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Define the recursive sequence $a_n$ by putting $a_1 =0 $ and $ a_n+1 = (frac14)^a_n $, $nge 1$
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Define the recursive sequence $a_n$ by putting $a_1 =0 $ and $ a_n+1 = (frac14)^a_n $, $nge 1$
my solution : $,a_n+1 = a_n = l $ ,$ l = (frac14)^l $,
$ l - (frac14)^l= 0 $
Now i can not proceed Further as how i can find the value of l ??
Pliz help me.....
Any hints/ solution will be appreciated
real-analysis
asked Jul 27 at 2:18
stupid
53518
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up vote
0
down vote
favorite
Define the recursive sequence $a_n$ by putting $a_1 =0 $ and $ a_n+1 = (frac14)^a_n $, $nge 1$
my solution : $,a_n+1 = a_n = l $ ,$ l = (frac14)^l $,
$ l - (frac14)^l= 0 $
Now i can not proceed Further as how i can find the value of l ??
Pliz help me.....
Any hints/ solution will be appreciated
real-analysis
asked Jul 27 at 2:18
stupid
53518
1
a_n+1 = a_n = lYou need to first prove that the limitlexists i.e. that the sequence converges.how i can find the value of lLook for a small semi-obviouslthat works, then prove it is the only solution.
– dxiv
Jul 27 at 2:21
okks @dxiv let me thinks more....
– stupid
Jul 27 at 2:24
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Define the recursive sequence $a_n$ by putting $a_1 =0 $ and $ a_n+1 = (frac14)^a_n $, $nge 1$
my solution : $,a_n+1 = a_n = l $ ,$ l = (frac14)^l $,
$ l - (frac14)^l= 0 $
Now i can not proceed Further as how i can find the value of l ??
Pliz help me.....
Any hints/ solution will be appreciated
real-analysis
asked Jul 27 at 2:18
stupid
53518
Define the recursive sequence $a_n$ by putting $a_1 =0 $ and $ a_n+1 = (frac14)^a_n $, $nge 1$
my solution : $,a_n+1 = a_n = l $ ,$ l = (frac14)^l $,
$ l - (frac14)^l= 0 $
Now i can not proceed Further as how i can find the value of l ??
Pliz help me.....
Any hints/ solution will be appreciated
real-analysis
asked Jul 27 at 2:18
stupid
53518
asked Jul 27 at 2:18
stupid
53518
asked Jul 27 at 2:18
stupid
53518
asked Jul 27 at 2:18
stupid
53518
53518
1
a_n+1 = a_n = lYou need to first prove that the limitlexists i.e. that the sequence converges.how i can find the value of lLook for a small semi-obviouslthat works, then prove it is the only solution.
– dxiv
Jul 27 at 2:21
okks @dxiv let me thinks more....
– stupid
Jul 27 at 2:24
add a comment |Â
1
a_n+1 = a_n = lYou need to first prove that the limitlexists i.e. that the sequence converges.how i can find the value of lLook for a small semi-obviouslthat works, then prove it is the only solution.
– dxiv
Jul 27 at 2:21
okks @dxiv let me thinks more....
– stupid
Jul 27 at 2:24
1
1
a_n+1 = a_n = l You need to first prove that the limit l exists i.e. that the sequence converges. how i can find the value of l Look for a small semi-obvious l that works, then prove it is the only solution.– dxiv
Jul 27 at 2:21
a_n+1 = a_n = l You need to first prove that the limit l exists i.e. that the sequence converges. how i can find the value of l Look for a small semi-obvious l that works, then prove it is the only solution.– dxiv
Jul 27 at 2:21
okks @dxiv let me thinks more....
– stupid
Jul 27 at 2:24
okks @dxiv let me thinks more....
– stupid
Jul 27 at 2:24
add a comment |Â
1 Answer
1
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Notice that the first terms of the sequence are $0,1,1/4,1/4^1/4,1/4^1/4^1/4, etc.$ I will prove it converges using a special case of Banach's fixed-point theorem, in other words, I'll state without proof that
Given $g$ continuously differentiable on $[a,b]$ such that $g([a,b]) subset [a,b]$ and $lambda=sup_aleq x leq b|g'(x)|<1$, then $x=g(x)$ has a unique solution $alpha$ in $[a,b]$ and for any choice of $x_0$ in $[a,b]$, with $x_n+1=g(x_n)$, $ngeq 0$, the sequence $x_n$ converges to $alpha$.
