Prove that $x_n+1=frac14-x_n$ converges [duplicate]

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  Prove recursively defined sequence converges
  
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Let $x_1 =3.$ Prove that $$x_n+1=frac14-x_n$$ converges.

The sequence seems to be converging to $1/4$ as the first few terms are $$3,1,frac13,frac311,frac1141,frac41153, cdots$$

In order to prove convergence I know I must make use of the definition that a sequence $(x_n)$ converges to $x$ if there exists an $N$ for all $n ge N Rightarrow |x_n-x|lt epsilon$. But how do I choose $N$ as a function of $epsilon$? What can I assume? Could someone show me what a proof for this looks like? Thanks.

real-analysis sequences-and-series

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asked Jul 21 at 17:23

Red

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marked as duplicate by Martin R, Parcly Taxel, Nosrati, rtybase, Andres Mejia Jul 21 at 18:34

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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  Also: math.stackexchange.com/q/1095493/42969, math.stackexchange.com/q/971104/42969, math.stackexchange.com/q/2195654/42969 – all found with Approach0
  – Martin R
  Jul 21 at 17:32
  
  

  
  
  
  

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up vote
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This question already has an answer here:

• 
  Prove recursively defined sequence converges
  
  1 answer
  
  

Let $x_1 =3.$ Prove that $$x_n+1=frac14-x_n$$ converges.

The sequence seems to be converging to $1/4$ as the first few terms are $$3,1,frac13,frac311,frac1141,frac41153, cdots$$

In order to prove convergence I know I must make use of the definition that a sequence $(x_n)$ converges to $x$ if there exists an $N$ for all $n ge N Rightarrow |x_n-x|lt epsilon$. But how do I choose $N$ as a function of $epsilon$? What can I assume? Could someone show me what a proof for this looks like? Thanks.

real-analysis sequences-and-series

share|cite|improve this question

asked Jul 21 at 17:23

Red

1,747733

marked as duplicate by Martin R, Parcly Taxel, Nosrati, rtybase, Andres Mejia Jul 21 at 18:34

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Also: math.stackexchange.com/q/1095493/42969, math.stackexchange.com/q/971104/42969, math.stackexchange.com/q/2195654/42969 – all found with Approach0
  – Martin R
  Jul 21 at 17:32
  
  

  
  
  
  

add a comment | 

up vote
1
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favorite

up vote
1
down vote

favorite

This question already has an answer here:

• 
  Prove recursively defined sequence converges
  
  1 answer
  
  

Let $x_1 =3.$ Prove that $$x_n+1=frac14-x_n$$ converges.

The sequence seems to be converging to $1/4$ as the first few terms are $$3,1,frac13,frac311,frac1141,frac41153, cdots$$

In order to prove convergence I know I must make use of the definition that a sequence $(x_n)$ converges to $x$ if there exists an $N$ for all $n ge N Rightarrow |x_n-x|lt epsilon$. But how do I choose $N$ as a function of $epsilon$? What can I assume? Could someone show me what a proof for this looks like? Thanks.

real-analysis sequences-and-series

share|cite|improve this question

asked Jul 21 at 17:23

Red

1,747733

This question already has an answer here:

• 
  Prove recursively defined sequence converges
  
  1 answer
  
  

Let $x_1 =3.$ Prove that $$x_n+1=frac14-x_n$$ converges.

The sequence seems to be converging to $1/4$ as the first few terms are $$3,1,frac13,frac311,frac1141,frac41153, cdots$$

In order to prove convergence I know I must make use of the definition that a sequence $(x_n)$ converges to $x$ if there exists an $N$ for all $n ge N Rightarrow |x_n-x|lt epsilon$. But how do I choose $N$ as a function of $epsilon$? What can I assume? Could someone show me what a proof for this looks like? Thanks.

This question already has an answer here:

• 
  Prove recursively defined sequence converges
  
  1 answer
  
  

real-analysis sequences-and-series

share|cite|improve this question

asked Jul 21 at 17:23

Red

1,747733

share|cite|improve this question

share|cite|improve this question

asked Jul 21 at 17:23

Red

1,747733

asked Jul 21 at 17:23

Red

1,747733

asked Jul 21 at 17:23

Red

1,747733

1,747733

marked as duplicate by Martin R, Parcly Taxel, Nosrati, rtybase, Andres Mejia Jul 21 at 18:34

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

marked as duplicate by Martin R, Parcly Taxel, Nosrati, rtybase, Andres Mejia Jul 21 at 18:34

