Show that $|frac{1}{2n}-frac{1}{2m}| frac{1}{epsilon}.$
In Example 1.5-9 of the book Functional Analysis by Kreyszig it claims that $|frac{1}{2n}-frac{1}{2m}| < epsilon$ holds for all $m, n > frac{1}{epsilon}.$
My calculations don't lead to that! :
1st objection: wlog let $m>n$ and $m=n+k$ so $$|frac{1}{2n}-frac{1}{2m}| < epsilon implies k-2epsilon(n)(n+k)<0.$$ Differentiation with respect to $k$ and desiring for the $k-2epsilon(n)(n+k) $ to decrease after its 'right-zero' gives $n > frac{1}{2epsilon} $ and thus $m, n > frac{1}{2epsilon}.$
2nd objection: $k-2epsilon(n)(n+k)= k-2epsilon n^2 -2epsilon nk <0$ holds for $n> dfrac{-2epsilon k + sqrt{4{epsilon}^2k^2+8epsilon k}}{4epsilon k}$. How change of $k$ doesn't effect $k-2epsilon n^2 -2epsilon nk <0$?
3rd objection: how intersection of two domains for $n,m$ obtained in 1st and 2nd objections is a superset of $n,m > frac{1}{epsilon}? $
real-analysis quadratics cauchy-sequences
edited Nov 29 '18 at 6:11
epimorphic
2,73631533
asked Nov 29 '18 at 3:08
72D
572116
add a comment |
In Example 1.5-9 of the book Functional Analysis by Kreyszig it claims that $|frac{1}{2n}-frac{1}{2m}| < epsilon$ holds for all $m, n > frac{1}{epsilon}.$
My calculations don't lead to that! :
1st objection: wlog let $m>n$ and $m=n+k$ so $$|frac{1}{2n}-frac{1}{2m}| < epsilon implies k-2epsilon(n)(n+k)<0.$$ Differentiation with respect to $k$ and desiring for the $k-2epsilon(n)(n+k) $ to decrease after its 'right-zero' gives $n > frac{1}{2epsilon} $ and thus $m, n > frac{1}{2epsilon}.$
2nd objection: $k-2epsilon(n)(n+k)= k-2epsilon n^2 -2epsilon nk <0$ holds for $n> dfrac{-2epsilon k + sqrt{4{epsilon}^2k^2+8epsilon k}}{4epsilon k}$. How change of $k$ doesn't effect $k-2epsilon n^2 -2epsilon nk <0$?
3rd objection: how intersection of two domains for $n,m$ obtained in 1st and 2nd objections is a superset of $n,m > frac{1}{epsilon}? $
real-analysis quadratics cauchy-sequences
edited Nov 29 '18 at 6:11
epimorphic
2,73631533
asked Nov 29 '18 at 3:08
72D
572116
1
How is thealgebra-precalculustag relevant when you've used differentiation?
– Shaun
Nov 29 '18 at 3:17
@Shaun, isn't it algebra minus precalculus? ;-)
– 72D
Nov 29 '18 at 3:19
Very clever, but no. The dash is used instead of a space because a space is used to end the current tag name when typing it in.
– Shaun
Nov 29 '18 at 3:21
1
@Shaun, I know I was kidding :-)) but I didn't know that algebra-precalculus shouldn't be used for differentiation content. I'll edit.
– 72D
Nov 29 '18 at 3:24
1
Yeah, I thought so. And thank you $ddotsmile$
– Shaun
Nov 29 '18 at 3:25
add a comment |
In Example 1.5-9 of the book Functional Analysis by Kreyszig it claims that $|frac{1}{2n}-frac{1}{2m}| < epsilon$ holds for all $m, n > frac{1}{epsilon}.$
My calculations don't lead to that! :
1st objection: wlog let $m>n$ and $m=n+k$ so $$|frac{1}{2n}-frac{1}{2m}| < epsilon implies k-2epsilon(n)(n+k)<0.$$ Differentiation with respect to $k$ and desiring for the $k-2epsilon(n)(n+k) $ to decrease after its 'right-zero' gives $n > frac{1}{2epsilon} $ and thus $m, n > frac{1}{2epsilon}.$
2nd objection: $k-2epsilon(n)(n+k)= k-2epsilon n^2 -2epsilon nk <0$ holds for $n> dfrac{-2epsilon k + sqrt{4{epsilon}^2k^2+8epsilon k}}{4epsilon k}$. How change of $k$ doesn't effect $k-2epsilon n^2 -2epsilon nk <0$?
