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Math Help - summation = constant, solve for unknown variable in summation

  1. #1
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    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
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  2. #2
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    Re: summation = constant, solve for unknown variable in summation

    Quote Originally Posted by kjs356 View Post
    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

    $x=\sqrt[3]{\dfrac{C_1}{C_2}}$
    Thanks from kjs356
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  3. #3
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    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.
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  4. #4
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    Re: summation = constant, solve for unknown variable in summation

    Quote Originally Posted by kjs356 View Post
    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....
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  5. #5
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    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.
    Thanks from kjs356
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  6. #6
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    Re: summation = constant, solve for unknown variable in summation

    Thanks! Sorry... for doubling the work.
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