Mixing
Recurrence relation - second order
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I read this book, and I do not understand one thing in proof.
The theorem is:
Let $s_1, s_2$ be number so that quadratic equation $x^2-s_1x-s_2 = 0$ has exactly one root, $r neq 0.$
Then every solution to the recurrence relation $$a_n = s_1a_{n-1} + s_2a_{n-2}$$ is of the form $$a_n = c_1r^n+c_2nr^n$$
Proof: Since the quadratic equation has a single (repeated) root, it must be of the form $(x-r)(x-r)=x^2-2rx+r^2$. Thus the recurrence must be $a_n = 2ra_{n-1}-r^na_{n-2}$.
And I stop here, because I do not understand the last sentence (in bold).
Thank you for explenation.
recurrence-relations proof-explanation
asked Jan 16 at 18:00
KapurKapur
556
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add a comment |
$begingroup$
I read this book, and I do not understand one thing in proof.
The theorem is:
Let $s_1, s_2$ be number so that quadratic equation $x^2-s_1x-s_2 = 0$ has exactly one root, $r neq 0.$
Then every solution to the recurrence relation $$a_n = s_1a_{n-1} + s_2a_{n-2}$$ is of the form $$a_n = c_1r^n+c_2nr^n$$
Proof: Since the quadratic equation has a single (repeated) root, it must be of the form $(x-r)(x-r)=x^2-2rx+r^2$. Thus the recurrence must be $a_n = 2ra_{n-1}-r^na_{n-2}$.
And I stop here, because I do not understand the last sentence (in bold).
Thank you for explenation.
recurrence-relations proof-explanation
asked Jan 16 at 18:00
KapurKapur
556
$endgroup$
add a comment |
$begingroup$
I read this book, and I do not understand one thing in proof.
The theorem is:
Let $s_1, s_2$ be number so that quadratic equation $x^2-s_1x-s_2 = 0$ has exactly one root, $r neq 0.$
Then every solution to the recurrence relation $$a_n = s_1a_{n-1} + s_2a_{n-2}$$ is of the form $$a_n = c_1r^n+c_2nr^n$$
Proof: Since the quadratic equation has a single (repeated) root, it must be of the form $(x-r)(x-r)=x^2-2rx+r^2$. Thus the recurrence must be $a_n = 2ra_{n-1}-r^na_{n-2}$.
And I stop here, because I do not understand the last sentence (in bold).
Thank you for explenation.
recurrence-relations proof-explanation
asked Jan 16 at 18:00
KapurKapur
556
$endgroup$
I read this book, and I do not understand one thing in proof.
The theorem is:
Let $s_1, s_2$ be number so that quadratic equation $x^2-s_1x-s_2 = 0$ has exactly one root, $r neq 0.$
Then every solution to the recurrence relation $$a_n = s_1a_{n-1} + s_2a_{n-2}$$ is of the form $$a_n = c_1r^n+c_2nr^n$$
Proof: Since the quadratic equation has a single (repeated) root, it must be of the form $(x-r)(x-r)=x^2-2rx+r^2$. Thus the recurrence must be $a_n = 2ra_{n-1}-r^na_{n-2}$.
And I stop here, because I do not understand the last sentence (in bold).
Thank you for explenation.
recurrence-relations proof-explanation
recurrence-relations proof-explanation
asked Jan 16 at 18:00
KapurKapur
556
asked Jan 16 at 18:00
KapurKapur
556
asked Jan 16 at 18:00
KapurKapur
556
asked Jan 16 at 18:00
KapurKapur
556
asked Jan 16 at 18:00
KapurKapur
556
556
add a comment |
add a comment |
1 Answer
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It could be either a typo in your post, on in the book, but the right expression is $a_n = 2r a_{n - 1} - r^{color{red}{2}} a_{n - 2}$. To show the statement in bold you can try this way. Imagine the recurrence
$$
a_n = 2r a_{n - 1} - r^{color{red}{2}} a_{n - 2} tag{1}
$$
and consider a solution of the form $a_n = c x^n$, if you replace this in Eqn (1) you get
$$
require{cancel}
cancel{c} x^n = 2 r (cancel{c} x^{n - 1}) - r^2 (cancel{c} x^{n - 2})
$$
Multiply everything by $x^{2-n}$, you will get
$$
x^2 = 2r x - r^2 ~~~Leftrightarrow (x-r)^2 = 0 tag{2}
$$
Or equivalently, an equation of the form $x^2 -2rx + r^2 = 0$ implies $a_n = 2r a_{n - 1} - r^{color{red}{2}} a_{n - 2}$
answered Jan 16 at 22:59
caveraccaverac
14.6k31130
$endgroup$
$begingroup$
Thank you, I rewrote it wrong. I think it is better. You put me on the different way on thinking about this.
