[tex] y_i=A{x_i}^b [/tex](adsbygoogle = window.adsbygoogle || []).push({});

When I solve for A two different ways I am getting different answers..so somewhere I'm doing something wrong. If someone could point out where I would be grateful :).

Using logs:

[tex] y_i=A{x_i}^b [/tex]

[tex] ln(y_i)=ln(A)+b*ln(x_i) [/tex]

[tex] ln(y_i)-(ln(A)+b*ln(x_i))=r_i [/tex] for least squares we want to minimize: [tex] S=\sum_i^{n}{r_i}^2 [/tex] which means the gradient has to be zero. I only care about finding A right now so I only have to deal with the partial with respect to A:

[tex] \frac{\partial S}{\partial A}\sum_i^{n}{r_i}^2 =2 \sum_i^{n}{r_i}\frac{\partial r_i}{\partial A}= 2\sum_i^{n}(ln(y_i)-ln(A)-b*ln(x_i))\frac{1}{A}=0 [/tex]

The numerator inside the sum has to be zero, and we can ignore the 2, so:

[tex]\sum_i^{n}(ln(y_i)-ln(A)-b*ln(x_i))=0[/tex]

[tex]\sum_i^{n}ln(y_i)-n*ln(A)-b\sum_i^{n}ln(x_i)=0[/tex]

[tex] ln(A)=\frac{\sum_i^{n}ln(y_i)-b\sum_i^{n}ln(x_i)}{n} [/tex]

(this is the derivation that I think is correct).

But when i solve for A without taking the logs of each side first I get something else:

[tex] y_i - A{x_i}^b = r_i [/tex]

[tex] \frac{\partial S}{\partial A}\sum_i^{n}{r_i}^2 =2 \sum_i^{n}{r_i}\frac{\partial r_i}{\partial A}= 2\sum_i^{n}(y_i - A{x_i}^b)*-{x_i}^b=0 [/tex]

[tex]\sum_i^{n}(-{x_i}{y_i} + A{x_i}^{2b})=0 [/tex]

[tex]-\sum_i^{n}{x_i}{y_i}+A\sum_i^{n}{x_i}^{2b}=0 [/tex]

[tex] A=\frac{\sum_i^{n}{x_i}{y_i}}{\sum_i^{n}{x_i}^{2b}} [/tex]

And if you take the ln of it to compare it with what we got before you get:

[tex] ln(A)= ln(\frac{\sum_i^{n}{x_i}{y_i}}{\sum_i^{n}{x_i}^{2b}}) =ln(\sum_i^{n}{x_i}{y_i})-ln(\sum_i^{n}{x_i}^{2b}) [/tex]

Which is not the same as:

[tex] ln(A)=\frac{\sum_i^{n}ln(y_i)-b\sum_i^{n}ln(x_i)}{n} [/tex]

as far as I can tell...

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# Solving for least square coefficients of power law

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