In our case, $g(x)=1/4^x$ and $[a,b]=[1/4,1]$ and it can be shown that all the hypotheses are met (please, verify this). Hence your sequence converges to the only solution of $1/4^x=x$ in the interval $[1/4,1]$, which by a short inspection turns out to be $x=1/2$.
answered Jul 27 at 4:45
SEBASTIAN VARGAS LOAIZA
1464
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Notice that the first terms of the sequence are $0,1,1/4,1/4^1/4,1/4^1/4^1/4, etc.$ I will prove it converges using a special case of Banach's fixed-point theorem, in other words, I'll state without proof that
Given $g$ continuously differentiable on $[a,b]$ such that $g([a,b]) subset [a,b]$ and $lambda=sup_aleq x leq b|g'(x)|<1$, then $x=g(x)$ has a unique solution $alpha$ in $[a,b]$ and for any choice of $x_0$ in $[a,b]$, with $x_n+1=g(x_n)$, $ngeq 0$, the sequence $x_n$ converges to $alpha$.
In our case, $g(x)=1/4^x$ and $[a,b]=[1/4,1]$ and it can be shown that all the hypotheses are met (please, verify this). Hence your sequence converges to the only solution of $1/4^x=x$ in the interval $[1/4,1]$, which by a short inspection turns out to be $x=1/2$.
answered Jul 27 at 4:45
SEBASTIAN VARGAS LOAIZA
1464
add a comment |Â
up vote
2
down vote
accepted
Notice that the first terms of the sequence are $0,1,1/4,1/4^1/4,1/4^1/4^1/4, etc.$ I will prove it converges using a special case of Banach's fixed-point theorem, in other words, I'll state without proof that
Given $g$ continuously differentiable on $[a,b]$ such that $g([a,b]) subset [a,b]$ and $lambda=sup_aleq x leq b|g'(x)|<1$, then $x=g(x)$ has a unique solution $alpha$ in $[a,b]$ and for any choice of $x_0$ in $[a,b]$, with $x_n+1=g(x_n)$, $ngeq 0$, the sequence $x_n$ converges to $alpha$.
In our case, $g(x)=1/4^x$ and $[a,b]=[1/4,1]$ and it can be shown that all the hypotheses are met (please, verify this). Hence your sequence converges to the only solution of $1/4^x=x$ in the interval $[1/4,1]$, which by a short inspection turns out to be $x=1/2$.
answered Jul 27 at 4:45
SEBASTIAN VARGAS LOAIZA
1464
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Notice that the first terms of the sequence are $0,1,1/4,1/4^1/4,1/4^1/4^1/4, etc.$ I will prove it converges using a special case of Banach's fixed-point theorem, in other words, I'll state without proof that
Given $g$ continuously differentiable on $[a,b]$ such that $g([a,b]) subset [a,b]$ and $lambda=sup_aleq x leq b|g'(x)|<1$, then $x=g(x)$ has a unique solution $alpha$ in $[a,b]$ and for any choice of $x_0$ in $[a,b]$, with $x_n+1=g(x_n)$, $ngeq 0$, the sequence $x_n$ converges to $alpha$.
In our case, $g(x)=1/4^x$ and $[a,b]=[1/4,1]$ and it can be shown that all the hypotheses are met (please, verify this). Hence your sequence converges to the only solution of $1/4^x=x$ in the interval $[1/4,1]$, which by a short inspection turns out to be $x=1/2$.
answered Jul 27 at 4:45
SEBASTIAN VARGAS LOAIZA
1464
Notice that the first terms of the sequence are $0,1,1/4,1/4^1/4,1/4^1/4^1/4, etc.$ I will prove it converges using a special case of Banach's fixed-point theorem, in other words, I'll state without proof that
Given $g$ continuously differentiable on $[a,b]$ such that $g([a,b]) subset [a,b]$ and $lambda=sup_aleq x leq b|g'(x)|<1$, then $x=g(x)$ has a unique solution $alpha$ in $[a,b]$ and for any choice of $x_0$ in $[a,b]$, with $x_n+1=g(x_n)$, $ngeq 0$, the sequence $x_n$ converges to $alpha$.
In our case, $g(x)=1/4^x$ and $[a,b]=[1/4,1]$ and it can be shown that all the hypotheses are met (please, verify this). Hence your sequence converges to the only solution of $1/4^x=x$ in the interval $[1/4,1]$, which by a short inspection turns out to be $x=1/2$.
answered Jul 27 at 4:45
SEBASTIAN VARGAS LOAIZA
1464
answered Jul 27 at 4:45
SEBASTIAN VARGAS LOAIZA
1464
answered Jul 27 at 4:45
SEBASTIAN VARGAS LOAIZA
1464
answered Jul 27 at 4:45
SEBASTIAN VARGAS LOAIZA
1464
1464
add a comment |Â
add a comment |Â
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1
a_n+1 = a_n = lYou need to first prove that the limitlexists i.e. that the sequence converges.how i can find the value of lLook for a small semi-obviouslthat works, then prove it is the only solution.– dxiv
Jul 27 at 2:21
okks @dxiv let me thinks more....
– stupid
Jul 27 at 2:24