## 100 Reputation

11 years, 140 days

## correct...

yes my sum and products are finite and i agree that i should test them. thanks guys for you time and help

## agreed too...

yes i noticed this bug but will it affect my differentiation or not?

i mean, are these results correct or this bug make them wrong?

## thankx alot...

i'm using maple 13 . thank you very much for your time. i still can't decide which mean should i consider so here you are the problem:-

enter this line to maple

w*(product(sum(p[i][j]*exp(-z[i][j]/t[j]), j = 1 .. m), i = 1 .. n))+(1-w)*[sum(sum(z[i][j], j = 1 .. m), i = 1 .. n

then differentiate w.r.t.     z[i][j]

then see the result and you will see the trailing tildes

## yes i intented...

yes i intented multiplications and lambda*b*j. i'll correct the code and try solve. if not i hope fsolve can do something

## help...

please please, i want to know which command (function) in maple should i use to solve these two equations. i posted the maple code of these equations before to copy them to your maple if you want. thanks

## @ Edgar...

thanks edgar. i used exp for exponential fn. and i will use * for multiply. but which maple function can solve them?

## yes robert that is what i'm...

yes robert that is what i'm saying. these are not differential equations.

here's the equations again. just copy them to maple and press enter

the first one:

n(-(2*(2*n+1))*exp(-2*`&lambda;a`(2*n+1))+(3*n+2)*exp(-`&lambda;a`(3*n+2))+(3*n+1)*exp(-`&lambda;a`(3*n+1))-(2*n+1)*e^(-`&lambda;a`(2*n+1)))+(2*(n-1))*exp(-`&lambda;a`(2*n+1))*n+(n-1)*exp(-`&lambda;a`(2*n+1))-(sum(-2*exp(-2*`&lambda;a`(2*n+i+1))*n-2*exp(-2*`&lambda;a`(2*n+i+1))*i-2*exp(-2*`&lambda;a`(2*n+i+1))+2*exp(-`&lambda;a`(2*n+i+2))*n+exp(-`&lambda;a`(2*n+i+2))*i+2*exp(-`&lambda;a`(2*n+i+2))+2*exp(-`&lambda;a`(2*n+i+1))*n+exp(-`&lambda;a`(2*n+i+1))*i+exp(-`&lambda;a`(2*n+i+1)), i = 1 .. n-1))-k(2*n+1)*(exp(-`&lambda;a`(2*n+1)))(exp(-`&lambda;b`(2*k+1))-exp(-`&lambda;b`(k+1))-exp(-`&lambda;bk`)+1)+(2*n+1)*exp(-`&lambda;a`(2*n+1))*(-(k-1)*exp(-`&lambda;bj`)+k-1+sum(exp(-`&lambda;b`(2*j+1))-exp(-`&lambda;b`(j+1)), j = 1 .. k-1))-(2*`&lambda;n`(n(e^(-2*`&lambda;a`(2*n+1))-e^(-`&lambda;a`(3*n+2))-e^(-`&lambda;a`(3*n+1))+e^(-`&lambda;a`(2*n+1)))-(n-1)*e^(-`&lambda;a`(2*n+1))-(sum(e^(-2*`&lambda;a`(n+i+1))-e^(-`&lambda;a`(2*n+i+2))-e^(-`&lambda;a`(2*n+i+1)), i = 1 .. n-1))+ke(e^(-`&lambda;b`(2*k+1))-e^`&lambda;b`(k+1)-e^(-`&lambda;kb`)+1)^(-`&lambda;a`(2*n+1))-e^(-`&lambda;a`(2*n+1))*(-(k-1)*e^(-`&lambda;jb`)+k-1+sum(e^(-`&lambda;b`(2*j+1))-e^`&lambda;b`(j+1), j = 1 .. k-1)))*(exp(-2*`&lambda;an`))(n(e^(-2*`&lambda;a`(2*n+1))-e^(-`&lambda;a`(3*n+2))-e^(-`&lambda;a`(3*n+1))+e^(-`&lambda;a`(2*n+1)))-(n-1)*e^(-`&lambda;a`(2*n+1))-(sum(e^(-2*`&lambda;a`(n+i+1))-e^(-`&lambda;a`(2*n+i+2))-e^(-`&lambda;a`(2*n+i+1)), i = 1 .. n-1))+ke(e^(-`&lambda;b`(2*k+1))-e^`&lambda;b`(k+1)-e^(-`&lambda;kb`)+1)^(-`&lambda;a`(2*n+1))-e^(-`&lambda;a`(2*n+1))*(-(k-1)*e^(-`&lambda;jb`)+k-1+sum(e^(-`&lambda;b`(2*j+1))-e^`&lambda;b`(j+1), j = 1 .. k-1)))+2*(lambda(n+1))(n(e^(-2*`&lambda;a`(2*n+1))-e^(-`&lambda;a`(3*n+2))-e^(-`&lambda;a`(3*n+1))+e^(-`&lambda;a`(2*n+1)))-(n-1)*e^(-`&lambda;a`(2*n+1))-(sum(e^(-2*`&lambda;a`(n+i+1))-e^(-`&lambda;a`(2*n+i+2))-e^(-`&lambda;a`(2*n+i+1)), i = 1 .. n-1))+ke(e^(-`&lambda;b`(2*k+1))-e^`&lambda;b`(k+1)-e^(-`&lambda;kb`)+1)^(-`&lambda;a`(2*n+1))-e^(-`&lambda;a`(2*n+1))*(-(k-1)*e^(-`&lambda;jb`)+k-1+sum(e^(-`&lambda;b`(2*j+1))-e^`&lambda;b`(j+1), j = 1 .. k-1)))*(exp(-2*`&lambda;a`(n+1)))(n(e^(-2*`&lambda;a`(2*n+1))-e^(-`&lambda;a`(3*n+2))-e^(-`&lambda;a`(3*n+1))+e^(-`&lambda;a`(2*n+1)))-(n-1)*e^(-`&lambda;a`(2*n+1))-(sum(e^(-2*`&lambda;a`(n+i+1))-e^(-`&lambda;a`(2*n+i+2))-e^(-`&lambda;a`(2*n+i+1)), i = 1 .. n-1))+ke(e^(-`&lambda;b`(2*k+1))-e^`&lambda;b`(k+1)-e^(-`&lambda;kb`)+1)^(-`&lambda;a`(2*n+1))-e^(-`&lambda;a`(2*n+1))*(-(k-1)*e^(-`&lambda;jb`)+k-1+sum(e^(-`&lambda;b`(2*j+1))-e^`&lambda;b`(j+1), j = 1 .. k-1)))-2*(lambda(2*n+1))(n(e^(-2*`&lambda;a`(2*n+1))-e^(-`&lambda;a`(3*n+2))-e^(-`&lambda;a`(3*n+1))+e^(-`&lambda;a`(2*n+1)))-(n-1)*e^(-`&lambda;a`(2*n+1))-(sum(e^(-2*`&lambda;a`(n+i+1))-e^(-`&lambda;a`(2*n+i+2))-e^(-`&lambda;a`(2*n+i+1)), i = 1 .. n-1))+ke(e^(-`&lambda;b`(2*k+1))-e^`&lambda;b`(k+1)-e^(-`&lambda;kb`)+1)^(-`&lambda;a`(2*n+1))-e^(-`&lambda;a`(2*n+1))*(-(k-1)*e^(-`&lambda;jb`)+k-1+sum(e^(-`&lambda;b`(2*j+1))-e^`&lambda;b`(j+1), j = 1 .. k-1)))*(exp(-2*`&lambda;a`(2*n+1)))(n(e^(-2*`&lambda;a`(2*n+1))-e^(-`&lambda;a`(3*n+2))-e^(-`&lambda;a`(3*n+1))+e^(-`&lambda;a`(2*n+1)))-(n-1)*e^(-`&lambda;a`(2*n+1))-(sum(e^(-2*`&lambda;a`(n+i+1))-e^(-`&lambda;a`(2*n+i+2))-e^(-`&lambda;a`(2*n+i+1)), i = 1 .. n-1))+ke(e^(-`&lambda;b`(2*k+1))-e^`&lambda;b`(k+1)-e^(-`&lambda;kb`)+1)^(-`&lambda;a`(2*n+1))-e^(-`&lambda;a`(2*n+1))*(-(k-1)*e^(-`&lambda;jb`)+k-1+sum(e^(-`&lambda;b`(2*j+1))-e^`&lambda;b`(j+1), j = 1 .. k-1))))/(1-(exp(-2*`&lambda;a`(n+1)))(1-exp(-2*`&lambda;an`))) = 0

