Proof by Induction of a Recursively Defined Sequence
$begingroup$
The question I'm currently struggling with is:
Consider the following recursively defined sequence:
$$ a_1 = 0, a_{n+1}= frac{a^2_n+4}{5} $$
a) Show by induction that $0le a_n le 1 $ for all n
b) Prove that if $a_n$ converges, so that $lim_{nto infty}a_n=A$ then $A^2−5A+4 =0$, and thus $A = 1$.
c) Prove by induction that $a_n$ is strictly increasing, $a_{n+1}>a_n$ for all n.
d) Conclude that $lim_{n toinfty}a_n$ exists, and thus that $lim_{n to infty}a_n=1$
I've managed to complete part a) where I did:
Basis - at $n=1$ $a_{1+1}=frac{0+4}{5}$
Inductive Hypothesis - for k as an element of the natural numbers, it is always true that $0 le a_{k+1} le 1$ where $a_{k+1}=frac{a_k+4}{5}$
Inductive Step - Here I tried to prove it by making $n=k+1$ which meant
$a_{k+2}= frac{a_k+1^2+4}{5}$
which I split in to $frac{a_k+1^2}{5}+frac{4}{5}$
and from the inductive hypothesis I know that $0 le a_{k+1} le 1$ and therefore it's evident that $frac{a_k+1^2}{5}$ can take a minimum value of $frac{0}{5}$ and a minimum value of $frac{1}{5}$
Part b) seems relatively simple as you can prove
$A^2−5A+4 =0$ quite easily but I'm unsure how to get the equation in the first place. I think it has something to do with rearranging:
$a_{k+1}= frac{a^2_k+4}{5} $
Part c) is the part I'm struggling with most and the answer in my notes doesn't really explain to me how it's done.
I did the basis and my inductive hypothesis and at the inductive step I got to:
$a_{k+1}^2-5a_{k+1}+4>0$ but I'm not sure what to do after this
Would anyone mind helping me with part c) and d)?
real-analysis sequences-and-series induction
New contributor
$endgroup$
add a comment |
$begingroup$
The question I'm currently struggling with is:
Consider the following recursively defined sequence:
$$ a_1 = 0, a_{n+1}= frac{a^2_n+4}{5} $$
a) Show by induction that $0le a_n le 1 $ for all n
b) Prove that if $a_n$ converges, so that $lim_{nto infty}a_n=A$ then $A^2−5A+4 =0$, and thus $A = 1$.
c) Prove by induction that $a_n$ is strictly increasing, $a_{n+1}>a_n$ for all n.
d) Conclude that $lim_{n toinfty}a_n$ exists, and thus that $lim_{n to infty}a_n=1$
I've managed to complete part a) where I did:
Basis - at $n=1$ $a_{1+1}=frac{0+4}{5}$
Inductive Hypothesis - for k as an element of the natural numbers, it is always true that $0 le a_{k+1} le 1$ where $a_{k+1}=frac{a_k+4}{5}$
Inductive Step - Here I tried to prove it by making $n=k+1$ which meant
$a_{k+2}= frac{a_k+1^2+4}{5}$
which I split in to $frac{a_k+1^2}{5}+frac{4}{5}$
and from the inductive hypothesis I know that $0 le a_{k+1} le 1$ and therefore it's evident that $frac{a_k+1^2}{5}$ can take a minimum value of $frac{0}{5}$ and a minimum value of $frac{1}{5}$
Part b) seems relatively simple as you can prove
$A^2−5A+4 =0$ quite easily but I'm unsure how to get the equation in the first place. I think it has something to do with rearranging:
$a_{k+1}= frac{a^2_k+4}{5} $
Part c) is the part I'm struggling with most and the answer in my notes doesn't really explain to me how it's done.
