Help with Recurrence Equation

Tygra
Messages
55
Reaction score
8
TL;DR Summary
Equation that calculates joint rotations in a multi-storey structure
Hi there,

I am going through a book on multi-storey steel structures and I have come to a chapter that gives approximate methods to calculate rotations at the joints (The intersecting members) of a rigid frame. There is a recurrence equation that computes the rotations and this is given below:

equation.png


The book mentions a few strategies to solve this equation. However, I am finding it quite difficult. The book mentions that you can use iterative methods and that you can start by approximating theta(i) as:

1716036446853.png


Is there anybody that can help with this, please?

Many thanks.
 
Mathematics news on Phys.org
I've never seen a recurrence formula like the one you showed. In all of the recurrence relations I've seen, to get the n-th value in the chain, you need one or more of the preceding values, plus a starting value or values. For example, a common recurrence relation for factorial numbers is given by ##n! = n \cdot (n - 1)!##, with ##1! = 1##.
Personally, I would use the iterative formula you wrote for ##\theta_i## to calculate as many terms in the sequence as you need.
 
  • Like
Likes FactChecker
Hi Mark44,

I have no experience with reccurence relations. Its just the book called it that.

Would you be so kind to a have a read through the section of the book to see if it becomes more clear to you. Here are some images of the section. Read from equation (2.23) for the first image.



20240518_160403.jpg


20240518_160456.jpg


20240518_160507.jpg


If you are strunggling to read the text I can manually type out the section if you like?

Many thanks
 
Mark44 said:
I've never seen a recurrence formula like the one you showed. In all of the recurrence relations I've seen, to get the n-th value in the chain, you need one or more of the preceding values, plus a starting value or values. For example, a common recurrence relation for factorial numbers is given by ##n! = n \cdot (n - 1)!##, with ##1! = 1##.
Personally, I would use the iterative formula you wrote for ##\theta_i## to calculate as many terms in the sequence as you need.
Good point! This looks more like the situation with computational fluid dynamics, where the value at each point (i'th) depends on all surrounding values. That requires iteratively converging to a solution that is all compatible and consistent. I have no experience with those techniques, but I am sure that the solution depends on the initial values at all the boundary points. In this case, it would include assumptions about the initial values at both the bottom and top story.
 
Last edited:
Hi FactChecker,

well it does say in the book at even numbered joints (joint 2, 4...etc) the assumption is made that theta(i-1) = theta(i) = theta(i+1) and that it followws that:

1716056009685.png



It goes on to say that the initial values are obtained from the above formula (at even numbered joints). Next, the odd numbered joints for the first approximations are calculated from:

1716056506111.png

Next, improved values can be calculated at even numbered joints from this equation. Lastly, it says the equation is then used alternately for the set of both odd and even numbered joints until a satisfactory degree of convergence is achieved.

This is quite double dutch to me. I really need some help with it.
 
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
Fermat's Last Theorem has long been one of the most famous mathematical problems, and is now one of the most famous theorems. It simply states that the equation $$ a^n+b^n=c^n $$ has no solutions with positive integers if ##n>2.## It was named after Pierre de Fermat (1607-1665). The problem itself stems from the book Arithmetica by Diophantus of Alexandria. It gained popularity because Fermat noted in his copy "Cubum autem in duos cubos, aut quadratoquadratum in duos quadratoquadratos, et...
I'm interested to know whether the equation $$1 = 2 - \frac{1}{2 - \frac{1}{2 - \cdots}}$$ is true or not. It can be shown easily that if the continued fraction converges, it cannot converge to anything else than 1. It seems that if the continued fraction converges, the convergence is very slow. The apparent slowness of the convergence makes it difficult to estimate the presence of true convergence numerically. At the moment I don't know whether this converges or not.
Back
Top