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Find the minimum sum of a series of equations

  1. Jun 23, 2008 #1
    I’m dealing with a series of equations to find out the values of x1 and x2 so that the sum of S0+S1+...+Sn will have the minimum value.
    The x1 and x2 values are limited to –1<x1<1 and –1<x2<1.

    S0 = 0
    S1 = a1 – [B(1 – x1) + a0* x1 – S0*x2]
    S2 = a2 – [B(1 – x1) + a1* x1 – S1*x2]
    S3 = a3 – [B(1 – x1) + a2* x1 – S2*x2]
    S4 = a4 – [B(1 – x1) + a3* x1 – S3*x2]
    S5 = a5 – [B(1 – x1) + a4* x1 – S4*x2]
    ...
    Sn = an – [B(1 – x1) + an* x1 – Sn-1*x2]
    n
    T = [tex]\Sigma[/tex]Sn
    n=1

    Where
    B is a constant.
    T is the minimum sum of the equations.
    Sn is the result of each equation.
    a0, a1, a2...,an are the coefficients of the equation.

    Hope you understand my question and thanks alot.
     
  2. jcsd
  3. Jun 23, 2008 #2
    I see. Let me see:

    S0 = 0
    S1 = a1 – [B(1 – x1) + a0* x1 – S0*x2] = a1 - [B(1 – x1) + a0* x1]=
    = (a1 - B) + (B - a0) * x1

    Assume n = 1.

    T = S1 = ((a1 - B) + (B - a0) * x1

    This is a polynomial of first degree in x1. It has a minimum at x1 = + or -1 depending on the sign of B - a0.

    Now calculate S2:

    S2 = a2 – [B(1 – x1) + a1* x1 – S1*x2] =
    = a2 – [B(1 – x1) + a1* x1 – ((a1 - B) + (B - a0) * x1)*x2] =
    = a2 - B + (B+a1)*x1 + (a1 - b) * x2 + (B - a0) * x1 * x2

    and for n = 2 T will be a polynom of second degree in x1 and x2. (the second degree comes from the product x1 * x2)

    Similarly for higher n. The result, T, is a polynom in x1 and x2. It is not that difficult to find a minimum of such a polynom.

    Anything else?
     
  4. Jun 23, 2008 #3
    hi Tomy,
    First of all, thanks for your prompt reply. I still not quite get what you meant, how do I know the values of x1 and x2 where (S0 + S1 + S2) is the minimum? Please show me a simple example, let n = 2
    S0 = 0
    S1 = (a1 - B) + (B - a0)*x1
    S2 = a2 - B + (B - a1)*x1 + (a1 - B)*x2 + (B - a0)*x1*x2

    S0+S1+S2 = (a1 - B) + (B - a0)*x1 + a2 - B + (B - a1)*x1 + (a1 - B)*x2 + (B - a0)*x1*x2

    Thanks
     
  5. Jun 24, 2008 #4
    Ok, so let me look at the case of n = 2. Consider function T of variables x1 and x2. From analysis of function of several variables we know that this function attains an extremum (that is either a minimum or maximum) only if the partial derivatives of T with respect of x1 and x2 vanish:

    dT/dx1 = 0
    dT/dx2 = 0

    In your case T is a polynom linear in x1 and x2 but it has a term that is x1*x2.

    To decide whether this extremum is a minimum or maximum we have to look at the second derivatives. Again from analysis of functions of several variables we know that function T has a minimum if see
    http://en.wikipedia.org/wiki/Second_partial_derivative_test
    .

    Can you continue from here?
     
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