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Trouble understanding the $ε$-$N$ proof for limit of sequence
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I'm currently stuck on a math problem proposed by my teacher.
For $varepsilon=0,001$, find $N=N_0$ such that $|a_n-L|lt varepsilon$ if $ngeqslant N$
$$a_n = frac(n+1)3n-1, rm and , L=1/3$$
According to him, the answer could be any $N_0$ greater than $left[frac23varepsilonright]+1$. I have an idea of how the proof works, but I can't figure out how he came up with that specific value for N$_0$. And why the final value has a $+1$.
calculus sequences-and-series limits
edited Sep 5 at 2:43
user21820
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asked Sep 4 at 19:24
Vinicius Alexandre Zonta
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I'm currently stuck on a math problem proposed by my teacher.
For $varepsilon=0,001$, find $N=N_0$ such that $|a_n-L|lt varepsilon$ if $ngeqslant N$
$$a_n = frac(n+1)3n-1, rm and , L=1/3$$
According to him, the answer could be any $N_0$ greater than $left[frac23varepsilonright]+1$. I have an idea of how the proof works, but I can't figure out how he came up with that specific value for N$_0$. And why the final value has a $+1$.
calculus sequences-and-series limits
edited Sep 5 at 2:43
user21820
36.2k440140
asked Sep 4 at 19:24
Vinicius Alexandre Zonta
161
New contributor
Vinicius Alexandre Zonta is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
In general, the purpose of a formula for $N_0$ is so that when you use that formula in the proof, the proof will work. The formula your teacher used gives a much larger than minimal value of $N_0,$ which is fine (you can never go wrong by setting $N_0$ larger than needed), but to know the reason for choosing that particular formula it might help to see the specific steps of the proof.
– David K
Sep 4 at 19:52
Note that the standard in the English speaking world is to use the period to denote decimals ($epsilon =.001$).
– Acccumulation
Sep 5 at 0:09
@Acccumulation Note that the standard is not to skip the leading zero as well, $epsilon = 0.001$.
– Roman Odaisky
Sep 5 at 1:32
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
I'm currently stuck on a math problem proposed by my teacher.
For $varepsilon=0,001$, find $N=N_0$ such that $|a_n-L|lt varepsilon$ if $ngeqslant N$
$$a_n = frac(n+1)3n-1, rm and , L=1/3$$
According to him, the answer could be any $N_0$ greater than $left[frac23varepsilonright]+1$. I have an idea of how the proof works, but I can't figure out how he came up with that specific value for N$_0$. And why the final value has a $+1$.
calculus sequences-and-series limits
edited Sep 5 at 2:43
user21820
36.2k440140
asked Sep 4 at 19:24
Vinicius Alexandre Zonta
161
New contributor
Vinicius Alexandre Zonta is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I'm currently stuck on a math problem proposed by my teacher.
For $varepsilon=0,001$, find $N=N_0$ such that $|a_n-L|lt varepsilon$ if $ngeqslant N$
$$a_n = frac(n+1)3n-1, rm and , L=1/3$$
According to him, the answer could be any $N_0$ greater than $left[frac23varepsilonright]+1$. I have an idea of how the proof works, but I can't figure out how he came up with that specific value for N$_0$. And why the final value has a $+1$.
calculus sequences-and-series limits
edited Sep 5 at 2:43
user21820
36.2k440140
asked Sep 4 at 19:24
Vinicius Alexandre Zonta
161
New contributor
Vinicius Alexandre Zonta is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Sep 5 at 2:43
user21820
36.2k440140
edited Sep 5 at 2:43
user21820
36.2k440140
edited Sep 5 at 2:43
user21820
36.2k440140
36.2k440140
asked Sep 4 at 19:24
Vinicius Alexandre Zonta
161
New contributor
Vinicius Alexandre Zonta is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Sep 4 at 19:24
Vinicius Alexandre Zonta
161
asked Sep 4 at 19:24
Vinicius Alexandre Zonta
161
161
New contributor
Vinicius Alexandre Zonta is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Vinicius Alexandre Zonta is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Vinicius Alexandre Zonta is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
In general, the purpose of a formula for $N_0$ is so that when you use that formula in the proof, the proof will work. The formula your teacher used gives a much larger than minimal value of $N_0,$ which is fine (you can never go wrong by setting $N_0$ larger than needed), but to know the reason for choosing that particular formula it might help to see the specific steps of the proof.
– David K
Sep 4 at 19:52
Note that the standard in the English speaking world is to use the period to denote decimals ($epsilon =.001$).
– Acccumulation
Sep 5 at 0:09
@Acccumulation Note that the standard is not to skip the leading zero as well, $epsilon = 0.001$.
– Roman Odaisky
Sep 5 at 1:32
add a comment |Â
In general, the purpose of a formula for $N_0$ is so that when you use that formula in the proof, the proof will work. The formula your teacher used gives a much larger than minimal value of $N_0,$ which is fine (you can never go wrong by setting $N_0$ larger than needed), but to know the reason for choosing that particular formula it might help to see the specific steps of the proof.