I did the basis and my inductive hypothesis and at the inductive step I got to:
$a_{k+1}^2-5a_{k+1}+4>0$ but I'm not sure what to do after this
Would anyone mind helping me with part c) and d)?
real-analysis sequences-and-series induction
New contributor
$endgroup$
$begingroup$
@saulspatz I already wrote that
$endgroup$
– greedoid
6 hours ago
add a comment |
$begingroup$
The question I'm currently struggling with is:
Consider the following recursively defined sequence:
$$ a_1 = 0, a_{n+1}= frac{a^2_n+4}{5} $$
a) Show by induction that $0le a_n le 1 $ for all n
b) Prove that if $a_n$ converges, so that $lim_{nto infty}a_n=A$ then $A^2−5A+4 =0$, and thus $A = 1$.
c) Prove by induction that $a_n$ is strictly increasing, $a_{n+1}>a_n$ for all n.
d) Conclude that $lim_{n toinfty}a_n$ exists, and thus that $lim_{n to infty}a_n=1$
I've managed to complete part a) where I did:
Basis - at $n=1$ $a_{1+1}=frac{0+4}{5}$
Inductive Hypothesis - for k as an element of the natural numbers, it is always true that $0 le a_{k+1} le 1$ where $a_{k+1}=frac{a_k+4}{5}$
Inductive Step - Here I tried to prove it by making $n=k+1$ which meant
$a_{k+2}= frac{a_k+1^2+4}{5}$
which I split in to $frac{a_k+1^2}{5}+frac{4}{5}$
and from the inductive hypothesis I know that $0 le a_{k+1} le 1$ and therefore it's evident that $frac{a_k+1^2}{5}$ can take a minimum value of $frac{0}{5}$ and a minimum value of $frac{1}{5}$
Part b) seems relatively simple as you can prove
$A^2−5A+4 =0$ quite easily but I'm unsure how to get the equation in the first place. I think it has something to do with rearranging:
$a_{k+1}= frac{a^2_k+4}{5} $
Part c) is the part I'm struggling with most and the answer in my notes doesn't really explain to me how it's done.
I did the basis and my inductive hypothesis and at the inductive step I got to:
$a_{k+1}^2-5a_{k+1}+4>0$ but I'm not sure what to do after this
Would anyone mind helping me with part c) and d)?
real-analysis sequences-and-series induction
New contributor
$endgroup$
The question I'm currently struggling with is:
Consider the following recursively defined sequence:
$$ a_1 = 0, a_{n+1}= frac{a^2_n+4}{5} $$
a) Show by induction that $0le a_n le 1 $ for all n
b) Prove that if $a_n$ converges, so that $lim_{nto infty}a_n=A$ then $A^2−5A+4 =0$, and thus $A = 1$.
c) Prove by induction that $a_n$ is strictly increasing, $a_{n+1}>a_n$ for all n.
d) Conclude that $lim_{n toinfty}a_n$ exists, and thus that $lim_{n to infty}a_n=1$
I've managed to complete part a) where I did:
Basis - at $n=1$ $a_{1+1}=frac{0+4}{5}$
Inductive Hypothesis - for k as an element of the natural numbers, it is always true that $0 le a_{k+1} le 1$ where $a_{k+1}=frac{a_k+4}{5}$
Inductive Step - Here I tried to prove it by making $n=k+1$ which meant
$a_{k+2}= frac{a_k+1^2+4}{5}$
which I split in to $frac{a_k+1^2}{5}+frac{4}{5}$
and from the inductive hypothesis I know that $0 le a_{k+1} le 1$ and therefore it's evident that $frac{a_k+1^2}{5}$ can take a minimum value of $frac{0}{5}$ and a minimum value of $frac{1}{5}$
Part b) seems relatively simple as you can prove
$A^2−5A+4 =0$ quite easily but I'm unsure how to get the equation in the first place. I think it has something to do with rearranging:
$a_{k+1}= frac{a^2_k+4}{5} $
Part c) is the part I'm struggling with most and the answer in my notes doesn't really explain to me how it's done.
