0
$\begingroup$

I got this kind of expression as a value of an infinite product:

$$\prod_{k=1}^{\infty} \left(1-\frac{A}{k^2}+\frac{B}{k^4} \right)$$

It's easy to see how it can be factored into a product of two sines.

$$\frac{1}{\pi^2 \sqrt{B}} \sin \sqrt{a-ib} \sin \sqrt{a+ib}$$

$$a=\frac{\pi^2A}{2}$$

$$b=\frac{\pi^2}{2}\sqrt{4B-A^2}$$

In my case, $4B>A^2$. However, it is obvious both by original expression and numerical computation, that the expression is real valued.

So how do I get rid of $i$ in this expression?

The only idea I have is series expansion. Either expand the sines and multiply the series or move to exponential form and expand the roots.

Is there another, easier way?

$\endgroup$
  • 2
    $\begingroup$ $\sqrt{a-ib} = \overline{\sqrt{a+ib}}$ probably. Otherwise we just get a factor of $-1$. Now, with $z = x+iy$, we have $$\sin z \sin \overline{z} = \sin z \, \overline{\sin z} = \lvert \sin (x+iy)\rvert^2 = \dotsc = \sin^2 x + \sinh^2 y.$$ $\endgroup$ – Daniel Fischer Feb 23 '16 at 22:49
  • $\begingroup$ Ah, so I have to use the exponential form for $a\pm ib$, then raise it to power $1/2$ and then use this formula? $\endgroup$ – Yuriy S Feb 23 '16 at 22:51
  • $\begingroup$ You don't need to. You can also get the square root in Cartesian form without too much ado. $\endgroup$ – Daniel Fischer Feb 23 '16 at 22:54
0
$\begingroup$

I figured it out. Thanks to @DanielFischer for the comment. I still don't know how to proceed without exponential form though.

Let $a,b$ be real positive numbers.

$$a+i b=\sqrt{a^2+b^2} \exp \left(i \arctan \frac{b}{a} \right)$$

$$\sqrt{a+i b}=\sqrt[4]{a^2+b^2} \exp \left(\frac{i}{2} \arctan \frac{b}{a} \right)=$$

I use the principal value of the root.

$$=\sqrt[4]{a^2+b^2}\left( \cos \left(\frac{1}{2} \arctan \frac{b}{a} \right)+i \sin \left(\frac{1}{2} \arctan \frac{b}{a} \right) \right)=$$

I leave out the trigonometric transforms.

$$=\frac{\sqrt[4]{a^2+b^2}}{\sqrt{2}} \left( \sqrt{1+\frac{a}{\sqrt{a^2+b^2}}}+i \frac{b}{\sqrt{a^2+b^2} \sqrt{1+\frac{a}{\sqrt{a^2+b^2}}}} \right)$$

Letting $b \to -b$ we similarly obtain:

$$\sqrt{a-i b}=\frac{\sqrt[4]{a^2+b^2}}{\sqrt{2}} \left( \sqrt{1+\frac{a}{\sqrt{a^2+b^2}}}-i \frac{b}{\sqrt{a^2+b^2} \sqrt{1+\frac{a}{\sqrt{a^2+b^2}}}} \right)$$

Now, it is convenient to transform the sine product in the following way:

$$\sin \sqrt{a+i b} \sin \sqrt{a-i b}=\frac{1}{2} \left( \cos (\sqrt{a+i b}-\sqrt{a-i b})-\cos (\sqrt{a+i b}+\sqrt{a-i b}) \right)$$

Let's denote:

$$X=Re (\sqrt{a+i b})=Re (\sqrt{a-i b})=\frac{\sqrt[4]{a^2+b^2}}{\sqrt{2}} \sqrt{1+\frac{a}{\sqrt{a^2+b^2}}}$$

$$Y=|Im (\sqrt{a+i b})|=|Im (\sqrt{a-i b})|=\frac{b}{2 X}$$

Finally we get:

$$\sin \sqrt{a+i b} \sin \sqrt{a-i b}=\frac{1}{2} \left( \cosh \left(\frac{b}{X} \right)-\cos (2X) \right)$$

$$X=\frac{1}{\sqrt{2}} \sqrt{a+\sqrt{a^2+b^2}}$$

Additionally, we obtain:

$$\prod_{k=1}^{\infty} \left(1-\frac{A}{k^2}+\frac{B}{k^4} \right)=\frac{\cosh (\pi \sqrt{2 \sqrt{B}-A})-\cos (\pi \sqrt{2 \sqrt{B}+A})}{2 \pi^2 \sqrt{B}}$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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