solving an equation not by sym

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Pooneh Shah Malekpoor
Pooneh Shah Malekpoor on 28 Oct 2022
Commented: Torsten on 31 Oct 2022
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?

Answers (4)

Walter Roberson
Walter Roberson on 28 Oct 2022
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.

John D'Errico
John D'Errico on 28 Oct 2022
Edited: John D'Errico on 28 Oct 2022
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.
  2 Comments
Walter Roberson
Walter Roberson on 28 Oct 2022
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
John D'Errico on 28 Oct 2022
Edited: John D'Errico on 28 Oct 2022
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.

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Torsten
Torsten on 28 Oct 2022
Edited: Torsten on 28 Oct 2022
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

Pooneh Shah Malekpoor
Pooneh Shah Malekpoor on 28 Oct 2022
Edited: Pooneh Shah Malekpoor on 28 Oct 2022
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
  9 Comments
Pooneh Shah Malekpoor
Pooneh Shah Malekpoor on 31 Oct 2022
thanks, whats the fun here?
Torsten
Torsten on 31 Oct 2022
Given x, the function "fun" calculates the resulting value for E(5).
"fzero" tries to adjust x so that E(5) becomes 0.

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