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#1 2011-11-13 08:17:42

wintersolstice
Real Member Registered: 2009-06-06
Posts: 128

sums of power sequences

many more to come:D

PS sorry about that last one I worked it out myself (the otehrs I got from a book years ago) I'm no good at factorising so if someone could do that for me I would be most greatful

Last edited by wintersolstice (2011-11-13 08:26:46)

Why did the chicken cross the Mobius Band?
To get to the other ...um...!!!

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#2 2011-11-14 05:02:34

bobbym
bumpkin From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606

Re: sums of power sequences

Hi;

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

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#3 2011-11-16 05:58:39

wintersolstice
Real Member Registered: 2009-06-06
Posts: 128

Re: sums of power sequences

Thanks bobbym

here is 5 as far as I know it is fully factorised

Why did the chicken cross the Mobius Band?
To get to the other ...um...!!!

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#4 2011-11-16 06:03:48

bobbym
bumpkin From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606

Re: sums of power sequences

Hi wintersolstice;

You can do just a bit better:

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

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#5 2011-11-17 03:25:29

benice
Member Registered: 2010-06-10
Posts: 117
Website

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#6 2012-05-31 14:32:03

knighthawk
Member
Registered: 2012-05-29
Posts: 7

Re: sums of power sequences

I think I've got a generic format for sums of power sequences. I'm not a math expert, so maybe somebody can express this better. First, is the constraint that
1 = 1, i.e. for any positive integer k

1
∑ i^k = 1
1

n
∑ i^0 = n^1
1

To go to sum of powers of 1...
Step 1) Multiply to clear the denominator of the first term
Step 2) Integrate the expression
Step 3) Add k*n such that (1)=1

n
∑ i^1 = (n^2)/2 + n/2
1

To go to sum of powers of 2...
Step 1) Multiply to clear the denominator of the first term
Step 2) Integrate the expression
Step 3) Add k*n such that (1)=1

n
∑ i^2 = (n^3)/3 + (n^2)/2 + n/6
1

To go to sum of powers of 3...
Step 1) Multiply to clear the denominator of the first term
Step 2) Integrate the expression
Step 3) Add k*n such that (1)=1

n
∑ i^3 = (n^4)/4 + (n^3)/2 + (n^2)/4
1

Rinse; Lather; Repeat. You can take this to any arbitrarily large integer. From this point onwards, it appears that every second co-efficient is zero.

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#7 2013-03-07 19:24:28

cmowla
Member
Registered: 2012-06-14
Posts: 70

Re: sums of power sequences

I know it's almost been a year since this thread's last post, but I made an "Adjusted Pascal's Triangle" to create these power sum formulas for positive integers, and I finally got around to posting it on this forum. Which I drew from the following formula I derived "from scratch":

(Putting in the n at the top of the triangle for sum of i^0 worked out perfectly, even though the formula itself cannot compute the correct value for R = 0).

Adjusting this triangle and using the method that knighthawk used in the previous post, I created an "Adjusted Pascal's Triangle" to compute the Bernoulli numbers which can be viewed here (it's too wide to post here, I think).

Basically, I found that

, and I used that to create that "Bernoulli Triangle".

I then wrote the following recursion formula from that Bernoulli triangle.

where

, where you can calculate a Bernoulli number in terms of its predecessor Bernoulli numbers.

Has anyone seen similar images like these "Adjusted Pascal Triangles" to compute the power sum and Bernoulli numbers visually?  I'm curious because I never heard of either, particularly the sum of power one, in high school or college math courses.

Last edited by cmowla (2013-03-08 12:26:05)

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#8 2013-03-08 00:04:21

Stangerzv
Member
Registered: 2012-01-30
Posts: 265

Re: sums of power sequences

Hi cmowla

I happened to read the article regarding the sums of power of integer using Bernoulli & Pascal long time ago. Could look similar but you can read it here www.sanjosemathcircle.org/handouts/2008-2009/20081112.pdf

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#9 2013-03-29 08:41:56

cmowla
Member
Registered: 2012-06-14
Posts: 70

Re: sums of power sequences

Stangerzv wrote:

Hi cmowla

I happened to read the article regarding the sums of power of integer using Bernoulli & Pascal long time ago. Could look similar but you can read it here www.sanjosemathcircle.org/handouts/2008-2009/20081112.pdf

When I was in the process of deriving the formula which I used to create the "adjusted pascal triangle" for the sum of power formulas, I actually went through the same process that Bernoulli did for a portion of my research.

