# I Differential Equations For Solving A Recursive equation

1. Feb 13, 2017

### MAGNIBORO

Hi, i have a question about a proof of some recursive equation,
the function is
$$c_{n}(a)=\int_{0}^{\pi } \frac{cos(nx)-cos(na)}{cos(x)-cos(a)}$$
whit $n\in \mathbb{N}$ and $a\in \mathbb{R}$ .

whit some algebra is easy to see $c_{0}(a)=0$ and $c_{1}(a)=\pi$

and the recursive equation

$$c_{n+1}(a)+c_{n-1}(a) -2 \, cos(a) \, c_{n}(a) =0$$

Then the author says that this is a differential equation and solves it (making $c_{k}(a)=Ce^{sk}$)

I dont understand , i mean this is a recursive relation not a differencial equation.
so every recursive relation can be solve with a differencial equation?
thanks

2. Feb 14, 2017

### Stephen Tashi

For a fixed value of $a$ you have one recursive equation. Each such recursive equation has a solution that is a function of "$a$".

Does that notation indicate that the right hand side of the equation is independent of $a$?

Whether it does or not, I can see that the author would use words to indicate that he is doing more than solving one recursive equation. We have one recursive equation for each different value of $a$. I agree that it's confusing to call this infinite family of recursive equations a "differential equation".

3. Feb 14, 2017

### MAGNIBORO

hi, i was searching and i see that the recursive equation can solve by the formula $c_{1} \, \alpha ^{n} + c_{2}\, \beta ^{n}$ where alpha and beta are roots of a polynomial. the autor did something similar but instead of making $c_{n} = h^{n}$ , he makes $c_{n} = C e^{sn}$.
I leave you the part of the paper
https://gyazo.com/472073d82147bdebf094c2c66cb12f16 [Broken]
thanks

Last edited by a moderator: May 8, 2017
4. Feb 14, 2017

### Stephen Tashi

He make $s$ a function of $a$.

He does not call the equation in question a "differential equation". He calls it a "difference equation". The terms "difference equation" and "linear recursive relation" refer to essentially the same types of equations. Some people might reserve the term "difference equation" for an equation that is stated using the "difference operator" $\triangle f(n) = f(n+1) - f(n)$ , but "difference equations" can be expressed as recursive relations.

Last edited by a moderator: May 8, 2017
5. Feb 14, 2017

### MAGNIBORO

oh my error, this happen when you read fast i guess, I will continue studying this recursive relations they show up in many places
thanks for the help =D

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