Solve the recurrence relation $n(n-1)a_n - (n-2)^2a_{n-2}= 0$, $a_0=0$, $a_1=1$
Solve the recurrence relation $$n(n-1)a_n - (n-2)^2a_{n-2}= 0,$$ where $a_0=0$, $a_1=1$.
I think I might need to use generating functions, but I'm still not sure how to get started with this problem. Typing this into software came back as no solution for some reason.
sequences-and-series combinatorics discrete-mathematics recurrence-relations
add a comment |
Solve the recurrence relation $$n(n-1)a_n - (n-2)^2a_{n-2}= 0,$$ where $a_0=0$, $a_1=1$.
I think I might need to use generating functions, but I'm still not sure how to get started with this problem. Typing this into software came back as no solution for some reason.
sequences-and-series combinatorics discrete-mathematics recurrence-relations
Are you sure about the definition of the $a_n$'s? With that definition, it is trivial that $a_n=0$ when $n$ is even.
– José Carlos Santos
Nov 24 at 21:05
@JoséCarlosSantos Yes, that is what I was given.
– cosmicbrownie
Nov 24 at 21:16
add a comment |
Solve the recurrence relation $$n(n-1)a_n - (n-2)^2a_{n-2}= 0,$$ where $a_0=0$, $a_1=1$.
I think I might need to use generating functions, but I'm still not sure how to get started with this problem. Typing this into software came back as no solution for some reason.
sequences-and-series combinatorics discrete-mathematics recurrence-relations
Solve the recurrence relation $$n(n-1)a_n - (n-2)^2a_{n-2}= 0,$$ where $a_0=0$, $a_1=1$.
I think I might need to use generating functions, but I'm still not sure how to get started with this problem. Typing this into software came back as no solution for some reason.
sequences-and-series combinatorics discrete-mathematics recurrence-relations
sequences-and-series combinatorics discrete-mathematics recurrence-relations
edited Nov 24 at 21:17
greedoid
37.7k114794
37.7k114794
asked Nov 24 at 21:00
cosmicbrownie
1016
1016
Are you sure about the definition of the $a_n$'s? With that definition, it is trivial that $a_n=0$ when $n$ is even.
– José Carlos Santos
Nov 24 at 21:05
@JoséCarlosSantos Yes, that is what I was given.
– cosmicbrownie
Nov 24 at 21:16
add a comment |
Are you sure about the definition of the $a_n$'s? With that definition, it is trivial that $a_n=0$ when $n$ is even.
– José Carlos Santos
Nov 24 at 21:05
@JoséCarlosSantos Yes, that is what I was given.
– cosmicbrownie
Nov 24 at 21:16
Are you sure about the definition of the $a_n$'s? With that definition, it is trivial that $a_n=0$ when $n$ is even.
– José Carlos Santos
Nov 24 at 21:05
Are you sure about the definition of the $a_n$'s? With that definition, it is trivial that $a_n=0$ when $n$ is even.
– José Carlos Santos
Nov 24 at 21:05
@JoséCarlosSantos Yes, that is what I was given.
– cosmicbrownie
Nov 24 at 21:16
@JoséCarlosSantos Yes, that is what I was given.
– cosmicbrownie
Nov 24 at 21:16
add a comment |
2 Answers
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As José said for $n$ even $a_n=0$. Now for $n$ odd it is easy to show (say with inudction) that $$a_n = {1cdot 3^2cdot 5^2cdot...cdot (n-2)^2over n!}$$ or
$$a_n = {((n-2)!!)^2over n!}$$
You may find this formula by calculating the values of $a_n$ for small $n$ and then prove it (as I said by induction).
add a comment |
$a_n=frac{(n-2)^2a_{n-2}}{n(n-1)}, ngeq2; a_0=0, a_1=1$
Clearly, all the even-index terms of the recurrence are $0$.
