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Here's something that's been bothering me for a while now, what I don't understand is, if I have a function and I wish to constrain it to specific values... And let's say I have three pairs of x,y values, why do I have to have three constraints? Can someone elaborate on that?

$$f(x) = k_1e^{k_2x} + k_3$$

where

$$f(0) = 0$$

$$f(0.1) = 1$$

$$f(1) = 100$$

For example, the $x = 0$ simply "cancels" the $e$ term. Therefore:

$k_1+k_3= f(0) = 0$

And so forth... But why do I need three of them to model it properly?

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1 Answer 1

up vote 3 down vote accepted

...let's say I have three pairs of x,y values, why do I have to have three constraints?

Solving a system of equations requires one to have as many variables as there are equations. If not, your problem is underdetermined (more unknowns than equations) or overdetermined (more equations than unknowns), and those are a bit more complicated to handle in general...

Getting back to the matter of fitting $k_1e^{k_2x} + k_3$, I presume you've already assembled your equations like so:

$$\begin{align*} 0&=k_1e^{0\cdot k_2} + k_3\\ 1&=k_1e^{0.1\cdot k_2} + k_3\\ 100&=k_1e^{1\cdot k_2} + k_3 \end{align*}$$

You said you've managed to turn the first equation into $k_1+k_3=0$. That's now one less degree of freedom. You can now replace all the $k_1$'s appearing in the other two equations with $-k_3$, or replace all the $k_3$'s with $-k_1$. You call the shots. The solution is really a matter of replacing your variables one-by-one with "known" quantities, up until the point where you have one equation in one unknown. That bit is usually easier to solve than many equations in many unknowns...

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Thank you for a quick response, could you perhaps reference someplace where I could find a bit more about this topic? I'm trying to understand exactly why does the number of equations have to equate the number of variables. I understand the relational under/over-determined reasoning, but something isn't clicking in my head. For example, if there's a lesser or bigger number of variables, is it still solvable? –  Curiosity Sep 22 '11 at 18:58
2  
I don't really have an off-hand reference. What it really all boils down to is that you know how to solve a single, solitary, equation in one unknown. If you have two equations in two unknowns, say $x$ and $y$, you can pick one of the two equations, solve for one of the variables, say $x$, which will be in terms of $y$. If you replace all the $x$'s in the other equation with the expression you got for $x$, you've just reduced your problem to "one equation in one unknown". And so on for $n$ equations in $n$ unknowns... –  J. M. Sep 22 '11 at 19:03
2  
For the linear case it's a question of linear algebra. For the differentiable case, it's the implicit function theorem. I'm not sure what machinery we invoke for the general case... –  Zhen Lin Sep 23 '11 at 0:54
1  
@Zhen: that last bit sounds awfully like a good question... :) –  J. M. Sep 23 '11 at 1:11

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