$endgroup$
– Kapur
Jan 17 at 14:02
add a comment |
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1 Answer
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$begingroup$
It could be either a typo in your post, on in the book, but the right expression is $a_n = 2r a_{n - 1} - r^{color{red}{2}} a_{n - 2}$. To show the statement in bold you can try this way. Imagine the recurrence
$$
a_n = 2r a_{n - 1} - r^{color{red}{2}} a_{n - 2} tag{1}
$$
and consider a solution of the form $a_n = c x^n$, if you replace this in Eqn (1) you get
$$
require{cancel}
cancel{c} x^n = 2 r (cancel{c} x^{n - 1}) - r^2 (cancel{c} x^{n - 2})
$$
Multiply everything by $x^{2-n}$, you will get
$$
x^2 = 2r x - r^2 ~~~Leftrightarrow (x-r)^2 = 0 tag{2}
$$
Or equivalently, an equation of the form $x^2 -2rx + r^2 = 0$ implies $a_n = 2r a_{n - 1} - r^{color{red}{2}} a_{n - 2}$
answered Jan 16 at 22:59
caveraccaverac
14.6k31130
$endgroup$
$begingroup$
Thank you, I rewrote it wrong. I think it is better. You put me on the different way on thinking about this.
$endgroup$
– Kapur
Jan 17 at 14:02
add a comment |
$begingroup$
It could be either a typo in your post, on in the book, but the right expression is $a_n = 2r a_{n - 1} - r^{color{red}{2}} a_{n - 2}$. To show the statement in bold you can try this way. Imagine the recurrence
$$
a_n = 2r a_{n - 1} - r^{color{red}{2}} a_{n - 2} tag{1}
$$
and consider a solution of the form $a_n = c x^n$, if you replace this in Eqn (1) you get
$$
require{cancel}
cancel{c} x^n = 2 r (cancel{c} x^{n - 1}) - r^2 (cancel{c} x^{n - 2})
$$
Multiply everything by $x^{2-n}$, you will get
$$
x^2 = 2r x - r^2 ~~~Leftrightarrow (x-r)^2 = 0 tag{2}
$$
Or equivalently, an equation of the form $x^2 -2rx + r^2 = 0$ implies $a_n = 2r a_{n - 1} - r^{color{red}{2}} a_{n - 2}$
answered Jan 16 at 22:59
caveraccaverac
14.6k31130
$endgroup$
$begingroup$
Thank you, I rewrote it wrong. I think it is better. You put me on the different way on thinking about this.
$endgroup$
– Kapur
Jan 17 at 14:02
add a comment |
$begingroup$
It could be either a typo in your post, on in the book, but the right expression is $a_n = 2r a_{n - 1} - r^{color{red}{2}} a_{n - 2}$. To show the statement in bold you can try this way. Imagine the recurrence
$$
a_n = 2r a_{n - 1} - r^{color{red}{2}} a_{n - 2} tag{1}
$$
and consider a solution of the form $a_n = c x^n$, if you replace this in Eqn (1) you get
$$
require{cancel}
cancel{c} x^n = 2 r (cancel{c} x^{n - 1}) - r^2 (cancel{c} x^{n - 2})
$$
Multiply everything by $x^{2-n}$, you will get
$$
x^2 = 2r x - r^2 ~~~Leftrightarrow (x-r)^2 = 0 tag{2}
$$
Or equivalently, an equation of the form $x^2 -2rx + r^2 = 0$ implies $a_n = 2r a_{n - 1} - r^{color{red}{2}} a_{n - 2}$
answered Jan 16 at 22:59
caveraccaverac
14.6k31130
$endgroup$
It could be either a typo in your post, on in the book, but the right expression is $a_n = 2r a_{n - 1} - r^{color{red}{2}} a_{n - 2}$. To show the statement in bold you can try this way. Imagine the recurrence
$$
a_n = 2r a_{n - 1} - r^{color{red}{2}} a_{n - 2} tag{1}
$$
and consider a solution of the form $a_n = c x^n$, if you replace this in Eqn (1) you get
$$
require{cancel}
cancel{c} x^n = 2 r (cancel{c} x^{n - 1}) - r^2 (cancel{c} x^{n - 2})
$$
Multiply everything by $x^{2-n}$, you will get
$$
x^2 = 2r x - r^2 ~~~Leftrightarrow (x-r)^2 = 0 tag{2}
$$
Or equivalently, an equation of the form $x^2 -2rx + r^2 = 0$ implies $a_n = 2r a_{n - 1} - r^{color{red}{2}} a_{n - 2}$
answered Jan 16 at 22:59
caveraccaverac
14.6k31130
answered Jan 16 at 22:59
caveraccaverac
14.6k31130
answered Jan 16 at 22:59
caveraccaverac
14.6k31130
answered Jan 16 at 22:59
caveraccaverac
14.6k31130
14.6k31130
$begingroup$
Thank you, I rewrote it wrong. I think it is better. You put me on the different way on thinking about this.
$endgroup$
– Kapur
Jan 17 at 14:02
add a comment |
$begingroup$
Thank you, I rewrote it wrong. I think it is better. You put me on the different way on thinking about this.
$endgroup$
– Kapur
Jan 17 at 14:02
$begingroup$
Thank you, I rewrote it wrong. I think it is better. You put me on the different way on thinking about this.
$endgroup$
– Kapur
Jan 17 at 14:02
$begingroup$
Thank you, I rewrote it wrong. I think it is better. You put me on the different way on thinking about this.
$endgroup$
– Kapur
Jan 17 at 14:02
add a comment |
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