Limit of Product Cosine | Double-Angle Formulas for Sine | (0,π/2) Range

[gop]

Homework Statement

Calculate for x\in(0,\pi/2)

\lim_{N\rightarrow\infty}\prod_{n=0}^{N}cos(\frac{x}{2^{n}})

Hint: Use the Double-Angle Formulas for the sine.

Homework Equations

The Attempt at a Solution

cos(x)\cdot cos(\frac{x}{2})\cdot cos(\frac{x}{4})\cdot...
\frac{\sin2x}{2\cdot\sin x}\cdot\frac{\sin x}{2\cdot\sin x/2}\cdot\frac{\sin x/2}{2\cdot\sin x/4}\cdot...
\frac{\sin2x}{2}\cdot\frac{1}{2}\cdot\frac{1}{2}\cdot...\cdot\frac{1}{2\cdot\sin x/N}
\frac{\sin2x}{\sin(x/{2^N})\cdot2^{N}}
However, now I have to resolve the 0*infinity in the denom. But how do I resolve that.

 

[Dick]

Shouldn't the N in sin(x/N) be like sin(x/(2^N))?

 

[gop]

yes, of course. I updated the information.

 

[Dick]

Then your denominator is a simple limit. It looks like limit c-> 0 of sin(c*x)/c, where c=1/(2^N).

 

[Dick]

BTW, make sure you've counted the 2's in the denominator correctly. I'm finding an extra one.

 

[gop]

that, of course, works. thanks you a lot.

 

[Gib Z]

An alternative method that I find quite nice is to use the fact that cos(x/2^n) is the real part of exp(ix/2^n). The product of the exponentials becomes an easy geometric series in the exponent =]

 

[Kummer]

You can also use a 'collapsing product'.

Note,
A_4 = \cos \frac{x}{2} \cos \frac{x}{4} \cos \frac{x}{8} \cos \frac{x}{16}
Then,
\sin \frac{x}{16} A_4 = \cos \frac{x}{2} \cos \frac{x}{4} \cos \frac{x}{8} \cos \frac{x}{16} \sin \frac{x}{16}
So,
2A_4\sin \frac{x}{16} = \cos \frac{x}{2} \cos \frac{x}{4} \cos \frac{x}{8}\sin \frac{x}{8}
Again,
4A_4\sin \frac{x}{16} = \cos \frac{x}{2} \cos \frac{x}{4} \sin \frac{x}{4}
Again,
8A_4\sin \frac{x}{16} = \cos \frac{x}{2} \sin \frac{x}{2}
Last time,
16A_4\sin \frac{x}{16} = \sin x
That means (since x\in (0,\pi/2))
A_4 = \frac{\sin x}{2^4 \sin \frac{x}{2^4}}
And in general,
A_n = \frac{\sin x}{2^n \sin \frac{x}{2^n}}

 

[Dick]

An alternative method that I find quite nice is to use the fact that cos(x/2^n) is the real part of exp(ix/2^n). The product of the exponentials becomes an easy geometric series in the exponent =]

Did you try doing it that way? I take the product of the exponentials and I get exp(i*2*x). Now what? The real part of that doesn't have much to do with the product of the real parts of the terms. That trick doesn't ALWAYS make things easier.

 

[Gib Z]

Damn It I assumed that the Real part of a product is equal to the product of the real part :( Never mind me =]