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I had a problem evaluating the series \begin{equation} S(\omega)=\sum_{k=0}^\infty (-1)^k {\alpha \choose k}\cos(k\omega),\quad 0<\alpha<2,\quad \omega\in(-\pi,\pi) \end{equation} where \begin{equation} {\alpha \choose k}=\frac{\Gamma(1+\alpha)}{\Gamma(\alpha-k+1)\Gamma(k+1)} \end{equation} is the binomial coefficient generalized to non-integer.

Seems it is a bit like Fourier series. However the coefficients are strange. I have drawn the curve of $S(\omega)$ vs. $\omega$ and through visualization I thought that $S(\omega)$ may be a well behaved function which has a simpler form.

Can you help me find a simple, equivalent expression to the above series? If that doesn't exist, is there an approximation to the sum?

Any answer would be appreciated.

P.S.:

Some answers given are concerned with complex numbers $(1-e^{i\omega})$. As far as I know, $(1-e^{i\omega})^\alpha$ is a multi-valued function. It is not convenient for evaluation in Matlab.

Is there an equivalent function that only involves real numbers?

Did
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ecook
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  • Note that $\cos(k\omega)$ is the real part of $e^{i k\omega}$ with $k$ an integer. Try thinking about this as a sum in $z=e^{i \omega}$... – Semiclassical Aug 23 '14 at 03:17
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    I started writing an answer as follows: $\displaystyle S(\omega) = \sum_{k=0}^\infty (-1)^k {\alpha \choose k}\cos(k\omega)$ $\displaystyle{} = \sum_{k=0}^\infty (-1)^k {\alpha \choose k}\operatorname{Re}(e^{ik\omega})$ $\displaystyle{} = \Re\left(\sum_{k=0}^\infty {\alpha \choose k} (-e^{i\omega})^k\right) = \operatorname{Re}\left( ( 1-e^{i\omega} )^\alpha \right)$. Then I thought there might be some complications and I didn't finish it. But I think it might work if pursued a bit further. ${}\qquad{}$ – Michael Hardy Aug 23 '14 at 03:22
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    The simplest way to finish the algebra in @MichaelHardy's approach is to factor out something that's obviously real (and being able to spot that is a good trick to know). – Semiclassical Aug 23 '14 at 03:27
  • @ Michael Hardy @ Semiclassical Is there a real function that equals? (I have modified my post) – ecook Aug 23 '14 at 03:41
  • Looking at it more carefully, it's more thorny than I gave credit for due to $\alpha$ not being an integer. Out of curiousity, what value of $\alpha$ did you use when plotting? – Semiclassical Aug 23 '14 at 03:55
  • @Semiclassical I have tested $\alpha=0.1,0.2,\cdots,1.9$. – ecook Aug 23 '14 at 03:57
  • Ok. Note that those are all rational values with denominator 10. That means that in every case $(1-e^{i\omega})^\alpha$ is at most 10-valued. I'm not certain, but I wouldn't be surprised that taking $\alpha$ to be real---say, $\pi/4$---gives results which are a bit hairier since $(1-e^{i\omega})^\alpha$ will be infinitely multi-valued. – Semiclassical Aug 23 '14 at 04:08
  • Yes. You are right. I referred to my textbook and found that the number of values is infinite if $\alpha$ is not rational. That is the question. Because I want to study the series in a simpler way, say, as a combination of elementary functions such as $x^a$, $e^x$, $\sin(x)$ etc. – ecook Aug 23 '14 at 07:13
  • @MichaelHardy The complications are involving convergence issue. The identity will be only valid if convergence is established. That can be done by Abel's theorem. – Sungjin Kim Aug 23 '14 at 07:26
  • @i707107 Can you say more about establishing the convergence? I am only familiar with Abel's theorem when the series is power series. – ecook Aug 23 '14 at 07:30
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    @ecook The power series involving binomial coefficient, has radius of convergence $1$. This can be achieved by ratio test. The convergence on the unit circle can be achieved by Abel's theorem. $\sum_{n\leq x} z^n$ is bounded when $z\neq 1$ and the binomial coefficient $\binom{\alpha}{k}$ converges to zero. Also, this power $\alpha$ function is actually not multi-valued. It is $\exp(\alpha\log (1-e^{iw}))$. – Sungjin Kim Aug 23 '14 at 07:35
  • @i707107 Can I conclude that the power series $\sum_{k=0}^\infty \left( \alpha \atop k \right)z^k$ converges for all $|z|\le1$? – ecook Aug 23 '14 at 12:28
  • @ecook As far as I know for $0<\alpha<2$ as in your assumption, the power series converges for all $|z|\leq 1$, $z\neq 1$. For $z=1$, the analysis should be more careful, but I did not work it out yet. Also, there would be more to be said about $\alpha$. Such as, what are the set of all $\alpha$'s such that $\sum_k\binom{\alpha}{k} z^k$ converges for all $|z|\leq 1$, $z\neq 1$? – Sungjin Kim Aug 23 '14 at 13:35
  • @i707107 Can the convergence at $z=1$ be evaluated as $\sum_{k=0}^\infty{\alpha \choose k}z^k=\sum_{k=0}^\infty{\alpha \choose k}=(1+1)^\alpha=2^\alpha$? Can you show how the series is evaluated at $|z|=1$ and $z\ne1$? – ecook Aug 24 '14 at 01:50
  • Got something from an answer below? – Did Aug 27 '14 at 14:06
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    For integer $\alpha$ it is useful to employ the generating function $g(z)=\sum _{n=0}^{\infty } z^n S(\omega ,n)=\frac{1-2 z \sin ^2\left(\frac{\omega }{2}\right)}{1-4 z (1-z) \sin ^2\left(\frac{\omega }{2}\right)}$ – Dr. Wolfgang Hintze Mar 18 '21 at 08:19
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    For half integer $\alpha$ I find after some simple but lengthy calculations this generating function $g\left(z,\alpha \to \frac{1}{2}\right)\text{:=}\sum _{m=0}^{\infty } z^m S\left(\omega ,\frac{m}{2}\right)\=\frac{1-z \sqrt{2 \sin \left(\frac{\omega }{2}\right)} \cos \left(\frac{\omega -\pi }{4}\right)}{1-2 \sqrt{2} z \sqrt{\sin \left(\frac{\omega }{2}\right)} \cos \left(\frac{\omega -\pi }{4}\right)+2 z^2 \sin \left(\frac{\omega }{2}\right)}$ – Dr. Wolfgang Hintze Mar 18 '21 at 09:25

