How Can You Express Fibonacci Terms f_{n+1} and f_{n} Using f_{n+2} and f_{n+3}?

  • Thread starter Thread starter morbello
  • Start date Start date
  • Tags Tags
    Relation
AI Thread Summary
To express Fibonacci terms f_{n+1} and f_{n} using f_{n+2} and f_{n+3}, a recurrence relation is needed. One proposed relation is f_{n+1} + f_{n+2} + f_{n+3} = f_{n}. Alternative methods mentioned do not incorporate f_{n+3}. There is some confusion regarding the proper formatting of the equations, particularly with LaTeX syntax. Clarification on the inclusion of an equals sign in the equations is also suggested. The discussion emphasizes the need for precise mathematical expressions to solve the problem effectively.
morbello
Messages
73
Reaction score
0
i have to find the recurrence relation to express both f_{n+1} and f_{n} with f_{n+2} and f_{n+3}

my answer is
f_{n+1}+f_{n+2}+f_{n+3}...f_{n}

ive got other ways off doing it in the book but they do not use f_{n+3}

these are

f_{n+2}+f_{f-1}-f_{n}


they all should be in lower case but i can not seam to get the post to do it i hope it does not make it hard to help me with the question.



The Attempt at a Solution






 
Physics news on Phys.org
hi morbello

you can click on any latex code to see how it is typed, easiest to put tex quotes around your whole equation


but I'm not too sure what you're trying to do... ;) shouldn't you have an equals sign somewhere?

something like
f^{n+1} = f(n+1) = f(n) + ...?
 
I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...
Essentially I just have this problem that I'm stuck on, on a sheet about complex numbers: Show that, for ##|r|<1,## $$1+r\cos(x)+r^2\cos(2x)+r^3\cos(3x)...=\frac{1-r\cos(x)}{1-2r\cos(x)+r^2}$$ My first thought was to express it as a geometric series, where the real part of the sum of the series would be the series you see above: $$1+re^{ix}+r^2e^{2ix}+r^3e^{3ix}...$$ The sum of this series is just: $$\frac{(re^{ix})^n-1}{re^{ix} - 1}$$ I'm having some trouble trying to figure out what to...
Back
Top