Show that Odd Euler Numbers are 0

[RJLiberator]

Homework Statement

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

Homework Equations

Not entirely sure what to put here for this one.

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.

 

[micromass]

What can you say about odd/evenness of the derivative of both sides?

 

[RJLiberator]

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?

 

[micromass]

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.

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 $f^{(k)}$.

 

[RJLiberator]

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?

 

[micromass]

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?

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.

 

[RJLiberator]

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?

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.

 

[RJLiberator]

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

 

[micromass]

What is $f(0)$ for an odd function?

 

[RJLiberator]

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.

 

[RJLiberator]

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?

 

[micromass]

But you can use it to show $E_1 = 0$, right? Now mimic the same argument for the other $E_n$.

 

[RJLiberator]

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 ?

 

[micromass]

Yes.

 

[RJLiberator]

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?

 

[micromass]

Yep, it appears you got it. Congratulations!

 

[RJLiberator]

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?

 

[micromass]

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.

 

[RJLiberator]

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.

 

[micromass]

How do you get an ln in a series expansion?

 

[RJLiberator]

Oi, you are absolutely right. My notes had an error.

I see: https://www.wolframalpha.com/input/?i=taylor+series+(1/cosh(z))
I have now solved the problem. You have been a great help in my understanding here.

Kind regards.