4 Answers4

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Assume that $\omega\ne0$. Then $1-\mathrm e^{\mathrm i\omega}\ne0$ and users @MichaelHardy and @i707107 explained in comments why $$ S(\omega) = \Re\exp(\alpha\cdot\mathrm{Log}(1-\mathrm e^{\mathrm i\omega})), $$ under the condition that $$ \left|\alpha\cdot\arg(1-\mathrm e^{\mathrm i\omega})\right| \lt \pi, $$ where $\arg$ denotes the principal argument, with values in $(-\pi,\pi]$, and $\mathrm{Log}$ denotes the principal branch of the complex logarithm, defined on $\mathbb C\setminus\mathbb R_-$ by the identity $$ \mathrm{Log}(r\mathrm e^{\mathrm it}) = \ln(r)+\mathrm i\mathrm t, \quad r\gt0, \quad |t|\lt\pi. $$ Thus, the main task is to identify $z=1-\mathrm e^{\mathrm i\omega}$ as $z=r\mathrm e^{\mathrm it}$ with $r\gt0$ and $|t|\lt\pi$, and to check the argument condition. To do so, note that $$ z = \mathrm e^{\mathrm i\omega/2}\cdot(\mathrm e^{-\mathrm i\omega/2}-\mathrm e^{\mathrm i\omega/2}) = -2\mathrm i\,\sin(\omega/2)\,\mathrm e^{\mathrm i\omega/2}, $$ hence $$ r = 2\,|\sin(\omega/2)|, \qquad \mathrm e^{\mathrm it} = -\mathrm i\,\mathrm{sgn}(\omega)\,\mathrm e^{\mathrm i\omega/2} = \mathrm e^{\mathrm i(\omega-\mathrm{sgn}(\omega)\pi)/2}. $$ This proves that $t=\frac12(\omega-\mathrm{sgn}(\omega)\pi)$ since this number is in the interval $(-\pi/\pi)$. Furthermore, separating the cases $\omega\gt0$ and $\omega\lt0$, one can see that $|t|\lt\pi/2$. Since $|\alpha|\lt2$, the argument condition holds and $$ S(\omega) = r^\alpha\,\Re\mathrm e^{\mathrm i\alpha t} = r^\alpha\,\cos(\alpha t), $$ that is,

$$ S(\omega) = 2^\alpha\,\left|\sin\left(\tfrac12\omega\right)\right|^\alpha\,\cos\left(\tfrac12\alpha(\omega-\mathrm{sgn}(\omega)\pi)\right). $$

Exercise: Extend this formula to the case $\omega=0$. Check that this defines an even function on $(-\pi,\pi)$. When $\alpha=1$, check that the RHS is $2\sin^2(\omega/2)$ for every $\omega$ in $(-\pi,\pi)$ and explain why it ought to.

