Filter löschen
Filter löschen

How can I create a ''fun=@x...'' function for integration by multipling ''fun= @x...'' functions that consists of the originals funs subarguments?

39 Ansichten (letzte 30 Tage)
I would like to illustrate the following code but an error occurs: this code works: fun=@(x)(exp(-hlpexpvar.*x)).*(1-(x./univar(1)))
but how can I do something like this:
funone=@x exp(-hlpexpvar.*x)
funtwo=@x 1-(x./univar(1))
fun= @x funone.*funtwo
I would like to thank you in advance for any kind of help to this problem.

Akzeptierte Antwort

Steven Lord
Steven Lord am 1 Jul. 2015
You were close. I'm going to define values for the variables for demonstration purposes
hlpexpvar = 1;
univar = 2;
Now to define the three functions
funone =@(x) exp(-hlpexpvar.*x);
funtwo =@(x) 1-(x./univar(1));
You can't multiply function handles. However you can multiply the values returned by evaluating the function handles.
fun = @(x) funone(x).*funtwo(x);
Let's check by comparing the integral of fun (using funone and funtwo) and the explicitly specified function fun2.
fun2 = @(x) exp(-hlpexpvar.*x).*(1-(x./univar(1)));
answer1 = integral(fun, 0, 1)
answer2 = integral(fun2, 0, 1)
  1 Kommentar
Konstantinos Dragonas
Konstantinos Dragonas am 1 Jul. 2015
Thank you very much for your help! This is the only problem i had in my effort to make a dynamic program that will take as input a rate transition matrix and and calculate the steady state probabilities by semi markov model.

Melden Sie sich an, um zu kommentieren.

Weitere Antworten (1)

chaman lal dewangan
chaman lal dewangan am 12 Mär. 2018
Thanks, I understood how to use "fun".

Kategorien

Mehr zu Programming finden Sie in Help Center und File Exchange

Community Treasure Hunt

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

Start Hunting!

Translated by