Proving the convergence of the sequence defined by $x_1=3$ and $x_{n+1}=frac{1}{4-x_n}$
$begingroup$
Consider the sequence defined by
$$x_1=3 quadtext{and}quad x_{n+1}=dfrac{1}{4-x_n}$$
I can calculate limit by assuming limit exist and solving quadratic equation, but I first wanted to give existence of limit.
I tried to show given sequence is decreasing and bounded below by 0.
I used derivative test as
$$f^prime(x)=frac{1}{(4-x)^2}$$
but form this I am not able to show
Also, I tried to shoe $x_{n+1}-x_n<0$ but that way also I am not succeed.
Please tell me how to approach such problem
real-analysis sequences-and-series convergence
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add a comment |
$begingroup$
Consider the sequence defined by
$$x_1=3 quadtext{and}quad x_{n+1}=dfrac{1}{4-x_n}$$
I can calculate limit by assuming limit exist and solving quadratic equation, but I first wanted to give existence of limit.
I tried to show given sequence is decreasing and bounded below by 0.
I used derivative test as
$$f^prime(x)=frac{1}{(4-x)^2}$$
but form this I am not able to show
Also, I tried to shoe $x_{n+1}-x_n<0$ but that way also I am not succeed.
Please tell me how to approach such problem
real-analysis sequences-and-series convergence
$endgroup$
add a comment |
$begingroup$
Consider the sequence defined by
$$x_1=3 quadtext{and}quad x_{n+1}=dfrac{1}{4-x_n}$$
I can calculate limit by assuming limit exist and solving quadratic equation, but I first wanted to give existence of limit.
I tried to show given sequence is decreasing and bounded below by 0.
I used derivative test as
$$f^prime(x)=frac{1}{(4-x)^2}$$
but form this I am not able to show
Also, I tried to shoe $x_{n+1}-x_n<0$ but that way also I am not succeed.
Please tell me how to approach such problem
real-analysis sequences-and-series convergence
$endgroup$
Consider the sequence defined by
$$x_1=3 quadtext{and}quad x_{n+1}=dfrac{1}{4-x_n}$$
I can calculate limit by assuming limit exist and solving quadratic equation, but I first wanted to give existence of limit.
I tried to show given sequence is decreasing and bounded below by 0.
I used derivative test as
$$f^prime(x)=frac{1}{(4-x)^2}$$
but form this I am not able to show
Also, I tried to shoe $x_{n+1}-x_n<0$ but that way also I am not succeed.
Please tell me how to approach such problem
real-analysis sequences-and-series convergence
real-analysis sequences-and-series convergence
edited Dec 2 '18 at 5:56
Blue
47.8k870152
47.8k870152
asked Dec 2 '18 at 5:48
MathLoverMathLover
49310
49310
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add a comment |
5 Answers
5
active
oldest
votes
$begingroup$
You will want to show $x>f(x)$ in order to establish that the sequence is decreasing. So consider $g(x) = x - dfrac{1}{4-x}.$
Since $g'(x) = 1-dfrac{1}{(4-x)^2} =dfrac{(x-5)(x-3)}{(4-x)^2},$ it suggest that your intuition true if $x_nleq 3$ for $n>1$, which can be easily established by induction.
$endgroup$
$begingroup$
1. The formula for $g'(x)$ is wrong. 2. Computing $g'(x)$ is not necessary at all to solve the question. (But two upvotes and an accepted answer in 1 hour...)
$endgroup$
– Did
Dec 2 '18 at 9:21
$begingroup$
Since my first comment, a silent correction of $g'(x)$ occurred. Point 2. still stands, fully.
$endgroup$
– Did
Dec 2 '18 at 10:59
$begingroup$
"This comment is rude or condescending. Learn more in our Code of Conduct."
$endgroup$
– Did
Dec 2 '18 at 21:22
add a comment |
$begingroup$
It can be approached in a graphical manner:
- Draw the graph of $y = frac{1}{4-x}$ to scale while marking the essentials.
- Asymptote at $x=4$; Value at $x = 3$ is $1$.
- Comparing it to the previous value of the sequence would require the plot of $y=x$ on the same axes.
- Mark the intersection as $x=2-sqrt3$ whereas $x=2+sqrt3$ is near $x=4$.
