How to speed up the calculation of jacobian?

Hi all,

I have a vector f with dimension 10000. I used jacobian(f,x) to compute its jacobian, but this turns out to be very slow. Is there any way that I can speed up or use parallel computation?

Many thanks.

[Walter Roberson]

To check, is your "f" a symbolic vector, or is it numeric only?

[Marrie]

thanks for your prompt reply. 'f' is a symbolic vector.

[Walter Roberson]

Is your x a single variable or is it a vector of symbolic variables? If it is a single variable, have you tried

diff(f,x)

as that should be equivalent ?

[Marrie]

Hi Walter,

x is a vector of symbolic variables. I tried:

   parfor j=1:1:N
      J(j)=jacobian(f(j),x);
   end

but this is more time-consuming.

[Matt J]

I don't do a lot of symbolic work, but 10000 symbolic functions seems like a lot to work with. What do the f(j) look like? How do they vary with j? Is there a compelling reason to choose symbolic manipulation over numerical manipulation?

[Marrie]

Hi Matt,

The purpose is to solve f(x)=0, where f and x are both vectors with dimension 10000. If I use fsolve, it is also very time-consuming, plus i need to solve f(x)=0 many many times (since x will be updated with time). For those reasons, I think computing the Jacobian for once would be the best choice.

[Walter Roberson]

Try

parfor xi = 1 : length(x)
  Jx(xi) = diff(f,x(xi));
end

[Matt J]

I think computing the Jacobian for once would be the best choice.

Yes, but it's still not clear why you have to do that using Symbolic Toolbox functions. Why is it so hard to derive an expression for the Jacobian on your own? If you can, then it will be much faster to use that expression. For example, given

 f=@(x) (x.^(1:N)).'

it would be much faster to compute Jacobian via

 Jac=@(x) spdiags((1:N).*(x.^(0:N-1)),0,N,N);

then going through jacobian(f,x).