Two for loops taking very long to get the complete solutions

The two for loops in the code take very long to complete

clear all
syms a
for k=0:0.1:pi
  for x=0:0.1:pi
    f = sin(k+a) - x*cos(x);
    y=vpasolve(f,a,[0 pi])
  end
end

How to fix this?

[Stephen23]

"How to fix this?"

Do not use vpasolve in a loop.

Seriously, symbolic maths is slow. If you want fast code, find the solution numerically.

[AtoZ]

@Stephen Thanks. I thought I was doing the solution numerically already. :-)

[Stephen23]

"I thought I was doing the solution numerically already. :-)"

Kind of, but you are finding a numeric solution using symbolic variables:

syms a

Symbolic maths is slow. Why not just write fast numeric code without symbolic variables?

[AtoZ]

Oh got it now. Thanks.

Answer 1

[madhan ravi]

EDITED

syms k a x
f = sin(k+a) - x*cos(x);
solution=solve(f,a)
x=0:0.1:pi;
k=x;
solutions=vpa(subs(solution),3)

[AtoZ]

@madhan thanks. Could you please also say something about `cell(1)` and ctr=1; what are these? Secondly,

if isempty(y)==0
      break

Is it excluding the regions with no solutions? Actually there should be some regions where there's no solution, so I want to also know those regions explicitly. and the output format is very long, I tried to put

format short

at different places in your code but it didn't work.

[madhan ravi]

 `cell(1)` and ctr=1;

its called as pre-allocation for increasing speed and efficiency of the loop car is just a basic counter, see edited answer

[madhan ravi]

of course it will take time because your loop is huge

[Torsten]

Solve your equation once outside the loop and only substitute k and x values into the solution inside the loop.

[Stephen23]

"its called as pre-allocation for increasing speed and efficiency ..."

Nope. Note that cell(1) only creates a 1x1 cell array, so does not preallocate the cell array to its final size, or anything close to it.

https://www.mathworks.com/help/matlab/matlab_prog/preallocate-memory-for-a-cell-array.html

Also note that isempty(y)==0 will always be true, so it serves no useful purpose.

[madhan ravi]

syms k a x

f = sin(k+a) - x*cos(x);

solution=solve(f,a)
x=0:0.1:pi;
k=x;
solutions=vpa(subs(solution),3)

[madhan ravi]

Thank you @Torsten

[AtoZ]

@madhan Thanks. This last one was quicker. What does 3 in

   solutions=vpa(subs(solution),3)

mean? Sorry but i am very new to Matlab :-)

[madhan ravi]

3 represents numbers to be displayed after the decimal point

[AtoZ]

How can we set a range on `a`? `a` has to be between 0<=a<=pi. just by

solution=solve(f,a,[0 pi])?

[Torsten]

I don't understand why you don't take the code I suggested under

https://de.mathworks.com/matlabcentral/answers/428022-numerical-solution-of-a-trigonometric-equation

It's efficient and takes into account the constraint on a.

[AtoZ]

I actually didn't understand why did we need to use all those conditions and why not just a straightforward solution like this one. I am actually trying to get the full solutions in the form of a table(x,k). I will try to do it if I can. Or if you know how, you could post there and I will accept. Thanks

[Torsten]

But the thing here is just the same.

You have to manually exclude complex solutions and real solutions that are not in the interval [0:pi]. Or did I get something wrong ?

[Stephen23]

"I actually didn't understand why did we need to use all those conditions and why not just a straightforward solution like this one."

Torsten's code is a straightforward solution. It is very simple!

You are failing to take into account the complexity of symbolic maths. Just because it is hidden inside two commands (e.g. syms and vpasolve) does not mean that symbolic maths is simple or fast. You are comparing a few simple lines of numeric code with a huge symbolic black-box, which is about as apples and oranges as you can get. Just because you are not looking at the black box does not mean that you can disregard its effects.

[Bruno Luong]

'I actually didn't understand why did we need to use all those conditions and why not just a straightforward solution like this one.' Thanks

The apparent complication because that fact that you deliberately restrict the solution a in [0,pi] as show in pour VPA code.

But Torsten code is the most efficient instead of using those horrible vpasove for something that can be computed from formula and few tests.

Torsten solution is fatest, simplest and mots reliable.

[madhan ravi]

I second Stephen and Brunos comment

[Bruno Luong]

If you need a vectorized Torsen code, here it is

x=0:0.1:pi;
k=(0:0.1:pi)';
b=asin(x.*cos(x));
b(imag(b)~=0)=NaN;
a1 = mod(b-k,2*pi);
a2 = mod(pi-b-k,2*pi);
a = a1;
outside = a1>pi & a2<=pi;
a(outside) = a2(outside);
a(a>pi) =NaN;

close all
surf(x,k,a);
xlabel('x');
ylabel('k');
zlabel('a');

[madhan ravi]

+1 @Bruno brilliant stuff right there

[AtoZ]

@Bruno this is really nice. Thanks :-)