small.

I think this problem is trickier than most other questions like it because in the original source there is comment saying "if you got $ 1+\frac{x^2}{6} $ [what I got] then think again!". My attempt was:

When $ x $ is small, $ \sin x \approx x $ so

$$ \frac{\sin^2{x}}{x^2 \sqrt{1-\frac{\sin^2 x}{3}}} = \frac{x^2}{x^2 \sqrt{1-\frac{x^2}{3}}} = \left ( 1 - \frac{x^2}{3} \right )^{-\frac{1}{2}} $$

Then using the binomial series approximation,

$$ \left ( 1 - \frac{x^2}{3} \right )^{-\frac{1}{2}} \approx 1 - \frac{1}{2}\left ( -\frac{x^2}{3} \right ) + ... = 1 + \frac{x^2}{6} $$

...and so it looks like I've fallen into whatever trap the question set.

Where is my error?

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$${\sin x\over x}\approx1-{1\over6}x^2$$

so

$${\sin^2x\over x^2}\approx\left(1-{1\over6}x^2\right)^2\approx1-{1\over3}x^2$$

not just $1$. We get

$${\sin^2x\over x^2\sqrt{1-{\sin^2x\over3}}}\approx\left(1-{1\over3}x^2\right)\left(1+{1\over6}x^2\right)\approx1-{1\over6}x^2$$

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    $\begingroup$ Is there any way to know when I should use more than one term to approximate sin(x) (like you did here)? $\endgroup$ – Nick_2440 Jul 17 at 21:04
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    $\begingroup$ @Nick_2440, it might be a good idea to carry asymptotics along explicitly in the form of $\sin x=x-{1\over6}x^3+O(x^5)$, etc. That way if you try $\sin x=x+O(x^3)$ you quickly see that ${\sin x\over x}=1+O(x^2)$ isn't good enough to give you an answer of the form $1-{1\over6}x^2+O(x^3)$. In fact it might be a good exercise to write out your own answer here doing that. $\endgroup$ – Barry Cipra Jul 17 at 21:21
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Suppose for simplicity you had two polynomials $P(z) = 1+ z + z^2 + 4z^4 + 7z^5$ and $Q(z) = 1 + 2z + 3z^2 + 4z^3$, and I asked you to calculate the product $P(z)Q(z)$... but not the entire thing. Suppose I only want the terms up to quadratic term; i.e if we write \begin{align} P(z)Q(z) &= a_0 + a_1z + a_2z^2 + a_3z^3 + a_4z^4 + a_5z^5 +a_6z^6 + a_7z^7 +a_8z^8 \end{align} then I'm asking you to find the coefficients $a_0,a_1,a_2$ (but for now, let's just say for some reason I'm interested in what happens when $|z|$ is very small up to an accuracy of quadratic order, so I don't really care about the rest of the terms ). Well, we just multiply everything out: \begin{align} P(z)Q(z) &= (1+ z + z^2 + 4z^4 + 7z^5)(1 + 2z + 3z^2 + 4z^3) \\ &= 1 + (1\cdot 2z + z \cdot 1) + (1\cdot 3z^2 + z\cdot 2z + z^2 \cdot 1) \\ &+ \text{(terms involving $z^3$ or higher, which I don't care about for now)} \\ &= 1 + 3z + 6z^2 + O(z^3) \end{align} In other words, because in my final product, I'm only interested in calculating up to the quadratic term, I can simply truncate the polynomials $P$ and $Q$ to quadratic order, and then multiply them out (and then again only keep terms up to quadratic order): \begin{align} P(z)Q(z) &= (1 + z + z^2 + \cdots)(1 + 2z + 3z^2 + \cdots) \\ &= 1 + 3z + 6z^2 + O(z^3) \end{align}

Again, because I'm only interested up to quadratic order, there's no need for me to keep any terms beyond that for $P(z)$ and $Q(z)$, because if I approximate $P(z) \approx 1+ z + z^2 + \color{red}{4z^4}$ (i.e I keep the $4^{th}$ order term) and I multiply with $Q(z) = 1+2z+3z^2 + 4z^3$, then the red term multiplied with anything in $Q(z)$ will yield terms which are $4^{th}$ order or higher (which I don't care about).

