Undergrad Differential Equations For Solving A Recursive equation

[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 don't understand , i mean this is a recursive relation not a differencial equation.
so every recursive relation can be solve with a differencial equation?
thanks

 

[Stephen Tashi]

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

I don't understand , i mean this is a recursive relation not a differencial equation.

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

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

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".

 

[MAGNIBORO]

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".

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
thanks

 

[Stephen Tashi]

I leave you the part of the paper
https://gyazo.com/472073d82147bdebf094c2c66cb12f16

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.

 

[MAGNIBORO]

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.

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

 

[EuclidsPoopyPants]

The recursive relation can be derived from the equation where you set a variable equal to the infinite sum of that variable's sequential derivatives: y = y' + y'' + y''' + ... + y^n'
Take the derivative of the above equation and subtract the derived equation from the first to obtain the recursive relation you stated: y = 2y'. Set y = c{n+1}(a) + c{n-1}(a), and y' then equals cos(a)*c{n}(a). I used the {} to indicate subscript because I'm a Luddite. The solution of the reduced form of the equation can be easily seen as: y=e^(x/2), or whatever variable you have in place of x as the independent, continuous parameter. Hope this helps! I've found that this infinite-order linear differential equation can be solved in this way by assuming that y^n' is equal to y^(n+1)' in the limit as n tends to infinity, due to the 1/2 in the exponent making both terms converge to zero in a manner similar to Cantor's theorem of fractals.