Prove the sequence $a_{1} = 4$, $a_{n + 1} = frac{a_{n}}{2} + frac{2}{a_{n}}$, $n = 1, 2, ldots$ satisfies...
$begingroup$
Prove the sequence $a_{1} = 4$, $a_{n + 1} = frac{a_{n}}{2} + frac{2}{a_{n}}$, $n = 1, 2, ldots$ satisfies $a_{n} > 2$
Let $x = a_{n}/2$.
Then $a_{n + 1} = x + 1/x$.
Define $f(x) = x + 1/x$ so that $f'(x) = -1/x^2 + 1 = 0 implies x = 1,$ meaning that $a_{n + 1}$ has a minimum at $1 + 1/1 = 2$.
This shows $a_{n + 1} geq 2$. But I want to show the strict bound $a_{n + 1} > 2$. Perhaps I can prove the case separately. I can't make any progress.
real-analysis calculus sequences-and-series inequality monotone-functions
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add a comment |
$begingroup$
Prove the sequence $a_{1} = 4$, $a_{n + 1} = frac{a_{n}}{2} + frac{2}{a_{n}}$, $n = 1, 2, ldots$ satisfies $a_{n} > 2$
Let $x = a_{n}/2$.
Then $a_{n + 1} = x + 1/x$.
Define $f(x) = x + 1/x$ so that $f'(x) = -1/x^2 + 1 = 0 implies x = 1,$ meaning that $a_{n + 1}$ has a minimum at $1 + 1/1 = 2$.
This shows $a_{n + 1} geq 2$. But I want to show the strict bound $a_{n + 1} > 2$. Perhaps I can prove the case separately. I can't make any progress.
real-analysis calculus sequences-and-series inequality monotone-functions
$endgroup$
1
$begingroup$
Prove $a_n>0$ and use $a+bgeq 2sqrt{ab}$.
$endgroup$
– Lau
Dec 12 '18 at 5:43
$begingroup$
All these and a lot more are covered here, including linked questions ...
$endgroup$
– rtybase
Dec 12 '18 at 9:22
add a comment |
$begingroup$
Prove the sequence $a_{1} = 4$, $a_{n + 1} = frac{a_{n}}{2} + frac{2}{a_{n}}$, $n = 1, 2, ldots$ satisfies $a_{n} > 2$
Let $x = a_{n}/2$.
Then $a_{n + 1} = x + 1/x$.
Define $f(x) = x + 1/x$ so that $f'(x) = -1/x^2 + 1 = 0 implies x = 1,$ meaning that $a_{n + 1}$ has a minimum at $1 + 1/1 = 2$.
This shows $a_{n + 1} geq 2$. But I want to show the strict bound $a_{n + 1} > 2$. Perhaps I can prove the case separately. I can't make any progress.
real-analysis calculus sequences-and-series inequality monotone-functions
$endgroup$
Prove the sequence $a_{1} = 4$, $a_{n + 1} = frac{a_{n}}{2} + frac{2}{a_{n}}$, $n = 1, 2, ldots$ satisfies $a_{n} > 2$
Let $x = a_{n}/2$.
Then $a_{n + 1} = x + 1/x$.
Define $f(x) = x + 1/x$ so that $f'(x) = -1/x^2 + 1 = 0 implies x = 1,$ meaning that $a_{n + 1}$ has a minimum at $1 + 1/1 = 2$.
This shows $a_{n + 1} geq 2$. But I want to show the strict bound $a_{n + 1} > 2$. Perhaps I can prove the case separately. I can't make any progress.
real-analysis calculus sequences-and-series inequality monotone-functions
real-analysis calculus sequences-and-series inequality monotone-functions
edited Dec 12 '18 at 6:31
stackofhay42
asked Dec 12 '18 at 5:37
stackofhay42stackofhay42
1956
1956
1
$begingroup$
Prove $a_n>0$ and use $a+bgeq 2sqrt{ab}$.
$endgroup$
– Lau
Dec 12 '18 at 5:43
$begingroup$
All these and a lot more are covered here, including linked questions ...
$endgroup$
– rtybase
Dec 12 '18 at 9:22
add a comment |
1
$begingroup$
Prove $a_n>0$ and use $a+bgeq 2sqrt{ab}$.
$endgroup$
– Lau
Dec 12 '18 at 5:43
$begingroup$
All these and a lot more are covered here, including linked questions ...
