solving an equation not by sym

2 次查看（过去 30 天）
Pooneh Shah Malekpoor
评论： Torsten 2022-10-31
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?

请先登录，再回答此问题。

回答（4 个）

Walter Roberson
Walter Roberson 2022-10-28
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 2022-10-28
编辑：John D'Errico 2022-10-28
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 个评论
Walter Roberson
Walter Roberson 2022-10-28
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 2022-10-28
编辑：John D'Errico 2022-10-28
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.

请先登录，再进行评论。

Torsten
Torsten 2022-10-28
编辑：Torsten 2022-10-28
Ran in:
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
编辑：Pooneh Shah Malekpoor 2022-10-28
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 个评论
显示 7更早的评论
Torsten
Torsten 2022-10-31
Given x, the function "fun" calculates the resulting value for E(5).
"fzero" tries to adjust x so that E(5) becomes 0.

请先登录，再进行评论。

类别

在 Help Center 和 File Exchange 中查找有关 Calculus 的更多信息

标签

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by