Unrecognized function or variable in a code

조회 수: 1 (최근 30일)
Mathew
Mathew 2024년 6월 19일
편집: Mathew 2024년 6월 19일
% param value
clear all; close all;
% p is constant and q varies
k0 = 0.0000169; u=0.6; p=536.2; q=0.0000376;
% your function
k = @(t,u) (k0 + (p*k0.^u-q*k0).*t.^a/gamma(a+1) +(p*k0.^u-q*k0).*(p*u*k0.^u-q).*t.^(2*a)/gamma(2*a+1)...
+(p*k0.^u-q*k0).*((p*u*k0.^u-q).^2-p*u.*(u-1)*k0.^(u-1)./2).*t.^(3*a)/gamma(3*a+1));
% grid
%t = linspace(0,0.1);
%u = linspace(0.3,0.9);
%[T,U] = meshgrid(t,q);
t = linspace(0,0.1,50);
a = linspace(0.3,0.9,50);
[T,A] = meshgrid(t,a);
% evaluate function
Z = k(T,A);
Unrecognized function or variable 'a'.

Error in solution>@(t,u)(k0+(p*k0.^u-q*k0).*t.^a/gamma(a+1)+(p*k0.^u-q*k0).*(p*u*k0.^u-q).*t.^(2*a)/gamma(2*a+1)+(p*k0.^u-q*k0).*((p*u*k0.^u-q).^2-p*u.*(u-1)*k0.^(u-1)./2).*t.^(3*a)/gamma(3*a+1)) (line 6)
k = @(t,u) (k0 + (p*k0.^u-q*k0).*t.^a/gamma(a+1) +(p*k0.^u-q*k0).*(p*u*k0.^u-q).*t.^(2*a)/gamma(2*a+1)...
% plot
figure
surf(T,A,Z)
%surf(T,Q,Z,'facecolor','none')
xlabel('t');
ylabel('\mu');
zlabel('k(t)')

채택된 답변

Cris LaPierre
Cris LaPierre 2024년 6월 19일
You create an anonymous function, k, that uses a variable a in the function. However, this variable was not defined at the time the anonymous function was defined.
  댓글 수: 4
Steven Lord
Steven Lord 2024년 6월 19일
% p is constant and q varies
k0 = 0.0000169; u=0.6; p=536.2; q=0.0000376;
k = @(t,a) (k0 + (p*k0.^u-q*k0).*t.^a./gamma(a+1) +(p*k0.^u-q*k0).*(p*u*k0.^u-q).*t.^(2*a)./gamma(2*a+1)...
+(p*k0.^u-q*k0).*((p*u*k0.^u-q).^2-p*u.*(u-1)*k0.^(u-1)./2).*t.^(3*a)./gamma(3*a+1));
% grid
t = linspace(0,1,50);
a = linspace(0.25,0.85,50);
[T,A] = meshgrid(t,a);
% evaluate function
Z = k(T,A);
% plot
figure
surf(T,A,Z)
xlabel('t');
ylabel('\alpha');
zlabel('k(t)')
Mathew
Mathew 2024년 6월 19일
편집: Mathew 2024년 6월 19일
Thanks @ Steven Lord and @ Cris LaPierre

댓글을 달려면 로그인하십시오.

추가 답변 (0개)

카테고리

Help CenterFile Exchange에서 Logical에 대해 자세히 알아보기

태그

제품


릴리스

R2023b

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by