– David K
Sep 4 at 19:52
Note that the standard in the English speaking world is to use the period to denote decimals ($epsilon =.001$).
– Acccumulation
Sep 5 at 0:09
@Acccumulation Note that the standard is not to skip the leading zero as well, $epsilon = 0.001$.
– Roman Odaisky
Sep 5 at 1:32
In general, the purpose of a formula for $N_0$ is so that when you use that formula in the proof, the proof will work. The formula your teacher used gives a much larger than minimal value of $N_0,$ which is fine (you can never go wrong by setting $N_0$ larger than needed), but to know the reason for choosing that particular formula it might help to see the specific steps of the proof.
– David K
Sep 4 at 19:52
In general, the purpose of a formula for $N_0$ is so that when you use that formula in the proof, the proof will work. The formula your teacher used gives a much larger than minimal value of $N_0,$ which is fine (you can never go wrong by setting $N_0$ larger than needed), but to know the reason for choosing that particular formula it might help to see the specific steps of the proof.
– David K
Sep 4 at 19:52
Note that the standard in the English speaking world is to use the period to denote decimals ($epsilon =.001$).
– Acccumulation
Sep 5 at 0:09
Note that the standard in the English speaking world is to use the period to denote decimals ($epsilon =.001$).
– Acccumulation
Sep 5 at 0:09
@Acccumulation Note that the standard is not to skip the leading zero as well, $epsilon = 0.001$.
– Roman Odaisky
Sep 5 at 1:32
@Acccumulation Note that the standard is not to skip the leading zero as well, $epsilon = 0.001$.
– Roman Odaisky
Sep 5 at 1:32
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
6
down vote
We have that
$$left|fracn+13n-1-frac13right|<epsilon iff frac43(3n-1)<epsilon. $$
Now
$$frac43(3n-1)<epsiloniff 9n-3>frac4epsiloniff 9n>frac4epsilon+3.$$ So we need
$$n>frac49epsilon+frac13.$$ So it is enough to have
$$n>left[frac49epsilonright]+left[frac13right]=left[frac49epsilonright]+1.$$
answered Sep 4 at 19:35
mfl
25.2k12141
Almost identical answers! You were quicker though. =)
– Sobi
Sep 4 at 19:37
add a comment |Â
up vote
6
down vote
We have
$$ a_n - L = fracn+13n-1- frac13 = frac3(n+1)-(3n-1)3(3n-1) = frac43(3n-1). $$
So we want
$$ frac43(3n-1) < epsilon quadLeftrightarrowquad 3n-1 > frac43epsilon quad Leftrightarrow quad n > frac49epsilon+frac13, $$
so we can take
$$ n_0 = left[ frac49epsilonright] + 1. $$
We have $+1$ here to be on the safe side, since we want the previous inequality to be strict.
In particular, if $epsilon = 0.001 = 10^-3$ this becomes
$$ n_0 = left[ frac40009right]+1 = left[ frac40009right]+1 = 445. $$
answered Sep 4 at 19:36
Sobi
2,653415
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
6
down vote
We have that
$$left|fracn+13n-1-frac13right|<epsilon iff frac43(3n-1)<epsilon. $$
Now
$$frac43(3n-1)<epsiloniff 9n-3>frac4epsiloniff 9n>frac4epsilon+3.$$ So we need
$$n>frac49epsilon+frac13.$$ So it is enough to have
$$n>left[frac49epsilonright]+left[frac13right]=left[frac49epsilonright]+1.$$
answered Sep 4 at 19:35
mfl
25.2k12141
Almost identical answers! You were quicker though. =)
– Sobi
Sep 4 at 19:37
add a comment |Â
up vote
6
down vote
We have that
$$left|fracn+13n-1-frac13right|<epsilon iff frac43(3n-1)<epsilon. $$
Now
$$frac43(3n-1)<epsiloniff 9n-3>frac4epsiloniff 9n>frac4epsilon+3.$$ So we need
$$n>frac49epsilon+frac13.$$ So it is enough to have
$$n>left[frac49epsilonright]+left[frac13right]=left[frac49epsilonright]+1.$$
answered Sep 4 at 19:35
mfl
25.2k12141
Almost identical answers! You were quicker though. =)
– Sobi
Sep 4 at 19:37
add a comment |Â
up vote
6
down vote
up vote
6
down vote
We have that
$$left|fracn+13n-1-frac13right|<epsilon iff frac43(3n-1)<epsilon. $$
Now
$$frac43(3n-1)<epsiloniff 9n-3>frac4epsiloniff 9n>frac4epsilon+3.$$ So we need
$$n>frac49epsilon+frac13.$$ So it is enough to have
$$n>left[frac49epsilonright]+left[frac13right]=left[frac49epsilonright]+1.$$
answered Sep 4 at 19:35
mfl
25.2k12141
We have that
$$left|fracn+13n-1-frac13right|<epsilon iff frac43(3n-1)<epsilon. $$
Now
$$frac43(3n-1)<epsiloniff 9n-3>frac4epsiloniff 9n>frac4epsilon+3.$$ So we need
$$n>frac49epsilon+frac13.$$ So it is enough to have
$$n>left[frac49epsilonright]+left[frac13right]=left[frac49epsilonright]+1.$$
answered Sep 4 at 19:35
mfl
25.2k12141
answered Sep 4 at 19:35
mfl
25.2k12141
answered Sep 4 at 19:35
mfl
25.2k12141
answered Sep 4 at 19:35
mfl
25.2k12141
25.2k12141
Almost identical answers! You were quicker though. =)
– Sobi
Sep 4 at 19:37
add a comment |Â
Almost identical answers! You were quicker though. =)
– Sobi
Sep 4 at 19:37
Almost identical answers! You were quicker though. =)
– Sobi
Sep 4 at 19:37
Almost identical answers! You were quicker though. =)
– Sobi
Sep 4 at 19:37
add a comment |Â
up vote
6
down vote
We have
$$ a_n - L = fracn+13n-1- frac13 = frac3(n+1)-(3n-1)3(3n-1) = frac43(3n-1). $$
So we want
$$ frac43(3n-1) < epsilon quadLeftrightarrowquad 3n-1 > frac43epsilon quad Leftrightarrow quad n > frac49epsilon+frac13, $$
so we can take
$$ n_0 = left[ frac49epsilonright] + 1. $$
We have $+1$ here to be on the safe side, since we want the previous inequality to be strict.
