Why is $\sin(x^{2})$ similar to $\sin(x) \cdot x$?

Question

Why is $\sin(x^2)$ similar of $\ x \sin(x)$?

I graphed it using desmos and when I look at it, the behavior as x approaches zero seems to be to oscillate less.

Yet as x approaches infinity and negative infinity $\sin(x^2)$ oscillates between y=1 and y=-1 while $\ x *sin(x)$ oscillates between y=x and y=-x.

I was wondering why these functions are so similar yet so different. I'm in 10th grade and I"m currently learning precalculus so if answers could be targeted to a precalculus level that would be great.

It's not at kind of "version." It is similar, but the notion of "version" here is mathematically useless. — Thomas Andrews, Nov 26 '15 at 23:39
For small values of $y$, $\sin(y) \approx y$ is what you get from linear approximation (a major topic of calculus). Therefore $\sin(x^2) \approx x^2$ and $\sin(x) \approx x$ for small $x$. Thus $\sin(x) \cdot x \approx x^2$ for small $x$ as well. For large $x$ they will be completely different, because $x \sin(x)$ will have growing amplitude. — Ian, Nov 26 '15 at 23:45
This works for many functions where $f(0)=0$. For example, consider $f(x)=2^x-1$. Try graphing $f(x^2)=2^{x^2}-1$ and $xf(x)=x(2^x-1)$. — Akiva Weinberger, Nov 27 '15 at 00:24

score 6 · Accepted Answer · answered Nov 26 '15 at 23:46

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When $x$ gets close to zero, $\sin x \approx x-\frac{x^3}{6}$. So $$\sin(x^2)\approx x^2-\frac{x^6}{6}\\x\sin(x)\approx x^2-\frac{x^4}{6}$$

Now, when $x$ is small, $x^4$ and $x^6$ are "very small." So the functions are dominated by $x^2$ near $x=0$. Indeed, if you graphed $y=x^2$ alongside, you'd see that both of your functions are close to but smaller than $y=x^2$.

Add in $y=\sin^2(x)$, and it will be similar, too.

When you get to calculus, this will be explored by studying "power series" for functions.

We also see from this approximation that since $\frac{x^6}{6}<\frac{x^4}{6}$ when $x$ near enough to zero (say $|x|<1$) we see that $\sin(x^2)>x\sin(x)$.

answered Nov 26 '15 at 23:46

Thomas Andrews

177,126

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Thank you! That makes sense. So they're not similar because they're closely related functions( sorry for the wording ) but because as x approaches 0 the small numbers mean that the functions are dominated by the same term in both (x^2 in this case). – SocketPhys Nov 26 '15 at 23:51
Yes. If $f(x)$ has a power series at $0$ and $f(0)=0$, then it is true that $xf(x)$ will be "close to" $f(x^2)$, and that will be as strong as above if $f(x)$ is an odd function. But that's because the functions will agree at the first term of the power series. – Thomas Andrews Nov 26 '15 at 23:53
@ThomasAndrews Can you help me rigorously understand the language that you use? If we were in a topology setting, and I said "for all $x$ near the point $y$, blah blah blah", using that language is another way of saying "for all $x$ in some open neighborhood of $y$, blah blah blah". I feel like it's the same story when you say "when $x$ is small, $x^{4}$ and $x^{6}$ are very small, so they are dominated by $x^{2}$". Can you help me interpret this formally/rigorously? I have had so much trouble recently interpreting this language. I hope my question is clear. – layman Nov 26 '15 at 23:57
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@user46944 I think it'd be:$$\lim_{x\to\color{Blue}0}\frac{x^4}{x^2}=\color{Red}0$$(or the equivalent for $x^6$). Topologically, this means that, if $f(x)=\dfrac{x^4}{x^2}$, and $A$ is a neighborhood of $\color{Red}0$, then $f^{-1}(A)$ is a deleted neighborhood of $\color{Blue}0$. (EDIT: Added color-coding) – Akiva Weinberger Nov 27 '15 at 00:02
Topology doesn't have a notion of distance or order, so it won't answer this. But what you can show is that for some $\delta_1>0$ and $C_1>0$ we have that $|\sin(x^2)-(x^2-x^6/6)|<C_1x^8$ when $|x|<\delta$. Similarly, for we have $\delta_2,C_2$ so that $|x\sin x-(x^2-x^4/6)|<C_2x^6$ for $|x|<\delta_2$. This is often written with "big-O" notation, since the above is verbose: $$\sin(x^2) = x^2 -x^6/6 + O(x^8)\x\sin x = x^2-x^4/6 + O(x^6)$$ The big-O notation statement just means the same thing, but is a powerful way to talk about approximations like this. – Thomas Andrews Nov 27 '15 at 00:02
For example, we can say $\sin(x^2)=x\sin(x) + O(x^4)=x^2+O(x^4)$ – Thomas Andrews Nov 27 '15 at 00:04
@ThomasAndrews You hit the nail on the head when introducing big-O notation. This notation really throws me off, because it doesn't make sense to me that we are saying $f(x) = g(x) + O(h(x))$ means $|f(x) - g(x)| < Ch(x)$. Why does the former involve an equality, and the latter involve an inequality? In my mind, $f(x) = g(x) + O(h(x))$ should mean that $f(x)$ is $g(x)$ plus some constant times $h(x)$, even though this isn't true. – layman Nov 27 '15 at 00:06
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@user46944 Added color-coding so you could see which $0$ corresponded to which. – Akiva Weinberger Nov 27 '15 at 00:07

