Intersection of piecewise defined functions.

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?

Answer 1

[Walter Roberson]

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 .

[Max]

Hello Walter,

thank you. I tried out your solution. But the code doesn´t work on my pc. Altough, the parameter deltal2_origin is known I get the message: Undefined function or variable 'deltal2_origin'.

[Max]

Sorry, I get the fault. You´re right.

[Walter Roberson]

probably a lower-case L vs digit-1 typo.

Answer 2

[John D'Errico]

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.

[Max]

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).