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bomba923
Oct2-04, 12:13 AM
What would be the sum formula for the summation in the attachment?

For any real constant 'c', what is the sum formula for

k
sigma (n^c) ?
n=1

bomba923
Oct4-04, 05:44 PM
oh sorry! :frown: the file needs mathtype to view....sorry!

This file is GIF image format--should be easier to open i hope :surprised
I attached the equation as a GIF image file...i hope it can be open....the GIF file that i attached (the equation is a GIF image file...yeah) (whtisthsum.gif)

(*Note: this is not really a power series--the exponent 'c' is a constant!! does not change!...so the sum really goes like
1+(2^c)+(3^c)+(4^c)+(5^c)+(6^c)+...+(k^c)

The exponent c does not change...it is the same for every term of the equation as u add them up...(2^c) and so on to (k^c)..the c exponent does not change, so it's not really a power series)

Gokul43201
Oct4-04, 10:14 PM
I don't believe there's a general closed form for that sum

There are formulas for specific whole number value of c. ie : for c=0,1,2,3 etc.

Also, it's not hard to find a formula, for a general positive integer value of c. This can be done by simply assuming the sum is a polynomial of degree c+1, and determining the coefficients by plugging in the first c+2 values of the sum.

Need to think more about a general method for real c.

Gokul43201
Oct4-04, 10:16 PM
Of course, that's other than using Mathematica, or Maple. In fact, this can be done quite easily using Excel, too.

bomba923
Oct5-04, 02:18 AM
Wait, but what would the formula be for any real c>0 ?

I tried solving it, but is there a formula for (a+b)^c , where c>0 but where 'c' could be real???...(not just natural). Let's just take the case where (a+b)>0 , because u cannot have a real root for an irrational power of a negative number...

So is there a formula for (a+b)^c where c is real and c>0 and (a+b)>0?

Look at the attachments...um, i posted three sorry :frown:
I needed to break one GIF file into two...so there are three (sorry)

Gokul43201
Oct5-04, 10:41 AM
I don't believe there is a general formula for real c. If you use the binomial expansion for (a+b)^c~,~c~\epsilon~\mathbb{R}, you will still have terms like a^c.

You don't have to be including attachments for math representations. You can simply use LaTeX typesetting, as I've done. Look at this thread for LaTeX :
http://www.physicsforums.com/showthread.php?t=8997

bomba923
Oct8-04, 11:46 PM
Hmm...i've found a solution elsewhere on some polysum tripod site:

< http://polysum.tripod.com/ >

What does it mean when an integrand is written only with a lower limit without an upper one?
Does that mean it applies from that lower limit to infinity? or something else?