How can I tell if a sequence can be a recursive sequence
Obviously 1, 1, 2, 3, 5, 8, etc.. is possibly a recursive sequence because we know it's the Fibonacci sequence and because it's really easy to see the pattern. It's recursive "function" for the form $a(n_{i - 1}) + b (n_{i - 2})$ where $a = 1$ and $b = 1$.
But how do we know for sure that 1, 2, 3, 4, _, _, etc.. can't have a recursive function? where there is a $a$ or $b$ that works for that sequence?
linear-algebra
asked Dec 3 '18 at 8:51
ming
3165
add a comment |
Obviously 1, 1, 2, 3, 5, 8, etc.. is possibly a recursive sequence because we know it's the Fibonacci sequence and because it's really easy to see the pattern. It's recursive "function" for the form $a(n_{i - 1}) + b (n_{i - 2})$ where $a = 1$ and $b = 1$.
But how do we know for sure that 1, 2, 3, 4, _, _, etc.. can't have a recursive function? where there is a $a$ or $b$ that works for that sequence?
linear-algebra
asked Dec 3 '18 at 8:51
ming
3165
Argh I don't know how to make more than one character go in a subscript can someone help me out? Thanks!
– ming
Dec 3 '18 at 8:52
to make multiple subscript charachters you can do : normal text_{your subscript text} which gives: $$normal text_{your subscript text}$$ Place all characters inside brackets
– TheD0ubleT
Dec 3 '18 at 8:59
Fancy, thanks!!
– ming
Dec 3 '18 at 9:40
add a comment |
Obviously 1, 1, 2, 3, 5, 8, etc.. is possibly a recursive sequence because we know it's the Fibonacci sequence and because it's really easy to see the pattern. It's recursive "function" for the form $a(n_{i - 1}) + b (n_{i - 2})$ where $a = 1$ and $b = 1$.
But how do we know for sure that 1, 2, 3, 4, _, _, etc.. can't have a recursive function? where there is a $a$ or $b$ that works for that sequence?
linear-algebra
asked Dec 3 '18 at 8:51
ming
3165
Obviously 1, 1, 2, 3, 5, 8, etc.. is possibly a recursive sequence because we know it's the Fibonacci sequence and because it's really easy to see the pattern. It's recursive "function" for the form $a(n_{i - 1}) + b (n_{i - 2})$ where $a = 1$ and $b = 1$.
But how do we know for sure that 1, 2, 3, 4, _, _, etc.. can't have a recursive function? where there is a $a$ or $b$ that works for that sequence?
linear-algebra
linear-algebra
asked Dec 3 '18 at 8:51
ming
3165
asked Dec 3 '18 at 8:51
ming
3165
edited Dec 3 '18 at 9:40
asked Dec 3 '18 at 8:51
ming
3165
asked Dec 3 '18 at 8:51
ming
3165
asked Dec 3 '18 at 8:51
ming
3165
3165
Argh I don't know how to make more than one character go in a subscript can someone help me out? Thanks!
– ming
Dec 3 '18 at 8:52
to make multiple subscript charachters you can do : normal text_{your subscript text} which gives: $$normal text_{your subscript text}$$ Place all characters inside brackets
– TheD0ubleT
Dec 3 '18 at 8:59
Fancy, thanks!!
– ming
Dec 3 '18 at 9:40
add a comment |
Argh I don't know how to make more than one character go in a subscript can someone help me out? Thanks!
– ming
Dec 3 '18 at 8:52
to make multiple subscript charachters you can do : normal text_{your subscript text} which gives: $$normal text_{your subscript text}$$ Place all characters inside brackets
– TheD0ubleT
Dec 3 '18 at 8:59
Fancy, thanks!!
– ming
Dec 3 '18 at 9:40
Argh I don't know how to make more than one character go in a subscript can someone help me out? Thanks!
– ming
Dec 3 '18 at 8:52
Argh I don't know how to make more than one character go in a subscript can someone help me out? Thanks!
– ming
Dec 3 '18 at 8:52
to make multiple subscript charachters you can do : normal text_{your subscript text} which gives: $$normal text_{your subscript text}$$ Place all characters inside brackets
– TheD0ubleT
Dec 3 '18 at 8:59
to make multiple subscript charachters you can do : normal text_{your subscript text} which gives: $$normal text_{your subscript text}$$ Place all characters inside brackets
– TheD0ubleT
Dec 3 '18 at 8:59
Fancy, thanks!!
– ming
Dec 3 '18 at 9:40
Fancy, thanks!!
– ming
Dec 3 '18 at 9:40
add a comment |
1 Answer
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You might want to look into : Solving homogeneous linear recurrence relations with constant coefficients.
We can easily find the solutions for this kind of linear recurrence relation:
If $x^2-ax-b = 0$ has 2 real solutions $r_1$ and $r_2$ then the sequence looks like : $u_n=A*r_1^n + B*r_2^n$
If $x^2-ax-b = 0$ has 1 real solution $r$ then the sequence looks like : $u_n=(n*A+B)*r^n $
If $x^2-ax-b = 0$ has 2 complex solutions $r_1$ and $r_2$ then the sequence looks like : $u_n=|r|^n*(Acos(ntheta)+Bsin(ntheta))$ with $theta = arg(r)$
As you can see there is always r^n which prevents linear sequences like $1,2,3,4,5...$ (except if $r = 1$)
In that case your characteristic polynomial looks like $(r-1)^2=0$, thus your recursive relation looks like $u_n=2u_{n-1}-u_{n-2}$
Also, you'll need $A = 1$ and $B = 0$.
that means $u_0=0$ and $u_1=1$
Note: I can make a more in depth answer but is is very well described in the article i have linked
answered Dec 3 '18 at 9:44
TheD0ubleT
39218
add a comment |
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You might want to look into : Solving homogeneous linear recurrence relations with constant coefficients.