the second one:-

n(e^(-2*`&lambda;a`(2*n+1))-e^(-`&lambda;a`(3*n+2))-e^(-`&lambda;a`(3*n+1))+e^(-`&lambda;a`(2*n+1)))-(n-1)*e^(-`&lambda;a`(2*n+1))-(sum(e^(-2*`&lambda;a`(n+i+1))-e^(-`&lambda;a`(2*n+i+2))-e^(-`&lambda;a`(2*n+i+1)), i = 1 .. n-1))+ke(e^(-`&lambda;b`(2*k+1))-e^`&lambda;b`(k+1)-e^(-`&lambda;kb`)+1)^(-`&lambda;a`(2*n+1))-e^(-`&lambda;a`(2*n+1))*(-(k-1)*e^(-`&lambda;bj`)+k-1+sum(e^(-`&lambda;b`(2*j+1))-e^(-`&lambda;b`(j+1)), j = 1 .. k-1))+be(k(-lambda(2*k+1)*e^(-`&lambda;b`(2*k+1))-lambda(k+1)*e^(-`&lambda;b`(k+1))+`&lambda;ke`^(-`&lambda;bk`))-(k-1)*`&lambda;je`^(-`&lambda;bj`)-(sum(-lambda(2*j+1)*e^(-`&lambda;b`(2*j+1))+lambda(j+1)*e^(-`&lambda;b`(j+1)), j = 1 .. k-1)))^(-`&lambda;a`(2*n+1)) = 0

and my variables are (a,b)

plz some one tell me how can i solve them.

## @ternox...

these equations are not differential equation and i don't have initial conditions how can i apply your code

## ok. applying.......

thanks  Robert  i'll consider what you said.

thanks  Ternox . i will need a while to understand the code and apply it on my equations but when i do it i'll tell you what did i get.

thanks again. later