If a sequence converges to a, prove the squares of its terms converge to a^2
$begingroup$
Given:
$$a_nto a$$
That means :
For all $epsilon^*:=frac{epsilon }{2|a|+1} $ we can find an N such that for all n>N $left |a_n-a right |<frac{epsilon }{2|a|+1}$
I have to prove
$$(a_n)^2 to a^2 $$
At first we guess
$left |a_n+a right |=left |a_n-a+a+a right |<frac{epsilon }{2|a|+1 }+2|a | $
Now we can prove
$$(a_n)^2 to a^2 $$
$left |(a_n)^2- a^2 right |=left |(a_n-a)(a_n+a)right |=left |a_n-a right |left |a_n+a right |<frac{epsilon }{2|a|+1}(frac{epsilon }{2|a|+1}+2|a |)<epsilon^2+epsilon$
I changed my epsilon and i think it works now??
any suggestion ?
real-analysis
asked Feb 1 '15 at 22:47
BeginnerBeginner
284
$endgroup$
|
show 2 more comments
$begingroup$
Given:
$$a_nto a$$
That means :
For all $epsilon^*:=frac{epsilon }{2|a|+1} $ we can find an N such that for all n>N $left |a_n-a right |<frac{epsilon }{2|a|+1}$
I have to prove
$$(a_n)^2 to a^2 $$
At first we guess
$left |a_n+a right |=left |a_n-a+a+a right |<frac{epsilon }{2|a|+1 }+2|a | $
Now we can prove
$$(a_n)^2 to a^2 $$
$left |(a_n)^2- a^2 right |=left |(a_n-a)(a_n+a)right |=left |a_n-a right |left |a_n+a right |<frac{epsilon }{2|a|+1}(frac{epsilon }{2|a|+1}+2|a |)<epsilon^2+epsilon$
I changed my epsilon and i think it works now??
any suggestion ?
real-analysis
asked Feb 1 '15 at 22:47
BeginnerBeginner
284
$endgroup$
$begingroup$
Your question isn't clear. Are you trying to prove that if some sequence $a_ninmathbb R$ converges to $a$, then its square converges to $a^2$?
$endgroup$
– fonini
Feb 1 '15 at 22:54
$begingroup$
i have to prove if a_n converges to a, then (a_n)^2 convergens to a^2
$endgroup$
– Beginner
Feb 1 '15 at 22:56
$begingroup$
$lim_{nto infty} a_n = a$ means that for all $epsilon gt 0$ there exists natural number $N$ such that for all $n gt N$, $|a_n - a| lt epsilon$.
$endgroup$
– hardmath
Feb 1 '15 at 22:58
$begingroup$
This is good, but to finish you need to show that you can make $epsilon^*$ as small as you want. It is not difficult, but it is the essence of the proof.
$endgroup$
– copper.hat
Feb 1 '15 at 23:16
$begingroup$
@Beginner how is $frac{epsilon}{2|a| + epsilon}(frac{epsilon}{2|a|} + 2|a|) < epsilon$? Putting aside the fact that $a$ can be zero (in which case $epsilon/|a|$ is undefined), if $epsilon = 2$ and $a = 1/4$, then the left hand side of the inequality is equal to $45/8$, but the right hand side is equal to $2$, which is less than $45/8$.
$endgroup$
– kobe
Feb 1 '15 at 23:52
|
show 2 more comments
$begingroup$
Given:
$$a_nto a$$
That means :
For all $epsilon^*:=frac{epsilon }{2|a|+1} $ we can find an N such that for all n>N $left |a_n-a right |<frac{epsilon }{2|a|+1}$
I have to prove
$$(a_n)^2 to a^2 $$
At first we guess
$left |a_n+a right |=left |a_n-a+a+a right |<frac{epsilon }{2|a|+1 }+2|a | $
Now we can prove
$$(a_n)^2 to a^2 $$
$left |(a_n)^2- a^2 right |=left |(a_n-a)(a_n+a)right |=left |a_n-a right |left |a_n+a right |<frac{epsilon }{2|a|+1}(frac{epsilon }{2|a|+1}+2|a |)<epsilon^2+epsilon$
I changed my epsilon and i think it works now??
