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Recurrence relation $a_n+1=a_n+(b-a_n)cdot x$

Let the sequence be defined recursively $X_n+1=pX_n+q$ for p nonzero and $X_1$ arbitrary. In what conditions does the sequence converge?Solving the recurrence relation $a_n = fracg1-ga_n-1$Solving a recurrence relation, $a_n = sqrtn(n+1)a_n-1 + n!(n+1)^3/2$recurrence relation related problemsWhy does this recurrence relation generate a sinusoidal curve?Linear Homogeneous Recurrence Relations and Inhomogenous Recurrence RelationsAmateur Math and a Linear Recurrence RelationSolving recurrence relation $f(n) = f(lfloorsqrt nrfloor) + 1; f(1) = 1, f(2) = 1$Recurrence with Polynomial Coefficients of $n$Solve complexity of recurrence relationSolving recurrence relation $a_n=3a_n-1-2a_n-2+2^n$

2

2

$begingroup$

I'd like to know how to find a general term for this recurrence:

$$a_n+1=a_n+(b-a_n)cdot x, textwhere b, x are positive constants$$

Background:

Linear interpolation equation for a starting point $a$ and ending point $b$ looks like this: $f(x)=(b-a)cdot x$. I assume $x in [0, 1]$. So for $x=1$ we are at the point $b$.

It is used in computer graphics to move things around :) Very often the parameters of the function are swaped, $x$ becomes constant and $a$ is taken from the previous iteration of a recursion. It gives an effect of homographic-function-like approaching to $b$ (there is a horizontal asymptote in $b$).

I want to find an exact solution for $a_n$. If I'm correct the relation is a following recurrence:

$$a_n+1=a_n+(b-a_n)cdot x$$

I'm curious about the method to get the solution. Wolfram alpha gave me $$a(n)=c_1(1-x)^n-1-b(1-x)^n+b$$

sequences-and-series discrete-mathematics recurrence-relations

share|cite|improve this question

edited yesterday

rtybase

11.4k31533

asked yesterday

LukeNLukeN

286

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Note that $a_n = -qa_n-1 + kq$. This is a linear, nonhomogenous recurrence relation, for which there are general methods to solve it. Some are done in this video - youtube.com/watch?v=EfF_XSEX1Sk. If you would like more specific help, you should edit your OP to include the relevant details - including context, what you know, your attempts at the problem, and what exactly is giving you trouble. (MSE dislikes being used as a site where people just expect us to give solutions to your homework and as posted this question looks exactly like that kind of scenario.)
  $endgroup$
  – Eevee Trainer
  yesterday
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Okay, thank you for your tip. I'm writing the context right now.
  $endgroup$
  – LukeN
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Try this here $$a_n=c_1 (-q)^n-1+frack q left(1-(-q)^nright)q+1$$
  $endgroup$
  – Dr. Sonnhard Graubner
  yesterday
  
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  @Dr.SonnhardGraubner The solution doesn't help if the OP wants to know how to get there.
  $endgroup$
  – Toby Mak
  yesterday
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Seems alright. At least it seems to sufficiently establish from the context that you're not exactly familiar with solving recurrence relations. I maintain that the video I linked would still be a good resource for the general method though, or whatever results you can dig up for solving (non)homogenous linear recurrences (might be a bit of a trawl if you're not familiar with them and depending on how much you need to learn). If I'm feeling up to it (and remember) and no one else has, I might present a write-up specific to this example tomorrow, but for now it's 5 AM for me so I should sleep :p
  $endgroup$
  – Eevee Trainer
  yesterday
  
  

  
  
  
  

 | 
show 2 more comments

2

2

$begingroup$

I'd like to know how to find a general term for this recurrence:

$$a_n+1=a_n+(b-a_n)cdot x, textwhere b, x are positive constants$$

Background:

Linear interpolation equation for a starting point $a$ and ending point $b$ looks like this: $f(x)=(b-a)cdot x$. I assume $x in [0, 1]$. So for $x=1$ we are at the point $b$.

It is used in computer graphics to move things around :) Very often the parameters of the function are swaped, $x$ becomes constant and $a$ is taken from the previous iteration of a recursion. It gives an effect of homographic-function-like approaching to $b$ (there is a horizontal asymptote in $b$).

