# Thread: Limit of an infinite product (1 + x^(2^n))

1. ## Limit of an infinite product (1 + x^(2^n))

I was sailing through my individual college stuff until I found a few infinite products...

I don't seem to even have a clue about appropriate methods here, so maybe anyone can give me an idea on this one:

I would greatly appreciate your help.

2. Originally Posted by Pranas
I was sailing through my individual college stuff until I found a few infinite products...

I don't seem to even have a clue about appropriate methods here, so maybe anyone can give me an idea on this one:

Notice that $\prod\limits_{n=0}^{N-1}\bigl(1+x^{2^n}\bigr) = (1+x)(1+x^2)\cdots(1+x^{2^{N-1}}) = \sum\limits_{k=0}^{2^N-1}x^k = \dfrac{1-x^{2^N}}{1-x}.$ Now let $N\to\infty.$

3. Originally Posted by Opalg
Notice that $\prod\limits_{n=0}^{N-1}\bigl(1+x^{2^n}\bigr) = (1+x)(1+x^2)\cdots(1+x^{2^{N-1}}) = \sum\limits_{k=0}^{2^N-1}x^k = \dfrac{1-x^{2^N}}{1-x}.$ Now let $N\to\infty.$
That is very slick, thank you!

However, I took another product (fortunately there are only two of them)

As you can see, I used your concept of taking N instead of infinity at first. I even applied some school mathematics and figured out an equivalent summation instead of product, but.. that just looks terrible to me. Or is it still a way to go even with trigonometric functions?

4. For the second one, note that, by repeat use of the sine double angle formula:

$\displaystyle sin(x) = 2sin(\frac{x}{2})cos(\frac{x}{2}) = 4sin(\frac{x}{4})cos(\frac{x}{2})cos(\frac{x}{4}) = \ldots = 2^Nsin(\frac{x}{2^N}) \cdot \prod_{n=1}^{N} cos(\frac{x}{2^n})$

5. Originally Posted by Defunkt
For the second one, note that, by repeat use of the sine double angle formula:

$\displaystyle sin(x) = 2sin(\frac{x}{2})cos(\frac{x}{2}) = 4sin(\frac{x}{4})cos(\frac{x}{2})cos(\frac{x}{4}) = \ldots = 2^Nsin(\frac{x}{2^N}) \cdot \prod_{n=1}^{N} cos(\frac{x}{2^n})$
And therefore $\prod _{n=1}^{N}\cos \left( {\frac {x}{{2}^{n}}} \right) =\sin \left(
x \right) \left( {2}^{N} \right) ^{-1} \left( \sin \left( {\frac {x}{
{2}^{N}}} \right) \right) ^{-1}
$

as $N\to\infty.$ it becomes ${\frac {\sin \left( x \right) }{x}}$

That's just brilliant! Thanks again.
However, it is my first year in college (more like a first month to be precise) and we are getting stuff virtually nobody is able to solve at home. Weird.

6. We had the second product punishing a basic limit over here few weeks ago.

Originally Posted by Pranas
However, it is my first year in college (more like a first month to be precise) and we are getting stuff virtually nobody is able to solve at home. Weird.
It means it's a good College. Stick to it.

7. Originally Posted by Pranas
I took another product (fortunately there are only two of them)

As you can see, I used your concept of taking N instead of infinity at first. I even applied some school mathematics and figured out an equivalent summation instead of product, but.. that just looks terrible to me. Or is it still a way to go even with trigonometric functions?
If you write that last equation as $\displaystyle \prod_{n=1}^N\cos\Bigl(\frac x{2^n}\Bigr) = \frac 1x\frac x{2^{N-1}}\sum_{n=1}^{2^{N-1}}\cos\biggl(\frac{\bigl(n-\frac12\bigr)x}{2^{N-1}}\biggr)$, then you can think of it as a Riemann approximating sum for the integral $\displaystyle \frac1x\int_0^x\!\!\!\cos t\,dt$. That gives you another way to get the answer $\frac{\sin x}x.$