any suggestion ?
real-analysis
asked Feb 1 '15 at 22:47
BeginnerBeginner
284
$endgroup$
Given:
$$a_nto a$$
That means :
For all $epsilon^*:=frac{epsilon }{2|a|+1} $ we can find an N such that for all n>N $left |a_n-a right |<frac{epsilon }{2|a|+1}$
I have to prove
$$(a_n)^2 to a^2 $$
At first we guess
$left |a_n+a right |=left |a_n-a+a+a right |<frac{epsilon }{2|a|+1 }+2|a | $
Now we can prove
$$(a_n)^2 to a^2 $$
$left |(a_n)^2- a^2 right |=left |(a_n-a)(a_n+a)right |=left |a_n-a right |left |a_n+a right |<frac{epsilon }{2|a|+1}(frac{epsilon }{2|a|+1}+2|a |)<epsilon^2+epsilon$
I changed my epsilon and i think it works now??
any suggestion ?
real-analysis
real-analysis
asked Feb 1 '15 at 22:47
BeginnerBeginner
284
asked Feb 1 '15 at 22:47
BeginnerBeginner
284
edited Feb 1 '15 at 23:58
Beginner
asked Feb 1 '15 at 22:47
BeginnerBeginner
284
asked Feb 1 '15 at 22:47
BeginnerBeginner
284
asked Feb 1 '15 at 22:47
BeginnerBeginner
284
284
$begingroup$
Your question isn't clear. Are you trying to prove that if some sequence $a_ninmathbb R$ converges to $a$, then its square converges to $a^2$?
$endgroup$
– fonini
Feb 1 '15 at 22:54
$begingroup$
i have to prove if a_n converges to a, then (a_n)^2 convergens to a^2
$endgroup$
– Beginner
Feb 1 '15 at 22:56
$begingroup$
$lim_{nto infty} a_n = a$ means that for all $epsilon gt 0$ there exists natural number $N$ such that for all $n gt N$, $|a_n - a| lt epsilon$.
$endgroup$
– hardmath
Feb 1 '15 at 22:58
$begingroup$
This is good, but to finish you need to show that you can make $epsilon^*$ as small as you want. It is not difficult, but it is the essence of the proof.
$endgroup$
– copper.hat
Feb 1 '15 at 23:16
$begingroup$
@Beginner how is $frac{epsilon}{2|a| + epsilon}(frac{epsilon}{2|a|} + 2|a|) < epsilon$? Putting aside the fact that $a$ can be zero (in which case $epsilon/|a|$ is undefined), if $epsilon = 2$ and $a = 1/4$, then the left hand side of the inequality is equal to $45/8$, but the right hand side is equal to $2$, which is less than $45/8$.
$endgroup$
– kobe
Feb 1 '15 at 23:52
|
show 2 more comments
$begingroup$
Your question isn't clear. Are you trying to prove that if some sequence $a_ninmathbb R$ converges to $a$, then its square converges to $a^2$?
$endgroup$
– fonini
Feb 1 '15 at 22:54
$begingroup$
i have to prove if a_n converges to a, then (a_n)^2 convergens to a^2
$endgroup$
– Beginner
Feb 1 '15 at 22:56
$begingroup$
$lim_{nto infty} a_n = a$ means that for all $epsilon gt 0$ there exists natural number $N$ such that for all $n gt N$, $|a_n - a| lt epsilon$.
$endgroup$
– hardmath
Feb 1 '15 at 22:58
$begingroup$
This is good, but to finish you need to show that you can make $epsilon^*$ as small as you want. It is not difficult, but it is the essence of the proof.