I want to find an exact solution for $a_n$. If I'm correct the relation is a following recurrence:

$$a_n+1=a_n+(b-a_n)cdot x$$

I'm curious about the method to get the solution. Wolfram alpha gave me $$a(n)=c_1(1-x)^n-1-b(1-x)^n+b$$

sequences-and-series discrete-mathematics recurrence-relations

share|cite|improve this question

edited yesterday

rtybase

11.4k31533

asked yesterday

LukeNLukeN

286

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Note that $a_n = -qa_n-1 + kq$. This is a linear, nonhomogenous recurrence relation, for which there are general methods to solve it. Some are done in this video - youtube.com/watch?v=EfF_XSEX1Sk. If you would like more specific help, you should edit your OP to include the relevant details - including context, what you know, your attempts at the problem, and what exactly is giving you trouble. (MSE dislikes being used as a site where people just expect us to give solutions to your homework and as posted this question looks exactly like that kind of scenario.)
  $endgroup$
  – Eevee Trainer
  yesterday
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Okay, thank you for your tip. I'm writing the context right now.
  $endgroup$
  – LukeN
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Try this here $$a_n=c_1 (-q)^n-1+frack q left(1-(-q)^nright)q+1$$
  $endgroup$
  – Dr. Sonnhard Graubner
  yesterday
  
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  @Dr.SonnhardGraubner The solution doesn't help if the OP wants to know how to get there.
  $endgroup$
  – Toby Mak
  yesterday
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Seems alright. At least it seems to sufficiently establish from the context that you're not exactly familiar with solving recurrence relations. I maintain that the video I linked would still be a good resource for the general method though, or whatever results you can dig up for solving (non)homogenous linear recurrences (might be a bit of a trawl if you're not familiar with them and depending on how much you need to learn). If I'm feeling up to it (and remember) and no one else has, I might present a write-up specific to this example tomorrow, but for now it's 5 AM for me so I should sleep :p
  $endgroup$
  – Eevee Trainer
  yesterday
  
  

  
  
  
  

 | 
show 2 more comments

$begingroup$

I'd like to know how to find a general term for this recurrence:

$$a_n+1=a_n+(b-a_n)cdot x, textwhere b, x are positive constants$$

Background:

Linear interpolation equation for a starting point $a$ and ending point $b$ looks like this: $f(x)=(b-a)cdot x$. I assume $x in [0, 1]$. So for $x=1$ we are at the point $b$.

It is used in computer graphics to move things around :) Very often the parameters of the function are swaped, $x$ becomes constant and $a$ is taken from the previous iteration of a recursion. It gives an effect of homographic-function-like approaching to $b$ (there is a horizontal asymptote in $b$).

I want to find an exact solution for $a_n$. If I'm correct the relation is a following recurrence:

$$a_n+1=a_n+(b-a_n)cdot x$$

I'm curious about the method to get the solution. Wolfram alpha gave me $$a(n)=c_1(1-x)^n-1-b(1-x)^n+b$$

sequences-and-series discrete-mathematics recurrence-relations

share|cite|improve this question

edited yesterday

rtybase

11.4k31533

asked yesterday

LukeNLukeN

286

$endgroup$

share|cite|improve this question

edited yesterday

rtybase

11.4k31533

asked yesterday

LukeNLukeN

286

share|cite|improve this question

edited yesterday

rtybase

11.4k31533

asked yesterday

LukeNLukeN

286

edited yesterday

rtybase

11.4k31533

edited yesterday

rtybase

11.4k31533

asked yesterday

LukeNLukeN

286

asked yesterday

LukeNLukeN

286

2 Answers
2

active

oldest

votes

1

$begingroup$

This is equivalent to
$$a_n+1=(1-x)a_n+bcdot x$$
which can be solved using characteristic polynomials. In fact, you can use this ready-to-use result for
$$a_n+1=pa_n+q$$
where $p=(1-x)$ and $q=bx$. The result is
$$a_n=A+Bcdot p^n=A+Bcdot (1-x)^n$$
where $A,B$ are some constants depending on $a_0$. Further in that link
$$a_n=-fracqp-1+left(a_1+fracqp-1right)p^n-1=\
fracbxx+left(a_1-fracbxxright)(1-x)^n-1=
colorredb+left(a_1-bright)(1-x)^n-1=\
b+left((1-x)a_0+bx-bright)(1-x)^n-1=colorredb+left(a_0-bright)(1-x)^n=\
a_0(1-x)(1-x)^n-1-b(1-x)^n+b$$
where $c_1=a_0(1-x)$-constant. I don't know how Wolfram alpha decides on these constants.

share|cite|improve this answer

edited yesterday

answered yesterday

rtybasertybase

11.4k31533

$endgroup$

add a comment | 

0

$begingroup$

This is a so-called arithmetico-geometric sequence.

Notice that

$$a_n+1-b=a_n-b+(b-a_n)x=(1-x)(a_n-b).$$

Then by induction,

$$a_n-b=(1-x)^n(a_0-b),$$

or

$$a_n=(1-x)^n(a_0-b)+b.$$

Except when $x=1$, $a_n$ tends to $b$.

