I need to get a hobby... :P

aleclarsen12 · 2011-01-01 16:14:58

This has been driving me crazy for the past week now. Consider the function:

Can

be written in a closed form?

Here is how I have gone about achieving this so far:
Let

We can express in terms of as .

has an odd property such that or alternatively .

Now, notice the special case where the summation is geometric:

That in mind, it is possible to calculate any

such that by starting with the special geometric case and using to raise the degree of the summation to any integer value for . Therefore, I can also find any integer value of too.

Okay, so here is the part that is about to make me rip my hair out:

I can't find a pattern in the coefficients! Do any of you see a pattern or some alternative way to implicitly define

?

Also, assuming I can find a closed integer definition for

, how would I go about increasing the domain of to ultimately include all real numbers?

Thank you!

bobbym · 2011-01-01 18:02:24

Hi;

Can
be written in a closed form?

In terms of elementary functions? I do not think so.

Also coming up with a relationship that describes the next numerator is difficult.

Wait got it! There is a relationship!

aleclarsen12 · 2011-01-02 07:23:28

What kind of relationship? Did you find a way of expressing

in closed form? I can't find the pattern in the coefficients for the continued derivatives... Do you see one?

bobbym · 2011-01-02 08:03:47

Hi aleclarsen12;

Did you find a way of expressing
in closed form?

Unfortunately there is no closed form in terms of elementary functions. But for the polynomial in the numerator yes and no.

The numerators are the well known Eulerian polynomials and their coefficients are the well known Eulerian numbers.

It is impressive that you found a recurrence based on differentiation. That was very good and as far as I know original!

The coefficients can be generated by a recurrence relation very similar to the ones that generate the binomial coefficients, as a matter of fact the two are related. But not in an easy computational way. I will give you what I know.

1) A generating function: If we take the Mclaurin series of the GF we get the Eulerian polynomials as coefficients.

If you take the coefficients of the numerator and arrange them in a triangle:

You will notice that each row adds up to a factorial. The best known way to calculate those numbers is to use the recurrence,

A(n,m) = (n − m)A(n − 1,m − 1) + (m + 1)A(n − 1,m)

There is no known closed form for this recurrence relation.
To see how this is used there is a good chart and a good example right here:
http://en.wikipedia.org/wiki/Eulerian_number

aleclarsen12 · 2011-01-02 08:49:01

So, as I see it,

has the relationship with Eulerian polynomials such that

Does this seem correct to you?

bobbym · 2011-01-02 08:55:10

Yes, that looks okay.

If you look at the triangle you will see 2 important properties of it.

1) It is Unimodal, meaning it has one hump.

2) The symmetry on both sides of that hump. Meaning the coefficients are the same. I was able to generate closed forms for the coefficients of the x,x^2 and x^3. Naturally by symmetry that means the second to last 3 also.
After that the forms became to complex too derive.

aleclarsen12 · 2011-01-02 09:06:20

Or in more standard terms:

(

denotes the n-th Eulerian polynomial)

Is this a common or known identity of Eulerian polynomials?

bobbym · 2011-01-02 09:15:04

Hi;

That I do not know. I would say it is probably a standard one. Check that site I sent you, you might find others.

aleclarsen12 · 2011-01-02 09:25:46

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?

bobbym · 2011-01-02 21:04:37

Hi aleclarsen12;

Hold on a second. Euler polys and Eulerian polys are not the same thing.

Here is a nice page for you on Eulerian polynomials.

http://www.luschny.de/math/euler/Euleri … mials.html

aleclarsen12 · 2011-01-03 15:29:26

Oh thanks you! Thats confusing how 2 completely different things share the same name.

On a somewhat interesting note,

.

relates to the Polylogarithm.

Using this I can define my original function

Thank you for all your help Bobbym!

Last edited by aleclarsen12 (2011-01-03 16:09:27)

aleclarsen12 · 2011-01-03 16:22:43

Or in integral form:

bobbym · 2011-01-03 17:30:12

Hi;

I do not know the answer to that one.

I am getting:

That is the Hurwitz Lerch Phi function.

aleclarsen12 · 2011-01-03 19:44:41

Your answer is equivalent to mine. I numerically tested both of them.

I prefer my answer better because it is written in a semi-friendly integral form accompanied by a fairly elementary function (the Gamma function). From the little research I have done, the Hurwitz Lerch Phi function (a function that I had never even heard of until just now) appears to be just be a generalized function defined only by a series expansion and quite complex integrals involving several non-elementary functions (see http://mathworld.wolfram.com/LerchTranscendent.html).

bobbym · 2011-01-04 01:24:07

Hi aleclarsen12;

Your answer is equivalent to mine. I numerically tested both of them.

That is not my answer. That is the answer Mathematica spits out. Be careful with using numerical experiments to verify that 2 different relationships are equal. There are known cases where they are very close but not the same. There is a famous one that only differs in the billionth place. I am working on one right now that only differs in the 1671 decimal place.

Nothing can replace a mathematical proof in this type of problem. I am just saying I do not have one for your relationship.

The two answers may be the same. But if you have to compute with Mathematica the Hurwitz Lerch Phi ( on the left ) or that integral...
I think the built in function ( Lerch Phi ) will outperform the integral formula.