If through, notice that starting the sequence from $x=3$ means that the next value is $1$ from the hyperbola which is well below the straight line. Now to get the next value put $x=1$ and get the next value from the hyperbola, which is again less than $1$ as the straight line depicts.
If you follow the pattern, you would tend to reach the intersection $x=2-sqrt3$ as the gap between both the curves decreases to zero which gives the limit of the sequence as $x=2-sqrt3$ (the limit only, not one of the terms of the sequence, since these are all rational numbers).
Also, one can thus say that if $x_1 in (2-sqrt3,2+sqrt3)$ then the sequence would be decreasing and would converge at $x=2-sqrt3$ and that all the terms $x in (2-sqrt3,2+sqrt3)$.
$endgroup$
1
$begingroup$
Why the downvotes? Because this post explains a simple, illuminating, rigorous, graphical approach which avoids basically any formula and is definitely worth knowing?
$endgroup$
– Did
Dec 2 '18 at 10:37
$begingroup$
The approved answer about which I had condemned in my answer was edited. People thought that it was correct already so they downvoted. I think it is worth knowing also because I feel the other answers are fishy!
$endgroup$
– Sameer Baheti
Dec 2 '18 at 10:38
$begingroup$
You mean, your answer was downvoted because you (correctly) pointed out a serious flaw in another post? This would be outrageous.
$endgroup$
– Did
Dec 2 '18 at 10:42
$begingroup$
It got edited before people saw the flaw I pointed. That would be the only reason.
$endgroup$
– Sameer Baheti
Dec 2 '18 at 10:44
add a comment |
$begingroup$
Suppose $p(y)=y^2-4y+1=0$ and let $x_n=y+z_n$ then $$y+z_{n+1}=frac 1{4-y-z_n}$$ so that $$-yz_n+(4-y)z_{n+1}=y^2-4y+1=0$$ and $$z_{n+1}=frac y{4-y}z_n$$
So if $0lt ylt 2$ the error term $z_n$ has the same sign and is decreasing in magnitude in constant proportion so tends to zero. It is easy to test that there is a root of the quadratic in the required range by noting $p(0)=1, p(2)=-3$
Note: the equations hold for the other possible value of $y$ too, but the inequality does not then apply to prove that the error term tends to zero. Also this was done without computing $y$.
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add a comment |
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To show that $x_n$ is decreasing, use induction to prove that $x_n> x_{n+1}$. To show that $x_n$ is bounded below by zero show a stronger assertion that $0<x_n<2+sqrt{3}$ by using induction.
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You do not need to invoke $2+sqrt{3}$. Use $x_1=3$ as the upper bound and accept $le$ for the comparison with this bound.
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– Oscar Lanzi
Dec 2 '18 at 10:49
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Yes, i realized it later. Thanks :).
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– user9077
Dec 2 '18 at 15:11
add a comment |
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Hint: after finding the limit $l$ from $l={1over 4-l}$ to make sure that the sequence tends to $l$, define $e_n=a_n-l$ and by substituting it in $a_{n+1}={1over 4-a_n}$ conclude that $e_n to 0$
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add a comment |
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You will want to show $x>f(x)$ in order to establish that the sequence is decreasing. So consider $g(x) = x - dfrac{1}{4-x}.$
Since $g'(x) = 1-dfrac{1}{(4-x)^2} =dfrac{(x-5)(x-3)}{(4-x)^2},$ it suggest that your intuition true if $x_nleq 3$ for $n>1$, which can be easily established by induction.
$endgroup$
$begingroup$
1. The formula for $g'(x)$ is wrong. 2. Computing $g'(x)$ is not necessary at all to solve the question. (But two upvotes and an accepted answer in 1 hour...)
$endgroup$
– Did
Dec 2 '18 at 9:21
$begingroup$
Since my first comment, a silent correction of $g'(x)$ occurred. Point 2. still stands, fully.
$endgroup$
– Did
Dec 2 '18 at 10:59
$begingroup$
"This comment is rude or condescending. Learn more in our Code of Conduct."
$endgroup$
– Did
Dec 2 '18 at 21:22
add a comment |
$begingroup$
You will want to show $x>f(x)$ in order to establish that the sequence is decreasing. So consider $g(x) = x - dfrac{1}{4-x}.$
Since $g'(x) = 1-dfrac{1}{(4-x)^2} =dfrac{(x-5)(x-3)}{(4-x)^2},$ it suggest that your intuition true if $x_nleq 3$ for $n>1$, which can be easily established by induction.
