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Let $$f(x)=\sum_{k=1}^{\infty}\sin\left(\frac{x}{2^k}\right)$$

Is $f(x)$ bounded?

EDIT:

I'm asking if there is $M \in R$ (M constant), s.t. $\forall x $ $|f(x)|<M$

EDIT2:

I deleted parts that caused confusion.

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I get you are not asking for convergence. You define $f(x)$ by the convergent series, and ask if the absolute function is bounded. Why did you tag it with divergent-series? –  Sasha Aug 28 '12 at 18:04
    
@Sasha: forget the older versions, this one is correct. –  zxc Aug 28 '12 at 18:06
    
I think Henning;s answer nails it. Please reconsider which answer you accept. –  Sasha Aug 28 '12 at 18:55
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3 Answers 3

up vote 10 down vote accepted

There is no bound, because $$ f(2^{3n}\cdot\frac{2\pi}{7}) = f(\frac{2\pi}{7}) + n\biggl(\sin(\tfrac17 2\pi) + \sin(\tfrac27 2\pi) + \sin(\tfrac47 2\pi)\biggr)$$ and the factor of $n$ on the right-hand side is nonzero (it is about 1.32).

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+1 Nice! ${}{}{}{}{}$ –  Sasha Aug 28 '12 at 18:30
    
In fact $\sin(2 \pi/7) + \sin(4 \pi/7) + \sin(8 \pi/7) = \sqrt{7}/2$. –  Robert Israel Aug 28 '12 at 19:06
    
@Robert: By Euclid's goatee, so it does. Something something cyclotomic something Galois? –  Henning Makholm Aug 28 '12 at 19:47
    
Also $\sin(2\pi/15) + \sin(4\pi/15) + \sin(8\pi/15) + \sin(16\pi/15) = \sqrt{15}/2$. On the other hand, $\sin(2\pi/31) + \ldots + \sin(32 \pi/31)$ is a root of $4 z^6-31 z^4+62 z^2-31$. –  Robert Israel Aug 28 '12 at 21:49
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No, it is unbounded.

Let $f_N(x) = \sum_{k=1}^N \sin(x/2^k)$. Then $|f(x) - f_N(x)| \le \sum_{k=N+1}^\infty |x|/2^k = |x|/2^N$. Now on $[0, 2^N \pi]$, $\sin(x/2^k)$ for $k=1,\ldots,N$ are orthogonal, so $\int_0^{2^N \pi} f_N(x)^2 = \sum_{k=1}^N \int_0^{2^N \pi} \sin^2(x/2^k)\ dx = N 2^{N-1} \pi$. Thus there must be some $x_N \in [0, 2^N \pi]$ with $|f_N(x_N)| > \sqrt{N/2}$, and $|f(x_N)| > \sqrt{N/2} - \pi$.

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Later edit: I see you've edited your question so that the part of my answer that's in quotation marks is no longer a verbatim quote from the question, and maybe the whole intent of the question has changed.
end of later edit

The statement that $\lim\limits_{x\to0}\sin x = x$ doesn't make sense if taken literally with the usual definitions. In the expression $\lim\limits_{x\to0}\sin x$, the variable $x$ is a bound variable, so no "$x$" should appear when the expression is evaluated.

"We can treat the series as $\sum\limits_k (x/2^k)$" can make sense if you're talking about whether it converges, but not if you're talking about what the sum is.

Convergence or divergence is not affected by changing finitely many terms. Those early terms where $\sin(x/2^k)$ is not close to $x/2^k$ therefore don't affect convergence or divergence. But they do affect the value of the sum.

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I agree with what you say, so I edited my question accordingly. –  zxc Aug 28 '12 at 18:07
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