Question

least square method

So ive got some indata
t=0 y0=1.77
t=1
t=2
t=3
t=3

Answer 1

t=0 y0=1.77
t=1 y1 = 2.35
t=2 y2 = 3.46
t=3 y3 = 5.12
t=4 y4 = 7.01

And my first task was to make it fit B(t) = at+b and I found out that the best fit was a= 1.36 and b = 1.22
and my second task was to make it fit B(t) = Ae^(kt) which is equivalent to ln (B(t) ) = kt + ln(A)

Answer 2

So my question from here is what to do? Do i have to take the logarithm of all the y values and then do the calculations all again?

Answer 3

\[Y=\left[\begin{matrix}\ln(1.77) \\ \ln(2.35) \\ \ln(3.46) \\ \ln(5.12) \\ \ln(7.01)\end{matrix}\right]\]

\[A = \left[\begin{matrix}0 & 1 \\ 1 & 1 \\ 2 & 1 \\ 3 & 1 \\ 4 & 1\end{matrix}\right]\]

\[\left(\begin{matrix}k \\ \ln A\end{matrix}\right)\]

Answer 4

\[A^T=\left[\begin{matrix}0 & 1 &2&3&4\\ 1 & 1&1&1&1\end{matrix}\right]\]
\[A^T*A=\left[\begin{matrix}30 & 10 \\ 10 & 5\end{matrix}\right]\]
\[A^T*Y=\left[\begin{matrix}0 & 1 & 2 & 3 & 4\\ 1 & 1 & 1 & 1 & 1\end{matrix}\right]*\left[\begin{matrix}\ln(1.77) \\ \ln(2.35) \\ \ln(3.46) \\ \ln(5.12) \\ \ln(7.01) \end{matrix}\right]=\left(\begin{matrix}16.0258 \\ 6.24716\end{matrix}\right)\]

Answer 5

\[A^T*A*X=AT*Y\] Normal equation
\[\left[\begin{matrix}30 & 10 \\ 10 & 5\end{matrix}\right]*\left(\begin{matrix}k \\ \ln A\end{matrix}\right)=\left(\begin{matrix}16.0258 \\ 6.24716\end{matrix}\right)\]

then we get the equation system
\[30k+10\ln(A) = 16.0258\]
\[10k + 5\ln(A) = 6.24716\]

Answer 6

@ParthKohli am I on the right track

Answer 7

I have not checked your details, but
the idea is you start with
\[ y= Ae^{kx} \\ \ln y = \ln A + k x \]
or if we rename the left-hand side
\[ \tilde y = k x + \ln A\]
you have values for x and ln y
and find the best values for ln A and k

I would then use those values to write the final "best fit"
\[ y = \ln A e^{kx} \]

Answer 8

* that last equation should be y = A e^kx
where A = e^lnA
and ln A is what you found in the linear least-fit.

Answer 9

@phi I have an answear B(t) = 0.543136 * e^0353148t

Answer 10

But my A seems to be wrong... cause it doesnt fit at all..

Answer 11

My friend got it to 1.7214 but I cant figure out how he did and he is not available now

Answer 12

found it! Im done :) thanks