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Show that Odd Euler Numbers are 0

  1. Jul 18, 2015 #1

    RJLiberator

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    1. The problem statement, all variables and given/known data
    Complex Analysis Problem:

    The euler Numbers E_n, n=0, 1, 2,..., are defined by 1/cosh(z) = the sum from n=0 to n=infinity of E_n/n! z^n (|z|<pi/2).

    show that E_n=0 for n odd. Calculus E_0, E_2, E_4, E_6

    2. Relevant equations
    Not entirely sure what to put here for this one.

    3. The attempt at a solution

    I've been stuck on this one for a little while now.

    I started out by just writing out the terms.
    E_0+E_1*z+E_2/2! z^2+....

    I know that the coshz is an even function, I know the first x amount of terms of Euler's Numbers, but I'm struggling on how to 'prove' that Odd Euler Numbers are = 0.

    Any guidance on getting me going here?
    Thank you kindly.
     
  2. jcsd
  3. Jul 18, 2015 #2

    micromass

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    What can you say about odd/evenness of the derivative of both sides?
     
  4. Jul 18, 2015 #3

    RJLiberator

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    Interesting, let's check it out.

    Well, we know that the derivative of 1/cosh(z) is 1/((sqrt(z+1))sqrt(z-1))
    My first thought was that it would be an odd function. But I'm not sure how to tell here.

    The derivative of the summation with respect to z is
    sum from n=0 to n=infinity of E_n/n! n*z^(n-1)

    so by taking the derivatives of both sides we are left with that equality.

    Since cosh^-1(z) is even function, the derivative here is odd?
     
  5. Jul 18, 2015 #4

    micromass

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    Do it in general. Let ##f## be an arbitrary even function (that is differentiable of course), prove that ##f^\prime## is odd. Do this over and over again to make a statement about odd/even of ##f^{(k)}##.
     
  6. Jul 18, 2015 #5

    RJLiberator

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    Certainly.

    Sin(x) is odd.
    d/dx of sin(x) is cos(x) which is even
    -sin(x) is even
    -cos(x) is odd
    sin(x) is even
    cos(x) is odd

    Here, we see the statement being that they switch off based on derivatives.
    Since 1/cosh(z) is odd, we take the derivative to get an even function. Is this the correct strategy?
     
  7. Jul 18, 2015 #6

    micromass

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    Yes. But can you prove that the derivative of an arbitrary odd function is even, and that the derivative of an arbitrary even function is odd? Write down the defining equation for odd/even and differentiate.
     
  8. Jul 18, 2015 #7

    RJLiberator

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    This makes some sense to me for sure: http://i.imgur.com/WnJ7v.png

    So now that we know this, I can attack with a strategy in mind.

    So now, I have taking the first few terms of the derivative summation to see

    [0+E_1+E_2*z+E_3/2 z^2+E_4/6 z^3+...]

    Let's see what to do with this.
     
  9. Jul 18, 2015 #8

    RJLiberator

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    Ugh, just not sure on how to 'show that E_n=0 for n odd.

    I have written out a few terms from both series.

    1/cosh(z) = [E_0+E_1z+E_2/2! z^2+E_3/3! z^3+...] EVEN
    d/dz 1/cosh(z) = [0+E_1+E_2z+E_3/2 z^2+...] ODD
     
  10. Jul 18, 2015 #9

    micromass

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    What is ##f(0)## for an odd function?
     
  11. Jul 18, 2015 #10

    RJLiberator

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    f(0)=0 for an odd function
    So we see that in the series representation of the odd function d/dz 1/cosh(z) the first term is 0.
     
  12. Jul 18, 2015 #11

    RJLiberator

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    So, by just showing that fact that f(0) =0 for the derivative there, that is not enough to then say E_1, E_3, E_5, and so on must be 0 then. Correct?
     
  13. Jul 18, 2015 #12

    micromass

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    But you can use it to show ##E_1 = 0##, right? Now mimic the same argument for the other ##E_n##.
     
  14. Jul 18, 2015 #13

    RJLiberator

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    Hm. We can use that to show that E_1 = 0?

    I'm not entirely sure how that follows.
    Does the argument go as follows:
    "Since 1/cosh(z) is an even function, it's derivative must be an odd function and therefore we see in the series representation : [0+E_1+E_2*x+...] that E_1 must be = 0 as f(0)=0 ?
     
  15. Jul 18, 2015 #14

    micromass

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  16. Jul 18, 2015 #15

    RJLiberator

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    From here, is it safe to make the jump:
    Since this is an odd function, we know that E_3=0 and E_n for all n odd = 0?

    This seemingly long question has been broken down to taking the derivative of an even function, using properties of odd/even functions to evaluate the series to show that hte euler numbers odd must be = 0?

    It was that simple, eh?
     
  17. Jul 18, 2015 #16

    micromass

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    Yep, it appears you got it. Congratulations!
     
  18. Jul 18, 2015 #17

    RJLiberator

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    Thanks to your help here. I do understand the concept now rather well.
    The key was the odd/even distinction.

    Now I am trying to finish it off by calculating the values of E_0, E_2, E_4, E_6.
    I know what they equal as that is 1, -1, 5, -61 respectively. I'm trying to figure out how to 'calculate' that.
    Does it have to do with the coefficient values in front of the z's?
     
  19. Jul 18, 2015 #18

    micromass

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    Yes, you need to find the coefficients in front of the ##z##'s. So you need to decompose ##1/\text{cosh}(z)## into a Taylor series.
     
  20. Jul 18, 2015 #19

    RJLiberator

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    The taylor series of cosh^-1(z) is ln(2z)-1/(4z^2)-3/(32z^4)-15/(288z^6)-....

    It doesn't seem to add up to what my series presented of:
    1/cosh(z) = [E_0+E_1z+E_2/2! z^2+E_3/3! z^3+...] EVEN

    Here, E_0 = ln(2z)
    E_2 = 2z^-4

    That doesn't seem to work. :|
    Perhaps I made a mistake in my series expansion initially.
     
  21. Jul 18, 2015 #20

    micromass

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    How do you get an ln in a series expansion?
     
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