I did the basis and my inductive hypothesis and at the inductive step I got to:
$a_{k+1}^2-5a_{k+1}+4>0$ but I'm not sure what to do after this
Would anyone mind helping me with part c) and d)?
real-analysis sequences-and-series induction
real-analysis sequences-and-series induction
New contributor
New contributor
edited 3 hours ago
greedoid
45k1157112
45k1157112
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asked 6 hours ago
kingking
235
235
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$begingroup$
@saulspatz I already wrote that
$endgroup$
– greedoid
6 hours ago
add a comment |
$begingroup$
@saulspatz I already wrote that
$endgroup$
– greedoid
6 hours ago
$begingroup$
@saulspatz I already wrote that
$endgroup$
– greedoid
6 hours ago
$begingroup$
@saulspatz I already wrote that
$endgroup$
– greedoid
6 hours ago
add a comment |
4 Answers
4
active
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votes
$begingroup$
It is $$a_{k+1}^2-5a_{k+1}+4>0$$ and not $a_{k+1}^2-5a_{k+1}-4>0$. So you have $$(a_{k+1}-1)(a_{k+1}-4)>0$$
which is true by a).
$endgroup$
$begingroup$
Ahh yes thank you that was a typo in my post but I see what you do with it now
$endgroup$
– king
5 hours ago
add a comment |
$begingroup$
The inductive step for the monotonicity (c) is just
$$
a_{n+1} > a_n implies frac{a^2_{n+1}+4}{5} > frac{a^2_n+4}{5} implies a_{n+2} > a_{n+1}
$$
The first implication holds because $a_n ge 0$.
The convergence (d) then follows from the monotone convergence theorem.
$endgroup$
add a comment |
$begingroup$
b) The argument is that at the fixed point, which has the value A,
$$a_{k+1}=a_{k}=A$$
So
$$a_{k+1}=frac{a_k^2+4}{5}$$
becomes
$$A=frac{A^2+4}{5}$$
$$5A=A^2+4$$
$$A^2+5A+4=0$$
New contributor
$endgroup$
add a comment |
$begingroup$
a) By the assumption of the induction we obtain:
$$0leq a_{n+1}=frac{a_n^2+4}{5}<frac{1^2+4}{5}=1.$$
b),c) $$a_{n+1}-a_n=frac{a_n^2+4}{5}-a_n=frac{(a_n-1)(a_n-4)}{5}>0,$$
which says that $a$ increases and bounded.
Thus, there is $limlimits_{nrightarrow+infty}a_n$ and let this limit be equal to $A$.
Thus, by theorems about limits we obtain: $$A^2-5A+4=0,$$ which gives $A=1$.
I like the following way:
$$|a_{n+1}-1|=left|frac{a_n^2+4}{5}-1right|=frac{a_n+1}{5}|a_n-1|<frac{2}{5}|a_n-1|<...<left(frac{2}{5}right)^n|a_1-1|rightarrow0,$$
Which says $$lim_{nrightarrow+infty}a_n=1.$$
$endgroup$
add a comment |
Your Answer
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
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votes
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votes
$begingroup$
It is $$a_{k+1}^2-5a_{k+1}+4>0$$ and not $a_{k+1}^2-5a_{k+1}-4>0$. So you have $$(a_{k+1}-1)(a_{k+1}-4)>0$$
which is true by a).
$endgroup$
$begingroup$
Ahh yes thank you that was a typo in my post but I see what you do with it now
$endgroup$
– king
5 hours ago
add a comment |
$begingroup$
It is $$a_{k+1}^2-5a_{k+1}+4>0$$ and not $a_{k+1}^2-5a_{k+1}-4>0$. So you have $$(a_{k+1}-1)(a_{k+1}-4)>0$$
which is true by a).
$endgroup$
$begingroup$
Ahh yes thank you that was a typo in my post but I see what you do with it now
$endgroup$
– king
5 hours ago
add a comment |
$begingroup$
It is $$a_{k+1}^2-5a_{k+1}+4>0$$ and not $a_{k+1}^2-5a_{k+1}-4>0$. So you have $$(a_{k+1}-1)(a_{k+1}-4)>0$$
which is true by a).
