how to evaluate a symbolic function in matlab
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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.
回答（4 个）
Youssef Khmou
2013-2-11
hi, you can use function_handle :
func=@(x) cos((pi*x).^4)
x=0:100; % example of vector x .
RHS=diff(func(x),2);
8 个评论
Kamuran
2013-2-12
Youssef Khmou
2013-2-12
编辑：Youssef Khmou
2013-2-12
Kamuran, that is normal routine for "diff" function , if you differentiate func with length p : diff(func,n) n times then the returned vector has p-n points .
if you put :
RHS4=diff(func(x),4);
Length of RHS4 is 101-4 .
diff produces a shift, you can truncate the 1st element of the RHS(1)=[];
Brian B
2013-2-12
diff operates differently on symbolic objects than on numeric vectors. You want (and have, as shown in your question) an analytical expression for the second derivative. You just have to substitute the numerical values of x at which you want the solution. Did you look at the functions suggested above by ChristianW?
Youssef Khmou
2013-2-12
The values are right Kamuran :
x=0:100;
func=@(x) x.^3;
f=x.^3;
df2=6.*x;
RHS=diff(func(x),2);
figure, plot(df2), hold on, plot(RHS,'r')
legend(' numerical d²f',' diff(function_handle,2)')
Youssef Khmou
2013-2-12
编辑：Youssef Khmou
2013-2-12
Brian : of course with linear x , what is your solution then? can you write a Diff function that gives the approximate der whatever the x is .
Youssef Khmou
2013-2-12
编辑：Youssef Khmou
2013-2-12
Well in this case you have to add diff(x) as DENOMINATOR :
Given your x :
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 .
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
2013-2-12
编辑：Youssef Khmou
2013-2-12
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
2013-2-12
syms x
func=cos((pi*x)^4);
RHS=diff(func,x,2);
xx=-1:0.1:1;
d2f = subs(RHS, xx)
2 个评论
Youssef Khmou
2013-2-13
right , that works better with "subs" function .
vikas singh
2023-3-12
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)
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2023-3-12
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