$endgroup$
– rtybase
Dec 12 '18 at 9:22
1
1
$begingroup$
Prove $a_n>0$ and use $a+bgeq 2sqrt{ab}$.
$endgroup$
– Lau
Dec 12 '18 at 5:43
$begingroup$
Prove $a_n>0$ and use $a+bgeq 2sqrt{ab}$.
$endgroup$
– Lau
Dec 12 '18 at 5:43
$begingroup$
All these and a lot more are covered here, including linked questions ...
$endgroup$
– rtybase
Dec 12 '18 at 9:22
$begingroup$
All these and a lot more are covered here, including linked questions ...
$endgroup$
– rtybase
Dec 12 '18 at 9:22
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
Use principle of mathematical induction,
Basis for induction
$$a_1>2$$
Induction Hypothesis
$$a_k>2$$ $$Longleftrightarrow frac{a_k}{2}>1$$
Inductive step
$$a_{k+1}=frac{a_k}{2}+frac{2}{a_k}>2$$
which is true by condition of equality in AM-GM inequality and induction hypothesis.
Hence proved
Hope it is helpful
$endgroup$
add a comment |
$begingroup$
AM-GM gives:
$$frac{a}{2} + frac{2}{a} stackrel{a>0, aneq 2}{>} 2sqrt{frac{a}{2}cdotfrac{2}{a}} = 2$$
Edit after comments:
- Note that $a^{star} = 2$ is a fixpoint of the iteration as $frac{a^{star}}{2} + frac{2}{a^{star}} = 2$.
- For any other starting value $a_0 > 0, a_0 neq 2$ AM-GM tells you, that the iteration $a_{n+1} = frac{a_n}{2} + frac{2}{a_n}$ produces $a_{n+1} > 2$.
So, also in your specific case of $a_1 = 4$ all members of the iteration are greater than $2$.
For a direct proof that all members of the iteration are greater than $2$ you may consider
- $f(x) = frac{x}{2} + frac{2}{x} Rightarrow f'(x) = frac{1}{2} - frac{2}{x^2}$
Now use MVT:
$$f(x) - 2 = f(x) - f(2) = f'(xi)(x - 2) = left(frac{1}{2} - frac{2}{{xi}^2}right) (x - 2) stackrel{2 < xi < x}{>} 0$$
$endgroup$
$begingroup$
how do i know $a neq 2$?
$endgroup$
– stackofhay42
Dec 12 '18 at 6:31
$begingroup$
@stackofhay42: If $a=2$, we t get the AM-GM $textbf{equality}$.
$endgroup$
– Yadati Kiran
Dec 12 '18 at 6:47
$begingroup$
Yes, if $a = 2$, we get AM-GM equality, meaning that we need to show that we cannot have $a = 2$. We never showed that it cannot be true that $a = 2$. I think that this is circular reasoning.
$endgroup$
– stackofhay42
Dec 12 '18 at 6:51
$begingroup$
@RakeshBhatt: By induction, we can say that all of the terms of the given sequence are positive. We know that $a_1>0$. Suppose $a_k>0 ,Longrightarrow a_{k+1}>0$ by the given relation.
$endgroup$
– Martund
Dec 12 '18 at 9:25
$begingroup$
@stackofhay42: We can also prove that by induction. $a_1neq 2, a_kneq 2 Longrightarrow a_{k+1}neq 2$ by the given relation.
$endgroup$
– Martund
Dec 12 '18 at 9:27
add a comment |
$begingroup$
Note $a_n >0$, $n=1,2,3,.....$.
Let $n ge 1$:
$a_{n+1}= dfrac{a_n^2 +4}{2a_n}=$
$dfrac{(a_n-2)^2+4a_n}{2a_n}=$
$dfrac{(a_n-2)^2}{2a_n} + 2 ge 2$.
(The first term $ge 0$, a square divided by a positive number.)
Hence
$a_{n+1} ge 2$, $n in mathbb{N}$.
$endgroup$
add a comment |
$begingroup$
Following with your reasoning, suppose $x + 1/x = 2$. Then:
$$x + 1/x = 2implies x^2 - 2x + 1 = 0implies x = 1.$$
But this is impossible because...
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What about $a_n<2$?
