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Does anyone know an infinite series summation that is equal to i?

  1. Feb 25, 2014 #1
    The title pretty much says it all, does anyone know of an infinite series summation that is equal to $$\sqrt{-1}$$?
  2. jcsd
  3. Feb 25, 2014 #2
    Let ∑an be convergent sequence with limit L. Then ∑i an/ L converges to i
  4. Feb 25, 2014 #3
    If that works then it is clever and unexpected (a wonderfully typical trait of PF:) although it will work for anything in place of 'i' correct?
    Being recursive is fine but I am looking for a 'unique' solution.

    I have one and have a hard time believing I am first to do so, so what else is out there?
  5. Feb 25, 2014 #4
    Yes, you may replace "i" with anything.
  6. Feb 25, 2014 #5
    Borek posted on a couple of my previous threads 'make an infinite series equal to 1 and multiply (insert the value I was attempting to get) by it.' At least this is a cleverly disguised version of the same thing, well done!

    Anyway, have you ever seen a unique infinite series equal to 'i'? Google searches and the such are turning up nothing. How cool would it be if this is a first?! :biggrin:
  7. Feb 26, 2014 #6

    Stephen Tashi

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    Science Advisor

    What do you mean by a "unique" infinite series. For that matter, what would a non-unique infinite series be?
  8. Feb 26, 2014 #7
    As in one where we don't use an infinite series equal to '1' and then multiply by 'i'. Apparently a fairly common suggestion here on PF :tongue2:
  9. Feb 26, 2014 #8
    I do...but it's not an infinite sum...it's an infinite product.
    Care to investigate this further? Or do you need a sum,
    and can't see how a product could come in handy?

    Because I have a product, ##log(I)## is a sum and is
    easy to get. And powers of "I" are also easy to get
    using the infinite product identity.

    Would that be of any help in your project?
  10. Feb 26, 2014 #9
    Honestly I have little experience with infinite products so far but I would love to see your work!
  11. Feb 26, 2014 #10
    Here is your homework assignment:

    Given the information in the new thread "A Cool Identity" in General Math,
    find an infinite number of infinite product identities for "I". Partial credit
    will be given if you can only find a finite number of identities.
    HINT: factor the factors into complex and real conjugates.

    This is not a "snipe" hunt, we are dealing with the mathematical
    certainty of prohibiting polynomial cross-products. Or stated
    differently, an infinite number of equations in an infinite number
    of unknowns, IS solvable.

    Math is very fun.
    Last edited: Feb 26, 2014
  12. Feb 26, 2014 #11

    Stephen Tashi

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    Science Advisor

    Are you asking for a series that sums to i whose individual terms cannot each be divided by i ? If the individual terms can be divided by i, this gives an infinite series that sums to 1.
  13. Feb 26, 2014 #12
    And I'll give you half credit if the only ones you find are an infinite number
    of products equal to unity and trivially multiply by "I".

    Math is very, very fun...or none of us would be here, n'est ce pas?

    You don't have to be "super-brilliant", just interested in puzzles,
    and then you become "super-brilliant" in some tiny, tiny niche of
    reality; you learn more and more about less and less.

    EDIT: another HINT: be sure to remember that "I" need only be
    multiplied by any one of the infinite number of factors.
    Last edited: Feb 26, 2014
  14. Feb 26, 2014 #13
    For those of you who do not like to cross threads,
    here is the super-cool identity:

    \frac {1} {1-x} =
    (1+x) \prod_{n=1} ^{\infty}
    \left( \frac {1+x^{2^n}} {1-x^{2^n}} \right)^{2^{-n}} , {\;}
    \text{for} {\;} 0\leq x<1

    It's in LaTex, so feel free to move it around.
  15. Feb 26, 2014 #14
    Just food for thought, which is better? Exposing "I", or hiding it?
  16. Feb 26, 2014 #15
    Sounds like we are on the same page, although you guys at PF are masters of semantics so I look forward to your response :)

    ClamShell, I wish I could do as you are suggesting but infinite products are not yet part of my vocabulary.
  17. Feb 26, 2014 #16
    Sorry to have imposed...I used to like sums because
    their so easy to differentiate. Now I suspect that it
    is just another fatal attraction. The answer is that
    it is better to hide "I" than to expose it.
    Disclaimer: But I do respect the opinion that "I"
    is a pretty fun object. Adios.
  18. Feb 26, 2014 #17
    Infinite products are the next step after infinite series, I can't wait to 'dig in' to these wonderful things. When I get up to speed I look forward to accepting your challenges!
  19. Feb 26, 2014 #18
    Keep in touch.
  20. Feb 26, 2014 #19
    Just can't keep my mouth shut...

    What about 1 = 1/2 + 1/4 + 1/8 + 1/16 + ... (Zeno?)?

    Each denominator is 2^n for n = 1 to infinity
    or each denominator is double the previous denominator.

    What happens if each denominator is (2i)^n ?

    This simple sum may reveal what happens to
    some class of other sums.

    Or what happens when ##i = i/2 + i/4 + i/8 + ...## ?
    and you need to find powers of ##i## by powering the
    right hand side?
  21. Feb 26, 2014 #20
    Ah hah! With that we could do this,

    $$i=.2-\sum_{n=0}^{\infty} \frac{1}{.4(2i)^n}$$

    I think that will work.

    Very good, thank you ClamShell!
    Anyone else have anything?
    Last edited: Feb 26, 2014
  22. Feb 26, 2014 #21
    I don't see it, please show your work...
  23. Feb 26, 2014 #22
    Hang on, I'm throwing it in wolfram...
  24. Feb 26, 2014 #23
    $$i=2-\sum_{n=0}^{\infty} \frac{1}{.4(2i)^n}$$

    That's better, now it works :biggrin:
  25. Feb 26, 2014 #24
    So the sum is ##(2 - i)## ?
  26. Feb 26, 2014 #25
    Apparently so!

    So what else can we do?
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