solving an equation not by sym

Hello

My equation is developed in a step-by-step manner as follows with some assumptions. x is the unknown in here. This is just a simplified form of my equation.

E(1,1)=0

E(2,1)=(2/x)+(4*E(1,1))

E(3,1)=(2/x)+(4*E(2,1))

.....

E(n+1,1)=(2/x)+(4*E(n,1))

I set E(n+1,1) equal to zero and find the x via sym. However, it takes a long time via sym. Is there any alternative solution?

Answer 1

[Walter Roberson]

https://www.mathworks.com/help/symbolic/compute-z-transforms-and-inverse-z-transforms.html

You appear to have a recurrence relationship. Those are potentially solvable with ztrans.

Answer 2

[John D'Errico]

You have this basic first order linear recurrence relation:

E(n+1,x) = 2/x + 4*E(n,x)

Where E(1,x) = 0 is a given, but also for some chosen value of N, you will then also want E(N,x) = 0.

Of course, the recurrence relation as shown has a trivial solution. 2/x is a effectively a constant with respect to n. Can we solve this problem using z-transforms? That seems the simplest way. The general solution would be to solve for E(n,x), by taking the z-transform of your recurrence, solving using the inverse z-transform, then set the nth term to zero, and solve for x.

[Walter Roberson]

This particular case can be resolved without ztrans by constructing a numerator (to be divided by x) in base 4. The sequence goes 0, 2, 22, 222, 2222 and so on. At the end, the base 4 numerator divided by x, has to equal 0 because the E(n+1,1) is said to be 0. The only solutions are x being ±infinity

[John D'Errico]

And why I did not actually write out a solution to this specific problem, because it was clearly not the problem of interest. I'm not even positive the question is about a simple linear one term recurrence. But a z-transform is probably the solution method to use, as we both suggrested.

Answer 3

[Torsten]

Linear system of equations in E(1,1),...,E(n+1) and 2/x.

Setup for n = 4 (solution variables are E(1,1),E(2,1),E(3,1), E(4,1), E(5,1) and 2/x in this order):

A = [1 0 0 0 0 0;-4 1 0 0 0 -1;0 -4 1 0 0 -1; 0 0 -4 1 0 -1; 0 0 0 -4 1 -1; 0 0 0 0 1 0];
b = [0 0 0 0 0 0].';
rank(A)

ans = 6

sol = A\b

sol = 6×1

0 0 0 0 0 0

x = 2/sol(6)

x = -Inf

Answer 4

[Pooneh Shah Malekpoor]

Thanks for yur response. Ok, let me explain in detail. In equation below, E(1,1)=0, E(2,1) has E(1,1) inside it, in the same line E(3,1) has E(2,1).... till E(n,1) has E(n-1,1) and finally E(n+1,1) which is equal to zero and from E(n+1,1)=0 , x can be found.

how to solve this equation as fast as possible?

E(i,1)=((CC(i-1)'.*l(i-1)./x)+(((tan(phii(i-1)*pi/180)')./x)*((rho*9.81*A(i-1)*(cos(beta(i-1))))+(-rho*9.81*KH*A(i-1)*(sin(beta(i-1))))+(E(i-1)*(sin(beta(i-1)-(L*j(i-1)))))))-(rho*9.81*A(i-1)*(sin(beta(i-1))))-(KH*rho*9.81*A(i-1)*cos(beta(i-1)))+(E(i-1)*(cos(beta(i-1)-(L*j(i-1))))))./(cos(beta(i-1)-(L*j(i)))+((sin(beta(i-1)-(L*j(i))))*(tan(phii(i-1)*pi/180)')./x));
%     

P.S. CC, l, phii, rho, A, beta, KH, are all known values

[Torsten]

How large is n in your computations ?

And all other arrays/constants except for E and x are known ?

[Pooneh Shah Malekpoor]

It may have 200 arrays. Other contants have the same size az n

For the time being, I am using n=10 as this is a trial one and I want to debug the code

Just rho=18. Except for x, everything is known (E only contains x as the unknown)

[Torsten]

The sizes of the arrays CC, phii, A, beta, j and E are 200x1 or do you have 200 arrays ?

And where do the equations stem from ? Maybe from the discretization of a boundary-value problem ?

[Pooneh Shah Malekpoor]

they are all (200*1) double arrays

These double arrays are the results from another function that I previously defined in the code and imported here

[Torsten]

For a given value of x, compute E1,...,En and return resid_(N+1) to a nonlinear solver, e.g. fzero or fsolve.

Here, resid_(N+1) is the residual of the (N+1)th equation where you want E(n+1) = 0.

[Pooneh Shah Malekpoor]

could you explain with a simple example?

[Torsten]

x1 = 4;
sol1 = fzero(@fun,x1)

sol1 = 3.1724

fun(sol1)

ans = -4.9058e-15

x2 = -2;
sol2 = fzero(@fun,x2)

sol2 = -2.7814

fun(sol2)

ans = 7.6050e-15

x = -10:0.01:10;
plot(x,fun(x),x1,sol1,'o',x2,sol2,'o')

%E(1) = 0;
%E(2) = 4*E(1) + 2/x;
%E(3) = 4*E(2) + 3/x;
%E(4) = 4*E(3) - 5*x + 2;
%E(5) = 4*E(4) + sin(x)
%E(5) = 0;
function res = fun(x)
  res = 0;
  res = 4*res + 2./x;
  res = 4*res + 3./x;
  res = 4*res - 5*x + 2;
  res = 4*res + sin(x);
end

[Pooneh Shah Malekpoor]

thanks, whats the fun here?

[Torsten]

Given x, the function "fun" calculates the resulting value for E(5).

"fzero" tries to adjust x so that E(5) becomes 0.