Intersection of piecewise defined functions.

10 views (last 30 days)
Hello,
I would like to calculate the value for that two functions are equal:
function no. 1:
F_2_origin_cond = (1-((1- wblcdf(t-delta12_origin,eta_hat_2,beta_hat))./(1-Age_t1_origin))).*(t>=2000);
F_2_origin_cond = 0.*(t<2000);
function no. 2:
y=0.876;
Does somebody know if it is possible in MATLAB?

Accepted Answer

Walter Roberson
Walter Roberson on 3 Feb 2016
By examination, the first function is 0 for all t < 2000 . That cannot possibly be 0.876. So the question comes down to finding (1-((1- wblcdf(t-delta12_origin,eta_hat_2,beta_hat))./(1-Age_t1_origin))) == 0.876 and if the answer is < 2000 then discard that answer.
F_2_origin_cond = @(t) (1-((1- wblcdf(t-delta12_origin,eta_hat_2,beta_hat))./(1-Age_t1_origin))) - 0.876;
t_target - fzero(F_2_origin_cond, [max(2000,deltal2_origin), inf])
The max() is there to select between the greatest of 2000 (the part at which the equation becomes valid) and delta12_origin, as wblcdf(t-delta12_origin,...) will be 0 for t < delta12_origin .
  3 Comments
Walter Roberson
Walter Roberson on 3 Feb 2016
probably a lower-case L vs digit-1 typo.

Sign in to comment.

More Answers (1)

John D'Errico
John D'Errico on 3 Feb 2016
Edited: John D'Errico on 3 Feb 2016
Of course it is possible.
Hint: subtract the two, then search for a root. This applies even if one of the functions is not constant, as it is here.
So, you could use a numerical solver. Or you could use a symbolic solver.
As far as the function being a piecewise one, that is not relevant, since for t < 2000, the function is trivially zero, therefore the two can never be equal in that domain. So all that matters is you find a root of the difference above 2000 for t.
  1 Comment
Max
Max on 3 Feb 2016
Hey John,
thanks for your answer. That what your have written is correct but what I´m searching for is how to do that in matlab (matlab source code).

Sign in to comment.

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by