Fixed point only solution to this sequence? [closed]

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I'm trying to show convergence of a sequence, and I reduced the problem to proving that the fixed point a = a'' = a''' = ... ad infimum is the only solution to this problem.

a' - (1+r)a = a'' - (1+r)a' = a''' - (1+r)a'' = ... ad infimum

If that helps, we can assume that if it's not a fixed point, it has to be monotonic. Any idea? I tried finding a contradiction for both cases a > a' > a'' > ... and the opposite. However, I can't.

Clearly, it is a solution, what I need to show is that it is unique

Any help appreciated. Thanks.

Edit: Delete this, it is clearly not unique...

convergence

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asked Jul 31 at 0:31

john john

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closed as unclear what you're asking by amWhy, Xander Henderson, Isaac Browne, Leucippus, Taroccoesbrocco Jul 31 at 8:31

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

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up vote
0
down vote

favorite

1

I'm trying to show convergence of a sequence, and I reduced the problem to proving that the fixed point a = a'' = a''' = ... ad infimum is the only solution to this problem.

a' - (1+r)a = a'' - (1+r)a' = a''' - (1+r)a'' = ... ad infimum

If that helps, we can assume that if it's not a fixed point, it has to be monotonic. Any idea? I tried finding a contradiction for both cases a > a' > a'' > ... and the opposite. However, I can't.

Clearly, it is a solution, what I need to show is that it is unique

Any help appreciated. Thanks.

Edit: Delete this, it is clearly not unique...

convergence

share|cite|improve this question

asked Jul 31 at 0:31

john john

1

closed as unclear what you're asking by amWhy, Xander Henderson, Isaac Browne, Leucippus, Taroccoesbrocco Jul 31 at 8:31

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

add a comment | 

up vote
0
down vote

favorite

1

up vote
0
down vote

favorite

1

1

I'm trying to show convergence of a sequence, and I reduced the problem to proving that the fixed point a = a'' = a''' = ... ad infimum is the only solution to this problem.

a' - (1+r)a = a'' - (1+r)a' = a''' - (1+r)a'' = ... ad infimum

If that helps, we can assume that if it's not a fixed point, it has to be monotonic. Any idea? I tried finding a contradiction for both cases a > a' > a'' > ... and the opposite. However, I can't.

Clearly, it is a solution, what I need to show is that it is unique

Any help appreciated. Thanks.

Edit: Delete this, it is clearly not unique...

convergence

share|cite|improve this question

asked Jul 31 at 0:31

john john

1

I'm trying to show convergence of a sequence, and I reduced the problem to proving that the fixed point a = a'' = a''' = ... ad infimum is the only solution to this problem.

a' - (1+r)a = a'' - (1+r)a' = a''' - (1+r)a'' = ... ad infimum

If that helps, we can assume that if it's not a fixed point, it has to be monotonic. Any idea? I tried finding a contradiction for both cases a > a' > a'' > ... and the opposite. However, I can't.

Clearly, it is a solution, what I need to show is that it is unique

Any help appreciated. Thanks.

Edit: Delete this, it is clearly not unique...

convergence

share|cite|improve this question

asked Jul 31 at 0:31

john john

1

share|cite|improve this question

share|cite|improve this question

asked Jul 31 at 0:31

john john

1

asked Jul 31 at 0:31

john john

1

asked Jul 31 at 0:31

john john

1

1

closed as unclear what you're asking by amWhy, Xander Henderson, Isaac Browne, Leucippus, Taroccoesbrocco Jul 31 at 8:31

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

closed as unclear what you're asking by amWhy, Xander Henderson, Isaac Browne, Leucippus, Taroccoesbrocco Jul 31 at 8:31

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

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Taking your first equality we see
$$a'-(1+r)a=a''-(1+r)a'\a''=(1+r)a-ra'$$
and similarly with more primes on the $a$s. It would be more conventional to write the variables as $A(n)$ where $n$ is the number of primes and you have a homogeneous recurrence relation
$$A(n+2)=-rA(n+1)+(1+r)A(n)$$
The general solution is to substitute in $A(n)=bk^n$ and get the characteristic equation $$k^2=-rk+(1-r)\k^2+rk-1-r=0\(k-1)(k-(r+1))=0$$
with the solution $$A(n)=c1^n+d(r+1)^n$$
If your initial conditions are carefully chosen, you will have $d=0$ and all the $A(n)=c$

