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bqllpd
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Define a shift polynomial sequence operator as
[tex]S=\sum_{i=0}^mc_i\mathbf E^i[/tex]
where [itex]\mathbf E[/itex] is the shift operator and [itex]c_i[/itex] are some constants, variables, functions, etc. When [itex]S[/itex] is applied to a sequence [itex]\{a_n\}[/itex], then [tex]S(a)_n=\sum_{i=0}^mc_i\mathbf E^ia_n[/tex].
If [itex]S[/itex] is composed with itself [itex]k[/itex] times with the sequence, then define
[itex]S^k(a)_n=\left[\sum_{i=0}^mc_i\mathbf E^i\right]^ka_n[/itex].
If the first elements from each new sequence for each [itex]k[/itex] are taken, define this as [itex]S^n(a)_0=b_n=T(a)_n[/itex]. This is a problem I'm having. I know what [itex]T^k(a)_n=b_n[/itex] and [itex]T^{-k}(b)_n=a_n[/itex] are when [itex]m=1[/itex], [itex]c_0=\pm k[/itex] when [itex]c_1=1[/itex] and [itex]c_0=\pm 1[/itex] when [itex]c_1=-1[/itex].
I want to find a general formula for [itex]T^k(a)_n=b_n[/itex] and it's inverse, [itex]T^{-k}(b)_n=a_n[/itex] where [itex]S[/itex] is any shift polynomial operator. Any help would be greatly appreciated.Thanks
bq
[tex]S=\sum_{i=0}^mc_i\mathbf E^i[/tex]
where [itex]\mathbf E[/itex] is the shift operator and [itex]c_i[/itex] are some constants, variables, functions, etc. When [itex]S[/itex] is applied to a sequence [itex]\{a_n\}[/itex], then [tex]S(a)_n=\sum_{i=0}^mc_i\mathbf E^ia_n[/tex].
If [itex]S[/itex] is composed with itself [itex]k[/itex] times with the sequence, then define
[itex]S^k(a)_n=\left[\sum_{i=0}^mc_i\mathbf E^i\right]^ka_n[/itex].
If the first elements from each new sequence for each [itex]k[/itex] are taken, define this as [itex]S^n(a)_0=b_n=T(a)_n[/itex]. This is a problem I'm having. I know what [itex]T^k(a)_n=b_n[/itex] and [itex]T^{-k}(b)_n=a_n[/itex] are when [itex]m=1[/itex], [itex]c_0=\pm k[/itex] when [itex]c_1=1[/itex] and [itex]c_0=\pm 1[/itex] when [itex]c_1=-1[/itex].
I want to find a general formula for [itex]T^k(a)_n=b_n[/itex] and it's inverse, [itex]T^{-k}(b)_n=a_n[/itex] where [itex]S[/itex] is any shift polynomial operator. Any help would be greatly appreciated.Thanks
bq
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