$endgroup$
It is $$a_{k+1}^2-5a_{k+1}+4>0$$ and not $a_{k+1}^2-5a_{k+1}-4>0$. So you have $$(a_{k+1}-1)(a_{k+1}-4)>0$$
which is true by a).
answered 6 hours ago
greedoidgreedoid
45k1157112
45k1157112
$begingroup$
Ahh yes thank you that was a typo in my post but I see what you do with it now
$endgroup$
– king
5 hours ago
add a comment |
$begingroup$
Ahh yes thank you that was a typo in my post but I see what you do with it now
$endgroup$
– king
5 hours ago
$begingroup$
Ahh yes thank you that was a typo in my post but I see what you do with it now
$endgroup$
– king
5 hours ago
$begingroup$
Ahh yes thank you that was a typo in my post but I see what you do with it now
$endgroup$
– king
5 hours ago
add a comment |
$begingroup$
The inductive step for the monotonicity (c) is just
$$
a_{n+1} > a_n implies frac{a^2_{n+1}+4}{5} > frac{a^2_n+4}{5} implies a_{n+2} > a_{n+1}
$$
The first implication holds because $a_n ge 0$.
The convergence (d) then follows from the monotone convergence theorem.
$endgroup$
add a comment |
$begingroup$
The inductive step for the monotonicity (c) is just
$$
a_{n+1} > a_n implies frac{a^2_{n+1}+4}{5} > frac{a^2_n+4}{5} implies a_{n+2} > a_{n+1}
$$
The first implication holds because $a_n ge 0$.
The convergence (d) then follows from the monotone convergence theorem.
$endgroup$
add a comment |
$begingroup$
The inductive step for the monotonicity (c) is just
$$
a_{n+1} > a_n implies frac{a^2_{n+1}+4}{5} > frac{a^2_n+4}{5} implies a_{n+2} > a_{n+1}
$$
The first implication holds because $a_n ge 0$.
The convergence (d) then follows from the monotone convergence theorem.
$endgroup$
The inductive step for the monotonicity (c) is just
$$
a_{n+1} > a_n implies frac{a^2_{n+1}+4}{5} > frac{a^2_n+4}{5} implies a_{n+2} > a_{n+1}
$$
The first implication holds because $a_n ge 0$.
The convergence (d) then follows from the monotone convergence theorem.
edited 5 hours ago
answered 6 hours ago
Martin RMartin R
29.4k33558
29.4k33558
add a comment |
add a comment |
$begingroup$
b) The argument is that at the fixed point, which has the value A,
$$a_{k+1}=a_{k}=A$$
So
$$a_{k+1}=frac{a_k^2+4}{5}$$
becomes
$$A=frac{A^2+4}{5}$$
$$5A=A^2+4$$
$$A^2+5A+4=0$$
New contributor
$endgroup$
add a comment |
$begingroup$
b) The argument is that at the fixed point, which has the value A,
$$a_{k+1}=a_{k}=A$$
So
$$a_{k+1}=frac{a_k^2+4}{5}$$
becomes
$$A=frac{A^2+4}{5}$$
$$5A=A^2+4$$
$$A^2+5A+4=0$$
New contributor
$endgroup$
add a comment |
$begingroup$
b) The argument is that at the fixed point, which has the value A,
$$a_{k+1}=a_{k}=A$$
So
$$a_{k+1}=frac{a_k^2+4}{5}$$
becomes
$$A=frac{A^2+4}{5}$$
$$5A=A^2+4$$
$$A^2+5A+4=0$$
New contributor
$endgroup$
b) The argument is that at the fixed point, which has the value A,
$$a_{k+1}=a_{k}=A$$
So
$$a_{k+1}=frac{a_k^2+4}{5}$$
becomes
$$A=frac{A^2+4}{5}$$
$$5A=A^2+4$$
$$A^2+5A+4=0$$
New contributor
New contributor
answered 6 hours ago
Martin HansenMartin Hansen
687
687
New contributor
New contributor
add a comment |
add a comment |
$begingroup$
a) By the assumption of the induction we obtain:
$$0leq a_{n+1}=frac{a_n^2+4}{5}<frac{1^2+4}{5}=1.$$
b),c) $$a_{n+1}-a_n=frac{a_n^2+4}{5}-a_n=frac{(a_n-1)(a_n-4)}{5}>0,$$
which says that $a$ increases and bounded.