3rd objection: how intersection of two domains for $n,m$ obtained in 1st and 2nd objections is a superset of $n,m > frac{1}{epsilon}? $
real-analysis quadratics cauchy-sequences
edited Nov 29 '18 at 6:11
epimorphic
2,73631533
asked Nov 29 '18 at 3:08
72D
572116
In Example 1.5-9 of the book Functional Analysis by Kreyszig it claims that $|frac{1}{2n}-frac{1}{2m}| < epsilon$ holds for all $m, n > frac{1}{epsilon}.$
My calculations don't lead to that! :
1st objection: wlog let $m>n$ and $m=n+k$ so $$|frac{1}{2n}-frac{1}{2m}| < epsilon implies k-2epsilon(n)(n+k)<0.$$ Differentiation with respect to $k$ and desiring for the $k-2epsilon(n)(n+k) $ to decrease after its 'right-zero' gives $n > frac{1}{2epsilon} $ and thus $m, n > frac{1}{2epsilon}.$
2nd objection: $k-2epsilon(n)(n+k)= k-2epsilon n^2 -2epsilon nk <0$ holds for $n> dfrac{-2epsilon k + sqrt{4{epsilon}^2k^2+8epsilon k}}{4epsilon k}$. How change of $k$ doesn't effect $k-2epsilon n^2 -2epsilon nk <0$?
3rd objection: how intersection of two domains for $n,m$ obtained in 1st and 2nd objections is a superset of $n,m > frac{1}{epsilon}? $
real-analysis quadratics cauchy-sequences
real-analysis quadratics cauchy-sequences
edited Nov 29 '18 at 6:11
epimorphic
2,73631533
asked Nov 29 '18 at 3:08
72D
572116
edited Nov 29 '18 at 6:11
epimorphic
2,73631533
asked Nov 29 '18 at 3:08
72D
572116
edited Nov 29 '18 at 6:11
epimorphic
2,73631533
edited Nov 29 '18 at 6:11
epimorphic
2,73631533
edited Nov 29 '18 at 6:11
epimorphic
2,73631533
2,73631533
asked Nov 29 '18 at 3:08
72D
572116
asked Nov 29 '18 at 3:08
72D
572116
asked Nov 29 '18 at 3:08
72D
572116
572116
1
How is thealgebra-precalculustag relevant when you've used differentiation?
– Shaun
Nov 29 '18 at 3:17
@Shaun, isn't it algebra minus precalculus? ;-)
– 72D
Nov 29 '18 at 3:19
Very clever, but no. The dash is used instead of a space because a space is used to end the current tag name when typing it in.
– Shaun
Nov 29 '18 at 3:21
1
@Shaun, I know I was kidding :-)) but I didn't know that algebra-precalculus shouldn't be used for differentiation content. I'll edit.
– 72D
Nov 29 '18 at 3:24
1
Yeah, I thought so. And thank you $ddotsmile$
– Shaun
Nov 29 '18 at 3:25
add a comment |
1
How is thealgebra-precalculustag relevant when you've used differentiation?
– Shaun
Nov 29 '18 at 3:17
@Shaun, isn't it algebra minus precalculus? ;-)
– 72D
Nov 29 '18 at 3:19
Very clever, but no. The dash is used instead of a space because a space is used to end the current tag name when typing it in.
– Shaun
Nov 29 '18 at 3:21
1
@Shaun, I know I was kidding :-)) but I didn't know that algebra-precalculus shouldn't be used for differentiation content. I'll edit.