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answered Jul 31 at 3:45

Ross Millikan

275k21184351

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
0
down vote

Taking your first equality we see
$$a'-(1+r)a=a''-(1+r)a'\a''=(1+r)a-ra'$$
and similarly with more primes on the $a$s. It would be more conventional to write the variables as $A(n)$ where $n$ is the number of primes and you have a homogeneous recurrence relation
$$A(n+2)=-rA(n+1)+(1+r)A(n)$$
The general solution is to substitute in $A(n)=bk^n$ and get the characteristic equation $$k^2=-rk+(1-r)\k^2+rk-1-r=0\(k-1)(k-(r+1))=0$$
with the solution $$A(n)=c1^n+d(r+1)^n$$
If your initial conditions are carefully chosen, you will have $d=0$ and all the $A(n)=c$

share|cite|improve this answer

answered Jul 31 at 3:45

Ross Millikan

275k21184351

add a comment | 

up vote
0
down vote

Taking your first equality we see
$$a'-(1+r)a=a''-(1+r)a'\a''=(1+r)a-ra'$$
and similarly with more primes on the $a$s. It would be more conventional to write the variables as $A(n)$ where $n$ is the number of primes and you have a homogeneous recurrence relation
$$A(n+2)=-rA(n+1)+(1+r)A(n)$$
The general solution is to substitute in $A(n)=bk^n$ and get the characteristic equation $$k^2=-rk+(1-r)\k^2+rk-1-r=0\(k-1)(k-(r+1))=0$$
with the solution $$A(n)=c1^n+d(r+1)^n$$
If your initial conditions are carefully chosen, you will have $d=0$ and all the $A(n)=c$

share|cite|improve this answer

answered Jul 31 at 3:45

Ross Millikan

275k21184351

add a comment | 

up vote
0
down vote

up vote
0
down vote

Taking your first equality we see
$$a'-(1+r)a=a''-(1+r)a'\a''=(1+r)a-ra'$$
and similarly with more primes on the $a$s. It would be more conventional to write the variables as $A(n)$ where $n$ is the number of primes and you have a homogeneous recurrence relation
$$A(n+2)=-rA(n+1)+(1+r)A(n)$$
The general solution is to substitute in $A(n)=bk^n$ and get the characteristic equation $$k^2=-rk+(1-r)\k^2+rk-1-r=0\(k-1)(k-(r+1))=0$$
with the solution $$A(n)=c1^n+d(r+1)^n$$
If your initial conditions are carefully chosen, you will have $d=0$ and all the $A(n)=c$

share|cite|improve this answer

answered Jul 31 at 3:45

Ross Millikan

275k21184351

Taking your first equality we see
$$a'-(1+r)a=a''-(1+r)a'\a''=(1+r)a-ra'$$
and similarly with more primes on the $a$s. It would be more conventional to write the variables as $A(n)$ where $n$ is the number of primes and you have a homogeneous recurrence relation
$$A(n+2)=-rA(n+1)+(1+r)A(n)$$
The general solution is to substitute in $A(n)=bk^n$ and get the characteristic equation $$k^2=-rk+(1-r)\k^2+rk-1-r=0\(k-1)(k-(r+1))=0$$
with the solution $$A(n)=c1^n+d(r+1)^n$$
If your initial conditions are carefully chosen, you will have $d=0$ and all the $A(n)=c$

share|cite|improve this answer

answered Jul 31 at 3:45

Ross Millikan

275k21184351

share|cite|improve this answer

share|cite|improve this answer

answered Jul 31 at 3:45

Ross Millikan

275k21184351

answered Jul 31 at 3:45

Ross Millikan

275k21184351

answered Jul 31 at 3:45

Ross Millikan

275k21184351

275k21184351

add a comment | 

add a comment |