For example try using your integral formula with x = 4. The built in Lerch Phi can get an answer, the integral formula cannot.

aleclarsen12 · 2011-01-04 14:10:00

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?

bobbym · 2011-01-04 14:24:06

The only identities I can find of the Hurwitz Lerch Phi (Lerch Transcendent?)

I thought you had access to mathematica?! Was it not you in the other thread that used the FindRoot command?
That is why I presented it in that form.

Even still in numerical work we would consider the integral, gamma combination useless.
Strictly formal. Because you cannot compute hardly any x with it. As I said try x = 4. That ought to be an easy but not for that formula. The integral is impossible to get in closed form and the gamma function is infinity for negative integers.

Does an integral representation of a function count as a closed form?

The general definition is that the answer should only contain elementary functions. To quote:

In mathematics, an expression is said to be a closed-form expression if, and only if, it can be expressed analytically in terms of a bounded number of certain "well-known" functions. Typically, these well-known functions are defined to be elementary functions constants, one variable x, elementary operations of arithmetic (+ × ÷), nth roots, exponent and logarithm (which thus also include trigonometric functions and inverse trigonometric functions).
Infinite series, integrals, limits, and infinite continued fractions are not permitted.

I did not realize that is what you were after. I would have tried to discourage you. Here is a practical rule of thumb. If Mathematica and Maple cannot get a closed form then unless you are Euler do not bother. I know this is heretical to human ears but it is good, practical advice. The chance that they missed something a human does not is too small to think about. You undoubtedly have better things to do with your math time.

aleclarsen12 · 2011-01-04 15:30:32

bobbym wrote:

I thought you had access to mathematica?! Was it not you in the other thread that used the FindRoot command?
That is why I presented it in that form.

Yes, that was me in the other form. No, I do not directly have access to Mathematica. In many cases Wolfram|Alpha can be manipulated to preform Mathematica commands. Although many Mathematica functions will not execute directly, if you rewrite them in a non-standard way, Wolfram|Alpha will correct the syntax and then execute the command. You can then use the "copyable plain-text" option to verify the command was executed as intended. Although it can be tedious at times, it beats paying $140 for a student license.

bobbym wrote:

Even still in numerical work we would consider the integral, gamma combination useless.
Strictly formal. Because you cannot compute hardly any x with it. As I said try x = 4. That ought to be an easy but not for that formula. The integral is impossible to get in closed form and the gamma function is infinity for negative integers.

An excellent point. I evaluated it numerically by expressing it in terms of the Poly-Logarithm.

bobbym wrote:

The general definition is that the answer should only contain elementary functions.

Alright fine; I admit it.... my answer amounts to nothing and is useless.

bobbym wrote:

I did not realize that is what you were after. I would have tried to discourage you. Here is a practical rule of thumb. If Mathematica and Maple cannot get a closed form then unless you are Euler do not bother.

Agreed. Although I recognize my answer amounts to nothing, it's not about the answer. For me its about the fun. I enjoy the process of trying to produce answers. Its strangely addicting. Even though I know every conceivable thing has already been done (or is not worth doing), I do it just to know that I can.

bobbym wrote:

You undoubtedly have better things to do with your math time.

Psssh... It's not like I was going to do my homework anyway.

Thank you for all your help! Your questions and corrections are extremely beneficial to my overall understanding of upper level math functions.

bobbym · 2011-01-04 15:37:21

Hi;

Alright fine; I admit it.... my answer amounts to nothing and is useless.

Nope! I do not agree. I am the last person in the world that can criticize someone else's useless math! I had fun and yes new math was learned. Please make the effort someday to utilize your student status and shell out the 140 bucks. It is 1850 if you are not a student! You will find it very useful.

Psssh... It's not like I was going to do my homework anyway

And please do your homework so someday you can end up in a spot where your good head can do some good.

Math Is Fun Forum

#1 2011-01-01 16:14:58

I need to get a hobby... :P

#2 2011-01-01 18:02:24

Re: I need to get a hobby... :P

#3 2011-01-02 07:23:28

Re: I need to get a hobby... :P

#4 2011-01-02 08:03:47

Re: I need to get a hobby... :P

#5 2011-01-02 08:49:01

Re: I need to get a hobby... :P

#6 2011-01-02 08:55:10

Re: I need to get a hobby... :P

#7 2011-01-02 09:06:20

Re: I need to get a hobby... :P

#8 2011-01-02 09:15:04

Re: I need to get a hobby... :P

#9 2011-01-02 09:25:46

Re: I need to get a hobby... :P

#10 2011-01-02 21:04:37

Re: I need to get a hobby... :P

#11 2011-01-03 15:29:26

Re: I need to get a hobby... :P

#12 2011-01-03 16:22:43

Re: I need to get a hobby... :P

#13 2011-01-03 17:30:12

Re: I need to get a hobby... :P

#14 2011-01-03 19:44:41

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#15 2011-01-04 01:24:07

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#16 2011-01-04 14:10:00

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#17 2011-01-04 14:24:06

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#18 2011-01-04 15:30:32

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#19 2011-01-04 15:37:21

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