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I have data which can be described by $y=f(x,z)$ where $z$ varies from 170 ~ 154. Now values given by $ks$ are known sample values that equals value given in the table header, $uks$ are unknown samples. $$ \begin{array} & z & x_{ks_{1}=0} & x_{ks_{2}=50} & x_{ks_{3} = 60} & \cdots & x_{uks_{1}}& \cdots & x_{uks_{n}}\newline 170 & 3.5 & 5 & 6 &\cdots & 5.3 & \cdots & 12 \newline 168 & 3.7 & 5.1 & 6.6 &\cdots & 5.6 & \cdots & 13.2 \newline \vdots & \vdots & \vdots & \vdots &\ddots & \vdots & \ddots & \vdots \newline 154 & 6.8 & 9.52 & 10 &\cdots & 11.5 & \cdots & 26.5 \newline \end{array} $$

Plotting values of $y$ vs. $x$ for various $z$, I get this plot:

enter image description here

and on a semi-log plot:

enter image description here

Note the values on $y-axis$ are the values given by the subscripts of $x$ in the table headers.

**My objective is to find the value of $y$ for the corresponding $x_{uks_{1}}, x_{uks_{2}}, \cdots, x_{uks_{n}}$ for different $z$.**There might be alternate methods, such methods are most welcome. And my approach is as follow:

  1. I try to fit a linear function of the form:

$$y = \frac{A_{i}}{ln[x]^{3}} + B_{i},\quad \text{i for each z}$$

this gives me plots as follows:

enter image description here

  1. Since $A_{i}$ and $B_{i}$ is different for $z$, I can fit a function. Therefore $A_{i} = f(z)$ and $B_{i} = f(z)$. For step 1, I used the following function:

$$A = a_1 z^3 + a_2 z^2 + a_3 z + a_4$$ $$B = b_1 z^2 + b_2 z + b_3$$

  1. Using the correlation:

$$y = \frac{a_1 z^3 + a_2 z^2 + a_3 z + a_4}{ln[x]^{3}} + b_1 z^2 + b_2 z + b_3$$

And substituting different values of $x_{uks}$ I get corresponding $y$, but when I plot them vs. $z$, I get plots of the form:

enter image description here

Now, my questions are as follows:

  1. How am I getting $y$ to be a linear function of $z$?
  2. Is this approach a valid one?
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  • $\begingroup$ $y$ seems to be inversely proportional to $x$, which is given an offset and is scaled. How would the first figure look like if you would plot $1/y$ instead of $y$? $\endgroup$ – fibonatic Apr 20 '15 at 12:37
  • $\begingroup$ You mean $z$, right? $y$ is inversely proportional to $z$. $\endgroup$ – nxkryptor Apr 20 '15 at 12:59
  • $\begingroup$ No $x$, $y$ to me seems to be close to linear proportional to $z$. $\endgroup$ – fibonatic Apr 20 '15 at 13:07
  • $\begingroup$ Yes, both $x,y$ are inversely proportional to $z$. $x$ is note linearly proportional to $z$, but exponential to $z$. $\endgroup$ – nxkryptor Apr 20 '15 at 13:09

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