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Also: math.stackexchange.com/q/1095493/42969, math.stackexchange.com/q/971104/42969, math.stackexchange.com/q/2195654/42969 – all found with Approach0
  – Martin R
  Jul 21 at 17:32
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Also: math.stackexchange.com/q/1095493/42969, math.stackexchange.com/q/971104/42969, math.stackexchange.com/q/2195654/42969 – all found with Approach0
  – Martin R
  Jul 21 at 17:32
  
  

  
  
  
  

Also: math.stackexchange.com/q/1095493/42969, math.stackexchange.com/q/971104/42969, math.stackexchange.com/q/2195654/42969 – all found with Approach0
– Martin R
Jul 21 at 17:32

Also: math.stackexchange.com/q/1095493/42969, math.stackexchange.com/q/971104/42969, math.stackexchange.com/q/2195654/42969 – all found with Approach0
– Martin R
Jul 21 at 17:32

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3 Answers
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1
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accepted

Step 1

Show that $$0<x_n leq 3tag1$$ for $n=1,2,cdots.$

Proof

Obviously, $(1)$ holds for $n=1$. Assume that $(1)$ holds for $n=k$, namely, $0<x_k leq 3$, then $$0<frac14<x_k+1=frac14-x_k leq 1 leq 3,$$which implies $(1)$ also holds for $n=k+1$. Thus, by mathematical induction, $(1)$ holds for all $n=1,2,cdots.$

Step 2

Show that $$x_n+1<x_ntag2$$ for $n=1,2,cdots.$

Proof

Notice that $x_1=3, x_2=1.$ Hence $x_2<x_1,$ which implies that $(2)$ holds for $n=1$. Assume that $(2)$ holds for $n=k$, namely, $x_k+1<x_k$, then $$x_k+2=frac14-x_k+1<frac14-x_k=x_k+1,$$which implies $(2)$ also holds for $n=k+1$. Thus, by mathematical induction, $(2)$ holds for all $n=1,2,cdots.$

Combining the two steps, by monotone convergence theorem, we may claim $x_n$ has a limit, which could be denoted as $x.$ Thus, taking the simultaneous limits of both sides of the recursion formula, we have$$x=frac14-x.$$Hence, $x=2 pm sqrt3.$ But $x_n leq 3$, hence $x leq 3,$ which implies that $2+sqrt3>3$ does not satisfy the requirement. As a result, $$x=2-sqrt3,$$ which is desired limit.

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answered Jul 21 at 18:30

mengdie1982

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First show that $x_n+1 < x_n$ and note that $x_n geq 0$. Since the sequence is bounded below and decreasing, it must converge to some number $x_infty$. We then note $$lim_n toinfty x_n+1 = lim_ntoinftyfrac14 - x_n iff x_infty = frac14 - x_infty iff x_infty(4-x_infty) = 1 iff x_infty = 2 pm sqrt3.$$

Since the sequence is decreasing and $x_2 = 1$, we must have that the sequence converges to $2 - sqrt3$.

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answered Jul 21 at 17:29

Marcus M

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If such a limit say $l$ exists we must have$$l=dfrac14-l$$or $$4l-l^2=1$$which gives us $$l=2pm sqrt 3$$since all the terms except the first 2 ones are below $1$ the limit $2+sqrt3$ is invalid, then we need to show that the sequence converges to $2-sqrt 3$. Define $$e_n=x_n-(2-sqrt 3)$$therefore$$e_n+1=x_n+1-2+sqrt 3=dfrac14-x_n-2+sqrt 3=dfrac1-4(2-sqrt 3)+x_n(2-sqrt 3)4-x_n=dfrac-7+4sqrt 3+x_n(2-sqrt 3)4-x_n=(2-sqrt 3)dfrace_n4-x_n$$since for $n>2$ we have $x_n<1$ by substitution we obtain$$|e_n+1|<dfrac2-sqrt 33|e_n|$$ which means that $e_nto 0$ or $x_nto 2-sqrt 3$

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answered Jul 21 at 17:38

Mostafa Ayaz

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3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
1
down vote



accepted

Step 1

Show that $$0<x_n leq 3tag1$$ for $n=1,2,cdots.$

Proof

Obviously, $(1)$ holds for $n=1$. Assume that $(1)$ holds for $n=k$, namely, $0<x_k leq 3$, then $$0<frac14<x_k+1=frac14-x_k leq 1 leq 3,$$which implies $(1)$ also holds for $n=k+1$. Thus, by mathematical induction, $(1)$ holds for all $n=1,2,cdots.$

