# Thread: summation = constant, solve for unknown variable in summation

1. ## summation = constant, solve for unknown variable in summation

Hello, I am new here... so here I go.

I am trying to figure out how to solve for a variable which I call x in the following equation

(D(i)^3-x^3)*N(i) + (D(i+1)^3-x^3)*N(i+1)... (D(n)^3-x^3)*N(n)= m

I know the values of D, N and I know the constant m.

I appreciate your help.

Thanks

2. ## Re: summation = constant, solve for unknown variable in summation

Originally Posted by kjs356
Hello, I am new here... so here I go.

I am trying to figure out how to solve for a variable which I call x in the following equation

(D(i)^3-x^3)*N(i) + (D(i+1)^3-x^3)*N(i+1)... (D(n)^3-x^3)*N(n)= m

I know the values of D, N and I know the constant m.

I appreciate your help.

Thanks
$\displaystyle \left(\sum_{i=1}^n \left(D_i^3-x^3\right)N_i\right)=m$

$\displaystyle \sum_{i=1}^n D_i^3 N_i - \sum_{i=1}^n N_i x^3 =m$

rewrite this a bit as

$\displaystyle C_1 = C_2 x^3$ where

$C_1 = \displaystyle \left(\left(\sum_{i=1}^n D_i^3 N_i\right)-m\right)$ and $\displaystyle C_2=\sum_{i-1}^n N_i$

finally

$x=\sqrt[3]{\dfrac{C_1}{C_2}}$

3. ## Re: summation = constant, solve for unknown variable in summation

Thanks!... but I made a small mistake and still can't quite get it.
any chance you solve again, but instead of (D(i)^3-x^3) it should be (D(i)-x)^3.

4. ## Re: summation = constant, solve for unknown variable in summation

Originally Posted by kjs356
Thanks!... but I made a small mistake and still can't quite get it.
any chance you solve again, but instead of (D(i)^3-x^3) it should be (D(i)-x)^3.
why did I know you were going to say this....

5. ## Re: summation = constant, solve for unknown variable in summation

$\displaystyle \left(\sum_{i=1}^n \left(D_i-x\right)^3 N_i\right)=m$

$\displaystyle \left(\sum_{i=1}^n \left(D_i^3-3 D_i^2 x+3 D_i x^2-x^3\right) N_i\right)=m$

$\displaystyle \left(\left(\sum_{i=1}^n D_i^3 N_i\right)-m\right)-\left(3\sum_{i=1}^n D_i^2 N_i\right)x+\left(3\sum_{i=1}^n D_i N_i\right)x^2-\left(\sum_{i=1}^n N_i\right)x^3=0$

$a x^3+b x^2 + c x + d =0$

$\displaystyle a=-\left(\sum_{i=1}^n N_i\right)$

$\displaystyle b=3\sum_{i=1}^n D_i N_i$

$\displaystyle c=-\left(3\sum_{i=1}^n D_i^2 N_i\right)$

$\displaystyle d=\left(\sum_{i=1}^n D_i^3 N_i\right)-m$

There is a formula for calculating the roots of a cubic polynomial.

I'm not going to reproduce it here.

You can find it here.

6. ## Re: summation = constant, solve for unknown variable in summation

Thanks! Sorry... for doubling the work.