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– Rakesh Bhatt
Dec 12 '18 at 10:14
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@RakeshBhatt, the OP has proved $a_n > 2implies a_{n+1}ge 2$. Only is required excluding the equality.
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– Martín-Blas Pérez Pinilla
Dec 12 '18 at 10:43
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Why is this impossible?
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– Ekesh Kumar
Dec 13 '18 at 23:45
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@Ekesh, because you have supposed $a_n > 2$ ($x > 1$).
$endgroup$
– Martín-Blas Pérez Pinilla
Dec 14 '18 at 7:12
add a comment |
Your Answer
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Use principle of mathematical induction,
Basis for induction
$$a_1>2$$
Induction Hypothesis
$$a_k>2$$ $$Longleftrightarrow frac{a_k}{2}>1$$
Inductive step
$$a_{k+1}=frac{a_k}{2}+frac{2}{a_k}>2$$
which is true by condition of equality in AM-GM inequality and induction hypothesis.
Hence proved
Hope it is helpful
$endgroup$
add a comment |
$begingroup$
Use principle of mathematical induction,
Basis for induction
$$a_1>2$$
Induction Hypothesis
$$a_k>2$$ $$Longleftrightarrow frac{a_k}{2}>1$$
Inductive step
$$a_{k+1}=frac{a_k}{2}+frac{2}{a_k}>2$$
which is true by condition of equality in AM-GM inequality and induction hypothesis.
Hence proved
Hope it is helpful
$endgroup$
add a comment |
$begingroup$
Use principle of mathematical induction,
Basis for induction
$$a_1>2$$
Induction Hypothesis
$$a_k>2$$ $$Longleftrightarrow frac{a_k}{2}>1$$
Inductive step
$$a_{k+1}=frac{a_k}{2}+frac{2}{a_k}>2$$
which is true by condition of equality in AM-GM inequality and induction hypothesis.
Hence proved
Hope it is helpful
$endgroup$
Use principle of mathematical induction,
Basis for induction
$$a_1>2$$
Induction Hypothesis
$$a_k>2$$ $$Longleftrightarrow frac{a_k}{2}>1$$
Inductive step
$$a_{k+1}=frac{a_k}{2}+frac{2}{a_k}>2$$
which is true by condition of equality in AM-GM inequality and induction hypothesis.
Hence proved
Hope it is helpful
answered Dec 12 '18 at 6:08
MartundMartund
1,623213
1,623213
add a comment |
add a comment |
$begingroup$
AM-GM gives:
$$frac{a}{2} + frac{2}{a} stackrel{a>0, aneq 2}{>} 2sqrt{frac{a}{2}cdotfrac{2}{a}} = 2$$
Edit after comments:
- Note that $a^{star} = 2$ is a fixpoint of the iteration as $frac{a^{star}}{2} + frac{2}{a^{star}} = 2$.
- For any other starting value $a_0 > 0, a_0 neq 2$ AM-GM tells you, that the iteration $a_{n+1} = frac{a_n}{2} + frac{2}{a_n}$ produces $a_{n+1} > 2$.
So, also in your specific case of $a_1 = 4$ all members of the iteration are greater than $2$.
For a direct proof that all members of the iteration are greater than $2$ you may consider
- $f(x) = frac{x}{2} + frac{2}{x} Rightarrow f'(x) = frac{1}{2} - frac{2}{x^2}$
Now use MVT:
$$f(x) - 2 = f(x) - f(2) = f'(xi)(x - 2) = left(frac{1}{2} - frac{2}{{xi}^2}right) (x - 2) stackrel{2 < xi < x}{>} 0$$
$endgroup$
$begingroup$
how do i know $a neq 2$?
$endgroup$
– stackofhay42
Dec 12 '18 at 6:31
$begingroup$
@stackofhay42: If $a=2$, we t get the AM-GM $textbf{equality}$.
$endgroup$
– Yadati Kiran
Dec 12 '18 at 6:47
$begingroup$
Yes, if $a = 2$, we get AM-GM equality, meaning that we need to show that we cannot have $a = 2$. We never showed that it cannot be true that $a = 2$. I think that this is circular reasoning.
$endgroup$
– stackofhay42
Dec 12 '18 at 6:51
$begingroup$
@RakeshBhatt: By induction, we can say that all of the terms of the given sequence are positive. We know that $a_1>0$. Suppose $a_k>0 ,Longrightarrow a_{k+1}>0$ by the given relation.
