Empty sym: 0-by-1
조회 수: 14 (최근 30일)
이전 댓글 표시
When I run this code, my answer for A comes out as 'Empty sym: 0-by-1'
clc;clear;format compact;
syms V
P=[0.98 1.97 4.93 9.86 49.36 98.69]; %atm
T=[573 573 573 573 573 573]; %K
R=461.5; %Pa*m3/mol*K
R=convpres(R,'Pa','atm');
R=R*1000; %L*atm/mol*K
GI=(P*V)-(R*T);
A=solve(GI,V)
댓글 수: 0
채택된 답변
Torsten
2023년 10월 12일
Use
A = double(arrayfun(@(i)solve(GI(i),V),1:numel(P)))
instead of
A = solve(GI,V)
댓글 수: 1
Dyuman Joshi
2023년 10월 12일
which can be simplified to
out = double(arrayfun(@(x) solve(x,V), GI))
추가 답변 (1개)
Steven Lord
2023년 10월 12일
Let's look at the equations you're trying to solve.
syms V
P=[0.98 1.97 4.93 9.86 49.36 98.69]; %atm
T=[573 573 573 573 573 573]; %K
R=461.5; %Pa*m3/mol*K
R=convpres(R,'Pa','atm');
R=R*1000; %L*atm/mol*K
GI=(P*V)-(R*T)
The first element of GI has a solution, as does the second element of GI. Is the solution for those two elements exactly the same? To ask that question another way, is there a value of V that satisfies all the elements in GI simultaneously?
A1=solve(GI(1), V)
A2 = solve(GI(2), V)
isAlways(A1 == A2)
No, there isn't. So your original solve call that's looking for that one value of V that satisfies all the elements in GI simultaneously is correct to return an empty sym.
Now if you wanted a vector of values of V that satisfy each corresponding element in GI, that's possible to compute.
A = zeros(size(GI), 'sym');
for whichTerm = 1:numel(GI)
A(whichTerm) = solve(GI(whichTerm), V);
end
A
Do they satisfy their equations?
result = zeros(size(GI), 'sym');
for whichEquation = 1:numel(GI)
result(whichEquation) = subs(GI(whichEquation), V, A(whichEquation));
end
result
Yes. Are they close to the same value?
vpa(A, 5)
Nope. Not even same order of magnitude.
댓글 수: 1
Walter Roberson
2023년 10월 12일
This is an important point: solve is always looking for simultaneous solution of all the inputs. It is designed for simultaneous equation solving, and does not have any option for solving each equation individually.
참고 항목
카테고리
Help Center 및 File Exchange에서 Number Theory에 대해 자세히 알아보기
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!