sum of an exponential series (i know this is cliche, but it's urgent)

**theclasher** · October 24th 2006, 06:07 AM

Okay, I'm brand new to this forum. I'm supposed to formulate a conjecture for the series (1^k)+(2^k)+(3^k)+(4^k)+...+(n^k)...

I've already found that the formula when k=1 is [n(n+1)/2]
when k=2 the sum of the series is (n/6)(n+1)(2n+1)
for k=3 it is [(n^2)/4](n+1)(n+1)
and for k=4, (n/30)(n+1)(2n+1)(3(n^2)+3n-1)

I don't know where to go from here. I've gone on to find the sums when k=5,6 but the question is based on the four sums I put above. Any ideas? Anyone done something like this before?

**ThePerfectHacker** · October 24th 2006, 06:25 AM

Originally Posted by theclasher

Okay, I'm brand new to this forum. I'm supposed to formulate a conjecture for the series (1^k)+(2^k)+(3^k)+(4^k)+...+(n^k)...

I've already found that the formula when k=1 is [n(n+1)/2]
when k=2 the sum of the series is (n/6)(n+1)(2n+1)
for k=3 it is [(n^2)/4](n+1)(n+1)
and for k=4, (n/30)(n+1)(2n+1)(3(n^2)+3n-1)

I don't know where to go from here. I've gone on to find the sums when k=5,6 but the question is based on the four sums I put above. Any ideas? Anyone done something like this before?

When I was younger I would to play around with taking differences between terms of sequences.

These where the interesting results:

A trivial sequence would be reffered to here as a sequence with constant results, i.e.,
1,1,1,1,1,....

A derivative is a sequence obtained from subtracting terms, i.e.
1,2,3,4,5,6....
Becomes,
1,1,1,.... (trivial)

1)A polynomial sequence of degree n has a trivial sequence after exactly n derivatives (and none before).

2)If a sequence is trivial after n derivatives (and none before) then it is definable in a polynomial of degree n.

3)Given the sequence, (called cyclotonomic sequence of degree n)
0^n,1^n,2^n,4^n,....
Then its trivial sequence is n! (factorial).

4)Given a polynomial $Ax^n+Bx^{n-1}+...+K$ its trivial sequence is, $A\cdot n!$ .

There are more facts but that should suffice.
-------------------------------------
Consider the sequence,
$0^m,1^m+0^m,2^m+1^m+0^m,....$
Its derivative is,
$1^m,2^m,3^m,....$
A polynomial sequence of degree $m$ .
Therefore the sequence above it is definable as a polynomial sequence of $m+1$ (because the derivative reduced the exponent).
Furthermore, the trivial sequence is $m!$ .

Which is why your "generalized exponent summation" can be expressed as,
$\frac{1}{m!}n^{m+1}+An^m+...+K$
---------
These coefficient follow something called Bernoulli numbers, search for them.

**theclasher** · October 24th 2006, 07:38 AM

Originally Posted by ThePerfectHacker

When I was younger I would to play around with taking differences between terms of sequences.

These where the interesting results:

A trivial sequence would be reffered to here as a sequence with constant results, i.e.,
1,1,1,1,1,....

A derivative is a sequence obtained from subtracting terms, i.e.
1,2,3,4,5,6....
Becomes,
1,1,1,.... (trivial)

1)A polynomial sequence of degree n has a trivial sequence after exactly n derivatives (and none before).

2)If a sequence is trivial after n derivatives (and none before) then it is definable in a polynomial of degree n.

3)Given the sequence, (called cyclotonomic sequence of degree n)
0^n,1^n,2^n,4^n,....
Then its trivial sequence is n! (factorial).

4)Given a polynomial $Ax^n+Bx^{n-1}+...+K$ its trivial sequence is, $A\cdot n!$ .

There are more facts but that should suffice.
-------------------------------------
Consider the sequence,
$0^m,1^m+0^m,2^m+1^m+0^m,....$
Its derivative is,
$1^m,2^m,3^m,....$
A polynomial sequence of degree $m$ .
Therefore the sequence above it is definable as a polynomial sequence of $m+1$ (because the derivative reduced the exponent).
Furthermore, the trivial sequence is $m!$ .

Which is why your "generalized exponent summation" can be expressed as,
$\frac{1}{m!}n^{m+1}+An^m+...+K$
---------
These coefficient follow something called Bernoulli numbers, search for them.

Thanks, this is really helpful. I don't completely understand it though. The part where you said "the sequence above is definable as a polynomial sequence of m+1 (because the derivative reduced the exponent)". I'm not sure how to explain what I don't get, but I had noticed, for all values of the exponent in the series, the power in the formula for the sum was raised was one higher. And in the last part where you have (1/m!)(n^m+1), i couldn't quite follow how you got that...or at least why you took the reciprocal of m!