$endgroup$
– Martund
Dec 12 '18 at 9:25
$begingroup$
@stackofhay42: We can also prove that by induction. $a_1neq 2, a_kneq 2 Longrightarrow a_{k+1}neq 2$ by the given relation.
$endgroup$
– Martund
Dec 12 '18 at 9:27
add a comment |
$begingroup$
AM-GM gives:
$$frac{a}{2} + frac{2}{a} stackrel{a>0, aneq 2}{>} 2sqrt{frac{a}{2}cdotfrac{2}{a}} = 2$$
Edit after comments:
- Note that $a^{star} = 2$ is a fixpoint of the iteration as $frac{a^{star}}{2} + frac{2}{a^{star}} = 2$.
- For any other starting value $a_0 > 0, a_0 neq 2$ AM-GM tells you, that the iteration $a_{n+1} = frac{a_n}{2} + frac{2}{a_n}$ produces $a_{n+1} > 2$.
So, also in your specific case of $a_1 = 4$ all members of the iteration are greater than $2$.
For a direct proof that all members of the iteration are greater than $2$ you may consider
- $f(x) = frac{x}{2} + frac{2}{x} Rightarrow f'(x) = frac{1}{2} - frac{2}{x^2}$
Now use MVT:
$$f(x) - 2 = f(x) - f(2) = f'(xi)(x - 2) = left(frac{1}{2} - frac{2}{{xi}^2}right) (x - 2) stackrel{2 < xi < x}{>} 0$$
$endgroup$
$begingroup$
how do i know $a neq 2$?
$endgroup$
– stackofhay42
Dec 12 '18 at 6:31
$begingroup$
@stackofhay42: If $a=2$, we t get the AM-GM $textbf{equality}$.
$endgroup$
– Yadati Kiran
Dec 12 '18 at 6:47
$begingroup$
Yes, if $a = 2$, we get AM-GM equality, meaning that we need to show that we cannot have $a = 2$. We never showed that it cannot be true that $a = 2$. I think that this is circular reasoning.
$endgroup$
– stackofhay42
Dec 12 '18 at 6:51
$begingroup$
@RakeshBhatt: By induction, we can say that all of the terms of the given sequence are positive. We know that $a_1>0$. Suppose $a_k>0 ,Longrightarrow a_{k+1}>0$ by the given relation.
$endgroup$
– Martund
Dec 12 '18 at 9:25
$begingroup$
@stackofhay42: We can also prove that by induction. $a_1neq 2, a_kneq 2 Longrightarrow a_{k+1}neq 2$ by the given relation.
$endgroup$
– Martund
Dec 12 '18 at 9:27
add a comment |
$begingroup$
AM-GM gives:
$$frac{a}{2} + frac{2}{a} stackrel{a>0, aneq 2}{>} 2sqrt{frac{a}{2}cdotfrac{2}{a}} = 2$$
Edit after comments:
- Note that $a^{star} = 2$ is a fixpoint of the iteration as $frac{a^{star}}{2} + frac{2}{a^{star}} = 2$.
- For any other starting value $a_0 > 0, a_0 neq 2$ AM-GM tells you, that the iteration $a_{n+1} = frac{a_n}{2} + frac{2}{a_n}$ produces $a_{n+1} > 2$.
So, also in your specific case of $a_1 = 4$ all members of the iteration are greater than $2$.
For a direct proof that all members of the iteration are greater than $2$ you may consider
- $f(x) = frac{x}{2} + frac{2}{x} Rightarrow f'(x) = frac{1}{2} - frac{2}{x^2}$
Now use MVT:
$$f(x) - 2 = f(x) - f(2) = f'(xi)(x - 2) = left(frac{1}{2} - frac{2}{{xi}^2}right) (x - 2) stackrel{2 < xi < x}{>} 0$$
$endgroup$
AM-GM gives:
$$frac{a}{2} + frac{2}{a} stackrel{a>0, aneq 2}{>} 2sqrt{frac{a}{2}cdotfrac{2}{a}} = 2$$
Edit after comments:
- Note that $a^{star} = 2$ is a fixpoint of the iteration as $frac{a^{star}}{2} + frac{2}{a^{star}} = 2$.
- For any other starting value $a_0 > 0, a_0 neq 2$ AM-GM tells you, that the iteration $a_{n+1} = frac{a_n}{2} + frac{2}{a_n}$ produces $a_{n+1} > 2$.
