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Math Help - infinite series problem?

  1. #1
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    infinite series problem?

    (1+x^2)(1+x^4)(1+x^8)...(goes on forever)

    |x|< 1

    all i am told to do is simplify and i really have no idea where to start on this. please help.
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  2. #2
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    nm, below
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  3. #3
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    Hello, oblixps!

    Simplify: . S \:=\:(1+x^2)(1+x^4)(1+x^8)(1 + x^{16}) \hdots\qquad |x|\,<\, 1

    We have: . S \;=\;(1+x^2)(1+x^4)(1+x^8)(1+x^{16}) \hdots


    Multiply both sides by (1-x^2)

    . (1-x^2)S \;=\;\underbrace{(1-x^2)(1+x^2)}(1+x^4)(1+x^8)(1+x^{16}) \hdots

    . . . . . . . . . . = \;\underbrace{(1-x^4)(1+x^4)}(1+x^8)(1+x^{16}) \hdots

    . . . . . . . . . . . . = \;\underbrace{(1 - x^8)(1 + x^8)}(1 + x^{16}) \hdots

    . . . . . . . . . . . . . . . = \;(1 - x^{16})(1 + x^{16})\hdots


    The right side is of the form: . \left(1 - x^{2^n}\right)

    Since |x|\,<\,1, then: . \lim_{n\to\infty}\left(1 - x^{2^n}\right) \;=\;1 - 0 \;=\;1


    Our equation becomes: . (1-x^2)S \;=\;1 \quad\Rightarrow\quad\boxed{ S \;=\;\frac{1}{1-x^2}}

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  4. #4
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    Let's be brave:

     f(x)  =  {\color{blue}\left(1+x\right)\left(1+x^2\right)}\l  eft(1+x^4\right)\left(1+x^8\right)\left(1+x^{16}\r  ight)\cdots

      = {\color{blue}\left(1 +x + x^2 + x^3 \right)}\left(1+x^4\right)\left(1+x^8\right)\left(  1+x^{16}\right)\cdots

    = \left(1 + x + x^2 + x^3 + x^4 + x^5 + x^6 + x^7\right)\left(1+x^8\right)\left(1 + x^{16}\right)\cdots

    see the pattern?


     = \sum_{ n = 0 }^{\infty} x^n

    which is the infinite series representation of the known function: f(x) = \frac{1}{1-x}
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  5. #5
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    oh wow thanks!! i was kinda weirded out by the going on forever part that i didn't even think about multiplying the first few terms together. this problem makes complete sense now! thanks a bunch guys!
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