$endgroup$
$begingroup$
1. The formula for $g'(x)$ is wrong. 2. Computing $g'(x)$ is not necessary at all to solve the question. (But two upvotes and an accepted answer in 1 hour...)
$endgroup$
– Did
Dec 2 '18 at 9:21
$begingroup$
Since my first comment, a silent correction of $g'(x)$ occurred. Point 2. still stands, fully.
$endgroup$
– Did
Dec 2 '18 at 10:59
$begingroup$
"This comment is rude or condescending. Learn more in our Code of Conduct."
$endgroup$
– Did
Dec 2 '18 at 21:22
add a comment |
$begingroup$
You will want to show $x>f(x)$ in order to establish that the sequence is decreasing. So consider $g(x) = x - dfrac{1}{4-x}.$
Since $g'(x) = 1-dfrac{1}{(4-x)^2} =dfrac{(x-5)(x-3)}{(4-x)^2},$ it suggest that your intuition true if $x_nleq 3$ for $n>1$, which can be easily established by induction.
$endgroup$
You will want to show $x>f(x)$ in order to establish that the sequence is decreasing. So consider $g(x) = x - dfrac{1}{4-x}.$
Since $g'(x) = 1-dfrac{1}{(4-x)^2} =dfrac{(x-5)(x-3)}{(4-x)^2},$ it suggest that your intuition true if $x_nleq 3$ for $n>1$, which can be easily established by induction.
edited Dec 2 '18 at 9:27
answered Dec 2 '18 at 6:09
dezdichadodezdichado
6,3161929
6,3161929
$begingroup$
1. The formula for $g'(x)$ is wrong. 2. Computing $g'(x)$ is not necessary at all to solve the question. (But two upvotes and an accepted answer in 1 hour...)
$endgroup$
– Did
Dec 2 '18 at 9:21
$begingroup$
Since my first comment, a silent correction of $g'(x)$ occurred. Point 2. still stands, fully.
$endgroup$
– Did
Dec 2 '18 at 10:59
$begingroup$
"This comment is rude or condescending. Learn more in our Code of Conduct."
$endgroup$
– Did
Dec 2 '18 at 21:22
add a comment |
$begingroup$
1. The formula for $g'(x)$ is wrong. 2. Computing $g'(x)$ is not necessary at all to solve the question. (But two upvotes and an accepted answer in 1 hour...)
$endgroup$
– Did
Dec 2 '18 at 9:21
$begingroup$
Since my first comment, a silent correction of $g'(x)$ occurred. Point 2. still stands, fully.
$endgroup$
– Did
Dec 2 '18 at 10:59
$begingroup$
"This comment is rude or condescending. Learn more in our Code of Conduct."
$endgroup$
– Did
Dec 2 '18 at 21:22
$begingroup$
1. The formula for $g'(x)$ is wrong. 2. Computing $g'(x)$ is not necessary at all to solve the question. (But two upvotes and an accepted answer in 1 hour...)
$endgroup$
– Did
Dec 2 '18 at 9:21
$begingroup$
1. The formula for $g'(x)$ is wrong. 2. Computing $g'(x)$ is not necessary at all to solve the question. (But two upvotes and an accepted answer in 1 hour...)
$endgroup$
– Did
Dec 2 '18 at 9:21
$begingroup$
Since my first comment, a silent correction of $g'(x)$ occurred. Point 2. still stands, fully.
$endgroup$
– Did
Dec 2 '18 at 10:59
$begingroup$
Since my first comment, a silent correction of $g'(x)$ occurred. Point 2. still stands, fully.
$endgroup$
– Did
Dec 2 '18 at 10:59
$begingroup$
"This comment is rude or condescending. Learn more in our Code of Conduct."
$endgroup$
– Did
Dec 2 '18 at 21:22
$begingroup$
"This comment is rude or condescending. Learn more in our Code of Conduct."
$endgroup$
– Did
Dec 2 '18 at 21:22
add a comment |
$begingroup$
It can be approached in a graphical manner:
- Draw the graph of $y = frac{1}{4-x}$ to scale while marking the essentials.
- Asymptote at $x=4$; Value at $x = 3$ is $1$.