**ThePerfectHacker** · October 24th 2006, 08:29 AM

Originally Posted by theclasher

Thanks, this is really helpful. I don't completely understand it though. The part where you said "the sequence above is definable as a polynomial sequence of m+1 (because the derivative reduced the exponent)". I'm not sure how to explain what I don't get, but I had noticed, for all values of the exponent in the series, the power in the formula for the sum was raised was one higher. And in the last part where you have (1/m!)(n^m+1), i couldn't quite follow how you got that...or at least why you took the reciprocal of m!

Because after you took the derivative you ended up with a polynomial sequence, right? Of degree $m$ .
Thus, after $m$ derivatives we have a trivial sequence, right? That means in total we used, $m+1$ derivative, right? Thus, that generalized sum is a polynomial of degree $m+1$ .

Now for the reciprocal of factorial, it is a little bit more difficult to explain without a full prove, but at least understand how it appears, because after 1 derivaitve you have a cyclotonomic sequence of degree $m$ which implies the trivial sequence will have all its values of $m!$ . Basically, how the reciprocal appears is that you do not want it in your formula, so to remove it you multiply by the reciprocal.

**CaptainBlack** · October 24th 2006, 09:05 AM

Originally Posted by theclasher

Okay, I'm brand new to this forum. I'm supposed to formulate a conjecture for the series (1^k)+(2^k)+(3^k)+(4^k)+...+(n^k)...

I've already found that the formula when k=1 is [n(n+1)/2]
when k=2 the sum of the series is (n/6)(n+1)(2n+1)
for k=3 it is [(n^2)/4](n+1)(n+1)
and for k=4, (n/30)(n+1)(2n+1)(3(n^2)+3n-1)

I don't know where to go from here. I've gone on to find the sums when k=5,6 but the question is based on the four sums I put above. Any ideas? Anyone done something like this before?

You need to ask Jakob Bernoulli about this, unfortunatly he is dead, but
you will find a summary of his relevant work here.

RonL

**theclasher** · October 24th 2006, 09:13 AM

Makes sense. Are you multiplying the whole polynomial by 1/m! or just the first term?

**ThePerfectHacker** · October 24th 2006, 09:38 AM

Originally Posted by theclasher

Makes sense. Are you multiplying the whole polynomial by 1/m! or just the first term?

Only the first term is like that.
---
I have spend many hours and days to solve this problem and find a pattern with these coefficients. Failed. It made me so angry when I read about a person solving this problems with bernoulli numbers. I only succeded in finding the first term.

I have another way to demonstrate the first term of this polynomial, using Riemann sums and fundamental theorem of calculus. Would thou like to see it?

**theclasher** · October 24th 2006, 09:46 AM

Haha, sure. (thanks both of you, by the way)

**theclasher** · October 24th 2006, 10:53 AM

Originally Posted by ThePerfectHacker

I am going to re-edit this post later on because.
I am having problems posting the images.
Check again tonight (after 9) to see if I responded back.

This is my 3

th Post!!!

Heh, congratulations.

**ThePerfectHacker** · October 24th 2006, 05:00 PM

Consider the real function, $m\in \mathbb{Z}^+$
$f=x^m$ on $[0,1]$

The the integral on this interval is,
$\int_0^1 x^m dx=\frac{1}{n+1}$

That means the Riemann sum partitioned in any way with the limit of the norm sequence zero much give this value.

Let us partition the interval by "right-enpoints".

The limit of the sum, will them be:
$\lim_{n\to \infty}\sum_{k=1}^n \left( \frac{k}{n} \right)^n \cdot \frac{1}{n}$
Some easy manipulations,
$\lim_{n\to \infty}\sum_{k=1}^n \frac{k^m}{n^{m+1}}$
But, the generalized sum takes on a polynomial,
$\sum_{k=1}^n k^m=An^{m+1}+Bn^{m}+...+Nn+M$
Thus, the Riemann sum is,
$\lim_{n\to \infty} \frac{An^{m+1}+Bn^{m}+...+Nn+M}{n^{m+1}}$
Thus,
$\lim_{n\to\infty} A+\frac{B}{n}+...+\frac{N}{n^m}+\frac{M}{n^{m+1}}$
Which is,
$\boxed{A}$
Thus, the Riemann sum and integral are equal,
$A=\frac{1}{m+1}$ .

I am sorry about what I said before about $(m+1)!$ . I made a mistake. When I did it that way it worked out to $\frac{m!}{(m+1)!}=\frac{1}{m+1}$ that is where factorial appeared and why I got confused.

Note what I have done. I have shown that the generalized sum is definable as a polynomial sequence (of 1 more degree) then I used integrations in two different ways to show what the leading coefficient has to be equal to. And also note I by no way shown what the other coeffients are. Hope you like that proof.

This is my 3

th Post!!!

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