Math Is Fun Forum

Here is a mathematical proof:

therefore,

With some algebraic multiplication:

I see your point: that particular integral is very difficult to preform. However, my goal was not to numerically evaluate the sum (the easiest way to do that is in the original form ). My goal was rather to express

in the simplest closed form possible. The only identities I can find of the Hurwitz Lerch Phi (Lerch Transcendent?) are defined as series expansions (the one I started with) and/or complex double integrals including the Dirichlet beta function. I consider my answer to be "simpler" in the sense that it can be fundamentally understood at a glance (although it may not be particularly simple to evaluate).

That being said, I had hoped that the answer would be more "elegant". To be honest, the simplest form of

(for computation or comprehension) is probably the native series expansion. However, being the weirdo I am, I felt the dire need to express the summation in a closed form. Now that I think about it I still may not have achieved this goal. Does an integral representation of a function count as a closed form?

Or in integral form:

The only solution such that |a| < 1 is:
a ~~ 0.6180339887498948482045868343656381177203...

(obtained using Mathematica command FindRoot[100a^8001 - 100a^7999 - a^2 - a + 1 == 0, {a,-1, 1,WorkingPrecision -> 39] )

The Eulerian polynomials can be canceled out using identity 16 from

http://mathworld.wolfram.com/EulerPolynomial.html

Any ideas about how to go about eliminating the Q(x+1,n) term?

So, as I see it,

has the relationship with Eulerian polynomials such that

Does this seem correct to you?

Correct me if I'm wrong but I do not believe there is a such thing as taking a fractional derivative.

That being said, assuming there is, I believe it would have to work something like this:

What your saying makes sense theoretically but I do not understand the idea of a "1/2 derivative". :\ Are there also "1/2 integrals"?

EDIT: I did some Googling; apparently there is a "half derivative". Sorry to question to concept.

Right. I'm not denying that the integral does diverge at infinity. I'm saying that it is not a counter example to my statement:

Because when you place the infinity in the bounds it is the limit causing divergence, not the integral itself. Using my previous statement we see that the integral in this case diverges at

because when we divide by 0 the infinitesimal is "overpowered". Thus we are left with a non-infinitesimal value causing the integral (at that point) to diverge.

We can see this holds true by evaluating the integral and substituting zero into the natural logarithm. It does in fact diverge toward -∞ (as my statement predicted).

I'm not questioning the validity of the Zeta Function.

I am saying that for any function

. The infinite iteration (fixed point?) will be the solution for

in the integrand:

.

This holds true with f(x)=cos(x). Consider that since an integral is just an infinite summation of infinitesimals in order for it to diverge toward negative infinity either one of the "pieces" is -∞ or each piece is not in fact infinitesimal.

We see that with this integral that if that since the dependent values are with in the denominator in oder to "overpower" the infinitesimal it must become 0. So what value for "t" causes the denominator to be 0?

We know this must be the value for Q because this is the only time one of the pieces can diverge. We don't have to worry about the lower bound of the integral because we know that the solution at that point is finite and is "sucked in" to the -∞.

I'm sorry my explanation is so bad. I have never formally taken a Calculus class. As a result virtually everything I know about upper level math beyond Algebra II I have had to teach myself from articles that I have read online or any text books I've been able to borrow. On top of that I have a hard time remembering the names of thermos, postulates or axioms (Geometry was a nightmare). So I end up inventing my own terminology. Don't hesitate to let me know if I need to clarify something or if I'm just out-right wrong.

Thanks!

Oh wow. I'm stupid I mis-evaluated it.

Thats an awsome defenition! How did you derive that?

Also, is there any way to use this identiy to solve my original problem (see my first post in this thread)?

(substituting n=3 into your equasion)

I do not belive this is the correct answer.

Hmm... Again I am receiving different values for n=3.