20 апр. 2015 г. · A nice proof is given by using generating functions. Say your recurrence is: ckan+k+ck−1an+k−1+⋯+c0an=0. with given values for a0,…,ak−1.

17 июл. 2017 г. · The formula for the determinant as a sum over permutations (or by induction over n using Laplace expansion).

29 дек. 2014 г. · The characteristic equation can only be formed when the differential equation (recursive sequence) is linear, homogeneous, and has constant ...

4 июл. 2012 г. · Here's a rote rule for doing so. Start with the recurrence: Un=3Un−1−Un−3. Convert each subscript to an exponent: Un=3Un−1−Un−3.

28 мар. 2014 г. · I'd like to proof: The caracteristic polynomial of A∈M(n×n,K) has the form: PA(λ)=(−1)nλn+(−1)n−1tr(A)λn−1+⋯+det(A). My proof idea:.

23 июн. 2012 г. · With substitution u=x, v=˙x we can rewrite this ODE as a system {˙u=v˙v=−Au−Bv. and hence with characteristic equation χ(λ)=det[λI−IAλI+B]. ...

20 мая 2024 г. · Derivation of the characteristic polynomial for homogenous difference equations ... an=c1an−1+c2an−2+⋯+cdan−d. ... an=k1np1rn+k2np2rn+⋯+knnpnrn.

18 нояб. 2015 г. · Think how you compute the characteristic polynomial. The highest order term will come from the product of the diagonal elements of A−λI. The ...

4 янв. 2018 г. · Let PA denote the characteristic polynomial for a 3×3 matrix A. Then PA(x)=16[Tr3(A)+2Tr(A3)−3Tr(A)Tr(A2)]−12[Tr2(A)−Tr(A2)]x+Tr(A)x2−x3.

7 окт. 2019 г. · Is there a more formally generalized process to explain what is going on when you derive a characteristic equation that would allow you to ...