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Is there an infinite product representation of e^(z)

  1. Jul 13, 2006 #1
    I was wondering if anyone knew where I might find, or formulate, the infinite product representation of the entire function f(z) = e^(z).

    Wikipedia says, and I qoute, "One important result concerning infinite products is that every entire function f(z) (i.e., every function that is holomorphic over the entire complex plane) can be factored into an infinite product of entire functions each with at most a single zero..."

    Source of quote: http://en.wikipedia.org/wiki/Infinite_product

    According to wikipedia, every entire function can be factored into an infinite product of entire functions with at most one zero.

    Now the sum of two entire functions is an entire function. So I was wondering, if one is given the infinite product representations of say,
    sin(pi*z) and e^(z), how might one construct a single product representation of
    g(z) = sin(pi*z) - e^(z)?

    Inquisitively,

    Edwin G. Schasteen
     
  2. jcsd
  3. Jul 13, 2006 #2

    shmoe

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    e^z has no zeros, nor does e^(g(z)). These are 'unbreakdownable' terms in the expansion. You can think of e^(g(z)) is your basic non-vanishing entire function, they all have this form (g analytic).

    Find the zeros and go from there. It will be icky. Product forms don't tend to add very well. Think what you do if you wanted to add two polynomials you had already factored. You'd have to multiply them out again. On the other hand, multiplication is the opposite- multiplying two series together is gross, multiplying two infinite products together is a pleasure.
     
  4. Jul 13, 2006 #3

    NateTG

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    Isn't it something like:
    [tex]\prod{1+\frac{z}{n}}[/tex]

    That's just a guess based on the formula for sine on the wikipedia site.
     
  5. Jul 13, 2006 #4

    mathwonk

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    you usually have to throw in some factors to make the product converge.
     
  6. Jul 13, 2006 #5

    shmoe

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    the negative integers aren't zeros of sin(pi*z)-exp(z), they're off a little bit. This is the ickiness I meant- you won't be finding this "little bit" nicely.

    They are close enough to the integers that the exp factors needed to ensure convergence will be the same form as the one they have in gamma, namely the exp(z/zero). sin(pi*z)-exp(z) is a function of order 1, so this "close enough" is justified. These factors also exist in the sine product form if you don't pair up the zeros like they did (and is usually done) (1-z/n)*(1+z/n)=1-(z/n)^2
     
  7. Jul 14, 2006 #6

    benorin

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    Last edited: Jul 14, 2006
  8. Jul 14, 2006 #7

    shmoe

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    You caused latex and the forum to explode.

    That is not a product form in the form of weiestrass/hadamard factorization. The factors are not analytic over C. This only claims to be valid for non-negative real values of x.

    Remember the overall goal of these weiestrass product forms is to show that the global behavior of analytic functions is essentially determined by their zeros and poles, just like you would have for polynomials (or rational functions). Any entire function with no zeros is of the form exp(g(z)) for some entire g(z). So if you broke exp(g(z)) into the product of two entire functions, it would have to look like exp(g(z))=exp(f(z))*exp(h(z)) where f and h are entire. So you'd have g(z)=f(z)+h(z)+some integer multiple of 2*Pi*i. This is what I mean by "unbreakdownable", any furthur factorization of exp(g(z)) is essentially trivial.
     
  9. Jul 14, 2006 #8

    benorin

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    It seems that though the above product was cited to hold for non-negative real values of x, that it ought to hold in the half-plane Re[x]>0.
     
  10. Jul 15, 2006 #9

    shmoe

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    I was wondering if that was'nt the case. Have you tried to prove this?

    The factors still won't be entire though, that root and having a zero will see to that.
     
  11. Jul 17, 2006 #10
    Code (Text):
    e^z has no zeros, nor does e^(g(z)). These are 'unbreakdownable' terms in the expansion. You can think of e^(g(z)) is your basic non-vanishing entire function, they all have this form (g analytic).
    e^(z) is probably not a good choice then. How about something like
    f(z) = sin(pi*z) - cos(pi*z)? Or, how do you go about forming the infinite product representation of entire functions with zeros, like sine or cosine functions, in general?

    Inquisitively,

    Edwin
     
  12. Jul 17, 2006 #11

    shmoe

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    Find the zeros. determine what factors are needed for convergece, determine the phi(z) (logarithmic differentiation can sometimes help here).

    Things are simplified somewhat if you have an integral function of finite order, like sine, or gamma, or zeta, they're all of order 1. We then know the convergence factors will just be of the form exp(z/zero) (just a linear poly in the exp) and phi(z) will be a linear polynomial also. So sine would have to be:

    [tex]\sin{\pi z}=ze^{a+bz}\prod_{n\neq 0}\left(1-\frac{z}{n}\right)e^{z/n}[/tex]

    Taking the limit of sin(pi z)/z as z->0 tells you e^a=pi. Pairing up the n terms with the -n terms then gives:

    [tex]\sin{\pi z}=z\pi e^{bz}\prod_{n=1}^\infty\left(1-\frac{z^2}{n^2}\right)[/tex]

    Since sin is an odd function, all the terms in the product are even, z is odd, so e^(bz) must be even as well, i.e. b=0.

    This kind of thing will be in complex analysis texts, and also many analytic number theory texts (usually concerned with the order 1 case).
     
  13. Jul 17, 2006 #12
    Thanks!

    Best Regards ,

    Edwin
     
  14. Jul 18, 2006 #13
    [tex]
    e^x = \lim_{n\rightarrow\infty} (1+x/n)^n
    [/tex]

    Not the most helpful product expansion in the world, but I believe it's the only one you could get for e, seeing as it only has one zero (- infinity).
     
  15. Jul 18, 2006 #14

    benorin

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    Read post #6.
     
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