# how to evaluate a symbolic function in matlab

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Kamuran on 11 Feb 2013
Commented: vikas singh on 12 Mar 2023
Hi,
I am trying to automate my code to get the derivative of a function an evaluate that in given points. For example,
syms x
func=cos((pi*x)^4);
RHS=diff(func,x,2);
% RHS will be ==> - 16*pi^8*x^6*cos(pi^4*x^4) - 12*pi^4*x^2*sin(pi^4*x^4)
x=[ -1
-0.978147600733806
-0.913545457642601
-0.809016994374947
-0.669130606358858
-0.5
-0.309016994374947
-0.104528463267653
0.104528463267653
0.309016994374947
0.5
0.669130606358858
0.809016994374947
0.913545457642601
0.978147600733806
1]
and I want RHS and Func in those z values as a vector.
I appreciate the help. Thank you.

ChristianW on 11 Feb 2013
• doc subs
• doc subexpr

Youssef Khmou on 11 Feb 2013
hi, you can use function_handle :
func=@(x) cos((pi*x).^4)
x=0:100; % example of vector x .
RHS=diff(func(x),2);
Youssef Khmou on 12 Feb 2013
Edited: Youssef Khmou on 12 Feb 2013
Well in this case you have to add diff(x) as DENOMINATOR :
func=@(x) x.^3;
f=x.^3;
df2=6.*x;
df2(1)=[];
d1=diff(func(x))./diff(x);
RHS=diff(d1)./diff(x(1:end-1));
figure, plot(df2), hold on, plot(RHS,'r')
legend(' numerical d²f',' diff(function_handle,2)')
you have to increase the Sample Rate in x to get better approximation .
Brian B on 12 Feb 2013
Edited: Brian B on 12 Feb 2013
Hi Youssef,
I'm not saying your solution is not valid. I was simply pointing out that the original question refers to symbolic calculation of an exact derivative. It is possible to evaluate the symbolic expression at any arbitrary set of points, without regard to interval size or ordering. That is what the subs command does, to which ChristianW referred. See my answer below.
regards,
Brian

Youssef Khmou on 12 Feb 2013
Edited: Youssef Khmou on 12 Feb 2013
Kamuran, to get better approximation you need to increase the sample rate in x and interpolate :
x=[ -1
-0.978147600733806
-0.913545457642601
-0.809016994374947
-0.669130606358858
-0.5
-0.309016994374947
-0.104528463267653
0.104528463267653
0.309016994374947
0.5
0.669130606358858
0.809016994374947
0.913545457642601
0.978147600733806
1];
x=x';
x=interp(x,5); % Example
func=@(x) x.^3;
f=x.^3;
df2=6.*x;
df2(1)=[];
d1=diff(func(x))./diff(x);
RHS=diff(d1)./diff(x(1:end-1));
figure, plot(df2), hold on, plot(RHS,'r')
legend(' numerical d²f',' diff(function_handle,2)')
Compare this result with the one given in the Comment with original x, there is an enhancement .
I hope that helps

Brian B on 12 Feb 2013
syms x
func=cos((pi*x)^4);
RHS=diff(func,x,2);
xx=-1:0.1:1;
d2f = subs(RHS, xx)
Youssef Khmou on 13 Feb 2013
right , that works better with "subs" function .
vikas singh on 12 Mar 2023
if I have function of two variable suppose x and t and I have to evaluate it on x=0:100:10000 and t= 0,1,5,10 15. how to do. suppose the fucnction is F(x,t)=sin(x)*exp(t)