Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

$$\sum_{i=1}^n \frac{\sin{(ix)}}{i} < 2\sqrt{\pi}$$

I have this answer, please let me know if there is a more beautiful proof.

My answer:

at first, we prove two inequalities:

  1. If $x\in (x,\pi)$ then $\sin x \leq x$
  2. When $x\in(0,\frac{\pi}{2})$, $\sin x \geq \frac{2x}{\pi}$

1) first, let $y = \sin x -x $

$y^{\prime} = \cos x -1 \leq 0$

so $\sin x - x \leq \sin 0 -0 = 0$

which can be rewritten as

$\sin x \leq x$

2) Let $y=\sin x - \frac{2x}{\pi}$

thus $y^{\prime} = \cos x - \frac{2}{\pi}$ because $x\in (0, \frac{\pi}{2})$

so y at first decreases and then increases on the boundary of $x \in (0,\frac{\pi}{2})$

so $ \sin x - \frac{2}{\pi}\leq \max \{{\sin 0 - \frac{2}{\pi}0, \sin (\frac{\pi}{2} - \frac{2}{\pi}\frac{\pi}{2}) \}}$

so $\sin x \leq \frac{2x}{\pi}$

Then select $M\in N$

$\frac{\sin (mx)}{m} + \frac{\sin ((m+1)x)}{m+1} + \ldots + \frac{\sin ((m+n)x)}{m+n} \leq \frac{\sin (mx)}{m} + \frac{\sin ((m+1)x)}{m} + \ldots + \frac{\sin ((m+n)x)}{m} $

=> $\frac{1}{2M} \times \frac{\sin ((m-\frac{1}{2})x) - \sin ((n+\frac{1}{2})x)}{\sin \frac{x}{2}} < \frac{1}{m \times \sin \frac{x}{2}} \times \sin x + \frac{\sin 2x}{2} + \ldots + \frac{\sin ((m-1)x)}{m-1} < x + \frac{2x}{2} + \ldots + \frac{(m-1)x}{m-1} $

so just need to prove that

$(m-1)x + \frac{1}{m \times \sin \frac{x}{2}} \leq 2\sqrt{\pi}$

select M which satisfies

$ \frac{\sqrt{\pi}}{x} \leq m < \frac{\sqrt{\pi}}{x} + 1 $

so $ (m-1)x < [ \frac{\sqrt{\pi}}{x} \times x = \sqrt{\pi} ] $

thus

$\frac {1}{m \times \sin(\frac{x}{2})}\leq[ \frac{1}{\sqrt{\pi}}\times \frac{2}{\frac{ \sin (0.5x)}{0.5x}} = \frac{1}{\sqrt{\pi}} \times \frac{2}{\frac{\sin 0.5x}{0.5x}} ]$

because $x\in (0, \pi)$ thus $\frac{x}{2} \in (0, \frac{\pi}{2})$

$ (m-1)x + \frac{1}{m \times \sin(0.5x)} \leq 2\sqrt{\pi} $

thanks, for viewing and commenting.

ps. I'm still learning latex and mathematics, so my answer isn't pretty to read, nor is the latex I wrote.

share|improve this question
4  
You might include a reference to where you've seen it done with algebra, so we might know what you don't want. –  Gerry Myerson Apr 11 '12 at 0:08
1  
What's the meaning of "with algebra", and of "without algebra"? –  Martin Argerami Apr 11 '12 at 1:56
    
@MartinArgerami When I was shown the answer, the guy just use basic algebra manipulation to show the result. –  yiyi Apr 11 '12 at 2:02
2  
@Mao Maybe you can use $$\int\limits_0^\infty \frac {\sin (xt)}{t} dt =\frac{\pi}{2};\forall x$$ –  Pedro Tamaroff Apr 11 '12 at 2:45
2  
Probably relevant: $\sum_{i=1}^n \frac{\sin(ix)}{i}$ is the $n$'th partial sum of the Fourier series for $f(x) = \pi/2 - x/2$ on $[0,2\pi]$. The fact that the maximum of this function is somewhat larger than $\pi/2$ is related to the Gibbs phenomenon. I'm not sure where $2 \sqrt{\pi}$ would come in though, it's certainly not a tight bound. –  Robert Israel Apr 11 '12 at 7:46
show 4 more comments

2 Answers 2

The best it is possible to state is: $$\left|\sum_{n=1}^N \frac{\sin(nx)}{n}\right|\leq\int_{0}^{1}\frac{\sin(\pi x)}{x}\,dx = 1.85194\ldots$$ Call $f_N(x)=\sum_{n=1}^N \frac{\sin(nx)}{n}$: it is a $2\pi$-periodic function converging to $\frac{\pi-x}{2}$ in $L_2\left([0,2\pi]\right)$. Since: $$\frac{d f_N(x)}{dx}= \frac{\cos\left(\frac{N+1}{2}x\right)\sin\left(\frac{N}{2}x\right)}{\sin\left(\frac{x}{2}\right)},$$ we know that $f_N(x)$ has $2n$ stationary points in $[0,2\pi]$, local maxima in $x=\frac{(2k+1)\pi}{N+1}$, the first one occurring in $x_N=\frac{\pi}{N+1}$. Once we prove that the value of $f_N(x)$ in any other local maximum is less than $f_N(x_N)$, and that $f_N(x_N)$ is an increasing sequence (I still must find a convincing proof of this two facts, but they look not too hard to deal with and strongly supported by computer inspection) the best bound we can hope in is: $$\left|\sum_{n=1}^N \frac{\sin(nx)}{n}\right|\leq\lim_{N\to +\infty}\sum_{n=1}^{N}\frac{\sin\left(\frac{\pi x}{N+1}\right)}{n},$$ where the RHS a Riemann sum associated with: $$\int_{0}^{1}\frac{\sin(\pi x)}{x}\,dx = 1.85194\ldots<\frac{13}{7},$$ QED.

share|improve this answer
add comment
up vote 1 down vote accepted

I was correct.

Read the comments for better ideas.

Answering this out of a need for my question to have an answer, and I wrote a correct answer in my post.

Yea, for me.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.