For the odd-index terms, you could do this:
$a_3=frac{1^2a_1}{2cdot3}\
a_5=frac{3^2a_3}{4cdot5}\
a_7=frac{5^2a_5}{6cdot7}\
cdot\
cdot\
cdot\
a_{2n+1}=frac{(2n-1)^2a_{2n-1}}{(2n)cdot(2n+1)}, ngeq1\
$
Multiplying all the terms
$thinspacethinspace
implies a_3cdot a_5cdot a_7 cdot cdot cdot a_{2n+1}=frac{1^2cdot3^2cdot5^2cdotcdotcdot(2n-1)^2}{(2n+1)!}a_1cdot a_3cdot a_5cdotcdotcdot a_{2n-1}\
implies a_{2n+1}=frac{(1cdot3cdot5cdotcdotcdot(2n-1))^2}{(2n+1)!}a_1\
implies a_{2n+1}=frac{(2n-1)!^2}{2^{2n-2} (n-1)!^2 (2n+1)!}\
implies a_{2n+1}=frac{(2n)!}{2^{2n} (2n+1) n!^2}; ngeq1
$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
As José said for $n$ even $a_n=0$. Now for $n$ odd it is easy to show (say with inudction) that $$a_n = {1cdot 3^2cdot 5^2cdot...cdot (n-2)^2over n!}$$ or
$$a_n = {((n-2)!!)^2over n!}$$
You may find this formula by calculating the values of $a_n$ for small $n$ and then prove it (as I said by induction).
add a comment |
As José said for $n$ even $a_n=0$. Now for $n$ odd it is easy to show (say with inudction) that $$a_n = {1cdot 3^2cdot 5^2cdot...cdot (n-2)^2over n!}$$ or
$$a_n = {((n-2)!!)^2over n!}$$
You may find this formula by calculating the values of $a_n$ for small $n$ and then prove it (as I said by induction).
add a comment |
As José said for $n$ even $a_n=0$. Now for $n$ odd it is easy to show (say with inudction) that $$a_n = {1cdot 3^2cdot 5^2cdot...cdot (n-2)^2over n!}$$ or
$$a_n = {((n-2)!!)^2over n!}$$
You may find this formula by calculating the values of $a_n$ for small $n$ and then prove it (as I said by induction).
As José said for $n$ even $a_n=0$. Now for $n$ odd it is easy to show (say with inudction) that $$a_n = {1cdot 3^2cdot 5^2cdot...cdot (n-2)^2over n!}$$ or
$$a_n = {((n-2)!!)^2over n!}$$
You may find this formula by calculating the values of $a_n$ for small $n$ and then prove it (as I said by induction).
edited Nov 24 at 21:19
answered Nov 24 at 21:12
greedoid
37.7k114794
37.7k114794
add a comment |
add a comment |
$a_n=frac{(n-2)^2a_{n-2}}{n(n-1)}, ngeq2; a_0=0, a_1=1$
Clearly, all the even-index terms of the recurrence are $0$.
For the odd-index terms, you could do this:
$a_3=frac{1^2a_1}{2cdot3}\
a_5=frac{3^2a_3}{4cdot5}\
a_7=frac{5^2a_5}{6cdot7}\
cdot\
cdot\
cdot\
a_{2n+1}=frac{(2n-1)^2a_{2n-1}}{(2n)cdot(2n+1)}, ngeq1\
$
Multiplying all the terms
$thinspacethinspace
implies a_3cdot a_5cdot a_7 cdot cdot cdot a_{2n+1}=frac{1^2cdot3^2cdot5^2cdotcdotcdot(2n-1)^2}{(2n+1)!}a_1cdot a_3cdot a_5cdotcdotcdot a_{2n-1}\
implies a_{2n+1}=frac{(1cdot3cdot5cdotcdotcdot(2n-1))^2}{(2n+1)!}a_1\
implies a_{2n+1}=frac{(2n-1)!^2}{2^{2n-2} (n-1)!^2 (2n+1)!}\
implies a_{2n+1}=frac{(2n)!}{2^{2n} (2n+1) n!^2}; ngeq1
$
add a comment |
$a_n=frac{(n-2)^2a_{n-2}}{n(n-1)}, ngeq2; a_0=0, a_1=1$
Clearly, all the even-index terms of the recurrence are $0$.