- Comparing it to the previous value of the sequence would require the plot of $y=x$ on the same axes.
- Mark the intersection as $x=2-sqrt3$ whereas $x=2+sqrt3$ is near $x=4$.
If through, notice that starting the sequence from $x=3$ means that the next value is $1$ from the hyperbola which is well below the straight line. Now to get the next value put $x=1$ and get the next value from the hyperbola, which is again less than $1$ as the straight line depicts.
If you follow the pattern, you would tend to reach the intersection $x=2-sqrt3$ as the gap between both the curves decreases to zero which gives the limit of the sequence as $x=2-sqrt3$ (the limit only, not one of the terms of the sequence, since these are all rational numbers).
Also, one can thus say that if $x_1 in (2-sqrt3,2+sqrt3)$ then the sequence would be decreasing and would converge at $x=2-sqrt3$ and that all the terms $x in (2-sqrt3,2+sqrt3)$.
$endgroup$
1
$begingroup$
Why the downvotes? Because this post explains a simple, illuminating, rigorous, graphical approach which avoids basically any formula and is definitely worth knowing?
$endgroup$
– Did
Dec 2 '18 at 10:37
$begingroup$
The approved answer about which I had condemned in my answer was edited. People thought that it was correct already so they downvoted. I think it is worth knowing also because I feel the other answers are fishy!
$endgroup$
– Sameer Baheti
Dec 2 '18 at 10:38
$begingroup$
You mean, your answer was downvoted because you (correctly) pointed out a serious flaw in another post? This would be outrageous.
$endgroup$
– Did
Dec 2 '18 at 10:42
$begingroup$
It got edited before people saw the flaw I pointed. That would be the only reason.
$endgroup$
– Sameer Baheti
Dec 2 '18 at 10:44
add a comment |
$begingroup$
It can be approached in a graphical manner:
- Draw the graph of $y = frac{1}{4-x}$ to scale while marking the essentials.
- Asymptote at $x=4$; Value at $x = 3$ is $1$.
- Comparing it to the previous value of the sequence would require the plot of $y=x$ on the same axes.
- Mark the intersection as $x=2-sqrt3$ whereas $x=2+sqrt3$ is near $x=4$.
If through, notice that starting the sequence from $x=3$ means that the next value is $1$ from the hyperbola which is well below the straight line. Now to get the next value put $x=1$ and get the next value from the hyperbola, which is again less than $1$ as the straight line depicts.
If you follow the pattern, you would tend to reach the intersection $x=2-sqrt3$ as the gap between both the curves decreases to zero which gives the limit of the sequence as $x=2-sqrt3$ (the limit only, not one of the terms of the sequence, since these are all rational numbers).
Also, one can thus say that if $x_1 in (2-sqrt3,2+sqrt3)$ then the sequence would be decreasing and would converge at $x=2-sqrt3$ and that all the terms $x in (2-sqrt3,2+sqrt3)$.
$endgroup$
1
$begingroup$
Why the downvotes? Because this post explains a simple, illuminating, rigorous, graphical approach which avoids basically any formula and is definitely worth knowing?
$endgroup$
– Did
Dec 2 '18 at 10:37
$begingroup$
The approved answer about which I had condemned in my answer was edited. People thought that it was correct already so they downvoted. I think it is worth knowing also because I feel the other answers are fishy!
$endgroup$
– Sameer Baheti
Dec 2 '18 at 10:38
$begingroup$
You mean, your answer was downvoted because you (correctly) pointed out a serious flaw in another post? This would be outrageous.
$endgroup$
– Did
Dec 2 '18 at 10:42
$begingroup$
It got edited before people saw the flaw I pointed. That would be the only reason.
$endgroup$
– Sameer Baheti
Dec 2 '18 at 10:44
add a comment |
$begingroup$
It can be approached in a graphical manner:
- Draw the graph of $y = frac{1}{4-x}$ to scale while marking the essentials.
- Asymptote at $x=4$; Value at $x = 3$ is $1$.
- Comparing it to the previous value of the sequence would require the plot of $y=x$ on the same axes.
- Mark the intersection as $x=2-sqrt3$ whereas $x=2+sqrt3$ is near $x=4$.
