Getting Empty sym: 0-by-1 when using solve function

2018 11 23
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更新时间：2018 11 24
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I am trying to find the value of x when the equation is equal to zero. I keep getting an empty sym 0-by-1. Any suggestions? Thanks!
I = (pi*.015^4)/64;
E = 200*10^9;
syms x
[x] = solve(0 == (1/(E*I))*((-199.46/2)*x^2 + (290/2)*(x-.25)^2 - (90.66/2)*(x-.8)^2 + 11.22),x)
x = double(x)

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Walter Roberson
Walter Roberson 2018-11-24
solve requests that an infinitely precise solution be found if possible . It makes little sense to use solve with floating point numbers as floating point numbers represent a semirandom point within an interval . It is a category error to use the two together . You should only use solve with infinitely precise inputs
For example is the 11.22 standing in for 101/9 ?
Mason Osborn
Mason Osborn 2018-11-24
The 11.22 is a constant of integration that I calculated by hand. All I need to do is find the value of x at which the equation is zero. I have used the solve function before but not for something like this. Is there another function that will solve for x?

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回答（1 个）

Star Strider
Star Strider 2018-11-23
编辑：Star Strider 2018-11-23

0 个投票

Your code works for me:
I = (pi*.015^4)/64;
E = 200*10^9;
syms x
[x] = solve(0 == (1/(E*I))*((-199.46/2)*x^2 + (290/2)*(x-.25)^2 - (90.66/2)*(x-.8)^2 + 11.22),x)
x = double(x)
producing:
x =
7/30 - (2^(1/2)*261763^(1/2)*1i)/60
(2^(1/2)*261763^(1/2)*1i)/60 + 7/30
x =
0.233333333333333 - 12.0591827620651i
0.233333333333333 + 12.0591827620651i

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Mason Osborn
Mason Osborn 2018-11-24
I get that too when I run the code after first opening MATLAB. I needed x to be a real number so I did...
I = (pi*.015^4)/64;
E = 200*10^9;
syms x real
[x] = solve(0 == (1/(E*I))*((-199.46/2)*x^2 + (290/2)*(x-.25)^2 - (90.66/2)*(x-.8)^2 + 11.22),x)
x = double(x)
And got results of...
x =
Empty sym: 0-by-1
x =
0×1 empty double column vector
when I changed it back to the original code ..... (without the real)
I = (pi*.015^4)/64;
E = 200*10^9;
syms x
[x] = solve(0 == (1/(E*I))*((-199.46/2)*x^2 + (290/2)*(x-.25)^2 - (90.66/2)*(x-.8)^2 + 11.22),x)
x = double(x)
I got the same results
x =
Empty sym: 0-by-1
x =
0×1 empty double column vector
I'm not sure why it will not go back to the original results when I first ran the program.
Walter Roberson
Walter Roberson 2018-11-24
you appear to have a quadratic . Expand it out and group terms and use the quadratic formula . II would be surprised if matlab is not calculating it correctly .
In other words either your equation is wrong or your expectations of aa real solution are wrong .
Star Strider
Star Strider 2018-11-24
Your function has no real roots. If you plot it, this quickly becomes evident:
syms x real
I = sym((pi*.015^4)/64);
E = sym(200E9);
Eqn = (1/(E*I))*((-199.46/2)*x^2 + (290/2)*(x-.25)^2 - (90.66/2)*(x-.8)^2 + 11.22);
figure
fplot(Eqn, [-10 10])
ylim([-0.03 0.001])
grid
You can expand the fplot limits as far as you like. It does not change the result.
I got the empty result with real, and without it (using vpasolve):
x =
7/30 - (523526^(1/2)*1i)/60
(523526^(1/2)*1i)/60 + 7/30
x =
0.23333333333333333333333333333333 - 12.059182762065134606377924566193i
0.23333333333333333333333333333333 + 12.059182762065134606377924566193i
I am using R2018b.
Walter Roberson
Walter Roberson 2018-11-24
Your equation has an expression in x in the numerator, and has E*I in the denominator. For any E*I that is non-zero and independent of x, then the roots of the equation are the same as the roots of the numerator. Therefore the values of E and I do not matter for the purpose of finding roots.
Your equation simplifies to
-600*x^2+280*x-87287 / (10000*E*I)
"a" and "c" are both negative, so b^2-4*a*c is subtracting a positive value from b^2, and -4*a*c is larger than b^2, so you would be taking sqrt() of a negative number. It isn't borderline either: it is solidly negative, around -209410400
Therefore no real roots.

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