$endgroup$
– copper.hat
Feb 1 '15 at 23:16
$begingroup$
@Beginner how is $frac{epsilon}{2|a| + epsilon}(frac{epsilon}{2|a|} + 2|a|) < epsilon$? Putting aside the fact that $a$ can be zero (in which case $epsilon/|a|$ is undefined), if $epsilon = 2$ and $a = 1/4$, then the left hand side of the inequality is equal to $45/8$, but the right hand side is equal to $2$, which is less than $45/8$.
$endgroup$
– kobe
Feb 1 '15 at 23:52
$begingroup$
Your question isn't clear. Are you trying to prove that if some sequence $a_ninmathbb R$ converges to $a$, then its square converges to $a^2$?
$endgroup$
– fonini
Feb 1 '15 at 22:54
$begingroup$
Your question isn't clear. Are you trying to prove that if some sequence $a_ninmathbb R$ converges to $a$, then its square converges to $a^2$?
$endgroup$
– fonini
Feb 1 '15 at 22:54
$begingroup$
i have to prove if a_n converges to a, then (a_n)^2 convergens to a^2
$endgroup$
– Beginner
Feb 1 '15 at 22:56
$begingroup$
i have to prove if a_n converges to a, then (a_n)^2 convergens to a^2
$endgroup$
– Beginner
Feb 1 '15 at 22:56
$begingroup$
$lim_{nto infty} a_n = a$ means that for all $epsilon gt 0$ there exists natural number $N$ such that for all $n gt N$, $|a_n - a| lt epsilon$.
$endgroup$
– hardmath
Feb 1 '15 at 22:58
$begingroup$
$lim_{nto infty} a_n = a$ means that for all $epsilon gt 0$ there exists natural number $N$ such that for all $n gt N$, $|a_n - a| lt epsilon$.
$endgroup$
– hardmath
Feb 1 '15 at 22:58
$begingroup$
This is good, but to finish you need to show that you can make $epsilon^*$ as small as you want. It is not difficult, but it is the essence of the proof.
$endgroup$
– copper.hat
Feb 1 '15 at 23:16
$begingroup$
This is good, but to finish you need to show that you can make $epsilon^*$ as small as you want. It is not difficult, but it is the essence of the proof.
$endgroup$
– copper.hat
Feb 1 '15 at 23:16
$begingroup$
@Beginner how is $frac{epsilon}{2|a| + epsilon}(frac{epsilon}{2|a|} + 2|a|) < epsilon$? Putting aside the fact that $a$ can be zero (in which case $epsilon/|a|$ is undefined), if $epsilon = 2$ and $a = 1/4$, then the left hand side of the inequality is equal to $45/8$, but the right hand side is equal to $2$, which is less than $45/8$.
$endgroup$
– kobe
Feb 1 '15 at 23:52
$begingroup$
@Beginner how is $frac{epsilon}{2|a| + epsilon}(frac{epsilon}{2|a|} + 2|a|) < epsilon$? Putting aside the fact that $a$ can be zero (in which case $epsilon/|a|$ is undefined), if $epsilon = 2$ and $a = 1/4$, then the left hand side of the inequality is equal to $45/8$, but the right hand side is equal to $2$, which is less than $45/8$.
$endgroup$
– kobe
Feb 1 '15 at 23:52
|
show 2 more comments
1 Answer
1
active
oldest
votes
$begingroup$
Given $varepsilon > 0$, let $eta := min{1, frac{varepsilon}{(1 + 2|a|)}} > 0$.
So since $a_n to a$, there exists a positive integer $N$ such that $|a_n - a| < eta$ for all $n ge N$.
Now if $n ge N$, then in particular, $|a_n - a| < 1$, which implies
$$
|a_n + a| = |(a_n - a) + 2a| le |a_n -a| + 2|a| < 1 + 2|a|
$$
Thus
$$
|a_n^2 - a^2| = |a_n + a||a_n - a| < (1 + 2|a|) frac{varepsilon}{1 + 2|a|} = varepsilon$$
Since $varepsilon$ was arbitrary, $~a_n^2 to a^2$.
edited Apr 6 '18 at 11:15
Viktor Glombik
1,0101527
answered Feb 1 '15 at 23:02
kobekobe
35k22248
$endgroup$
$begingroup$
Thanks for the answer and can you tell me please, is my solution correct?