Step 2

Show that $$x_n+1<x_ntag2$$ for $n=1,2,cdots.$

Proof

Notice that $x_1=3, x_2=1.$ Hence $x_2<x_1,$ which implies that $(2)$ holds for $n=1$. Assume that $(2)$ holds for $n=k$, namely, $x_k+1<x_k$, then $$x_k+2=frac14-x_k+1<frac14-x_k=x_k+1,$$which implies $(2)$ also holds for $n=k+1$. Thus, by mathematical induction, $(2)$ holds for all $n=1,2,cdots.$

Combining the two steps, by monotone convergence theorem, we may claim $x_n$ has a limit, which could be denoted as $x.$ Thus, taking the simultaneous limits of both sides of the recursion formula, we have$$x=frac14-x.$$Hence, $x=2 pm sqrt3.$ But $x_n leq 3$, hence $x leq 3,$ which implies that $2+sqrt3>3$ does not satisfy the requirement. As a result, $$x=2-sqrt3,$$ which is desired limit.

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answered Jul 21 at 18:30

mengdie1982

2,912216

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up vote
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down vote



accepted

Step 1

Show that $$0<x_n leq 3tag1$$ for $n=1,2,cdots.$

Proof

Obviously, $(1)$ holds for $n=1$. Assume that $(1)$ holds for $n=k$, namely, $0<x_k leq 3$, then $$0<frac14<x_k+1=frac14-x_k leq 1 leq 3,$$which implies $(1)$ also holds for $n=k+1$. Thus, by mathematical induction, $(1)$ holds for all $n=1,2,cdots.$

Step 2

Show that $$x_n+1<x_ntag2$$ for $n=1,2,cdots.$

Proof

Notice that $x_1=3, x_2=1.$ Hence $x_2<x_1,$ which implies that $(2)$ holds for $n=1$. Assume that $(2)$ holds for $n=k$, namely, $x_k+1<x_k$, then $$x_k+2=frac14-x_k+1<frac14-x_k=x_k+1,$$which implies $(2)$ also holds for $n=k+1$. Thus, by mathematical induction, $(2)$ holds for all $n=1,2,cdots.$

Combining the two steps, by monotone convergence theorem, we may claim $x_n$ has a limit, which could be denoted as $x.$ Thus, taking the simultaneous limits of both sides of the recursion formula, we have$$x=frac14-x.$$Hence, $x=2 pm sqrt3.$ But $x_n leq 3$, hence $x leq 3,$ which implies that $2+sqrt3>3$ does not satisfy the requirement. As a result, $$x=2-sqrt3,$$ which is desired limit.

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answered Jul 21 at 18:30

mengdie1982

2,912216

add a comment | 

up vote
1
down vote



accepted

up vote
1
down vote



accepted

Step 1

Show that $$0<x_n leq 3tag1$$ for $n=1,2,cdots.$

Proof

Obviously, $(1)$ holds for $n=1$. Assume that $(1)$ holds for $n=k$, namely, $0<x_k leq 3$, then $$0<frac14<x_k+1=frac14-x_k leq 1 leq 3,$$which implies $(1)$ also holds for $n=k+1$. Thus, by mathematical induction, $(1)$ holds for all $n=1,2,cdots.$

Step 2

Show that $$x_n+1<x_ntag2$$ for $n=1,2,cdots.$

Proof

Notice that $x_1=3, x_2=1.$ Hence $x_2<x_1,$ which implies that $(2)$ holds for $n=1$. Assume that $(2)$ holds for $n=k$, namely, $x_k+1<x_k$, then $$x_k+2=frac14-x_k+1<frac14-x_k=x_k+1,$$which implies $(2)$ also holds for $n=k+1$. Thus, by mathematical induction, $(2)$ holds for all $n=1,2,cdots.$

Combining the two steps, by monotone convergence theorem, we may claim $x_n$ has a limit, which could be denoted as $x.$ Thus, taking the simultaneous limits of both sides of the recursion formula, we have$$x=frac14-x.$$Hence, $x=2 pm sqrt3.$ But $x_n leq 3$, hence $x leq 3,$ which implies that $2+sqrt3>3$ does not satisfy the requirement. As a result, $$x=2-sqrt3,$$ which is desired limit.

