Best fit k1,k2,k3 for w[i]-w[i-1]=k1*(c[i]-(k2*w[i-1]+k3)) with c[i], w[i] known At the end of each day "i", I record my weight "w[i]", and the number 
of calories I've consumed that day "c[i]". 
I believe my weight change "w[i]-w[i-1]" is proportional to my net 
calorie intake: "w[i]-w[i-1] = k1*(net calorie intake)". 
The number of calories I burn is a linear function of my prior day's 
weight: "k2*w[i-1]+k3" 
My net calorie intake is thus "c[i]-(k2*w[i-1]+k3)". 
My weight change is thus: "w[i]-w[i-1] = k1*(c[i]-(k2*w[i-1]+k3))" 
Given that I know "w[i]" and "c[i]" for each i, how do I find the best 
fit values of k1, k2, and k3? 
Note: I'm trying to solve this as a data-fitting problem. I realize it 
doesn't map reality perfectly. 
 A: This is a linear model, so the easiest (but not completely kosher) way to fit it is using ordinary least squares (OLS) regression. It's easier if you rewrite the model a bit,
$$w_i = \beta_0 + \beta_1 w_{i-1} + \beta_2 c_i.$$
You should be able to convince yourself that this is essentially the same model, and that you can solve backwards to find your $k_1, k_2, k_3$ in terms of $\beta_0, \beta_1, \beta_2$. We can fit it in R as follows (with some random data):
> w <- rnorm(101,180,2)
> c <- rnorm(100,2000,200)
> model <- lm(w[2:101] ~ w[1:100] + c)
> summary(model)

Call:
lm(formula = w[2:101] ~ w[1:100] + c)

Residuals:
    Min      1Q  Median      3Q     Max 
-5.5455 -1.4231  0.1289  1.1868  5.3232 

Coefficients:
             Estimate Std. Error t value Pr(>|t|)    
(Intercept) 1.429e+02  1.784e+01   8.013 2.51e-12 ***
w[1:100]    1.925e-01  9.919e-02   1.941   0.0552 .  
c           1.095e-03  1.180e-03   0.928   0.3559    
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 

Residual standard error: 2.126 on 97 degrees of freedom
Multiple R-squared: 0.04784,    Adjusted R-squared: 0.02821 
F-statistic: 2.437 on 2 and 97 DF,  p-value: 0.09277 

Since the data are completely random, we don't actually expect there to be any relationship (other than the intercept), and those p-values are actually surprisingly low. You can do a similar fit in any statistical software.
If you want to actually do it properly (as opposed to an easy and not-very-accurate hack), this type of analysis is called time series. MIT's OCW has a course about it, though I can't vouch for it personally (and I'm usually a little skeptical of stats courses from economists).