$endgroup$
– Beginner
Feb 1 '15 at 23:11
$begingroup$
It's not correct. For one, $a$ can be zero, so $varepsilon/a$ may be undefined.
$endgroup$
– kobe
Feb 1 '15 at 23:15
$begingroup$
For all $frac{epsilon }{2|a|+frac{epsilon}}$ we can find an N such that for all n>N $left |a_n-a right |<frac{epsilon }{2|a|+frac{epsilon}}$
$endgroup$
– Beginner
Feb 1 '15 at 23:19
$begingroup$
Your statement does not make sense. Even if you just have $|a_n - a| < varepsilon/(text{something})$ for all $n > N$, you cannot find an upper bound for $|a_n + a|$ that is independent of $varepsilon$ for $n > N$.
$endgroup$
– kobe
Feb 1 '15 at 23:25
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Given $varepsilon > 0$, let $eta := min{1, frac{varepsilon}{(1 + 2|a|)}} > 0$.
So since $a_n to a$, there exists a positive integer $N$ such that $|a_n - a| < eta$ for all $n ge N$.
Now if $n ge N$, then in particular, $|a_n - a| < 1$, which implies
$$
|a_n + a| = |(a_n - a) + 2a| le |a_n -a| + 2|a| < 1 + 2|a|
$$
Thus
$$
|a_n^2 - a^2| = |a_n + a||a_n - a| < (1 + 2|a|) frac{varepsilon}{1 + 2|a|} = varepsilon$$
Since $varepsilon$ was arbitrary, $~a_n^2 to a^2$.
edited Apr 6 '18 at 11:15
Viktor Glombik
1,0101527
answered Feb 1 '15 at 23:02
kobekobe
35k22248
$endgroup$
$begingroup$
Thanks for the answer and can you tell me please, is my solution correct?
$endgroup$
– Beginner
Feb 1 '15 at 23:11
$begingroup$
It's not correct. For one, $a$ can be zero, so $varepsilon/a$ may be undefined.
$endgroup$
– kobe
Feb 1 '15 at 23:15
$begingroup$
For all $frac{epsilon }{2|a|+frac{epsilon}}$ we can find an N such that for all n>N $left |a_n-a right |<frac{epsilon }{2|a|+frac{epsilon}}$
$endgroup$
– Beginner
Feb 1 '15 at 23:19
$begingroup$
Your statement does not make sense. Even if you just have $|a_n - a| < varepsilon/(text{something})$ for all $n > N$, you cannot find an upper bound for $|a_n + a|$ that is independent of $varepsilon$ for $n > N$.
$endgroup$
– kobe
Feb 1 '15 at 23:25
add a comment |
$begingroup$
Given $varepsilon > 0$, let $eta := min{1, frac{varepsilon}{(1 + 2|a|)}} > 0$.
So since $a_n to a$, there exists a positive integer $N$ such that $|a_n - a| < eta$ for all $n ge N$.
Now if $n ge N$, then in particular, $|a_n - a| < 1$, which implies
$$
|a_n + a| = |(a_n - a) + 2a| le |a_n -a| + 2|a| < 1 + 2|a|
$$
Thus
$$
|a_n^2 - a^2| = |a_n + a||a_n - a| < (1 + 2|a|) frac{varepsilon}{1 + 2|a|} = varepsilon$$
Since $varepsilon$ was arbitrary, $~a_n^2 to a^2$.
edited Apr 6 '18 at 11:15
Viktor Glombik
1,0101527
answered Feb 1 '15 at 23:02
kobekobe
35k22248
$endgroup$
$begingroup$
Thanks for the answer and can you tell me please, is my solution correct?
