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Can anybody help me to solve this?

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ASWATHI V
ASWATHI V on 15 Aug 2018
Edited: ASWATHI V on 18 Aug 2018
When I am running this code it shows this error
Undefined function 'times' for input arguments of type 'function_handle'.
Error in withoutpt_fd1 (line 41)
A2 = 1-(a2_1.*a2_5);
a2_1 = lamda_su1*lamda_u1p*(phi_1*k*I)/(p_smax*lamda_u1u2);
a2_2 = @(z) exp((-lamda_u1p-((sig_2.*lamda_u1u2)/(k.*I))).*z);
a2_3 = @(z) z+((lamda_su1.*phi_1.*k.*I)./(p_smax.*lamda_u1u2));
a2_4 = @(z) a2_2(z).*a2_3(z);
a2_5 = @(z) integral(a2_4,0,Inf);
A2 = 1-(a2_1.*a2_5);

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Accepted Answer

ADragon
ADragon on 15 Aug 2018
Hi ASWATHI, when you call an anonymous function, you need to provide inputs just like a .m function. So, assignment A2 and a2_5 should be written as
a2_5 = @(z) integral(a2_4(z),0,Inf);
A2 = 1 - (a2-1(<z input>).*a2_5(<z input>));
You will need to replace "<z input>" with a numeric value or variable with a numeric values. Hope this helps
AD

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ADragon
ADragon on 15 Aug 2018
*You will need to replace "<z input>" with a numeric value or variable with a numeric value*.
I don't know what your code does, but you made all your equations a function of z. What do you want z to be?
AD
ADragon
ADragon on 15 Aug 2018
I'm sorry. I see what you are doing... you are integrating a2_5 for z=0 to Inf. Modify a2_5 to the code below. a2_5 is not a function of z after you integrate. Everything else looks ok.
a2_5 = integral(a2_4,0,Inf);
The result should be the same as the solution provided by Star Strider.
ASWATHI V
ASWATHI V on 18 Aug 2018
Thank you so much sir

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