Non linear equation in STATA

[Lars Pete]

Hello,

I am looking for the code to solve for t(p):

e^t + t = p

for values p=1 and p = 4.

Thanks in advance!

[William Lisowski]

So you want to solve the equations

et + t = 1

and

et + t = 4

For the first, the solution is t=0.

For the second, the solution is t=1.0737289375564990871

Both solutions were obtained using the WolframAlpha site, by typing the equation into the query box, and in the second case, clicking on the "approximate form" and "more digits" buttons in the "real solution" results box.

[Lars Pete]

So you want to solve the equations

et + t = 1

and

et + t = 4

For the first, the solution is t=0.

For the second, the solution is t=1.0737289375564990871

Both solutions were obtained using the WolframAlpha site, by typing the equation into the query box, and in the second case, clicking on the "approximate form" and "more digits" buttons in the "real solution" results box.

William,

Thank you. I am aware of that site and other ways to solve the equation. I was looking for ways to solve it using STATA program such as nl. But I am completely clueless about how to do that.

Best regards.

[Clyde Schechter]

There is nothing built-in in Stata for this, as far as I know. However, for a simple function that has a simple derivative, like this one, it is easy to iterate an approximate solution using the Newton-Raphson method:

. clear*

.
. scalar t = 1

. scalar y = exp(t) + t - 4

.
. while abs(y) > 1e-6 {
  2.      scalar dydt = exp(t) + 1
  3.     scalar t = t - y/dydt
  4.     scalar y = exp(t) + t - 4
  5.     display t, y
  6. }
1.0757657 .0080029
1.0737305 6.069e-06
1.0737289 3.495e-12

There is nothing magical about 1e-6 as the degree of approximation. Pick what you like (within reason)

[William Lisowski]

Note that help nl tells us

nl fits an arbitrary nonlinear regression function by least squares.

whereas as both Clyde and I note, this is an equation that admits to an exact solution. Clyde's solution is a straightforward application of freshman calculus.

In general least squares is not helpful when there is no "error" to minimize, so nl would seem to be of little use.

. set obs 1
number of observations (_N) was 0, now 1

. generate p = 4

. generate e = exp(1)

. nl (p = e^{t} + {t})
(obs = 1)

p has zero variance
r(498);

But as a learning exercise, if we choose two distinct values of p close to each other, we can persuade nl to give us an approximation.

. set obs 2
number of observations (_N) was 0, now 2

. generate p = 3.995 + .010*(_n==2)

. generate e = exp(1)

. nl (p = e^{t} + {t})
(obs = 2)

Iteration 0:  residual SS =  7.815957
Iteration 1:  residual SS =  .1346935
Iteration 2:  residual SS =  .0001234
Iteration 3:  residual SS =    .00005
Iteration 4:  residual SS =    .00005


      Source |      SS            df       MS
-------------+----------------------------------    Number of obs =          2
       Model |          0          0           .    R-squared     =     0.0000
    Residual |     .00005          1  .000050002    Adj R-squared =     0.0000
-------------+----------------------------------    Root MSE      =   .0070712
       Total |     .00005          1  .000050002    Res. dev.     =  -15.51742

------------------------------------------------------------------------------
           p |      Coef.   Std. Err.      t    P>|t|     [95% Conf. Interval]
-------------+----------------------------------------------------------------
          /t |   1.073729   .0012735   843.13   0.001     1.057548     1.08991
------------------------------------------------------------------------------
  Parameter t taken as constant term in model & ANOVA table

Or we can pose it as an optimization problem to be solved using mata to minimize the difference between p and e^t+t, which we do by maximizing the negative of the square of the difference.

clear all
mata:
void myeval(todo, t,  y, g, H)
 {
         y = - ( exp(t) + t - 4)^2
 }
S = optimize_init()
optimize_init_evaluator(S, &myeval())
optimize_init_params(S, 0)
x = optimize(S)
x
end

. clear all

. mata:
------------------------------------------------- mata (type end to exit) -----------------------
: void myeval(todo, t,  y, g, H)
>  {
>          y = - ( exp(t) + t - 4)^2
>  }
note: argument todo unused
note: argument g unused
note: argument H unused

: S = optimize_init()

: optimize_init_evaluator(S, &myeval())

: optimize_init_params(S, 0)

: x = optimize(S)
Iteration 0:   f(p) =         -9  
Iteration 1:   f(p) = -3.9270875  
Iteration 2:   f(p) = -.00854249  
Iteration 3:   f(p) = -.00038131  
Iteration 4:   f(p) = -1.687e-06  
Iteration 5:   f(p) = -1.550e-09  
Iteration 6:   f(p) = -2.843e-11  

: x
  1.07372758

: end
-------------------------------------------------------------------------------------------------

.

[Joro Kolev]

You might also look up this FAQ: https://www.stata.com/support/faqs/p...ear-equations/

[Lars Pete]

Thank you very much Clyde. This is very helpful.

[Lars Pete]

Thank you so much William. I tried using the optimization method. However, I was using it mindlessly, without understanding what it actually does. This makes it clearer.

[Lars Pete]

You might also look up this FAQ: https://www.stata.com/support/faqs/p...ear-equations/

Thanks Joro. I did refer to these. But I couldn't get the solution.