– 72D
Nov 29 '18 at 3:24
1
Yeah, I thought so. And thank you $ddotsmile$
– Shaun
Nov 29 '18 at 3:25
1
1
How is the
algebra-precalculus tag relevant when you've used differentiation?– Shaun
Nov 29 '18 at 3:17
How is the
algebra-precalculus tag relevant when you've used differentiation?– Shaun
Nov 29 '18 at 3:17
@Shaun, isn't it algebra minus precalculus? ;-)
– 72D
Nov 29 '18 at 3:19
@Shaun, isn't it algebra minus precalculus? ;-)
– 72D
Nov 29 '18 at 3:19
Very clever, but no. The dash is used instead of a space because a space is used to end the current tag name when typing it in.
– Shaun
Nov 29 '18 at 3:21
Very clever, but no. The dash is used instead of a space because a space is used to end the current tag name when typing it in.
– Shaun
Nov 29 '18 at 3:21
1
1
@Shaun, I know I was kidding :-)) but I didn't know that algebra-precalculus shouldn't be used for differentiation content. I'll edit.
– 72D
Nov 29 '18 at 3:24
@Shaun, I know I was kidding :-)) but I didn't know that algebra-precalculus shouldn't be used for differentiation content. I'll edit.
– 72D
Nov 29 '18 at 3:24
1
1
Yeah, I thought so. And thank you $ddotsmile$
– Shaun
Nov 29 '18 at 3:25
Yeah, I thought so. And thank you $ddotsmile$
– Shaun
Nov 29 '18 at 3:25
add a comment |
2 Answers
2
active
oldest
votes
$m,n>{1overepsilon}$ implies that ${1over 2n}<epsilon/2$. This implies that $|{1over 2n}-{1over 2m}|leq |{1over 2n}|+|{1over 2m}|<epsilon/2+epsilon/2=epsilon$.
answered Nov 29 '18 at 3:14
Tsemo Aristide
56.1k11444
add a comment |
You can do a little better.
I write $c$ for $epsilon$
because lazy.
Since
$m, n > frac1{c}
$,
$0 le frac1{m}, frac1{n}
le c
$.
If
$m = n$
then
$|frac1{m}-frac1{n}|
= 0 < c$.
If
$m > n$
then
$|frac1{m}-frac1{n}|
=frac1{n}-frac1{m}
lt frac1{n}
< c$.
If
$m < n$
then
$|frac1{m}-frac1{n}|
=frac1{m}-frac1{n}
lt frac1{m}
< c$.
In all cases,
$|frac1{m}-frac1{n}|
< c$
so that
$|frac1{2m}-frac1{2n}|
< frac{c}{2}$.
answered Nov 29 '18 at 5:38
marty cohen
72.6k549127
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$m,n>{1overepsilon}$ implies that ${1over 2n}<epsilon/2$. This implies that $|{1over 2n}-{1over 2m}|leq |{1over 2n}|+|{1over 2m}|<epsilon/2+epsilon/2=epsilon$.
answered Nov 29 '18 at 3:14
Tsemo Aristide
56.1k11444
add a comment |
$m,n>{1overepsilon}$ implies that ${1over 2n}<epsilon/2$. This implies that $|{1over 2n}-{1over 2m}|leq |{1over 2n}|+|{1over 2m}|<epsilon/2+epsilon/2=epsilon$.
answered Nov 29 '18 at 3:14
Tsemo Aristide
56.1k11444
add a comment |
$m,n>{1overepsilon}$ implies that ${1over 2n}<epsilon/2$. This implies that $|{1over 2n}-{1over 2m}|leq |{1over 2n}|+|{1over 2m}|<epsilon/2+epsilon/2=epsilon$.
answered Nov 29 '18 at 3:14
Tsemo Aristide
56.1k11444
$m,n>{1overepsilon}$ implies that ${1over 2n}<epsilon/2$. This implies that $|{1over 2n}-{1over 2m}|leq |{1over 2n}|+|{1over 2m}|<epsilon/2+epsilon/2=epsilon$.
answered Nov 29 '18 at 3:14
Tsemo Aristide
56.1k11444
answered Nov 29 '18 at 3:14
Tsemo Aristide
56.1k11444
answered Nov 29 '18 at 3:14
Tsemo Aristide
56.1k11444
answered Nov 29 '18 at 3:14
Tsemo Aristide
56.1k11444
56.1k11444
add a comment |
add a comment |
You can do a little better.