Akiva Weinberger · Answer 2 · 2015-11-26T23:58:44.690

In calculus, you learn that many functions can be written as "infinite polynomials." For example, there's this: $$\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\dotsb$$ Yes, those are factorials. A consequence of this is that, since $\sin\pi=0$, we have: $$0=\pi-\frac{\pi^3}{3!}+\frac{\pi^5}{5!}-\frac{\pi^7}{7!}+\dotsb$$ This is not obvious.

Note that it begins with $x$, rather than $2x$ or $\frac x2$ or whatever. One consequence of this is that, when $x\approx0$, we have $\sin x\approx x$ (since all of the other terms are much smaller than $x$ when $x$ is small). This is actually much easier to prove than the rest of the series; the usual proof involves some geometry and the unit circle. However, for what follows, the exact coefficient of $x$ doesn't really matter.

In any case, multiplying by $x$, we get: $$x\sin x=x^2-\frac{x^4}{3!}+\frac{x^6}{5!}-\frac{x^8}{7!}+\dotsb$$ Or, replacing the $x$s with $x^2$s: $$\sin x^2=x^2-\frac{x^6}{3!}+\frac{x^{10}}{5!}-\frac{x^{14}}{7!}+\dotsb$$ We can see that these both begin with $x^2$. As $x$ gets smaller, these other terms are much less significant than the $x^2$ term. This means that, when $x\approx0$, we have $x^2\approx x\sin x\approx\sin x^2$.

score 0 · Answer 3 · answered Nov 26 '15 at 23:48

Try finding the $ \lim_{x \rightarrow 0} \frac{\sin(x^2)}{x \sin(x)}$, you will see it is finite, thus they behave simiraly near $0$, which is important for approximations near $0$. Although $\sin(x^2)$ is bounded, and $x \sin(x)$ is not, they have local minima and maxima at the same points.

score 0 · Answer 4 · answered Nov 27 '15 at 10:49

This is not an answer but it is too long for a comment.

You received the explanation of what you observed.

I give you another one which is also amazing : function $y=\sin^x(x)$ in the range $(2\pi,3\pi)$ almost looks like a gaussian. Doing things similar as in the answers, consider Taylor expansion around $x=\frac{5\pi} 2$ $$x\log(\sin(x))=-\frac{5\pi}{4} \left(x-\frac{5 \pi }{2}\right)^2+O\left(\left(x-\frac{5 \pi }{2}\right)^3\right)$$ $$\sin^x(x)\simeq e^{-\frac{5\pi}{4} \left(x-\frac{5 \pi }{2}\right)^2}$$ Pu the two functions on the same plot and enjoy.

Why is $\sin(x^{2})$ similar to $\sin(x) \cdot x$?

4 Answers4