share|cite|improve this answer

answered Jul 21 at 18:30

mengdie1982

2,912216

Step 1

Show that $$0<x_n leq 3tag1$$ for $n=1,2,cdots.$

Proof

Obviously, $(1)$ holds for $n=1$. Assume that $(1)$ holds for $n=k$, namely, $0<x_k leq 3$, then $$0<frac14<x_k+1=frac14-x_k leq 1 leq 3,$$which implies $(1)$ also holds for $n=k+1$. Thus, by mathematical induction, $(1)$ holds for all $n=1,2,cdots.$

Step 2

Show that $$x_n+1<x_ntag2$$ for $n=1,2,cdots.$

Proof

Notice that $x_1=3, x_2=1.$ Hence $x_2<x_1,$ which implies that $(2)$ holds for $n=1$. Assume that $(2)$ holds for $n=k$, namely, $x_k+1<x_k$, then $$x_k+2=frac14-x_k+1<frac14-x_k=x_k+1,$$which implies $(2)$ also holds for $n=k+1$. Thus, by mathematical induction, $(2)$ holds for all $n=1,2,cdots.$

Combining the two steps, by monotone convergence theorem, we may claim $x_n$ has a limit, which could be denoted as $x.$ Thus, taking the simultaneous limits of both sides of the recursion formula, we have$$x=frac14-x.$$Hence, $x=2 pm sqrt3.$ But $x_n leq 3$, hence $x leq 3,$ which implies that $2+sqrt3>3$ does not satisfy the requirement. As a result, $$x=2-sqrt3,$$ which is desired limit.

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answered Jul 21 at 18:30

mengdie1982

2,912216

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share|cite|improve this answer

answered Jul 21 at 18:30

mengdie1982

2,912216

answered Jul 21 at 18:30

mengdie1982

2,912216

answered Jul 21 at 18:30

mengdie1982

2,912216

2,912216

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add a comment | 

up vote
3
down vote

First show that $x_n+1 < x_n$ and note that $x_n geq 0$. Since the sequence is bounded below and decreasing, it must converge to some number $x_infty$. We then note $$lim_n toinfty x_n+1 = lim_ntoinftyfrac14 - x_n iff x_infty = frac14 - x_infty iff x_infty(4-x_infty) = 1 iff x_infty = 2 pm sqrt3.$$

Since the sequence is decreasing and $x_2 = 1$, we must have that the sequence converges to $2 - sqrt3$.

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answered Jul 21 at 17:29

Marcus M

8,1731847

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up vote
3
down vote

First show that $x_n+1 < x_n$ and note that $x_n geq 0$. Since the sequence is bounded below and decreasing, it must converge to some number $x_infty$. We then note $$lim_n toinfty x_n+1 = lim_ntoinftyfrac14 - x_n iff x_infty = frac14 - x_infty iff x_infty(4-x_infty) = 1 iff x_infty = 2 pm sqrt3.$$

Since the sequence is decreasing and $x_2 = 1$, we must have that the sequence converges to $2 - sqrt3$.

share|cite|improve this answer

answered Jul 21 at 17:29

Marcus M

8,1731847

add a comment | 

up vote
3
down vote

up vote
3
down vote

First show that $x_n+1 < x_n$ and note that $x_n geq 0$. Since the sequence is bounded below and decreasing, it must converge to some number $x_infty$. We then note $$lim_n toinfty x_n+1 = lim_ntoinftyfrac14 - x_n iff x_infty = frac14 - x_infty iff x_infty(4-x_infty) = 1 iff x_infty = 2 pm sqrt3.$$

Since the sequence is decreasing and $x_2 = 1$, we must have that the sequence converges to $2 - sqrt3$.

share|cite|improve this answer

answered Jul 21 at 17:29

Marcus M

8,1731847

First show that $x_n+1 < x_n$ and note that $x_n geq 0$. Since the sequence is bounded below and decreasing, it must converge to some number $x_infty$. We then note $$lim_n toinfty x_n+1 = lim_ntoinftyfrac14 - x_n iff x_infty = frac14 - x_infty iff x_infty(4-x_infty) = 1 iff x_infty = 2 pm sqrt3.$$

Since the sequence is decreasing and $x_2 = 1$, we must have that the sequence converges to $2 - sqrt3$.