In particular, if $epsilon = 0.001 = 10^-3$ this becomes
$$ n_0 = left[ frac40009right]+1 = left[ frac40009right]+1 = 445. $$
answered Sep 4 at 19:36
Sobi
2,653415
add a comment |Â
up vote
6
down vote
We have
$$ a_n - L = fracn+13n-1- frac13 = frac3(n+1)-(3n-1)3(3n-1) = frac43(3n-1). $$
So we want
$$ frac43(3n-1) < epsilon quadLeftrightarrowquad 3n-1 > frac43epsilon quad Leftrightarrow quad n > frac49epsilon+frac13, $$
so we can take
$$ n_0 = left[ frac49epsilonright] + 1. $$
We have $+1$ here to be on the safe side, since we want the previous inequality to be strict.
In particular, if $epsilon = 0.001 = 10^-3$ this becomes
$$ n_0 = left[ frac40009right]+1 = left[ frac40009right]+1 = 445. $$
answered Sep 4 at 19:36
Sobi
2,653415
add a comment |Â
up vote
6
down vote
up vote
6
down vote
We have
$$ a_n - L = fracn+13n-1- frac13 = frac3(n+1)-(3n-1)3(3n-1) = frac43(3n-1). $$
So we want
$$ frac43(3n-1) < epsilon quadLeftrightarrowquad 3n-1 > frac43epsilon quad Leftrightarrow quad n > frac49epsilon+frac13, $$
so we can take
$$ n_0 = left[ frac49epsilonright] + 1. $$
We have $+1$ here to be on the safe side, since we want the previous inequality to be strict.
In particular, if $epsilon = 0.001 = 10^-3$ this becomes
$$ n_0 = left[ frac40009right]+1 = left[ frac40009right]+1 = 445. $$
answered Sep 4 at 19:36
Sobi
2,653415
We have
$$ a_n - L = fracn+13n-1- frac13 = frac3(n+1)-(3n-1)3(3n-1) = frac43(3n-1). $$
So we want
$$ frac43(3n-1) < epsilon quadLeftrightarrowquad 3n-1 > frac43epsilon quad Leftrightarrow quad n > frac49epsilon+frac13, $$
so we can take
$$ n_0 = left[ frac49epsilonright] + 1. $$
We have $+1$ here to be on the safe side, since we want the previous inequality to be strict.
In particular, if $epsilon = 0.001 = 10^-3$ this becomes
$$ n_0 = left[ frac40009right]+1 = left[ frac40009right]+1 = 445. $$
answered Sep 4 at 19:36
Sobi
2,653415
answered Sep 4 at 19:36
Sobi
2,653415
answered Sep 4 at 19:36
Sobi
2,653415
answered Sep 4 at 19:36
Sobi
2,653415
2,653415
add a comment |Â
add a comment |Â
Vinicius Alexandre Zonta is a new contributor. Be nice, and check out our Code of Conduct.
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In general, the purpose of a formula for $N_0$ is so that when you use that formula in the proof, the proof will work. The formula your teacher used gives a much larger than minimal value of $N_0,$ which is fine (you can never go wrong by setting $N_0$ larger than needed), but to know the reason for choosing that particular formula it might help to see the specific steps of the proof.
– David K
Sep 4 at 19:52
Note that the standard in the English speaking world is to use the period to denote decimals ($epsilon =.001$).
– Acccumulation
Sep 5 at 0:09
@Acccumulation Note that the standard is not to skip the leading zero as well, $epsilon = 0.001$.
– Roman Odaisky
Sep 5 at 1:32