Question about the proof of Linear Homogeneous Recurrance Relations.
$begingroup$
In Rosen's book, there is a proof of a theorem on Linear Homogeneous Recurrance Relations.
To prove the only if part of this theorem, why all we need to prove is that
there are constants $α_1$ and $α_2$ such that the sequence ${a_n}$
with $a_n = α_1r_1^n + α_2r_2^n$ satisfies these same initial
conditions
? (the first sentence of the second picture)
Why we do not need to prove that $a_n = α_1r_1^n + α_2r_2^n$ holds for all $n>=2$?
Does the initial conditions hold implies that for all $n >= 2$, $a_n$ will equals to $α_1r_1^n + α_2r_2^n$?
Thanks!
discrete-mathematics
asked Dec 9 '18 at 3:45
cmalcmal
1104
$endgroup$
add a comment |
$begingroup$
In Rosen's book, there is a proof of a theorem on Linear Homogeneous Recurrance Relations.
To prove the only if part of this theorem, why all we need to prove is that
there are constants $α_1$ and $α_2$ such that the sequence ${a_n}$
with $a_n = α_1r_1^n + α_2r_2^n$ satisfies these same initial
conditions
? (the first sentence of the second picture)
Why we do not need to prove that $a_n = α_1r_1^n + α_2r_2^n$ holds for all $n>=2$?
Does the initial conditions hold implies that for all $n >= 2$, $a_n$ will equals to $α_1r_1^n + α_2r_2^n$?
Thanks!
discrete-mathematics
asked Dec 9 '18 at 3:45
cmalcmal
1104
$endgroup$
$begingroup$
If $(a_n)$ and $(b_n)$ satisfy the same second-order recurrence and $a_0=b_0$ and $a_1=b_1$ then a trivial induction shows that $a_n=b_n$ for all $n$.
$endgroup$
– David C. Ullrich
Dec 9 '18 at 15:58
add a comment |
$begingroup$
In Rosen's book, there is a proof of a theorem on Linear Homogeneous Recurrance Relations.
To prove the only if part of this theorem, why all we need to prove is that
there are constants $α_1$ and $α_2$ such that the sequence ${a_n}$
with $a_n = α_1r_1^n + α_2r_2^n$ satisfies these same initial
conditions
? (the first sentence of the second picture)
Why we do not need to prove that $a_n = α_1r_1^n + α_2r_2^n$ holds for all $n>=2$?
Does the initial conditions hold implies that for all $n >= 2$, $a_n$ will equals to $α_1r_1^n + α_2r_2^n$?
Thanks!
discrete-mathematics
asked Dec 9 '18 at 3:45
cmalcmal
1104
$endgroup$
In Rosen's book, there is a proof of a theorem on Linear Homogeneous Recurrance Relations.
To prove the only if part of this theorem, why all we need to prove is that
there are constants $α_1$ and $α_2$ such that the sequence ${a_n}$
with $a_n = α_1r_1^n + α_2r_2^n$ satisfies these same initial
conditions
? (the first sentence of the second picture)
Why we do not need to prove that $a_n = α_1r_1^n + α_2r_2^n$ holds for all $n>=2$?
Does the initial conditions hold implies that for all $n >= 2$, $a_n$ will equals to $α_1r_1^n + α_2r_2^n$?
Thanks!
discrete-mathematics
discrete-mathematics
asked Dec 9 '18 at 3:45
cmalcmal
1104
asked Dec 9 '18 at 3:45
cmalcmal
1104
edited Dec 9 '18 at 6:22
cmal
asked Dec 9 '18 at 3:45
cmalcmal
1104
asked Dec 9 '18 at 3:45
cmalcmal
1104
asked Dec 9 '18 at 3:45
cmalcmal
1104
1104
$begingroup$
If $(a_n)$ and $(b_n)$ satisfy the same second-order recurrence and $a_0=b_0$ and $a_1=b_1$ then a trivial induction shows that $a_n=b_n$ for all $n$.
$endgroup$
– David C. Ullrich
Dec 9 '18 at 15:58
add a comment |
$begingroup$
If $(a_n)$ and $(b_n)$ satisfy the same second-order recurrence and $a_0=b_0$ and $a_1=b_1$ then a trivial induction shows that $a_n=b_n$ for all $n$.
$endgroup$
– David C. Ullrich
Dec 9 '18 at 15:58
$begingroup$
If $(a_n)$ and $(b_n)$ satisfy the same second-order recurrence and $a_0=b_0$ and $a_1=b_1$ then a trivial induction shows that $a_n=b_n$ for all $n$.
$endgroup$
– David C. Ullrich
Dec 9 '18 at 15:58
$begingroup$
If $(a_n)$ and $(b_n)$ satisfy the same second-order recurrence and $a_0=b_0$ and $a_1=b_1$ then a trivial induction shows that $a_n=b_n$ for all $n$.
$endgroup$
– David C. Ullrich
Dec 9 '18 at 15:58
add a comment |
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$begingroup$
If $(a_n)$ and $(b_n)$ satisfy the same second-order recurrence and $a_0=b_0$ and $a_1=b_1$ then a trivial induction shows that $a_n=b_n$ for all $n$.
$endgroup$
– David C. Ullrich
Dec 9 '18 at 15:58