For the odd-index terms, you could do this:
$a_3=frac{1^2a_1}{2cdot3}\
a_5=frac{3^2a_3}{4cdot5}\
a_7=frac{5^2a_5}{6cdot7}\
cdot\
cdot\
cdot\
a_{2n+1}=frac{(2n-1)^2a_{2n-1}}{(2n)cdot(2n+1)}, ngeq1\
$
Multiplying all the terms
$thinspacethinspace
implies a_3cdot a_5cdot a_7 cdot cdot cdot a_{2n+1}=frac{1^2cdot3^2cdot5^2cdotcdotcdot(2n-1)^2}{(2n+1)!}a_1cdot a_3cdot a_5cdotcdotcdot a_{2n-1}\
implies a_{2n+1}=frac{(1cdot3cdot5cdotcdotcdot(2n-1))^2}{(2n+1)!}a_1\
implies a_{2n+1}=frac{(2n-1)!^2}{2^{2n-2} (n-1)!^2 (2n+1)!}\
implies a_{2n+1}=frac{(2n)!}{2^{2n} (2n+1) n!^2}; ngeq1
$
add a comment |
$a_n=frac{(n-2)^2a_{n-2}}{n(n-1)}, ngeq2; a_0=0, a_1=1$
Clearly, all the even-index terms of the recurrence are $0$.
For the odd-index terms, you could do this:
$a_3=frac{1^2a_1}{2cdot3}\
a_5=frac{3^2a_3}{4cdot5}\
a_7=frac{5^2a_5}{6cdot7}\
cdot\
cdot\
cdot\
a_{2n+1}=frac{(2n-1)^2a_{2n-1}}{(2n)cdot(2n+1)}, ngeq1\
$
Multiplying all the terms
$thinspacethinspace
implies a_3cdot a_5cdot a_7 cdot cdot cdot a_{2n+1}=frac{1^2cdot3^2cdot5^2cdotcdotcdot(2n-1)^2}{(2n+1)!}a_1cdot a_3cdot a_5cdotcdotcdot a_{2n-1}\
implies a_{2n+1}=frac{(1cdot3cdot5cdotcdotcdot(2n-1))^2}{(2n+1)!}a_1\
implies a_{2n+1}=frac{(2n-1)!^2}{2^{2n-2} (n-1)!^2 (2n+1)!}\
implies a_{2n+1}=frac{(2n)!}{2^{2n} (2n+1) n!^2}; ngeq1
$
$a_n=frac{(n-2)^2a_{n-2}}{n(n-1)}, ngeq2; a_0=0, a_1=1$
Clearly, all the even-index terms of the recurrence are $0$.
For the odd-index terms, you could do this:
$a_3=frac{1^2a_1}{2cdot3}\
a_5=frac{3^2a_3}{4cdot5}\
a_7=frac{5^2a_5}{6cdot7}\
cdot\
cdot\
cdot\
a_{2n+1}=frac{(2n-1)^2a_{2n-1}}{(2n)cdot(2n+1)}, ngeq1\
$
Multiplying all the terms
$thinspacethinspace
implies a_3cdot a_5cdot a_7 cdot cdot cdot a_{2n+1}=frac{1^2cdot3^2cdot5^2cdotcdotcdot(2n-1)^2}{(2n+1)!}a_1cdot a_3cdot a_5cdotcdotcdot a_{2n-1}\
implies a_{2n+1}=frac{(1cdot3cdot5cdotcdotcdot(2n-1))^2}{(2n+1)!}a_1\
implies a_{2n+1}=frac{(2n-1)!^2}{2^{2n-2} (n-1)!^2 (2n+1)!}\
implies a_{2n+1}=frac{(2n)!}{2^{2n} (2n+1) n!^2}; ngeq1
$
edited Nov 24 at 21:44
answered Nov 24 at 21:31
Shubham Johri
3,826716
3,826716
add a comment |
add a comment |
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Are you sure about the definition of the $a_n$'s? With that definition, it is trivial that $a_n=0$ when $n$ is even.
– José Carlos Santos
Nov 24 at 21:05
@JoséCarlosSantos Yes, that is what I was given.
– cosmicbrownie
Nov 24 at 21:16