Thus, there is $limlimits_{nrightarrow+infty}a_n$ and let this limit be equal to $A$.
Thus, by theorems about limits we obtain: $$A^2-5A+4=0,$$ which gives $A=1$.
I like the following way:
$$|a_{n+1}-1|=left|frac{a_n^2+4}{5}-1right|=frac{a_n+1}{5}|a_n-1|<frac{2}{5}|a_n-1|<...<left(frac{2}{5}right)^n|a_1-1|rightarrow0,$$
Which says $$lim_{nrightarrow+infty}a_n=1.$$
$endgroup$
add a comment |
$begingroup$
a) By the assumption of the induction we obtain:
$$0leq a_{n+1}=frac{a_n^2+4}{5}<frac{1^2+4}{5}=1.$$
b),c) $$a_{n+1}-a_n=frac{a_n^2+4}{5}-a_n=frac{(a_n-1)(a_n-4)}{5}>0,$$
which says that $a$ increases and bounded.
Thus, there is $limlimits_{nrightarrow+infty}a_n$ and let this limit be equal to $A$.
Thus, by theorems about limits we obtain: $$A^2-5A+4=0,$$ which gives $A=1$.
I like the following way:
$$|a_{n+1}-1|=left|frac{a_n^2+4}{5}-1right|=frac{a_n+1}{5}|a_n-1|<frac{2}{5}|a_n-1|<...<left(frac{2}{5}right)^n|a_1-1|rightarrow0,$$
Which says $$lim_{nrightarrow+infty}a_n=1.$$
$endgroup$
add a comment |
$begingroup$
a) By the assumption of the induction we obtain:
$$0leq a_{n+1}=frac{a_n^2+4}{5}<frac{1^2+4}{5}=1.$$
b),c) $$a_{n+1}-a_n=frac{a_n^2+4}{5}-a_n=frac{(a_n-1)(a_n-4)}{5}>0,$$
which says that $a$ increases and bounded.
Thus, there is $limlimits_{nrightarrow+infty}a_n$ and let this limit be equal to $A$.
Thus, by theorems about limits we obtain: $$A^2-5A+4=0,$$ which gives $A=1$.
I like the following way:
$$|a_{n+1}-1|=left|frac{a_n^2+4}{5}-1right|=frac{a_n+1}{5}|a_n-1|<frac{2}{5}|a_n-1|<...<left(frac{2}{5}right)^n|a_1-1|rightarrow0,$$
Which says $$lim_{nrightarrow+infty}a_n=1.$$
$endgroup$
a) By the assumption of the induction we obtain:
$$0leq a_{n+1}=frac{a_n^2+4}{5}<frac{1^2+4}{5}=1.$$
b),c) $$a_{n+1}-a_n=frac{a_n^2+4}{5}-a_n=frac{(a_n-1)(a_n-4)}{5}>0,$$
which says that $a$ increases and bounded.
Thus, there is $limlimits_{nrightarrow+infty}a_n$ and let this limit be equal to $A$.
Thus, by theorems about limits we obtain: $$A^2-5A+4=0,$$ which gives $A=1$.
I like the following way:
$$|a_{n+1}-1|=left|frac{a_n^2+4}{5}-1right|=frac{a_n+1}{5}|a_n-1|<frac{2}{5}|a_n-1|<...<left(frac{2}{5}right)^n|a_1-1|rightarrow0,$$
Which says $$lim_{nrightarrow+infty}a_n=1.$$
edited 2 hours ago
answered 5 hours ago
Michael RozenbergMichael Rozenberg
105k1893198
105k1893198
add a comment |
add a comment |
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$begingroup$
@saulspatz I already wrote that
$endgroup$
– greedoid
6 hours ago