If through, notice that starting the sequence from $x=3$ means that the next value is $1$ from the hyperbola which is well below the straight line. Now to get the next value put $x=1$ and get the next value from the hyperbola, which is again less than $1$ as the straight line depicts.
If you follow the pattern, you would tend to reach the intersection $x=2-sqrt3$ as the gap between both the curves decreases to zero which gives the limit of the sequence as $x=2-sqrt3$ (the limit only, not one of the terms of the sequence, since these are all rational numbers).
Also, one can thus say that if $x_1 in (2-sqrt3,2+sqrt3)$ then the sequence would be decreasing and would converge at $x=2-sqrt3$ and that all the terms $x in (2-sqrt3,2+sqrt3)$.
$endgroup$
It can be approached in a graphical manner:
- Draw the graph of $y = frac{1}{4-x}$ to scale while marking the essentials.
- Asymptote at $x=4$; Value at $x = 3$ is $1$.
- Comparing it to the previous value of the sequence would require the plot of $y=x$ on the same axes.
- Mark the intersection as $x=2-sqrt3$ whereas $x=2+sqrt3$ is near $x=4$.
If through, notice that starting the sequence from $x=3$ means that the next value is $1$ from the hyperbola which is well below the straight line. Now to get the next value put $x=1$ and get the next value from the hyperbola, which is again less than $1$ as the straight line depicts.
If you follow the pattern, you would tend to reach the intersection $x=2-sqrt3$ as the gap between both the curves decreases to zero which gives the limit of the sequence as $x=2-sqrt3$ (the limit only, not one of the terms of the sequence, since these are all rational numbers).
Also, one can thus say that if $x_1 in (2-sqrt3,2+sqrt3)$ then the sequence would be decreasing and would converge at $x=2-sqrt3$ and that all the terms $x in (2-sqrt3,2+sqrt3)$.
edited Dec 2 '18 at 11:06
answered Dec 2 '18 at 7:21
Sameer BahetiSameer Baheti
5168
5168
1
$begingroup$
Why the downvotes? Because this post explains a simple, illuminating, rigorous, graphical approach which avoids basically any formula and is definitely worth knowing?
$endgroup$
– Did
Dec 2 '18 at 10:37
$begingroup$
The approved answer about which I had condemned in my answer was edited. People thought that it was correct already so they downvoted. I think it is worth knowing also because I feel the other answers are fishy!
$endgroup$
– Sameer Baheti
Dec 2 '18 at 10:38
$begingroup$
You mean, your answer was downvoted because you (correctly) pointed out a serious flaw in another post? This would be outrageous.
$endgroup$
– Did
Dec 2 '18 at 10:42
$begingroup$
It got edited before people saw the flaw I pointed. That would be the only reason.
$endgroup$
– Sameer Baheti
Dec 2 '18 at 10:44
add a comment |
1
$begingroup$
Why the downvotes? Because this post explains a simple, illuminating, rigorous, graphical approach which avoids basically any formula and is definitely worth knowing?
$endgroup$
– Did
Dec 2 '18 at 10:37
$begingroup$
The approved answer about which I had condemned in my answer was edited. People thought that it was correct already so they downvoted. I think it is worth knowing also because I feel the other answers are fishy!
$endgroup$
– Sameer Baheti
Dec 2 '18 at 10:38
$begingroup$
You mean, your answer was downvoted because you (correctly) pointed out a serious flaw in another post? This would be outrageous.
$endgroup$
– Did
Dec 2 '18 at 10:42
$begingroup$
It got edited before people saw the flaw I pointed. That would be the only reason.
$endgroup$
– Sameer Baheti
Dec 2 '18 at 10:44
1
1
$begingroup$
Why the downvotes? Because this post explains a simple, illuminating, rigorous, graphical approach which avoids basically any formula and is definitely worth knowing?
$endgroup$
– Did
Dec 2 '18 at 10:37
$begingroup$
Why the downvotes? Because this post explains a simple, illuminating, rigorous, graphical approach which avoids basically any formula and is definitely worth knowing?
$endgroup$
– Did
Dec 2 '18 at 10:37
$begingroup$
The approved answer about which I had condemned in my answer was edited. People thought that it was correct already so they downvoted. I think it is worth knowing also because I feel the other answers are fishy!