Leaving out all of the ugly trial and error that I did (it was helpful to have a CAS handy through it all), here is my full derivation of my formula

which I used to create that "adjusted Pascal's Triangle" for the power sum polynomial formulas.

[Step 1]:  I derived a formula for sum i^2 from scratch.

We need to show that

[Step 2]:  Extrapolate

By applying the same pattern

to different power sums, I found that a general formula for sum i^R is simply:

, which has been known for a while, but I didn't know that unfortunately until after I found it.

This is the only step which is not fully justified, should one consider this entire derivation to possibly be a single proof, and is why I said I derived it "from scratch" (emphasizing with the quotes that there was one point of intuition).  However, one could simply find a proof for that formula, since it has existed for a while, and combine it with the remaining to have a complete proof that my "Adjusted Pascal's Triangle" is indeed correct to construct the power sum polynomials for positive integers.

[Step 3] Make a recurrence relation.

The above general formula isn't useful unless we can somehow use it to calculate the power sum sequence for the next power using all previous formulas.  The following is the recurrence relation I came up with which proved to be useful.  If the expression in the summation of the formula below is evaluated at any positive integer R>0, then the greatest exponent of i is R-1, i.e., i^(R-1), which is precisely what we need in a recurrence relation.

So we need to show that:

[Step 4]: Transforming into a pattern which can be generated by a diagram (which turned out to be an "adjusted Pascal Triangle")

This was the trickiest part.  I probably spent more time doing this step than anything else because I didn't even know if it was possible to create a formula which could generate an intuitive diagram.

So we show that

Last edited by cmowla (2013-03-29 09:15:03)

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#10 2015-06-12 04:59:55

cmowla
Member
Registered: 2012-06-14
Posts: 70

Re: sums of power sequences

Yesterday I created and uploaded a three part video series on the content I have posted in this thread previously (with the exception of the Bernoulli number stuff).

I present the method in "Part 1", abstract it in "Part 2", and prove it in "Part 3".  Enjoy!

I realized that I made a few mistakes in the proof(s) in the messages/spoilers in this post, but my video proof is correct.  In addition, I have created my own proof of

Therefore my proof is now both complete and correct.

Let me know what you guys think! Last edited by cmowla (2015-06-12 05:01:03)

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#11 2017-08-16 17:30:18

almighty100
Member
Registered: 2017-08-16
Posts: 5

Re: sums of power sequences

In 2005 I wanted to actually derive all of the formulas for the sum of integers, squares, cubes, etc.  I arrayed them vertically in a way and found that:

.
.
.

For each of them, the summation on the right hand side was distributed to each term, and after rearranging, collecting like-terms, I was able to derive the formulas for each case.

After finding the sum of the fourth powers, and recognizing the binomial coefficients, I had discovered more generally:

which I have never seen anywhere else before today (which I recognized as I watched your video on youtube for the proof of your theorem).  That video had a link directly to this page, so I signed up and here I am As I continued to find the formulas, I discovered another pattern and derived the following:

Where:

and

is a binomial coefficient.

After discovering this/these formulas, I began to research the sums of powers.  It was then that I learned about Jacob Bernoulli's work on the problem and Bernoulli Numbers.

Once I had learned about them, I found out that I can extract a recursive formula for Bernoulli Numbers out of my formula:

Using this formula, the first Bernoulli number is positive 1/2.  I was proud of myself for discovering these nasty looking formulas.  Although, there are much more efficient methods for arriving with the summation of powers and, of course, Bernoulli Numbers.
The most beautiful that I have seen for the summation of powers is:

or more simply:

where

is the
Bernoulli Number. i.e. the exponents of
act as subscripts.

For example:

This interesting relation of exponents acting as subscripts is referred to as "Umbral Calculus" which, other than this marvelous integral, I have never used before.

At any rate, I'm glad to have found this site.  Today is my first day here.  this is my first (mathematical) post.  I hope you guys are still active.

Last edited by almighty100 (2017-08-17 04:57:35)

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#12 2017-08-16 20:16:31

zetafunc
Moderator Registered: 2014-05-21
Posts: 2,376
Website

Re: sums of power sequences

Hi almighty100,

Welcome to the forum! Thank you for your excellent first contribution.

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#13 2017-08-17 17:08:43

ganesh Registered: 2005-06-28
Posts: 35,520

Re: sums of power sequences

Hi almighty100,

This is a Brilliant Post!

