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Автор fakeac, история, 8 лет назад, По-английски
Let's say we have 2 recurrence relations -
summation(a[i]*f[i] for i=1 to n1)+summation(b[i]*g[i] for i =1 to n2) = 0 --------> Equation 1
summation(c[i]*f[i] for i=1 to n3)+summation(d[i]*g[i] for i =1 to n4) = 0 --------> Equation 2
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`-----------------------------------------------------------------------------------> Equation m
WHERE a[i], b[i], c[i] AND d[i] are constants.

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In what cases is this above system of recurrence relations separable?
I want to express: f[i] in terms of only f terms and g[i] in terms of only g terms.
Is there a method to do it? If not, what are some of the specific cases under which this is doable?

I know of 1 such method when:
f[i]=a*f[i-1]+b*g[i-2] --------------------------------------------------Equation A
g[i]=c*f[i-5]+d*g[i-4] --------------------------------------------------Equation B
-> g[i-2]=(f[i]-a*f[i-1])/b -------------------------------------------- by rearranging A
-> f[i-5]=(g[i]-d*g[i-4])/c -------------------------------------------- by rearranging B
But now once you have say g[i-2]=(f[i]-a*f[i-1])/b, then by replacing i with i+2, we get,
-> g[i+2-2]=(f[i+2]-a*f[i+2-1])/b
-> g[i]=(f[i+2]-a*f[i+1])/b
Similarly g[i-4]=(f[i-2]-a*f[i-3])/b (from previous result)
now substitute these in Equation B.
Solve similarly for f.
Thus, separated.
##################################################################################################
Any general method?
Thanks in advance.
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8 лет назад, # |
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It is better to

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, like -

And you said "f[i] in terms of only f terms and g[i] in terms of only g terms", But the example you given, there you expressed f[i] in terms of g and g[i] in terms of f. Confusing.

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    8 лет назад, # ^ |
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    I forgot to show the rest of the steps :P But now once you have say g[i-2]=(f[i]-a*f[i-1])/b, then by replacing i with i+2, we get, -> g[i+2-2]=(f[i+2]-a*f[i+2-1])/b -> g[i]=(f[i+2]-a*f[i+1])/b Similarly g[i-4]=(f[i-2]-a*f[i-3])/b (from previous result) now substitute these in Equation B. Read blog above, I have modified it now. Thanks.

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8 лет назад, # |
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