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answered Jul 21 at 17:29

Marcus M

8,1731847

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share|cite|improve this answer

answered Jul 21 at 17:29

Marcus M

8,1731847

answered Jul 21 at 17:29

Marcus M

8,1731847

answered Jul 21 at 17:29

Marcus M

8,1731847

8,1731847

add a comment | 

add a comment | 

up vote
1
down vote

If such a limit say $l$ exists we must have$$l=dfrac14-l$$or $$4l-l^2=1$$which gives us $$l=2pm sqrt 3$$since all the terms except the first 2 ones are below $1$ the limit $2+sqrt3$ is invalid, then we need to show that the sequence converges to $2-sqrt 3$. Define $$e_n=x_n-(2-sqrt 3)$$therefore$$e_n+1=x_n+1-2+sqrt 3=dfrac14-x_n-2+sqrt 3=dfrac1-4(2-sqrt 3)+x_n(2-sqrt 3)4-x_n=dfrac-7+4sqrt 3+x_n(2-sqrt 3)4-x_n=(2-sqrt 3)dfrace_n4-x_n$$since for $n>2$ we have $x_n<1$ by substitution we obtain$$|e_n+1|<dfrac2-sqrt 33|e_n|$$ which means that $e_nto 0$ or $x_nto 2-sqrt 3$

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answered Jul 21 at 17:38

Mostafa Ayaz

8,5773630

add a comment | 

up vote
1
down vote

If such a limit say $l$ exists we must have$$l=dfrac14-l$$or $$4l-l^2=1$$which gives us $$l=2pm sqrt 3$$since all the terms except the first 2 ones are below $1$ the limit $2+sqrt3$ is invalid, then we need to show that the sequence converges to $2-sqrt 3$. Define $$e_n=x_n-(2-sqrt 3)$$therefore$$e_n+1=x_n+1-2+sqrt 3=dfrac14-x_n-2+sqrt 3=dfrac1-4(2-sqrt 3)+x_n(2-sqrt 3)4-x_n=dfrac-7+4sqrt 3+x_n(2-sqrt 3)4-x_n=(2-sqrt 3)dfrace_n4-x_n$$since for $n>2$ we have $x_n<1$ by substitution we obtain$$|e_n+1|<dfrac2-sqrt 33|e_n|$$ which means that $e_nto 0$ or $x_nto 2-sqrt 3$

share|cite|improve this answer

answered Jul 21 at 17:38

Mostafa Ayaz

8,5773630

add a comment | 

up vote
1
down vote

up vote
1
down vote

If such a limit say $l$ exists we must have$$l=dfrac14-l$$or $$4l-l^2=1$$which gives us $$l=2pm sqrt 3$$since all the terms except the first 2 ones are below $1$ the limit $2+sqrt3$ is invalid, then we need to show that the sequence converges to $2-sqrt 3$. Define $$e_n=x_n-(2-sqrt 3)$$therefore$$e_n+1=x_n+1-2+sqrt 3=dfrac14-x_n-2+sqrt 3=dfrac1-4(2-sqrt 3)+x_n(2-sqrt 3)4-x_n=dfrac-7+4sqrt 3+x_n(2-sqrt 3)4-x_n=(2-sqrt 3)dfrace_n4-x_n$$since for $n>2$ we have $x_n<1$ by substitution we obtain$$|e_n+1|<dfrac2-sqrt 33|e_n|$$ which means that $e_nto 0$ or $x_nto 2-sqrt 3$

share|cite|improve this answer

answered Jul 21 at 17:38

Mostafa Ayaz

8,5773630

If such a limit say $l$ exists we must have$$l=dfrac14-l$$or $$4l-l^2=1$$which gives us $$l=2pm sqrt 3$$since all the terms except the first 2 ones are below $1$ the limit $2+sqrt3$ is invalid, then we need to show that the sequence converges to $2-sqrt 3$. Define $$e_n=x_n-(2-sqrt 3)$$therefore$$e_n+1=x_n+1-2+sqrt 3=dfrac14-x_n-2+sqrt 3=dfrac1-4(2-sqrt 3)+x_n(2-sqrt 3)4-x_n=dfrac-7+4sqrt 3+x_n(2-sqrt 3)4-x_n=(2-sqrt 3)dfrace_n4-x_n$$since for $n>2$ we have $x_n<1$ by substitution we obtain$$|e_n+1|<dfrac2-sqrt 33|e_n|$$ which means that $e_nto 0$ or $x_nto 2-sqrt 3$

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answered Jul 21 at 17:38

Mostafa Ayaz

8,5773630

share|cite|improve this answer

share|cite|improve this answer

answered Jul 21 at 17:38

Mostafa Ayaz

8,5773630

answered Jul 21 at 17:38

Mostafa Ayaz

8,5773630

answered Jul 21 at 17:38

Mostafa Ayaz

8,5773630

8,5773630

add a comment | 

add a comment |