$endgroup$
– Beginner
Feb 1 '15 at 23:11
$begingroup$
It's not correct. For one, $a$ can be zero, so $varepsilon/a$ may be undefined.
$endgroup$
– kobe
Feb 1 '15 at 23:15
$begingroup$
For all $frac{epsilon }{2|a|+frac{epsilon}}$ we can find an N such that for all n>N $left |a_n-a right |<frac{epsilon }{2|a|+frac{epsilon}}$
$endgroup$
– Beginner
Feb 1 '15 at 23:19
$begingroup$
Your statement does not make sense. Even if you just have $|a_n - a| < varepsilon/(text{something})$ for all $n > N$, you cannot find an upper bound for $|a_n + a|$ that is independent of $varepsilon$ for $n > N$.
$endgroup$
– kobe
Feb 1 '15 at 23:25
add a comment |
$begingroup$
Given $varepsilon > 0$, let $eta := min{1, frac{varepsilon}{(1 + 2|a|)}} > 0$.
So since $a_n to a$, there exists a positive integer $N$ such that $|a_n - a| < eta$ for all $n ge N$.
Now if $n ge N$, then in particular, $|a_n - a| < 1$, which implies
$$
|a_n + a| = |(a_n - a) + 2a| le |a_n -a| + 2|a| < 1 + 2|a|
$$
Thus
$$
|a_n^2 - a^2| = |a_n + a||a_n - a| < (1 + 2|a|) frac{varepsilon}{1 + 2|a|} = varepsilon$$
Since $varepsilon$ was arbitrary, $~a_n^2 to a^2$.
edited Apr 6 '18 at 11:15
Viktor Glombik
1,0101527
answered Feb 1 '15 at 23:02
kobekobe
35k22248
$endgroup$
Given $varepsilon > 0$, let $eta := min{1, frac{varepsilon}{(1 + 2|a|)}} > 0$.
So since $a_n to a$, there exists a positive integer $N$ such that $|a_n - a| < eta$ for all $n ge N$.
Now if $n ge N$, then in particular, $|a_n - a| < 1$, which implies
$$
|a_n + a| = |(a_n - a) + 2a| le |a_n -a| + 2|a| < 1 + 2|a|
$$
Thus
$$
|a_n^2 - a^2| = |a_n + a||a_n - a| < (1 + 2|a|) frac{varepsilon}{1 + 2|a|} = varepsilon$$
Since $varepsilon$ was arbitrary, $~a_n^2 to a^2$.
edited Apr 6 '18 at 11:15
Viktor Glombik
1,0101527
answered Feb 1 '15 at 23:02
kobekobe
35k22248
edited Apr 6 '18 at 11:15
Viktor Glombik
1,0101527
edited Apr 6 '18 at 11:15
Viktor Glombik
1,0101527
edited Apr 6 '18 at 11:15
Viktor Glombik
1,0101527
1,0101527
answered Feb 1 '15 at 23:02
kobekobe
35k22248
answered Feb 1 '15 at 23:02
kobekobe
35k22248
answered Feb 1 '15 at 23:02
kobekobe
35k22248
35k22248
$begingroup$
Thanks for the answer and can you tell me please, is my solution correct?
$endgroup$
– Beginner
Feb 1 '15 at 23:11
$begingroup$
It's not correct. For one, $a$ can be zero, so $varepsilon/a$ may be undefined.
$endgroup$
– kobe
Feb 1 '15 at 23:15
$begingroup$
For all $frac{epsilon }{2|a|+frac{epsilon}}$ we can find an N such that for all n>N $left |a_n-a right |<frac{epsilon }{2|a|+frac{epsilon}}$
$endgroup$
– Beginner
Feb 1 '15 at 23:19
$begingroup$
Your statement does not make sense. Even if you just have $|a_n - a| < varepsilon/(text{something})$ for all $n > N$, you cannot find an upper bound for $|a_n + a|$ that is independent of $varepsilon$ for $n > N$.
$endgroup$
– kobe
Feb 1 '15 at 23:25
add a comment |
$begingroup$
Thanks for the answer and can you tell me please, is my solution correct?
