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Approximate $F$ by a function of the type $f(t) = ae^{bt}$ by the least squares method

F(t) - 132 65 31 17 8

t - ----1---2 -3--4 - 5

This is what my teacher did (sketches because I don't know all the logical steps that he did): $$\min ||F(t)-f(t) || = \min_{a,b} \sum_{i=1}^5 (F(t_i)-ae^{bt_i})^2$$

$$F -ae^{bt}\implies \ln F - \ln ae^{bt} = \ln a + bt $$

$$\min\sum(\ln F(t_i)-\ln a + bt_i)^2$$ $$ \begin{bmatrix} \langle 1,1\rangle & \langle 1,t\rangle \\ \langle t,1\rangle & \langle t,t\rangle \end{bmatrix}\begin{bmatrix} \alpha \\ \beta \end{bmatrix} = \begin{bmatrix} 5 & 15\\ 15 & 55 \end{bmatrix} \begin{bmatrix} \alpha \\ \beta \end{bmatrix} = \begin{bmatrix} \langle \ln F, 1\rangle \\ \langle \ln F,t\rangle \end{bmatrix} $$

I think that that matrix of inner products came from this:

We're trying to find $\min ||F(t) - f(t)||$. This is the same as projecting $F(t)-f(t)$ onto the subspace generated by $f(t)$, so we should have:

$$\langle \ln F-\ln a - bt,1\rangle=0\\ \langle \ln F-\ln a - bt,t\rangle=0$$ which implies

$$\langle \ln F,1\rangle = \langle\ln a,1\rangle + \langle bt,1\rangle\\\langle \ln F,t\rangle = \langle\ln a,t\rangle + \langle bt,t\rangle$$

which is the same as

$$\langle \ln F,1\rangle = \ln a\langle 1,1\rangle + b\langle t,1\rangle\\\langle \ln F,t\rangle = \ln a\langle1,t\rangle + b\langle t,t\rangle$$ which is the system we have above.

Now the question is: which inner product he used to get that matrix? I don't have idea on how to take inner product of those things.

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In linear least squares, you are using the little 'el two' $l^2$ inner product since you have discrete data points.

$$<f,g> = \sum_{i=1}^{5} f_ig_i.$$

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