So, also in your specific case of $a_1 = 4$ all members of the iteration are greater than $2$.
For a direct proof that all members of the iteration are greater than $2$ you may consider
- $f(x) = frac{x}{2} + frac{2}{x} Rightarrow f'(x) = frac{1}{2} - frac{2}{x^2}$
Now use MVT:
$$f(x) - 2 = f(x) - f(2) = f'(xi)(x - 2) = left(frac{1}{2} - frac{2}{{xi}^2}right) (x - 2) stackrel{2 < xi < x}{>} 0$$
edited Dec 12 '18 at 9:35
answered Dec 12 '18 at 5:49
trancelocationtrancelocation
10.8k1723
10.8k1723
$begingroup$
how do i know $a neq 2$?
$endgroup$
– stackofhay42
Dec 12 '18 at 6:31
$begingroup$
@stackofhay42: If $a=2$, we t get the AM-GM $textbf{equality}$.
$endgroup$
– Yadati Kiran
Dec 12 '18 at 6:47
$begingroup$
Yes, if $a = 2$, we get AM-GM equality, meaning that we need to show that we cannot have $a = 2$. We never showed that it cannot be true that $a = 2$. I think that this is circular reasoning.
$endgroup$
– stackofhay42
Dec 12 '18 at 6:51
$begingroup$
@RakeshBhatt: By induction, we can say that all of the terms of the given sequence are positive. We know that $a_1>0$. Suppose $a_k>0 ,Longrightarrow a_{k+1}>0$ by the given relation.
$endgroup$
– Martund
Dec 12 '18 at 9:25
$begingroup$
@stackofhay42: We can also prove that by induction. $a_1neq 2, a_kneq 2 Longrightarrow a_{k+1}neq 2$ by the given relation.
$endgroup$
– Martund
Dec 12 '18 at 9:27
add a comment |
$begingroup$
how do i know $a neq 2$?
$endgroup$
– stackofhay42
Dec 12 '18 at 6:31
$begingroup$
@stackofhay42: If $a=2$, we t get the AM-GM $textbf{equality}$.
$endgroup$
– Yadati Kiran
Dec 12 '18 at 6:47
$begingroup$
Yes, if $a = 2$, we get AM-GM equality, meaning that we need to show that we cannot have $a = 2$. We never showed that it cannot be true that $a = 2$. I think that this is circular reasoning.
$endgroup$
– stackofhay42
Dec 12 '18 at 6:51
$begingroup$
@RakeshBhatt: By induction, we can say that all of the terms of the given sequence are positive. We know that $a_1>0$. Suppose $a_k>0 ,Longrightarrow a_{k+1}>0$ by the given relation.
$endgroup$
– Martund
Dec 12 '18 at 9:25
$begingroup$
@stackofhay42: We can also prove that by induction. $a_1neq 2, a_kneq 2 Longrightarrow a_{k+1}neq 2$ by the given relation.
$endgroup$
– Martund
Dec 12 '18 at 9:27
$begingroup$
how do i know $a neq 2$?
$endgroup$
– stackofhay42
Dec 12 '18 at 6:31
$begingroup$
how do i know $a neq 2$?
$endgroup$
– stackofhay42
Dec 12 '18 at 6:31
$begingroup$
@stackofhay42: If $a=2$, we t get the AM-GM $textbf{equality}$.
$endgroup$
– Yadati Kiran
Dec 12 '18 at 6:47
$begingroup$
@stackofhay42: If $a=2$, we t get the AM-GM $textbf{equality}$.
$endgroup$
– Yadati Kiran
Dec 12 '18 at 6:47
$begingroup$
Yes, if $a = 2$, we get AM-GM equality, meaning that we need to show that we cannot have $a = 2$. We never showed that it cannot be true that $a = 2$. I think that this is circular reasoning.
$endgroup$
– stackofhay42
Dec 12 '18 at 6:51
$begingroup$
Yes, if $a = 2$, we get AM-GM equality, meaning that we need to show that we cannot have $a = 2$. We never showed that it cannot be true that $a = 2$. I think that this is circular reasoning.
$endgroup$
– stackofhay42
Dec 12 '18 at 6:51
$begingroup$
@RakeshBhatt: By induction, we can say that all of the terms of the given sequence are positive. We know that $a_1>0$. Suppose $a_k>0 ,Longrightarrow a_{k+1}>0$ by the given relation.
