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Math Help - Limit of an infinite product (1 + x^(2^n))

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    Member Pranas's Avatar
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    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.
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  2. #2
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    Quote Originally Posted by Pranas View Post
    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.
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    Member Pranas's Avatar
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    Quote Originally Posted by Opalg View Post
    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?
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    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})
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    Member Pranas's Avatar
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    Quote Originally Posted by Defunkt View Post
    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( <br />
x \right)  \left( {2}^{N} \right) ^{-1} \left( \sin \left( {\frac {x}{<br />
{2}^{N}}} \right)  \right) ^{-1}<br />

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

    Quote Originally Posted by Pranas View Post
    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.
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  7. #7
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    Quote Originally Posted by Pranas View Post
    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.
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