Proof

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July 5th 2006, 05:02 PM
Nichelle14

Proof

I think this goes in this section.

Let x be a real number. Prove that

abs(sin nx) <= n*abs(sinx) for all positive integers n.
July 5th 2006, 05:41 PM
ThePerfectHacker

Quote:

Originally Posted by Nichelle14

I think this goes in this section.

Let x be a real number. Prove that

abs(sin nx) <= n*abs(sinx) for all positive integers n.

I did not try to do the problem yet but you can use the fact that,
$\sqrt{x^2}=|x|$
Thus, you need to prove that,
$\sqrt{\sin ^2 nx}\leq n\sqrt{\sin^2 x}$
If and only if,
$\sin^2 nx\leq n^2\sin ^2 x$
July 8th 2006, 06:25 PM
Nichelle14

I still don't understand what to do.
July 8th 2006, 08:42 PM
ThePerfectHacker

Mathematical induction is the tool here.
It is true for $n=1$ thus, there is a $k$ such as, $|\sin kx|\leq k|\sin x|$
Thus,
$0\leq |\sin kx|\leq k|\sin x|$
But,
$0\leq |\cos x|\leq 1$
Thus, (mutiply inequalities notice they are non-negative),
$|\sin kx||\cos x|\leq k|\sin x|$ ......(1)
Also,
$0\leq |\sin x|\leq |\sin x|$
And,
$0\leq |\cos kx|\leq 1$
Thus, (mutiply inequalites notice they are non-negative),
$|\cos kx||\sin x|\leq |\sin x|$ ......(2)
Using the property $|x||y|=|xy|$ on (1) and (2) we have,
$|\sin kx \cos x|\leq k|\sin x|$
$|\cos kx \sin x|\leq |\sin x|$
Now, add these inequalities,
$|\sin kx \cos x|+|\cos kx \sin x|\leq k|\sin x|+|\sin x|$
Now, by triangular inequality,
$|\sin kx \cos x|+|\cos kx \sin x|\geq |\sin kx \cos x+\cos kx \sin x|$
By transitivity ( $a<b \mbox{ and }b<c \rightarrow a<c$ ),
$|\sin kx \cos x+\cos kx \sin x|\leq k|\sin x|+|\sin x|$
Thus, recognizing the sum for sine and simply the right side,
$|\sin (k+1) x|\leq (k+1)|\sin x|$
Proof is complete.
July 9th 2006, 04:58 AM
Quick

Is it possible to say $5\leq5$
(because it seems somewhat ridiculous)
July 9th 2006, 06:47 AM
ThePerfectHacker

Quote:

Originally Posted by Quick

Is it possible to say $5\leq5$
(because it seems somewhat ridiculous)

Of course.

$a\leq b$ is defined to be true whenever,
$a<b$ OR $a=b$ .
Furthermore, in mathematics this expression is used a lot. There is one powerful theorem in set theory (Zorn's Lemma) which is based on the fact that $a\leq a$ .