Recursion equation - problem with one same root

Question

i have problem with Recursion equation. It look like this: $x(k+2) + 2x(k+1) + x(k) = 0 $ , $x(0) = 3$, $x(1) = 6$

then i' m gettin: $x^2 + 2x + 1 = 0$

zero value are: $\Delta = 0$ , $x_{0} = -1$

$x(k) = C_{1}x_{1}^{k} + C_{2}x_{2}^{k}$

for $x(0) = 3$ i get

$C_{1} + C_{2} = 3$

The question is: Am i doing it right? In formula above $x_{1} , x_{2}$ are not the same. The question is how should i do it for $x(1)$ because it by putt for both $x_{1} , x_{2}$ $-1$ i will got $0=0$. What am i doing wrong?

//------------SECOND CASE-------------------------

$x(k+2) + 3x(k+1) + x(k) = 0, x(0) = 1, x(1) = 3$

$x^{2} + 3x + 1 = 0$

$\Delta = 1$

$r_{1} = -2$ $r_{2} = -1$

In this case i'm using this:

$x(k) = C_{1}r_{1}^{k} + C_{2}r_{2}^{k}$

right??

here, you have $x(k) = C_0x_0^k$ which is a solution (no $x_1, x_2$ roots). In the case of double zero there is one more: $C'_0.k.x_0^k$ — Fabrice NEYRET, Oct 26 '15 at 20:47
I'm lost you are using : $x(k) = C_{0}x_{0}^{k}$, Brian wrote $x_{k} = C_{1}r^{k} + C_{2}kr^{k}$ Is those formulas arent different? — JohneQue, Oct 26 '15 at 21:34
" In the case of double zero there is one more:xxx". so the general solution is the linear combination of both. — Fabrice NEYRET, Oct 26 '15 at 22:03
Ok, i think is clear now :D but sill this part of math is form me black magic :D — JohneQue, Oct 26 '15 at 22:09
yep; at least you can verify it works :-) . I don't remember the demonstration (and quick googling gave me results without demonstration). — Fabrice NEYRET, Oct 26 '15 at 22:21

score 0 · Accepted Answer · answered Oct 26 '15 at 20:39

0

When you have a double root $r$, the general form is

$$ x_k = C_1 r^k + C_2 kr^k $$

In this case, we have $r = -1$, so

$$ 3 = x_0 = C_1 $$ $$ 6 = x_1 = -C_1 - C_2 $$

which yields $C_1 = 3, C_2 = -9$, and

$$ x_k = 3(-1)^k-9k(-1)^k $$

Substitution yields $\{x_k\}_{k=0}^\infty = 3, 6, -15, 24, -33, \ldots$ which satisfies the recurrence.

Further multiplicities yield terms of the form $k^2 r^k, k^3 r^k, \ldots$ You may see a resemblance to differential equations (if you're familiar with those), in which a double root in the characteristic equation yields a solution of the form $f(x) = C_1 e^{rx} + C_2 xe^{rx}$ (and further multiplicities yield similar terms).

answered Oct 26 '15 at 20:39

Brian Tung

34,160

So the solution which i show above with two different root is correct? – JohneQue Oct 26 '15 at 20:44
I didn't see a solution you provided with two different roots; I only see a solution with a double root $r = -1$. What are your two different roots? (The characteristic equation does indeed have only one root of multiplicity $2$.) – Brian Tung Oct 26 '15 at 20:47
I just ask if this $x(k) = C_{1}r_{1}^{k} + C_{2}r_{2}^{k}$ is correct for 2 different roots just to be sure. :) – JohneQue Oct 26 '15 at 20:49
no: it is not correct. You prove there was one root, not two. It's what you do with this single root that differs in the case of double root. Read carefully our answers: there is a k factor. – Fabrice NEYRET Oct 26 '15 at 20:49
Look if you can, i edited question -> secon case. Am i on good track to solve this equation? – JohneQue Oct 26 '15 at 20:54
Yes, you are using the right approach on the second part. – Brian Tung Oct 26 '15 at 21:05
Thank you very much :D this stack is amazing (y) :D – JohneQue Oct 26 '15 at 21:30

Recursion equation - problem with one same root

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