$endgroup$
– Beginner
Feb 1 '15 at 23:11
$begingroup$
It's not correct. For one, $a$ can be zero, so $varepsilon/a$ may be undefined.
$endgroup$
– kobe
Feb 1 '15 at 23:15
$begingroup$
For all $frac{epsilon }{2|a|+frac{epsilon}}$ we can find an N such that for all n>N $left |a_n-a right |<frac{epsilon }{2|a|+frac{epsilon}}$
$endgroup$
– Beginner
Feb 1 '15 at 23:19
$begingroup$
Your statement does not make sense. Even if you just have $|a_n - a| < varepsilon/(text{something})$ for all $n > N$, you cannot find an upper bound for $|a_n + a|$ that is independent of $varepsilon$ for $n > N$.
$endgroup$
– kobe
Feb 1 '15 at 23:25
$begingroup$
Thanks for the answer and can you tell me please, is my solution correct?
$endgroup$
– Beginner
Feb 1 '15 at 23:11
$begingroup$
Thanks for the answer and can you tell me please, is my solution correct?
$endgroup$
– Beginner
Feb 1 '15 at 23:11
$begingroup$
It's not correct. For one, $a$ can be zero, so $varepsilon/a$ may be undefined.
$endgroup$
– kobe
Feb 1 '15 at 23:15
$begingroup$
It's not correct. For one, $a$ can be zero, so $varepsilon/a$ may be undefined.
$endgroup$
– kobe
Feb 1 '15 at 23:15
$begingroup$
For all $frac{epsilon }{2|a|+frac{epsilon}}$ we can find an N such that for all n>N $left |a_n-a right |<frac{epsilon }{2|a|+frac{epsilon}}$
$endgroup$
– Beginner
Feb 1 '15 at 23:19
$begingroup$
For all $frac{epsilon }{2|a|+frac{epsilon}}$ we can find an N such that for all n>N $left |a_n-a right |<frac{epsilon }{2|a|+frac{epsilon}}$
$endgroup$
– Beginner
Feb 1 '15 at 23:19
$begingroup$
Your statement does not make sense. Even if you just have $|a_n - a| < varepsilon/(text{something})$ for all $n > N$, you cannot find an upper bound for $|a_n + a|$ that is independent of $varepsilon$ for $n > N$.
$endgroup$
– kobe
Feb 1 '15 at 23:25
$begingroup$
Your statement does not make sense. Even if you just have $|a_n - a| < varepsilon/(text{something})$ for all $n > N$, you cannot find an upper bound for $|a_n + a|$ that is independent of $varepsilon$ for $n > N$.
$endgroup$
– kobe
Feb 1 '15 at 23:25
add a comment |
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Your question isn't clear. Are you trying to prove that if some sequence $a_ninmathbb R$ converges to $a$, then its square converges to $a^2$?
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– fonini
Feb 1 '15 at 22:54
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i have to prove if a_n converges to a, then (a_n)^2 convergens to a^2
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– Beginner
Feb 1 '15 at 22:56
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$lim_{nto infty} a_n = a$ means that for all $epsilon gt 0$ there exists natural number $N$ such that for all $n gt N$, $|a_n - a| lt epsilon$.
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– hardmath
Feb 1 '15 at 22:58
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This is good, but to finish you need to show that you can make $epsilon^*$ as small as you want. It is not difficult, but it is the essence of the proof.
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– copper.hat
Feb 1 '15 at 23:16
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@Beginner how is $frac{epsilon}{2|a| + epsilon}(frac{epsilon}{2|a|} + 2|a|) < epsilon$? Putting aside the fact that $a$ can be zero (in which case $epsilon/|a|$ is undefined), if $epsilon = 2$ and $a = 1/4$, then the left hand side of the inequality is equal to $45/8$, but the right hand side is equal to $2$, which is less than $45/8$.
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– kobe
Feb 1 '15 at 23:52