It appears to me that if one wants to make progress in mathematics, one should study the masters and not the pupils. - Niels Henrik Abel.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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#14 2017-08-18 10:19:30

almighty100
Member
Registered: 2017-08-16
Posts: 5

Re: sums of power sequences

Thank you both for the reply.  cmowla replied on his youtube video, as the links he was trying to post for me (to his work on the problem) apparently caused him to be banned... for a only few days, hopefully?

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#15 2017-08-18 10:42:52

zetafunc
Moderator Registered: 2014-05-21
Posts: 2,376
Website

Re: sums of power sequences

Occasionally if a member attempts to post a link which has a certain word in it too many times, that person's IP will automatically be banned from the site as a precaution -- this has happened to me a few times in the past, but it does also prevent a lot of spammers from flooding the forum. I will inform MathsIsFun about cmowla's situation.

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#16 2017-08-25 03:34:48

cmowla
Member
Registered: 2012-06-14
Posts: 70

Re: sums of power sequences

almighty100 wrote:

Thank you both for the reply.  cmowla replied on his youtube video, as the links he was trying to post for me (to his work on the problem) apparently caused him to be banned... for a only few days, hopefully?

I guess the spam filter block time is up.

For those interested, I replied to almighty100 with this pdf in one the comments in the third video of my video series.  In the hyperlinked thread which contained my first online post about this topic, if you go both backward and forwards from that post, you can see other proofs of specific powers of i.  (For example, if you go ahead, you will find how I proved Sum i^1 by writing a linear equation from two "data points", and you can see a system of equations approach for calculating i^2 from someone else a few posts before that.)

Last edited by cmowla (2017-08-25 03:43:07)

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#17 2019-03-23 22:27:52

Eulero
Member
Registered: 2016-09-14
Posts: 11

Re: sums of power sequences

cmowla wrote:

I know it's almost been a year since this thread's last post, but I made an "Adjusted Pascal's Triangle" to create these power sum formulas for positive integers, and I finally got around to posting it on this forum.

http://i.minus.com/i67c1SgPQAHxV.png

Which I drew from the following formula I derived "from scratch":

(Putting in the n at the top of the triangle for sum of i^0 worked out perfectly, even though the formula itself cannot compute the correct value for R = 0).

Adjusting this triangle and using the method that knighthawk used in the previous post, I created an "Adjusted Pascal's Triangle" to compute the Bernoulli numbers which can be viewed here (it's too wide to post here, I think).

Basically, I found that

, and I used that to create that "Bernoulli Triangle".

I then wrote the following recursion formula from that Bernoulli triangle.

where

, where you can calculate a Bernoulli number in terms of its predecessor Bernoulli numbers.

Has anyone seen similar images like these "Adjusted Pascal Triangles" to compute the power sum and Bernoulli numbers visually?  I'm curious because I never heard of either, particularly the sum of power one, in high school or college math courses.

http://www.mathisfunforum.com/viewtopic.php?id=23646          same result bro .

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#18 2019-03-23 22:38:56

cmowla
Member
Registered: 2012-06-14
Posts: 70

Re: sums of power sequences

Eulero wrote:

http://www.mathisfunforum.com/viewtopic.php?id=23646          same result bro .

I couldn't follow your proof.  Did you see my video series?  The third video is my complete direct proof.

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#19 2021-05-02 10:56:42

cmowla
Member
Registered: 2012-06-14
Posts: 70

Re: sums of power sequences

I know that I'm bumping an old thread, but I recently found a video where a guy explains (proves) why the integration method that knighthawk posted about works.

Here is the video where he shows the method knighthawk posted about.
Here is the video where he (apparently) proves the method.

I actually commented on the video (showing the method), but he was "kind enough" to delete my comment (wherein I just mentioned that I saw this method posted here 10 years ago and also asked him what he thought of my Pascal's Triangle method).  So he's nice on screen, but maybe a little insecure . . .

Last edited by cmowla (2021-05-02 10:57:49)

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#20 2021-11-18 16:15:04

anan21
Member
Registered: 2021-11-18
Posts: 1

Re: sums of power sequences

wintersolstice wrote:

many more to come:D

PS sorry about that last one I worked it out myself (the otehrs I got from a book years ago) I'm no good at factorising so if someone could do that for me I would be most greatful

I agree with you.

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