But what you should not do is truncate $P(z)$ and $Q(z)$ up to linear order, and say that \begin{align} P(z)Q(z) & \approx (1+z)(1+2z) = 1 + 3z + 2z^2 \end{align} Because in this way, you're missing out other second order contributions (by multiplying constant term of $P$ with quadratic term of $Q$ and vice-versa).


This is how you know how many terms you need to use in your approximation. In your case, you want to approximate \begin{align} f(x) &= \dfrac{\sin^2x}{x^2\sqrt{1 - \frac{\sin^2x}{3}}} \end{align} up to $2^{nd}$ order. So, what do we do? We write things as a product first: \begin{align} f(x) &= \left(\dfrac{\sin x}{x}\right)\cdot\left(\dfrac{\sin x}{x}\right) \cdot \left(\dfrac{1}{\sqrt{1- \frac{\sin^2x}{3}}}\right)\tag{$1$} \end{align} Now, we have to expand each bracketed term up to atleast second order in $x$ and then multiply the result together. First: \begin{align} \dfrac{\sin x}{x} &= \dfrac{x - \dfrac{x^3}{6} + O(x^4)}{x} = 1 - \dfrac{x^2}{6} + O(x^3) \tag{$2$} \end{align} Next, we recall that \begin{align} \dfrac{1}{\sqrt{1-z}} &= 1+ \dfrac{z}{2} + \dfrac{3z^2}{8} + O(z^3) \end{align} Now, plug in $z= \frac{\sin^2x}{3} = x + O(x^3)$, to get \begin{align} \dfrac{1}{\sqrt{1-\frac{\sin^2x}{3}}} &= 1+ \dfrac{1}{2}\left(\dfrac{\sin^2x}{3}\right) + \dfrac{3}{8}\left(\frac{\sin^2x}{3}\right)^2 + O((\sin^2 x)^3) \\ &= 1 + \dfrac{1}{2}\left(\dfrac{x^2 + O(x^4)}{3}\right) + O(x^4) + O(x^6) \\ &= 1 + \dfrac{1}{6}x^2 + O(x^4)\tag{$3$}, \end{align} where in the second line, hopefully it's clear how I got the various terms: for example $\sin x = x + O(x^3)$, so $\left(\frac{\sin^2x}{3}\right)^2 = \frac{1}{9}\sin^4x = \frac{1}{9} (x + O(x^3))^4 = O(x^4)$. Therefore, the final answer is obtained by plugging $(2)$ and $(3)$ into $(1)$ : \begin{align} f(x) &= \left(1 - \dfrac{x^2}{6} + O(x^3)\right)^2 \cdot \left(1 + \dfrac{1}{6}x^2 + O(x^4)\right) \\ &= 1 - \dfrac{x^2}{6} + O(x^4) \end{align}


Long story short, if your end goal is to calculate up to second order, then at each stage of your algebra make sure you're keeping terms atleast up to $x^2$.

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    $\begingroup$ +1 for a fantastic answer. While experienced users may not really want to read the long story, it's a boon for beginners. $\endgroup$ – Paramanand Singh Jul 18 at 0:49
  • $\begingroup$ I wish I could choose 2 answers as accepted...this was very well explained, thanks :) $\endgroup$ – Nick_2440 Jul 18 at 11:19
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You can also note the the singularity at $x=0$ is removable and that $f$ is in fact at least 4 times differentiable. Taylor's expansion gives you the answer...

$$ f(x)=f(0)+f'(0)x + \frac 12 f''(0) x^2 + O(x^3) $$

where

$$ f(0)=\lim_{x\to 0}f(0) = 1, \quad f'(0) = \lim_{x\to 0}f'(x)=0, \quad f''(0)=\lim_{x\to 0}f''(x)= -\frac 13, \quad f'''(0)=0 $$

yielding

$$ f(x)=1-\frac 16 x^2 + O(x^4)\approx 1-\frac 16 x^2 (\textrm{for small } x ). $$

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  • $\begingroup$ Typo in last line. $\endgroup$ – spalein Jul 17 at 21:47
  • $\begingroup$ @spalein thanks. $\endgroup$ – PierreCarre Jul 18 at 10:35

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