Summation Question: Substituting y=ai+b in c=Σ(i2*yi)?

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The discussion centers on the substitution of a linear polynomial fit, y=ai+b, into the summation c=Σ(i^2*yi). It clarifies that substituting Y(i) for yi results in an approximation C=Σ(i^2*Y(i)), rather than the exact value of c. The conversation highlights the importance of calculating polynomial coefficients accurately, especially when using methods like least-squares or linear programming. Additionally, it addresses the possibility of determining coefficients a and b that satisfy the equation s_3*a + s_2*b = c, while also fitting the data. Overall, constrained optimization methods, such as the EXCEL Solver, can be employed to find suitable solutions for higher-degree polynomials.
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Homework Statement



I have a set of data (i, yi). A polynomial fit of 1st degree would be y=ai+b, right?
If I have c=Σ(i2*yi) is it correct to substitute y=ai+b inside the summation?

Homework Equations


The Attempt at a Solution

 
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lilly92 said:

Homework Statement



I have a set of data (i, yi). A polynomial fit of 1st degree would be y=ai+b, right?
If I have c=Σ(i2*yi) is it correct to substitute y=ai+b inside the summation?

Homework Equations


The Attempt at a Solution


You are being careless with notation, and it is landing you in trouble. You have data ##\{ (i, y_i)
\}## and fit a formula of the form ##Y(x) = ax + b## to the data; that is, you are approximating ##y_i## by the value ##Y(i) = ai + b##. Hopefully, the approximation is good in some sense, but that is another, separate issue. Anyway, you have a quantity ##c = \sum i^2 y_i##. When you substitute ##Y(i)## instead of ##y_i## you are computing an approximation ##C = \sum i^2 Y(i)## instead of the exact value of ##c##.
 
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I don't care about the exact approximation because I test various polynomials to figure out for which c approximates a specific known value. But my problem is what to do about the coefficients of the polynomials. Is there a way to calculate them in order to calculate c?
 
lilly92 said:
I don't care about the exact approximation because I test various polynomials to figure out for which c approximates a specific known value. But my problem is what to do about the coefficients of the polynomials. Is there a way to calculate them in order to calculate c?

How do you perform the fit to the data? If you use the least-squares method there are formulas for the coefficients. If you use some other method, there may not be formulas---only algorithms. For example, if you do a least average absolute-deviation fit, you can set up the problem as a linear program and solve it using a standard package (such as the EXCEL Solver). The solution of the linear program will include values of the coefficients.
 
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Okay I understand that, thank you. But what if I want to test with polynomials of second degree or higher?
My y data can be manually changed and c takes a specific value. What I'm trying to do is figure out which ys to change to make c take that value and/or by how much. Is that possible?
 
lilly92 said:
Okay I understand that, thank you. But what if I want to test with polynomials of second degree or higher?
My y data can be manually changed and c takes a specific value. What I'm trying to do is figure out which ys to change to make c take that value and/or by how much. Is that possible?

Are you asking whether we can find numbers ##a## and ##b## that give
\sum_{i=1}^n i^2 (ai+b) = c
then the answer is an obvious yes. If we let ##s_3 = \sum_{i=1}^n i^3## and ##s_1 = \sum_{i=1}^n i^2## then the equation just says that ##s_3 a + s_2 b = c## and there are lots of ##(a,b)## combinations that satisfy that. If you also want the form ##Y(i) = ai + b## to be a (hopefully good) fit to some data ##\{ i, y_i \}##, then you just have a constrained version of the usual data-fitting methods. The standard fitting formulas may no longer apply--- because of your specified constraint ##s_3 a + s_2 b = c##---but you can use a constrained optimization method to get a numerical solution. For example, you can do it using the EXCEL Solver.
 
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