Menu Home Action My entries Defined browse Select Select in the list MathematicsPhysics Then Select Select in the list Then Select Select in the list Search

recurrence relation

 Definition/Summary A recurrence relation is an equation which defines each term of a sequence as a function of preceding terms. The most well-known are those defining the Fibonacci numbers and the binomial coefficients. An ordinary differential equation can be considered as a recurrence relation on the sequence of nth derivatives of a function (in which the 0th derivative is the function itself), and if linear can be solved using the characteristic polynomial method.

 Equations Fibonacci numbers: Each Fibonacci number is the sum of the two preceding numbers, starting with 1 and 1: $$F_n\ =\ F_{n-1}\ +\ F_{n-2}$$ 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89 ... Binomial coefficients (Pascal's triangle): Each binomial coefficient is the sum of the two coefficients "diagonally below", starting with 1: $$\left(\begin{array}{c}n\\k\end{array}\right)\ =\ \left(\begin{array}{c}n - 1\\k\end{array}\right)\ +\ \left(\begin{array}{c}n-1\\k-1\end{array}\right)$$ 1; 1 1; 1 2 1; 1 3 3 1; 1 4 6 4 1; 1 5 10 10 5 1; ...

 Scientists Leonardo of Pisa (Fibonacci) (c. 1170 – c. 1250) Blaise Pascal (1623 - 1662)

 Recent forum threads on recurrence relation

 Breakdown Mathematics > Calculus/Analysis >> Calculus

 Extended explanation Characteristic polynomial: This is not the same as the characteristic polynomial of a matrix or matroid In a linear differential equation, the derivative may be replaced by an operator, D, giving a polynomial equation in D: $$\sum_{n\,=\,0}^m\,a_n\,\frac{d^ny}{dx^n}\ =\ 0\ \mapsto\ \left(\sum_{n\,=\,0}^m\,a_n\,D^n\right)y\ =\ 0$$ Similarly, in a linear recurrence relation, the nth term may be replaced by an nth power: $$\sum_{n\,=\,0}^m\,a_{k+n}\,P_{k+n}\ =\ 0\ \mapsto\ \sum_{n\,=\,0}^m\,a_n\,x^n\ =\ 0$$ If this polynomial has distinct (different) roots $r_1,\dots,r_m$: $$\prod_{n\,=\,1}^m(D\,-\,r_n)\ =\ 0$$ or $$\prod_{n\,=\,1}^m(x\,-\,r_n)\ =\ 0$$ then the general solution is a linear combination of the solutions of each of the equations: $$\left(D\,-\,r_n\right)y\ =\ 0$$ or $$P_{k+1}\ -\ r_nP_k\ =\ 0$$ which are the same as $$\frac{dy}{dx}\ =\ r_n\,y$$ or $$\frac{P_{k+1}}{P_k}\ =\ r_n$$ and so is of the form: $$y\ =\ \sum_{n\,=\,1}^m\,C_n\,e^{r_nx}$$ or $$P_k\ =\ \sum_{n\,=\,1}^m\,C_n\,(r_n)^k$$ For a pair of complex roots (they always come in conjugate pairs) $p\ \pm\ iq$ or $r\,e^{\pm is}$, a pair of $$C_ne^{r_nx}$$ or $C_n\,(r_n)^k$may be replaced by $$e^{px}(A\,cos(qx)\,+\,iB\,sin(qx))$$ or $$r^k(A\,cos(sk)\,+\,iB\,sin(sk))$$ However, if the polynomial has some repeated roots: $$\prod_{p\,=\,1}^q\prod_{n\,=\,1}^p(D\,-\,r_n)^p\,y\ =\ 0$$ or $$\prod_{p\,=\,1}^q\prod_{n\,=\,1}^p(x\,-\,r_n)^p\ =\ 0$$ then the general solution is of the form: $$y\ =\ \sum_{p\,=\,1}^q\sum_{n\,=\,1}^pC_{n,p}\,x^{p-1}\,e^{r_nx}$$ or $$P_k\ =\ \sum_{p\,=\,1}^q\sum_{n\,=\,1}^pC_{n,p}\,n^{p-1}\,(r_n)^k$$