Dear all,

I have a tricky problem; I need to solve this equation for

Is it possible in some way?

The only constraints are ,

Thanks a lot

Simo

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- Nov 28th 2008, 06:39 AMSimonot-so-simple equation to solve
Dear all,

I have a tricky problem; I need to solve this equation for

Is it possible in some way?

The only constraints are ,

Thanks a lot

Simo - Nov 28th 2008, 07:54 AMshawsend
Going gets tough, tough resort to numerical methods. The following is Mathematica code to calculate the value of n for the supplied values of R and L. First out is the expression for the sum, second out is the numerically calculated value of n which makes the sum equal to zero, and the third out is a back-substitution of the the root for a check.

Code:`In[234]:=`

R = 3;

L = 5;

formula1 = Sum[Binomial[R, r]*(-1)^(R - r)*

(r/L)^n*Log[r/L], {r, 1, R}]

root = n /. FindRoot[formula1 == 0,

{n, 1.2}]

formula1 /. n -> root

Out[236]= -(3/5)^n Log[5/3] + 3 (2/5)^n Log[5/2] - 3 5^-n Log[5]

Out[237]= 1.38015

Out[238]= -5.55112*10^-17

- Nov 28th 2008, 08:07 AMSimo
Ok,

I already have a numerical solution like that, but it is not interesting for me.

Is not possible to find an analytical solution, without resort to numerical methods?