I came across the recurrence relation $T(k)-7T(k-2)+6T(k-3)=0$ where $T(0)=8,T(1)=6 $and $T(2)=22$.
I found the roots of the characteristic equation to be $1,-3,2$ and the constants to be $1,2,5$. So, the solution $T(k)=5(1)^k+2(-3)^k+1(2)^k$. But my book states the solution to be $ T(k)=5+2(2)^k+(-3)^k$. I can notice that this has been obtained by just assigning different constants to different roots. So, my question is how do we determine which constants are to be assigned to which root?
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Ayan Shah
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1Have you checked whether (i) your formula works for $k=0,1,2$, (ii) the book's formula works for $k=0,1,2$? – Angina Seng Feb 11 '18 at 06:38
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Mine doesn't work for T=1 ...but why??? – Ayan Shah Feb 11 '18 at 06:43
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Probably an arithmetic error? – Angina Seng Feb 11 '18 at 06:43
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I don't think it's an arithmetic error, for I got the same constants and roots as the book. – Ayan Shah Feb 11 '18 at 06:44
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You have to pair each "constant" with the correct root. – Angina Seng Feb 11 '18 at 06:46
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It's like they have assigned a different serial number to the roots from me – Ayan Shah Feb 11 '18 at 06:46
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What makes them "correct"?? – Ayan Shah Feb 11 '18 at 06:47
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What makes 1 the first root, 2 the second root and -3 the third root??? – Ayan Shah Feb 11 '18 at 06:48
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You properly found the roots of the characteristic equation. So $$T_k=c_1 (1)^k+c_2 (2)^k+c_3 (-3)^k=c_1 +c_2 (2)^k+c_3 (-3)^k$$ Now apply the conditions $$8=c_1+c_2+c_3$$ $$6=c_1+2c_2-3c_3$$ $$22=c_1+4c_2+9c_3$$
This seems to be easy to solve.
Claude Leibovici
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