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

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Konstantinos Dragonas
回答： chaman lal dewangan 2018-3-12
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.

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采纳的回答

Steven Lord
Steven Lord 2015-7-1
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)
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Konstantinos Dragonas
Konstantinos Dragonas 2015-7-1
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.

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更多回答（1 个）

chaman lal dewangan
Thanks, I understood how to use "fun".

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