We can easily find the solutions for this kind of linear recurrence relation:
If $x^2-ax-b = 0$ has 2 real solutions $r_1$ and $r_2$ then the sequence looks like : $u_n=A*r_1^n + B*r_2^n$
If $x^2-ax-b = 0$ has 1 real solution $r$ then the sequence looks like : $u_n=(n*A+B)*r^n $
If $x^2-ax-b = 0$ has 2 complex solutions $r_1$ and $r_2$ then the sequence looks like : $u_n=|r|^n*(Acos(ntheta)+Bsin(ntheta))$ with $theta = arg(r)$
As you can see there is always r^n which prevents linear sequences like $1,2,3,4,5...$ (except if $r = 1$)
In that case your characteristic polynomial looks like $(r-1)^2=0$, thus your recursive relation looks like $u_n=2u_{n-1}-u_{n-2}$
Also, you'll need $A = 1$ and $B = 0$.
that means $u_0=0$ and $u_1=1$
Note: I can make a more in depth answer but is is very well described in the article i have linked
answered Dec 3 '18 at 9:44
TheD0ubleT
39218
add a comment |
You might want to look into : Solving homogeneous linear recurrence relations with constant coefficients.
We can easily find the solutions for this kind of linear recurrence relation:
If $x^2-ax-b = 0$ has 2 real solutions $r_1$ and $r_2$ then the sequence looks like : $u_n=A*r_1^n + B*r_2^n$
If $x^2-ax-b = 0$ has 1 real solution $r$ then the sequence looks like : $u_n=(n*A+B)*r^n $
If $x^2-ax-b = 0$ has 2 complex solutions $r_1$ and $r_2$ then the sequence looks like : $u_n=|r|^n*(Acos(ntheta)+Bsin(ntheta))$ with $theta = arg(r)$
As you can see there is always r^n which prevents linear sequences like $1,2,3,4,5...$ (except if $r = 1$)
In that case your characteristic polynomial looks like $(r-1)^2=0$, thus your recursive relation looks like $u_n=2u_{n-1}-u_{n-2}$
Also, you'll need $A = 1$ and $B = 0$.
that means $u_0=0$ and $u_1=1$
Note: I can make a more in depth answer but is is very well described in the article i have linked
answered Dec 3 '18 at 9:44
TheD0ubleT
39218
add a comment |
You might want to look into : Solving homogeneous linear recurrence relations with constant coefficients.
We can easily find the solutions for this kind of linear recurrence relation:
If $x^2-ax-b = 0$ has 2 real solutions $r_1$ and $r_2$ then the sequence looks like : $u_n=A*r_1^n + B*r_2^n$
If $x^2-ax-b = 0$ has 1 real solution $r$ then the sequence looks like : $u_n=(n*A+B)*r^n $
If $x^2-ax-b = 0$ has 2 complex solutions $r_1$ and $r_2$ then the sequence looks like : $u_n=|r|^n*(Acos(ntheta)+Bsin(ntheta))$ with $theta = arg(r)$
As you can see there is always r^n which prevents linear sequences like $1,2,3,4,5...$ (except if $r = 1$)
In that case your characteristic polynomial looks like $(r-1)^2=0$, thus your recursive relation looks like $u_n=2u_{n-1}-u_{n-2}$
Also, you'll need $A = 1$ and $B = 0$.
that means $u_0=0$ and $u_1=1$
Note: I can make a more in depth answer but is is very well described in the article i have linked
answered Dec 3 '18 at 9:44
TheD0ubleT
39218
You might want to look into : Solving homogeneous linear recurrence relations with constant coefficients.
We can easily find the solutions for this kind of linear recurrence relation:
If $x^2-ax-b = 0$ has 2 real solutions $r_1$ and $r_2$ then the sequence looks like : $u_n=A*r_1^n + B*r_2^n$
If $x^2-ax-b = 0$ has 1 real solution $r$ then the sequence looks like : $u_n=(n*A+B)*r^n $
If $x^2-ax-b = 0$ has 2 complex solutions $r_1$ and $r_2$ then the sequence looks like : $u_n=|r|^n*(Acos(ntheta)+Bsin(ntheta))$ with $theta = arg(r)$
As you can see there is always r^n which prevents linear sequences like $1,2,3,4,5...$ (except if $r = 1$)
In that case your characteristic polynomial looks like $(r-1)^2=0$, thus your recursive relation looks like $u_n=2u_{n-1}-u_{n-2}$
Also, you'll need $A = 1$ and $B = 0$.
that means $u_0=0$ and $u_1=1$
Note: I can make a more in depth answer but is is very well described in the article i have linked
answered Dec 3 '18 at 9:44
TheD0ubleT
39218
answered Dec 3 '18 at 9:44
TheD0ubleT
39218
answered Dec 3 '18 at 9:44
TheD0ubleT
39218
answered Dec 3 '18 at 9:44
TheD0ubleT
39218
39218
add a comment |
add a comment |
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Argh I don't know how to make more than one character go in a subscript can someone help me out? Thanks!
– ming
Dec 3 '18 at 8:52
to make multiple subscript charachters you can do : normal text_{your subscript text} which gives: $$normal text_{your subscript text}$$ Place all characters inside brackets
– TheD0ubleT
Dec 3 '18 at 8:59
Fancy, thanks!!
– ming
Dec 3 '18 at 9:40