# My diff function won't work and I'm not sure why

조회 수: 39(최근 30일)
Sarah Johnson 2020년 2월 12일
편집: Dimitris Kalogiros 2020년 2월 13일
My goal is to give an initial x, and return the function values at that x
function [f, g, h] = myfunction(x)
syms x
q = exp(2*sin(x)) - x;
f = q(x);
dq = diff(q);
g = dq(x);
dq2 = diff(dq);
h = dq2(x);
end
So what I've been struggling with is that if I assign my equation as above, then I can find "f = q(x)" but there is an error for the "diff". If I change the syntax to
q = @(x) exp(2*sin(x)) - x;
then I'm able to calculate using "diff" but then the "f = q(x)" doesn't work. I'm pretty new to MATLAB and not great at it so any advice would be great!
Update: This is the error message I am recieving for the first code
Error using sym/subsindex (line 855)
Invalid indexing or function definition. Indexing must follow MATLAB indexing. Function arguments must be symbolic variables, and function body must be sym expression.
Error in sym/subsref (line 900)
R_tilde = builtin('subsref',L_tilde,Idx);
Error in myfunction (line 4)
f = q(x);
And for the other code it is:
Undefined function 'diff' for input arguments of type 'function_handle'.
Error in myfunction2 (line 4)
dq = diff(q);
##### 댓글 수: 1표시숨기기 없음
Steven Lord 2020년 2월 12일
What is the full and exact text of the error messages you receive in each of the circumstances? Please show us all the text displayed in red and/or orange in the Command Window when you try to run this code.

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### 답변(3개)

Dimitris Kalogiros 2020년 2월 12일
Try this :
clearvars; close all; clc;
syms x
q(x) = exp(2*sin(x)) - x;
f(x) = q(x)
dq(x) = diff(q(x));
g(x) = dq(x)
dq2(x) = diff(dq(x));
h(x) = dq2(x)
or this one:
clearvars; close all; clc;
syms x
q(x) = exp(2*sin(x)) - x;
f = q(x)
dq(x) = diff(q(x));
g = dq(x)
dq2(x) = diff(dq(x));
h = dq2(x)
##### 댓글 수: 1표시숨기기 없음
Sarah Johnson 2020년 2월 12일
Both of these still are only producing the functions themselves and not the value. I'm not sure what's wrong. 댓글을 달려면 로그인하십시오.

David Goodmanson 2020년 2월 12일
편집: David Goodmanson 2020년 2월 12일
Hi Julia,
if you want to use this for a variety of functions, then
fun = @(x) exp(2*sin(x)) - x; % one example
[f g h] = newfuns(fun)
function [a b c] = newfuns(fun)
syms x
a = fun(x);
b = diff(a);
c = diff(b);
end
f =
exp(2*sin(x)) - x
g =
2*exp(2*sin(x))*cos(x) - 1
h =
4*exp(2*sin(x))*cos(x)^2 - 2*exp(2*sin(x))*sin(x)
##### 댓글 수: 1표시숨기기 없음
Sarah Johnson 2020년 2월 12일
My goal is to find the function values at the initally given x. Not find the differentials themselves. I have done that but I'm having trouble plugging in the inital x to give me an answer. For example, if I plug in x = 0 and then call "myfunction(x)", I would want my f to be 1

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Dimitris Kalogiros 2020년 2월 13일
편집: Dimitris Kalogiros 2020년 2월 13일
I think , the answer to your problem is the following piece of code:
clearvars; clc; close all;
% call of the function
myVal=5;
[f, g, h] = myfunction(myVal);
fprintf('f = %d g = %d h = %d', f, g, h);
% definition of the function
function [f, g, h] = myfunction(y)
syms x
q = exp(2*sin(x)) - x;
f = double( vpa( subs( q, y) ) );
dq = diff(q);
g = double( vpa(subs( dq , y)) );
dq2 = diff(dq);
h = double( vpa(subs( dq2, y)) );
end
A much more usuful version , is the following. Where you can pass your mathematical function as an argument:
clearvars; clc; close all;
syms x
myF=exp(2*sin(x)) - x;
% define the evaluation point
myVal=5;
[f, g, h] = myFunctionCalculator(myVal, myF, x);
% print results
fprintf('f = %d g = %d h = %d', f, g, h);
function [f, g, h] = myFunctionCalculator(xo, q, x)
f = double( vpa( subs( q, x, xo) ) );
dq = diff(q);
g = double( vpa(subs( dq , x, xo)) );
dq2 = diff(dq);
h = double( vpa(subs( dq2, x, xo)) );
end
.

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