$endgroup$
– Sameer Baheti
Dec 2 '18 at 10:38
$begingroup$
The approved answer about which I had condemned in my answer was edited. People thought that it was correct already so they downvoted. I think it is worth knowing also because I feel the other answers are fishy!
$endgroup$
– Sameer Baheti
Dec 2 '18 at 10:38
$begingroup$
You mean, your answer was downvoted because you (correctly) pointed out a serious flaw in another post? This would be outrageous.
$endgroup$
– Did
Dec 2 '18 at 10:42
$begingroup$
You mean, your answer was downvoted because you (correctly) pointed out a serious flaw in another post? This would be outrageous.
$endgroup$
– Did
Dec 2 '18 at 10:42
$begingroup$
It got edited before people saw the flaw I pointed. That would be the only reason.
$endgroup$
– Sameer Baheti
Dec 2 '18 at 10:44
$begingroup$
It got edited before people saw the flaw I pointed. That would be the only reason.
$endgroup$
– Sameer Baheti
Dec 2 '18 at 10:44
add a comment |
$begingroup$
Suppose $p(y)=y^2-4y+1=0$ and let $x_n=y+z_n$ then $$y+z_{n+1}=frac 1{4-y-z_n}$$ so that $$-yz_n+(4-y)z_{n+1}=y^2-4y+1=0$$ and $$z_{n+1}=frac y{4-y}z_n$$
So if $0lt ylt 2$ the error term $z_n$ has the same sign and is decreasing in magnitude in constant proportion so tends to zero. It is easy to test that there is a root of the quadratic in the required range by noting $p(0)=1, p(2)=-3$
Note: the equations hold for the other possible value of $y$ too, but the inequality does not then apply to prove that the error term tends to zero. Also this was done without computing $y$.
$endgroup$
add a comment |
$begingroup$
Suppose $p(y)=y^2-4y+1=0$ and let $x_n=y+z_n$ then $$y+z_{n+1}=frac 1{4-y-z_n}$$ so that $$-yz_n+(4-y)z_{n+1}=y^2-4y+1=0$$ and $$z_{n+1}=frac y{4-y}z_n$$
So if $0lt ylt 2$ the error term $z_n$ has the same sign and is decreasing in magnitude in constant proportion so tends to zero. It is easy to test that there is a root of the quadratic in the required range by noting $p(0)=1, p(2)=-3$
Note: the equations hold for the other possible value of $y$ too, but the inequality does not then apply to prove that the error term tends to zero. Also this was done without computing $y$.
$endgroup$
add a comment |
$begingroup$
Suppose $p(y)=y^2-4y+1=0$ and let $x_n=y+z_n$ then $$y+z_{n+1}=frac 1{4-y-z_n}$$ so that $$-yz_n+(4-y)z_{n+1}=y^2-4y+1=0$$ and $$z_{n+1}=frac y{4-y}z_n$$
So if $0lt ylt 2$ the error term $z_n$ has the same sign and is decreasing in magnitude in constant proportion so tends to zero. It is easy to test that there is a root of the quadratic in the required range by noting $p(0)=1, p(2)=-3$
Note: the equations hold for the other possible value of $y$ too, but the inequality does not then apply to prove that the error term tends to zero. Also this was done without computing $y$.
$endgroup$
Suppose $p(y)=y^2-4y+1=0$ and let $x_n=y+z_n$ then $$y+z_{n+1}=frac 1{4-y-z_n}$$ so that $$-yz_n+(4-y)z_{n+1}=y^2-4y+1=0$$ and $$z_{n+1}=frac y{4-y}z_n$$
So if $0lt ylt 2$ the error term $z_n$ has the same sign and is decreasing in magnitude in constant proportion so tends to zero. It is easy to test that there is a root of the quadratic in the required range by noting $p(0)=1, p(2)=-3$
Note: the equations hold for the other possible value of $y$ too, but the inequality does not then apply to prove that the error term tends to zero. Also this was done without computing $y$.
edited Dec 2 '18 at 6:56
answered Dec 2 '18 at 6:49
Mark BennetMark Bennet
80.8k981179
80.8k981179
add a comment |
add a comment |
$begingroup$
To show that $x_n$ is decreasing, use induction to prove that $x_n> x_{n+1}$. To show that $x_n$ is bounded below by zero show a stronger assertion that $0<x_n<2+sqrt{3}$ by using induction.
