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I was trying to prove that for $n \geq 1$ and $n = 0$, $$\lvert \sin(nx)\rvert \leq n \cdot \lvert \sin(x) \rvert$$

Turns out it's actually not true...but I don't know why. My proof seems reasonable to me, please help me find the problem(s).

EDIT: This is for real numbers $n$

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Proof:

Since we're interested in the absolute value of $\sin(nx)$ we only have to consider $nx$ over the interval $[0, \frac{\pi}{2}]$, because any angle $kz \gt \frac{\pi}{2}$ can be expressed as $nx$ such that $\lvert \sin(nx) \rvert = \lvert \sin(kz) \rvert$.

We know that the rate of change of $\sin(x)$ with respect to $x$ decreases continuously as $x$ increases, for $x$ over $[0, \frac{\pi}{2}]$.

We also know that any angle $nx$ can be broken into the sum of $x$ and $(nx - x)$.

Therefore we know that $$\frac{\sin(x)}{x} \geq \frac{\sin(nx) - \sin(x)}{nx - x}$$

In other words, since $\frac{\mathrm{d}}{\mathrm{d}x} \sin(x)$ is continuously decreasing over the interval $[0, \frac{\pi}{2}]$, the slope over the interval $[0,x]$ is always greater than or equal to the slope over $[x, nx]$.

From this, we get

$$nx \cdot \sin(x) - x \cdot \sin(x) \geq x \cdot \sin(nx) - x \cdot \sin(x)$$

$$nx \cdot \sin(x) \geq x \cdot \sin(nx)$$

$$n \cdot \sin(x) \geq \sin(nx)$$

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I'm pretty sure what's wrong is my first assumption that: Since we're interested in the absolute value of $\sin(nx)$ we only have to consider $nx$ over the interval $[0, \frac{\pi}{2}]$, because any angle $kz \gt (\frac{\pi}{2})$ can be expressed as $nx$ such that $\lvert \sin(nx) \rvert = \lvert \sin(kz) \rvert$.

But I'm not sure exactly why. I'd really appreciate an explanation of why this part or others of my proof are wrong.

EDIT: A second question I have is whether my proof would actually work if we did indeed constain $nx$ to $[0, \frac{\pi}{2}]$. What about constraining just $x$ to $[0, \frac{\pi}{2}]$

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  • $\begingroup$ @MattSamuel, thank you but that was just a typo, though. I meant $pi/2$ as in the rest of the proof. $\endgroup$ – jeremy radcliff Oct 9 '15 at 22:22
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    $\begingroup$ $\lvert \sin(nx)\rvert \leq n \cdot \lvert \sin(x) \rvert$ is true (for nonnegative integers $n$ and real numbers $x$). I cannot see from your question why you think that it is false. $\endgroup$ – Martin R Oct 9 '15 at 22:25
  • $\begingroup$ @MartinR Well, if we let $n$ be a non-integer it's definitely false, and nowhere does the OP specify that $n$ has to be an integer . . . $\endgroup$ – Noah Schweber Oct 9 '15 at 22:57
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    $\begingroup$ Th line starting "Therefore we know that..." Um, how? $\endgroup$ – Thomas Andrews Oct 9 '15 at 22:58
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    $\begingroup$ If $n$ is not an integer, you can't restrict yourself to $[0,\pi/2]$ because $|\sin n(x+\pi)|\neq |\sin nx|$. $\endgroup$ – Thomas Andrews Oct 9 '15 at 22:59
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$\lvert \sin(nx)\rvert \leq n \cdot \lvert \sin(x) \rvert$ holds for all non-negative integers $n$ and all real numbers $x$. It can be proven easily using induction since $$ \lvert \sin((n+1)x)\rvert = \lvert \sin(nx + x)\rvert = \lvert \sin(nx) \cos(x) + \cos(nx) \sin(x) \rvert \\ \le \lvert \sin(nx) \cos(x) \rvert + \lvert \cos(nx)\sin(x) \rvert \le \lvert \sin(nx) \rvert + \lvert \sin(x) \rvert \, . $$

If $n$ is not an integer then the inequality cannot hold for all $x \in \mathbb R$, it is false e.g. for $x = \pi$.

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  • $\begingroup$ Thank you but $n$ is not supposed to be an integer, that was the problem. I can find a counter example for when $n$ is not an integer, but I don't understand what is wrong with my assumption... $\endgroup$ – jeremy radcliff Oct 9 '15 at 23:41
  • $\begingroup$ @jeremyradcliff: If $n$ is not an integer then $x=\pi$ is a counter example, and your inequality $\frac{\sin(x)}{x} \geq \frac{\sin(nx) - \sin(x)}{nx - x}$ already fails for $x=\pi$ because the lhs is zero. $\endgroup$ – Martin R Oct 9 '15 at 23:50
  • $\begingroup$ @jeremyradcliff: There already seems to be a flaw in your initial argument that you only have to consider $nx$ over the interval $[0, \frac{\pi}{2}]$. From $\lvert \sin(nx) \rvert = \lvert \sin(kz) \rvert$ and $\lvert \sin(nx)\rvert \leq n \cdot \lvert \sin(x) \rvert$ it does not follow that $\lvert \sin(kz)\rvert \leq k \cdot \lvert \sin(z) \rvert$. $\endgroup$ – Martin R Oct 10 '15 at 0:15
  • $\begingroup$ Thank you for following up. I had a hunch that was the problem. I can see it now algebraically, but I'm not sure I have the intuition as to why it isn't true. I just have to think about it until it clicks. $\endgroup$ – jeremy radcliff Oct 10 '15 at 1:13

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