Did
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  • Thank you very much for your great answer. – ecook Aug 23 '14 at 10:41
  • Hmmm... Surely you recognized that sgn(ω) denotes the sign of ω? – Did Aug 23 '14 at 11:16
  • I have answered your question below. By the way, I have another question. Why only considering principal branch? – ecook Aug 23 '14 at 11:29
  • Because this is the only way to make that the real part of the complex function one considers, indeed coincides with $S(\omega)$ in the question. – Did Aug 23 '14 at 11:38
  • Still I am not quite clear. If all branches are considered, $(1-e^{i\omega})^\alpha=r^\alpha e^{i\alpha t}\cdot e^{i\alpha 2k\pi}$, $k=0, \pm 1,\pm 2,\cdots$ and $r$, $t$ are the same as you give in the answer. Why only considering the case $k=0$ ? – ecook Aug 23 '14 at 12:12
  • This is entirely answered by my LAST comment. – Did Aug 23 '14 at 12:40
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Hint

If you replace $\cos(k\omega)$ by $\dfrac{{e^{ik\omega}+e^{-ik\omega}}}{2}$ or if you compute $$S(\omega)=\sum_{k=0}^\infty (-1)^k {\alpha \choose k}\cos(k\omega)$$ as being the real part of $$T(\omega)=\sum_{k=0}^\infty (-1)^k {\alpha \choose k}e^{i k \omega}$$ using the generalised binomial theorem, you end with $$S(\omega)=\sum_{k=0}^\infty (-1)^k {\alpha \choose k}\cos(k\omega)=\frac{1}{2} \left(\left(1-e^{i \omega}\right)^\alpha+\left(1-e^{-i \omega}\right)^\alpha\right)$$ which a real valued function.

Michael Hardy
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Claude Leibovici
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Use the identity

$$\cos(k\omega) = \frac{e^{ik\omega}+e^{-ik\omega}}{2}$$

and the Newton's generalised binomial theorem as

$$ S_1 = \sum_{k=0}^{\infty}(-1)^k {\alpha\choose k}e^{ik\omega} = (1-e^{i\omega})^{\alpha}. $$

You do the rest of the problem.

Mhenni Benghorbal
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  • Thanks. I have modified my question. Do you know what is the real function equivalent to the series? – ecook Aug 23 '14 at 03:44
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    The first identity has a typo. The approach using the second identity was underlined by others in comments. The whole is far from giving a full solution. – Did Aug 23 '14 at 09:10
  • @Did: the typo was corrected. Thanks for the comment. – Mhenni Benghorbal Aug 23 '14 at 15:33
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Thank you very much for the answer given by @Did.

In this reply, I'll try to answer the exercise he proposes.

  1. The formula \begin{equation} S(\omega) = 2^\alpha|\sin(\omega/2)|^\alpha \cos\left[\frac{\alpha}{2}(\omega-\text{sgn}(\omega)\pi)\right] \end{equation} can be extended by defining $\text{sgn}(0) = \frac{1}{\alpha}$ such that $\omega=0$ is also included.

  2. The formula is even because \begin{equation} \cos\left[\frac{\alpha}{2}(\omega-\text{sgn}(\omega)\pi)\right] = \cos(\alpha\omega/2)\cos(\alpha\text{sgn}(\omega)\pi/2)+\sin(\alpha\omega/2)\sin(\alpha\text{sgn}(\omega)\pi/2) \end{equation} and
    \begin{equation} \begin{split} &\cos(\alpha\text{sgn}(\omega)\pi/2)=\cos(\alpha\pi/2) ,\quad \omega>0\\ &\cos(\alpha\text{sgn}(\omega)\pi/2)=\cos(\alpha\pi/2) ,\quad \omega<0\\ &\sin(\alpha\text{sgn}(\omega)\pi/2)=\sin(\alpha\pi/2) ,\quad \omega>0\\ &\sin(\alpha\text{sgn}(\omega)\pi/2)=-\sin(\alpha\pi/2) ,\quad \omega<0\\ \end{split} \end{equation}

  3. When $\alpha=1$, the formula becomes \begin{equation} S(\omega)=2|\sin(\omega/2)| \cos\left[(\omega-\text{sgn}(\omega)\pi)/2\right] \end{equation} and since \begin{equation} \begin{split} &\cos\left[(\omega-\text{sgn}(\omega)\pi)/2\right] = \cos(\omega/2-\pi/2)=\sin(\omega/2),\quad \omega>0\\ &\cos\left[(\omega-\text{sgn}(\omega)\pi)/2\right] = \cos(\omega/2+\pi/2)=-\sin(\omega/2),\quad \omega<0\\ \end{split} \end{equation} We conclude \begin{equation} S(\omega)=2\sin^2(\omega/2) \end{equation}

ecook
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