$endgroup$
– Martund
Dec 12 '18 at 9:25
$begingroup$
@RakeshBhatt: By induction, we can say that all of the terms of the given sequence are positive. We know that $a_1>0$. Suppose $a_k>0 ,Longrightarrow a_{k+1}>0$ by the given relation.
$endgroup$
– Martund
Dec 12 '18 at 9:25
$begingroup$
@stackofhay42: We can also prove that by induction. $a_1neq 2, a_kneq 2 Longrightarrow a_{k+1}neq 2$ by the given relation.
$endgroup$
– Martund
Dec 12 '18 at 9:27
$begingroup$
@stackofhay42: We can also prove that by induction. $a_1neq 2, a_kneq 2 Longrightarrow a_{k+1}neq 2$ by the given relation.
$endgroup$
– Martund
Dec 12 '18 at 9:27
add a comment |
$begingroup$
Note $a_n >0$, $n=1,2,3,.....$.
Let $n ge 1$:
$a_{n+1}= dfrac{a_n^2 +4}{2a_n}=$
$dfrac{(a_n-2)^2+4a_n}{2a_n}=$
$dfrac{(a_n-2)^2}{2a_n} + 2 ge 2$.
(The first term $ge 0$, a square divided by a positive number.)
Hence
$a_{n+1} ge 2$, $n in mathbb{N}$.
$endgroup$
add a comment |
$begingroup$
Note $a_n >0$, $n=1,2,3,.....$.
Let $n ge 1$:
$a_{n+1}= dfrac{a_n^2 +4}{2a_n}=$
$dfrac{(a_n-2)^2+4a_n}{2a_n}=$
$dfrac{(a_n-2)^2}{2a_n} + 2 ge 2$.
(The first term $ge 0$, a square divided by a positive number.)
Hence
$a_{n+1} ge 2$, $n in mathbb{N}$.
$endgroup$
add a comment |
$begingroup$
Note $a_n >0$, $n=1,2,3,.....$.
Let $n ge 1$:
$a_{n+1}= dfrac{a_n^2 +4}{2a_n}=$
$dfrac{(a_n-2)^2+4a_n}{2a_n}=$
$dfrac{(a_n-2)^2}{2a_n} + 2 ge 2$.
(The first term $ge 0$, a square divided by a positive number.)
Hence
$a_{n+1} ge 2$, $n in mathbb{N}$.
$endgroup$
Note $a_n >0$, $n=1,2,3,.....$.
Let $n ge 1$:
$a_{n+1}= dfrac{a_n^2 +4}{2a_n}=$
$dfrac{(a_n-2)^2+4a_n}{2a_n}=$
$dfrac{(a_n-2)^2}{2a_n} + 2 ge 2$.
(The first term $ge 0$, a square divided by a positive number.)
Hence
$a_{n+1} ge 2$, $n in mathbb{N}$.
answered Dec 12 '18 at 11:01
Peter SzilasPeter Szilas
11.1k2821
11.1k2821
add a comment |
add a comment |
$begingroup$
Following with your reasoning, suppose $x + 1/x = 2$. Then:
$$x + 1/x = 2implies x^2 - 2x + 1 = 0implies x = 1.$$
But this is impossible because...
$endgroup$
$begingroup$
What about $a_n<2$?
$endgroup$
– Rakesh Bhatt
Dec 12 '18 at 10:14
$begingroup$
@RakeshBhatt, the OP has proved $a_n > 2implies a_{n+1}ge 2$. Only is required excluding the equality.
$endgroup$
– Martín-Blas Pérez Pinilla
Dec 12 '18 at 10:43
$begingroup$
Why is this impossible?
$endgroup$
– Ekesh Kumar
Dec 13 '18 at 23:45
$begingroup$
@Ekesh, because you have supposed $a_n > 2$ ($x > 1$).
$endgroup$
– Martín-Blas Pérez Pinilla
Dec 14 '18 at 7:12
add a comment |
$begingroup$
Following with your reasoning, suppose $x + 1/x = 2$. Then:
$$x + 1/x = 2implies x^2 - 2x + 1 = 0implies x = 1.$$
But this is impossible because...
$endgroup$
$begingroup$
What about $a_n<2$?
