Least Squares Fit for h(x)=ae^x+be^(-x) Homework

rey242
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Homework Statement


For the following data, find the least squares fit of the given form
x=1,2,3,4,6
y=14,10,8,6,5
h(x)=ae^x+be^(-x)

Homework Equations





The Attempt at a Solution


So I tried to linearize the equation by taking the natural log of everything
ln(h)=ln(ae^-x)+ln(be^ex)
ln(h)=ln(a)+ln(b)+x-x

that when I run into a problem, I eliminate the x's.
My question is, is there another way to linearize the equation or should I continue though?
 
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That's not right. In particular,
\ln (ae^{-x} + be^x) \neq \ln (ae^{-x}) + \ln (be^x)
 
There is no reason to linearize anything. Nor are you trying to fit the curve to those points- just the "least squares" fit.

If h(x)= ae^{x}+ be^{-x} then h(1)= ae+ be^{-1}. Since you are told that y= 14 when x= 1, the "error" is ae+ be^{-1}- 14 and the "error squared" is (ae+ be^{-1})^2. Similarly, for x= 2, y= 10, the "error squared" is (ae^2+ be^{-2}- 10)^2.

For all 5 x,y pairs, the sum of errors squared is (ae+ be^{-1}- 14)^2+ (ae^2+ be^{-2}- 10)^2+ (ae^3+ be^{-3}- 8)^2+ (ae^4+ be^{-4}- 6)^2+ (ae^6+ be^{-6}-5)^2. Find a and b to minimize that.
 
What procedure would I use to find a and b? I know about the error...would I convert the sum of the errors squared into normal equations?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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