Find the minimum sum of a series of equations

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Homework Help Overview

The discussion revolves around finding the minimum sum of a series of equations involving variables x1 and x2, constrained within specific limits. The equations define a sequence of sums S0, S1, S2, ..., Sn, where S0 is initialized to zero and subsequent Sn values are expressed in terms of previous sums and coefficients a0, a1, ..., an, along with a constant B.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants explore the formulation of the equations and their polynomial nature. Some discuss the implications of the degree of the polynomial in relation to the variables x1 and x2. Others seek clarification on how to determine the values of x1 and x2 that minimize the sum, particularly through examples.

Discussion Status

The conversation is ongoing, with participants providing insights into the mathematical properties of the equations. Some have offered guidance on the nature of the polynomial and the conditions for finding extrema, while others are still seeking concrete examples and further clarification on the minimization process.

Contextual Notes

Participants are working within the constraints of the problem, specifically the limits on x1 and x2. There is also mention of the need to analyze partial derivatives to find extrema, indicating a focus on calculus principles in the discussion.

alexleong
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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 = \SigmaSn
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 a lot.
 
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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?
 
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
 
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 "[URL
http://en.wikipedia.org/wiki/Second_partial_derivative_test.[/URL]

Can you continue from here?
 
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