share|cite|improve this answer

edited yesterday

answered yesterday

Yves DaoustYves Daoust

130k676227

$endgroup$

add a comment | 

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1

$begingroup$

This is equivalent to
$$a_n+1=(1-x)a_n+bcdot x$$
which can be solved using characteristic polynomials. In fact, you can use this ready-to-use result for
$$a_n+1=pa_n+q$$
where $p=(1-x)$ and $q=bx$. The result is
$$a_n=A+Bcdot p^n=A+Bcdot (1-x)^n$$
where $A,B$ are some constants depending on $a_0$. Further in that link
$$a_n=-fracqp-1+left(a_1+fracqp-1right)p^n-1=\
fracbxx+left(a_1-fracbxxright)(1-x)^n-1=
colorredb+left(a_1-bright)(1-x)^n-1=\
b+left((1-x)a_0+bx-bright)(1-x)^n-1=colorredb+left(a_0-bright)(1-x)^n=\
a_0(1-x)(1-x)^n-1-b(1-x)^n+b$$
where $c_1=a_0(1-x)$-constant. I don't know how Wolfram alpha decides on these constants.

share|cite|improve this answer

edited yesterday

answered yesterday

rtybasertybase

11.4k31533

$endgroup$

add a comment | 

1

$begingroup$

This is equivalent to
$$a_n+1=(1-x)a_n+bcdot x$$
which can be solved using characteristic polynomials. In fact, you can use this ready-to-use result for
$$a_n+1=pa_n+q$$
where $p=(1-x)$ and $q=bx$. The result is
$$a_n=A+Bcdot p^n=A+Bcdot (1-x)^n$$
where $A,B$ are some constants depending on $a_0$. Further in that link
$$a_n=-fracqp-1+left(a_1+fracqp-1right)p^n-1=\
fracbxx+left(a_1-fracbxxright)(1-x)^n-1=
colorredb+left(a_1-bright)(1-x)^n-1=\
b+left((1-x)a_0+bx-bright)(1-x)^n-1=colorredb+left(a_0-bright)(1-x)^n=\
a_0(1-x)(1-x)^n-1-b(1-x)^n+b$$
where $c_1=a_0(1-x)$-constant. I don't know how Wolfram alpha decides on these constants.

share|cite|improve this answer

edited yesterday

answered yesterday

rtybasertybase

11.4k31533

$endgroup$

add a comment | 

$begingroup$

This is equivalent to
$$a_n+1=(1-x)a_n+bcdot x$$
which can be solved using characteristic polynomials. In fact, you can use this ready-to-use result for
$$a_n+1=pa_n+q$$
where $p=(1-x)$ and $q=bx$. The result is
$$a_n=A+Bcdot p^n=A+Bcdot (1-x)^n$$
where $A,B$ are some constants depending on $a_0$. Further in that link
$$a_n=-fracqp-1+left(a_1+fracqp-1right)p^n-1=\
fracbxx+left(a_1-fracbxxright)(1-x)^n-1=
colorredb+left(a_1-bright)(1-x)^n-1=\
b+left((1-x)a_0+bx-bright)(1-x)^n-1=colorredb+left(a_0-bright)(1-x)^n=\
a_0(1-x)(1-x)^n-1-b(1-x)^n+b$$
where $c_1=a_0(1-x)$-constant. I don't know how Wolfram alpha decides on these constants.

share|cite|improve this answer

edited yesterday

answered yesterday

rtybasertybase

11.4k31533

$endgroup$

share|cite|improve this answer

edited yesterday

answered yesterday

rtybasertybase

11.4k31533

answered yesterday

rtybasertybase

11.4k31533

answered yesterday

rtybasertybase

11.4k31533

0

$begingroup$

This is a so-called arithmetico-geometric sequence.

Notice that

$$a_n+1-b=a_n-b+(b-a_n)x=(1-x)(a_n-b).$$

Then by induction,

$$a_n-b=(1-x)^n(a_0-b),$$

or

$$a_n=(1-x)^n(a_0-b)+b.$$

Except when $x=1$, $a_n$ tends to $b$.

share|cite|improve this answer

edited yesterday

answered yesterday

Yves DaoustYves Daoust

130k676227

$endgroup$

add a comment | 

0

$begingroup$

This is a so-called arithmetico-geometric sequence.

Notice that

$$a_n+1-b=a_n-b+(b-a_n)x=(1-x)(a_n-b).$$

Then by induction,

$$a_n-b=(1-x)^n(a_0-b),$$

or

$$a_n=(1-x)^n(a_0-b)+b.$$

Except when $x=1$, $a_n$ tends to $b$.

share|cite|improve this answer

edited yesterday

answered yesterday

Yves DaoustYves Daoust

130k676227

$endgroup$

add a comment | 

$begingroup$

This is a so-called arithmetico-geometric sequence.

Notice that

$$a_n+1-b=a_n-b+(b-a_n)x=(1-x)(a_n-b).$$

Then by induction,

$$a_n-b=(1-x)^n(a_0-b),$$

or

$$a_n=(1-x)^n(a_0-b)+b.$$

Except when $x=1$, $a_n$ tends to $b$.

share|cite|improve this answer

edited yesterday

answered yesterday

Yves DaoustYves Daoust

130k676227

$endgroup$

share|cite|improve this answer

edited yesterday

answered yesterday

Yves DaoustYves Daoust

130k676227

answered yesterday

Yves DaoustYves Daoust

130k676227

answered yesterday

Yves DaoustYves Daoust

130k676227