where are the errors in my code? please help me!!
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Urgent Help ,please!
I would like to Solve an equation for a variable (y) while setting another variable (r) across a range and then substituting the solution into the upper and lower limits of an integral (q). Eventually, plotting the relationship between the intgeral (q) and the variable (r) with the range.
I know that there are multiple errors! but I can not figure out where are they!!
here is the code:
r=linspace(0,0.09);
syms y(r)
eqn= (cosh(10^10.*y./0.5)).^2 == 5./16.*r;
V1 = double(vpasolve(eqn,[y,r],[0 2]));
V2 = double(vpasolve(eqn,[y,r],[-2 0]));
fun=@(x)0.0018./((5./((cosh(10^10.*x./0.5)).^2)-r.*16).^0.5);
q = integral(fun,V2,V1)
plot(r,q)
采纳的回答
Torsten
2022-5-25
syms r z
eqn = ((z+1/z)/2)^2 == 5/16*r;
S = solve(eqn,z);
y = log(S)*0.5/10^10
solves the equation
(cosh(10^10.*y./0.5)).^2 == 5./16.*r
for y.
Deduce the conditions on r that V1 and V2 in the subsequent integration are real-valued.
20 个评论
Hi,
Thank you ver much for response!
Here what I wrote:
syms r z
eqn = ((z+1/z)/2)^2 == 5/16*r;
S =solve(eqn,z);
y = log(S)*0.5/10^10
V1 = double(vpasolve(y,r,[0 0.09]));
V2 = double(vpasolve(y,r,[0 0.09]));
fun=@(x)0.0018./((5./((cosh(10^10.*x./0.5)).^2)-r.*16).^0.5);
q = integral(fun,V2,V1)
plot(r,q)
Maybe, I do not understand what you meant. could you please write the full code? becuase my code still gives me error!
Torsten
2022-5-26
And did you find out which of the four solutions for y give a positive real number under the log(...) and what condition on r is necessary for that this is the case ?
I mean the expression sqrt(5*r-16) gives one necessary condition that your subsequent integration is senseful, namely r >= 16/5.
Thus
V2 = log(sqrt(5*r)/4 - sqrt(5*r-16)/4)/2e10
V1 = log(sqrt(5*r)/4 + sqrt(5*r-16)/4)/2e10
under the assumption
r >= 16/5
Abdallah Qaswal
2022-5-26
Abdallah Qaswal
2022-5-26
Well, MATLAB told you what V1 and V2 are depending on r - and these expressions are complex numbers if r < 16/5.
And that's logical since cosh(x) >= 1 for real x, thus the right-hand side of your equation
(cosh(10^10.*y./0.5)).^2 == 5./16.*r;
must be >=1, thus
5/16*r >= 1
thus
r >= 16/5
Walter Roberson
2022-5-26
y = log(S)*0.5/10^10
If that is still how you are defining y then y is a vector of symbolic values. You cannot solve() something with respect to y when y is not a symbolic variable.
subs(eqn, r, y)
y = log(S)*0.5/10^10
is the symbolic vector-solution of the equation
eqn= (cosh(10^10.*y./0.5)).^2 == 5./16.*r;
got over the deviation to solve
eqn = ((z+1/z)/2)^2 == 5/16*r;
S = solve(eqn,z);
y = log(S)*0.5/10^10
first for z.
Or can MATLAB directly
syms r y
eqn = (cosh(10^10.*y./0.5)).^2 == 5./16.*r;
S = solve(eqn,y)
?
At least, both results should match.
Abdallah Qaswal
2022-5-26
Abdallah Qaswal
2022-5-26
I already told you that this will not work.
This is code you can use:
% Prescribe solutions V2, V1 for (cosh(10^10.*y./0.5)).^2 == 5./16.*r
V2 = @(r) log(sqrt(5*r)/4 - sqrt(5*r-16)/4)/2e10
V1 = @(r) log(sqrt(5*r)/4 + sqrt(5*r-16)/4)/2e10
% Define function to be integrated
fun = @(x,r)0.0018./((5./((cosh(10^10.*x./0.5)).^2)-r.*16).^0.5);
% Define integral
q = @(r) integral(@(x)fun(x,r),V2(r),V1(r));
% Evaluate integral
r = 16/5:1/5:20;
Q = arrayfun(q,r)
% Plot result
plot(r,[real(Q);imag(Q)])
Abdallah Qaswal
2022-5-26
Torsten
2022-5-26
Is there something incorrect with the description in your question ?
You want to get the two real-valued solutions V1 and V2 with V1 > V2 of the equation
(cosh(10^10.*y./0.5)).^2 == 5./16.*r;
and use these solutions as lower and upper limit to evaluate the integral of the function
fun=@(x)0.0018./((5./((cosh(10^10.*x./0.5)).^2)-r.*16).^0.5);
for different values of r.
If this is correct, then the code I posted solves this problem.
Abdallah Qaswal
2022-5-26
Abdallah Qaswal
2022-5-26
Abdallah Qaswal
2022-5-26
If you choose
r = linspace(0,0.09)
you must know that V1 and V2 are complex numbers and that you integrate over a line in the complex domain, not over an interval in the reals.
But in case this doesn't matter for you, it's now also questionable which two of the four solutions of the equation
(cosh(10^10.*y./0.5)).^2 == 5./16.*r;
to take as lower and upper limits of the integral.
I assumed taking the real solutions under the assumption that r >= 16/5.
Abdallah Qaswal
2022-5-26
Abdallah Qaswal
2022-5-26
Torsten
2022-5-26
Yes, the little things ...
更多回答（1 个）
Walter Roberson
2022-5-26
Your first solve() returns multiple roots.
Your vpasolve() is then trying to process all of the roots at the same time .
Remember that solve() and vpasolve() are trying to solve simultaneous equations. If you pass in four equations in one variable, it will not attempt to solve each of the equations independently: it will try to find a single value of the variable that satisfies all of the equations at the same time. (Imagine, for example,that you had a series of trig expressions in a single variable, then you might be trying to find the right period that solves all of the equations at once.)
If you want to solve each equation independently, then use arrayfun() to run solve.
arrayfun(@(Y) vpasolve(Y, r, [0 0.09]), Y, 'uniform', 0)
The 'uniform', 0 is there because vpasolve() might decide there are no solutions.
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