I write $c$ for $epsilon$
because lazy.
Since
$m, n > frac1{c}
$,
$0 le frac1{m}, frac1{n}
le c
$.
If
$m = n$
then
$|frac1{m}-frac1{n}|
= 0 < c$.
If
$m > n$
then
$|frac1{m}-frac1{n}|
=frac1{n}-frac1{m}
lt frac1{n}
< c$.
If
$m < n$
then
$|frac1{m}-frac1{n}|
=frac1{m}-frac1{n}
lt frac1{m}
< c$.
In all cases,
$|frac1{m}-frac1{n}|
< c$
so that
$|frac1{2m}-frac1{2n}|
< frac{c}{2}$.
answered Nov 29 '18 at 5:38
marty cohen
72.6k549127
add a comment |
You can do a little better.
I write $c$ for $epsilon$
because lazy.
Since
$m, n > frac1{c}
$,
$0 le frac1{m}, frac1{n}
le c
$.
If
$m = n$
then
$|frac1{m}-frac1{n}|
= 0 < c$.
If
$m > n$
then
$|frac1{m}-frac1{n}|
=frac1{n}-frac1{m}
lt frac1{n}
< c$.
If
$m < n$
then
$|frac1{m}-frac1{n}|
=frac1{m}-frac1{n}
lt frac1{m}
< c$.
In all cases,
$|frac1{m}-frac1{n}|
< c$
so that
$|frac1{2m}-frac1{2n}|
< frac{c}{2}$.
answered Nov 29 '18 at 5:38
marty cohen
72.6k549127
add a comment |
You can do a little better.
I write $c$ for $epsilon$
because lazy.
Since
$m, n > frac1{c}
$,
$0 le frac1{m}, frac1{n}
le c
$.
If
$m = n$
then
$|frac1{m}-frac1{n}|
= 0 < c$.
If
$m > n$
then
$|frac1{m}-frac1{n}|
=frac1{n}-frac1{m}
lt frac1{n}
< c$.
If
$m < n$
then
$|frac1{m}-frac1{n}|
=frac1{m}-frac1{n}
lt frac1{m}
< c$.
In all cases,
$|frac1{m}-frac1{n}|
< c$
so that
$|frac1{2m}-frac1{2n}|
< frac{c}{2}$.
answered Nov 29 '18 at 5:38
marty cohen
72.6k549127
You can do a little better.
I write $c$ for $epsilon$
because lazy.
Since
$m, n > frac1{c}
$,
$0 le frac1{m}, frac1{n}
le c
$.
If
$m = n$
then
$|frac1{m}-frac1{n}|
= 0 < c$.
If
$m > n$
then
$|frac1{m}-frac1{n}|
=frac1{n}-frac1{m}
lt frac1{n}
< c$.
If
$m < n$
then
$|frac1{m}-frac1{n}|
=frac1{m}-frac1{n}
lt frac1{m}
< c$.
In all cases,
$|frac1{m}-frac1{n}|
< c$
so that
$|frac1{2m}-frac1{2n}|
< frac{c}{2}$.
answered Nov 29 '18 at 5:38
marty cohen
72.6k549127
answered Nov 29 '18 at 5:38
marty cohen
72.6k549127
answered Nov 29 '18 at 5:38
marty cohen
72.6k549127
answered Nov 29 '18 at 5:38
marty cohen
72.6k549127
72.6k549127
add a comment |
add a comment |
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1
How is the
algebra-precalculustag relevant when you've used differentiation?– Shaun
Nov 29 '18 at 3:17
@Shaun, isn't it algebra minus precalculus? ;-)
– 72D
Nov 29 '18 at 3:19
Very clever, but no. The dash is used instead of a space because a space is used to end the current tag name when typing it in.
– Shaun
Nov 29 '18 at 3:21
1
@Shaun, I know I was kidding :-)) but I didn't know that algebra-precalculus shouldn't be used for differentiation content. I'll edit.
– 72D
Nov 29 '18 at 3:24
1
Yeah, I thought so. And thank you $ddotsmile$
– Shaun
Nov 29 '18 at 3:25