An Inequality?

**elim** · March 4th 2010, 06:24 PM

Prove or disproe:
For any positive integer n and $x \in [0,\pi], \quad \sum_{k=1}^n \frac{1}{k} \sin kx \geq 0$

**Sudharaka** · March 4th 2010, 10:49 PM

Originally Posted by elim

Prove or disproe:
For any positive integer n and $x \in [0,\pi], \quad \sum_{k=1}^n \frac{1}{k} \sin kx \geq 0$

Dear elim,

You could show that this statement is correct using mathamatical induction. Can you give it a try? If you need more help please don't hesitate to reply back.

**elim** · March 5th 2010, 07:39 AM

Say n (>1) is the smallest integer such that for some minimal point $x_0 \in (0,\pi)$ ,
$\sum_{k=1}^n \frac{1}{k} \sin kx_0 \leq 0$ ,
then $\sin nx_0 < 0$

On the other hand the critical points for the function $f_n(x) = \sum_{k=1}^n \frac{1}{k}\sin kx$ are
$x = \frac{(2m+1)\pi}{n+1}, \quad m=0,\cdots,\left[\frac{n-1}{2}\right]$
They are the points in $(0,\pi)$ such that
$\sum_{k=1}^n \cos kx = \frac{\cos \frac{(n+1)x}{2} \sin \frac{nx}{2}}{\sin \frac{x}{2}} = 0$
(note that $\sin \frac{nx_0}{2} \neq 0$ by our finding above)
But $\sin \frac{n(2m+1)\pi}{n+1} = \sin \left( 2m\pi+\frac{n-2m}{n+1}\right) = \sin \frac{n-2m}{n+1} > 0$

Thus we get a contradiction and we actually proved that
$\forall n \in \mathbf{N}^+, \forall x\in (0,\pi) \left( \sum_{k=1}^n \frac{1}{k}\sin kx > 0\right)$

This is a bit stronger statement.

Thanks Sudharaka for the direction. I thought this problem needs more advanced tech to solve.

**Sudharaka** · March 5th 2010, 08:19 AM

Originally Posted by elim

It seems that I still not be able to figure out the way of using induction.
say n (>1) is the smallest integer such that for some $x_0 \in (0,\pi)$ ,
we have $\sum_{k=1}^n \frac{1}{k} \sin kx \leq 0$ ,
then what?

I did get the critical points for the function $f_n(x) = \sum_{k=1}^n \frac{1}{k}\sin kx$ .
They are $x = \frac{(2m+1)\pi}{n+1}, \quad m=0,\cdots,\left[\frac{n-1}{2}\right]$
They are the points in $(0,\pi)$ such that
$\sum_{k=1}^n \cos kx = \frac{\cos \frac{(n+1)x}{2} \sin \frac{nx}{2}}{\sin \frac{x}{2}} = 0$

Dear elim,

We have to show that for any positive integer n and $<br /> x \in [0,\pi], \quad \sum_{k=1}^n \frac{1}{k} \sin kx \geq 0$

When n=1,

$\sum_{k=1}^1 \frac{1}{k} \sin kx =sinx\geq{0}~Since,~x \in [0,\pi]$

Therefore the expression is true for n=1

Suppose the expression is true for n=p where $p\in{Z^{+}}$

Then, $\sum_{k=1}^p \frac{1}{k} \sin kx\geq0$

Now, $\frac{1}{(p+1)} \sin (p+1)x\geq{0}~Since,~p+1>0~and~x\in[0,\pi]$

Therefore, $\sum_{k=1}^p \frac{1}{k} \sin kx+\frac{1}{(p+1)} \sin (p+1)x\geq{0}$

Hence, $\sum_{k=1}^{p+1} \frac{1}{k} \sin kx\geq0$

Therefore our the expression is true for $n=p+1~where~p\in{Z^+}$

Hence by mathamatical induction the expression is true for all $n\in{Z^+}$

Hope this will help you.

**elim** · March 5th 2010, 10:29 AM

Originally Posted by Sudharaka

Now, $\frac{1}{(p+1)} \sin (p+1)x\geq{0}~Since,~p+1>0~and~x\in[0,\pi]$

This is not true. And is the hard part.

On the other hand, actually I did used math induction to prove the inequality is true for all positive integer n and all $x \in [0,\pi]$ by showing that no such an integer n>1 and $x_0 \in (0,\pi)$ to let
$\sum_{k=1}^n \frac{1}{k} \sin kx_0 \leq 0$ happen assuming

(**) $\sum_{k=1}^m \frac{1}{k} \sin kx_0 > 0$ for $1 \leq m < n, \quad x \in (0,\pi)$ which is true for m=1

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