$endgroup$
$begingroup$
You do not need to invoke $2+sqrt{3}$. Use $x_1=3$ as the upper bound and accept $le$ for the comparison with this bound.
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– Oscar Lanzi
Dec 2 '18 at 10:49
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Yes, i realized it later. Thanks :).
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– user9077
Dec 2 '18 at 15:11
add a comment |
$begingroup$
To show that $x_n$ is decreasing, use induction to prove that $x_n> x_{n+1}$. To show that $x_n$ is bounded below by zero show a stronger assertion that $0<x_n<2+sqrt{3}$ by using induction.
$endgroup$
$begingroup$
You do not need to invoke $2+sqrt{3}$. Use $x_1=3$ as the upper bound and accept $le$ for the comparison with this bound.
$endgroup$
– Oscar Lanzi
Dec 2 '18 at 10:49
$begingroup$
Yes, i realized it later. Thanks :).
$endgroup$
– user9077
Dec 2 '18 at 15:11
add a comment |
$begingroup$
To show that $x_n$ is decreasing, use induction to prove that $x_n> x_{n+1}$. To show that $x_n$ is bounded below by zero show a stronger assertion that $0<x_n<2+sqrt{3}$ by using induction.
$endgroup$
To show that $x_n$ is decreasing, use induction to prove that $x_n> x_{n+1}$. To show that $x_n$ is bounded below by zero show a stronger assertion that $0<x_n<2+sqrt{3}$ by using induction.
edited Dec 2 '18 at 7:37
answered Dec 2 '18 at 6:25
user9077user9077
1,279612
1,279612
$begingroup$
You do not need to invoke $2+sqrt{3}$. Use $x_1=3$ as the upper bound and accept $le$ for the comparison with this bound.
$endgroup$
– Oscar Lanzi
Dec 2 '18 at 10:49
$begingroup$
Yes, i realized it later. Thanks :).
$endgroup$
– user9077
Dec 2 '18 at 15:11
add a comment |
$begingroup$
You do not need to invoke $2+sqrt{3}$. Use $x_1=3$ as the upper bound and accept $le$ for the comparison with this bound.
$endgroup$
– Oscar Lanzi
Dec 2 '18 at 10:49
$begingroup$
Yes, i realized it later. Thanks :).
$endgroup$
– user9077
Dec 2 '18 at 15:11
$begingroup$
You do not need to invoke $2+sqrt{3}$. Use $x_1=3$ as the upper bound and accept $le$ for the comparison with this bound.
$endgroup$
– Oscar Lanzi
Dec 2 '18 at 10:49
$begingroup$
You do not need to invoke $2+sqrt{3}$. Use $x_1=3$ as the upper bound and accept $le$ for the comparison with this bound.
$endgroup$
– Oscar Lanzi
Dec 2 '18 at 10:49
$begingroup$
Yes, i realized it later. Thanks :).
$endgroup$
– user9077
Dec 2 '18 at 15:11
$begingroup$
Yes, i realized it later. Thanks :).
$endgroup$
– user9077
Dec 2 '18 at 15:11
add a comment |
$begingroup$
Hint: after finding the limit $l$ from $l={1over 4-l}$ to make sure that the sequence tends to $l$, define $e_n=a_n-l$ and by substituting it in $a_{n+1}={1over 4-a_n}$ conclude that $e_n to 0$
$endgroup$
add a comment |
$begingroup$
Hint: after finding the limit $l$ from $l={1over 4-l}$ to make sure that the sequence tends to $l$, define $e_n=a_n-l$ and by substituting it in $a_{n+1}={1over 4-a_n}$ conclude that $e_n to 0$
$endgroup$
add a comment |
$begingroup$
Hint: after finding the limit $l$ from $l={1over 4-l}$ to make sure that the sequence tends to $l$, define $e_n=a_n-l$ and by substituting it in $a_{n+1}={1over 4-a_n}$ conclude that $e_n to 0$
$endgroup$
Hint: after finding the limit $l$ from $l={1over 4-l}$ to make sure that the sequence tends to $l$, define $e_n=a_n-l$ and by substituting it in $a_{n+1}={1over 4-a_n}$ conclude that $e_n to 0$
answered Dec 3 '18 at 16:25
Mostafa AyazMostafa Ayaz
15.2k3939
15.2k3939
add a comment |
add a comment |
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