$endgroup$
– Rakesh Bhatt
Dec 12 '18 at 10:14
$begingroup$
@RakeshBhatt, the OP has proved $a_n > 2implies a_{n+1}ge 2$. Only is required excluding the equality.
$endgroup$
– Martín-Blas Pérez Pinilla
Dec 12 '18 at 10:43
$begingroup$
Why is this impossible?
$endgroup$
– Ekesh Kumar
Dec 13 '18 at 23:45
$begingroup$
@Ekesh, because you have supposed $a_n > 2$ ($x > 1$).
$endgroup$
– Martín-Blas Pérez Pinilla
Dec 14 '18 at 7:12
add a comment |
$begingroup$
Following with your reasoning, suppose $x + 1/x = 2$. Then:
$$x + 1/x = 2implies x^2 - 2x + 1 = 0implies x = 1.$$
But this is impossible because...
$endgroup$
Following with your reasoning, suppose $x + 1/x = 2$. Then:
$$x + 1/x = 2implies x^2 - 2x + 1 = 0implies x = 1.$$
But this is impossible because...
edited Dec 12 '18 at 10:44
answered Dec 12 '18 at 7:44
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
34.3k42871
34.3k42871
$begingroup$
What about $a_n<2$?
$endgroup$
– Rakesh Bhatt
Dec 12 '18 at 10:14
$begingroup$
@RakeshBhatt, the OP has proved $a_n > 2implies a_{n+1}ge 2$. Only is required excluding the equality.
$endgroup$
– Martín-Blas Pérez Pinilla
Dec 12 '18 at 10:43
$begingroup$
Why is this impossible?
$endgroup$
– Ekesh Kumar
Dec 13 '18 at 23:45
$begingroup$
@Ekesh, because you have supposed $a_n > 2$ ($x > 1$).
$endgroup$
– Martín-Blas Pérez Pinilla
Dec 14 '18 at 7:12
add a comment |
$begingroup$
What about $a_n<2$?
$endgroup$
– Rakesh Bhatt
Dec 12 '18 at 10:14
$begingroup$
@RakeshBhatt, the OP has proved $a_n > 2implies a_{n+1}ge 2$. Only is required excluding the equality.
$endgroup$
– Martín-Blas Pérez Pinilla
Dec 12 '18 at 10:43
$begingroup$
Why is this impossible?
$endgroup$
– Ekesh Kumar
Dec 13 '18 at 23:45
$begingroup$
@Ekesh, because you have supposed $a_n > 2$ ($x > 1$).
$endgroup$
– Martín-Blas Pérez Pinilla
Dec 14 '18 at 7:12
$begingroup$
What about $a_n<2$?
$endgroup$
– Rakesh Bhatt
Dec 12 '18 at 10:14
$begingroup$
What about $a_n<2$?
$endgroup$
– Rakesh Bhatt
Dec 12 '18 at 10:14
$begingroup$
@RakeshBhatt, the OP has proved $a_n > 2implies a_{n+1}ge 2$. Only is required excluding the equality.
$endgroup$
– Martín-Blas Pérez Pinilla
Dec 12 '18 at 10:43
$begingroup$
@RakeshBhatt, the OP has proved $a_n > 2implies a_{n+1}ge 2$. Only is required excluding the equality.
$endgroup$
– Martín-Blas Pérez Pinilla
Dec 12 '18 at 10:43
$begingroup$
Why is this impossible?
$endgroup$
– Ekesh Kumar
Dec 13 '18 at 23:45
$begingroup$
Why is this impossible?
$endgroup$
– Ekesh Kumar
Dec 13 '18 at 23:45
$begingroup$
@Ekesh, because you have supposed $a_n > 2$ ($x > 1$).
$endgroup$
– Martín-Blas Pérez Pinilla
Dec 14 '18 at 7:12
$begingroup$
@Ekesh, because you have supposed $a_n > 2$ ($x > 1$).
$endgroup$
– Martín-Blas Pérez Pinilla
Dec 14 '18 at 7:12
add a comment |
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$begingroup$
Prove $a_n>0$ and use $a+bgeq 2sqrt{ab}$.
$endgroup$
– Lau
Dec 12 '18 at 5:43
$begingroup$
All these and a lot more are covered here, including linked questions ...
$endgroup$
– rtybase
Dec 12 '18 at 9:22