Mixing
Show that if $f''(0)$ exists and $f(0)=f'(0)=0$, then $sum f(frac1n)$ converges
$begingroup$
I need help with this problem:
Let $f$ be a continous function over an interval that contains $0$. Let $a_n = f(frac 1 n)$ (for n large enough).
- Show that if $f''(0)$ exists and $f(0)=f'(0)=0$, then $sum_{n=1}^infty a_n$ converges.
I've already shown that if $sum_{n=1}^infty a_n$ converges, then $f(0)=0$ and that if $f'(0)$ exists and $sum_{n=1}^infty a_n$ converges, then $f'(0)=0$.
How do I show this one?
sequences-and-series convergence
asked Feb 1 at 5:49
davidllerenavdavidllerenav
3128
$endgroup$
add a comment |
$begingroup$
I need help with this problem:
Let $f$ be a continous function over an interval that contains $0$. Let $a_n = f(frac 1 n)$ (for n large enough).
- Show that if $f''(0)$ exists and $f(0)=f'(0)=0$, then $sum_{n=1}^infty a_n$ converges.
I've already shown that if $sum_{n=1}^infty a_n$ converges, then $f(0)=0$ and that if $f'(0)$ exists and $sum_{n=1}^infty a_n$ converges, then $f'(0)=0$.
How do I show this one?
sequences-and-series convergence
asked Feb 1 at 5:49
davidllerenavdavidllerenav
3128
$endgroup$
add a comment |
$begingroup$
I need help with this problem:
Let $f$ be a continous function over an interval that contains $0$. Let $a_n = f(frac 1 n)$ (for n large enough).
- Show that if $f''(0)$ exists and $f(0)=f'(0)=0$, then $sum_{n=1}^infty a_n$ converges.
I've already shown that if $sum_{n=1}^infty a_n$ converges, then $f(0)=0$ and that if $f'(0)$ exists and $sum_{n=1}^infty a_n$ converges, then $f'(0)=0$.
How do I show this one?
sequences-and-series convergence
asked Feb 1 at 5:49
davidllerenavdavidllerenav
3128
$endgroup$
I need help with this problem:
Let $f$ be a continous function over an interval that contains $0$. Let $a_n = f(frac 1 n)$ (for n large enough).
- Show that if $f''(0)$ exists and $f(0)=f'(0)=0$, then $sum_{n=1}^infty a_n$ converges.
I've already shown that if $sum_{n=1}^infty a_n$ converges, then $f(0)=0$ and that if $f'(0)$ exists and $sum_{n=1}^infty a_n$ converges, then $f'(0)=0$.
How do I show this one?
sequences-and-series convergence
sequences-and-series convergence
asked Feb 1 at 5:49
davidllerenavdavidllerenav
3128
asked Feb 1 at 5:49
davidllerenavdavidllerenav
3128
edited Feb 1 at 16:46
davidllerenav
asked Feb 1 at 5:49
davidllerenavdavidllerenav
3128
asked Feb 1 at 5:49
davidllerenavdavidllerenav
3128
asked Feb 1 at 5:49
davidllerenavdavidllerenav
3128
3128
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
We can use comparison test based on the following fact
$$
lim_{xto 0} frac{f(x)}{x^2}=lim_{xto 0}frac{f'(x)}{2x}=frac{f''(0)}{2}
$$ where L'Hopital's rule is used in the first equation. Thus for all sufficiently large $nge N$,
$$
n^2|a_n| = n^2left|fleft(frac1nright)right|le |f''(0)|
$$ hence
$$
sum_{nge 1}|a_n| le sum_{n=1}^{N-1} |a_n|+sum_{nge N} frac{|f''(0)|}{n^2}<infty.
$$ The series converges absolutely, which implies that the series converges.
answered Feb 1 at 7:20
SongSong
18.6k21651
$endgroup$
$begingroup$
Why did you compare it to $x^2$?
$endgroup$
– davidllerenav
Feb 1 at 16:33
$begingroup$
@davidllerenav It is a natural choice in view of the Taylor-Mclaurin expansion $f(x) = f(0)+f'(0)x +frac{f''(0)}2 x^2 +cdots cong frac{f''(0)}2 x^2$. Also, $2$ is the first exponent $rin Bbb N$ such that $sum_{nge 1} frac1{n^r}<infty$. (Thus we find that $sum_{nge 1}f(1/n)cong sum_{nge 1} frac{f'(0)}{n}$ does not converge if $f'(0)ne 0$.)
$endgroup$
– Song
Feb 1 at 16:59
$begingroup$
Ok, I think I got it. Can't I also do it by the limit comparison? Because I already proved that if $f'(0)$ exists and $sum_{n=1}^infty a_n$ converges then $f'(0)=0$. I did it by using the derivative definition and supposing that $f'(0)$ isn't 0, ending up with $lim_ntoinfty nf(frac 1 n)=t$. Then by the limit comparison test, I reached a contradiction, thus $f'(0)=0$. Can't I do it that way instead?
$endgroup$
– davidllerenav
Feb 1 at 17:08
$begingroup$
@davidllerenav Yes, you can also do it by limit comparison test. Indeed, what I wrote in my answer can be viewed as a logic behind applying limit comparison test.
$endgroup$
– Song
Feb 1 at 17:10
$begingroup$
So if I just write the same thing but with $f''(0)$ and show that $f''(0)=0$, then $sum_{n=1}^infty a_n$ must converge?
$endgroup$
– davidllerenav
Feb 1 at 17:13
|
show 2 more comments
$begingroup$
By Taylor's Formula with remainder $|f(x)| leq cx^{2}$ for some constant $C$ for $x$ in a neighborhoodod $0$. Hence $sum f(frac 1 n)$ is dominated by a constant times $sum frac 1 {n^{2}}$.
answered Feb 1 at 6:00
Kavi Rama MurthyKavi Rama Murthy
73.7k53170
$endgroup$
$begingroup$
Why is the remainder $cx^2$? Is there another way of doing it?
$endgroup$
– davidllerenav
Feb 1 at 16:31
$begingroup$
$frac {f'(x)|} {|x|}=frac {|f'(x)-f'(0)|} {|x|} to f''(0)$. From this it follows that $|f'(x)| <(1+|f''(0)|) |x|$ if $|x|$ is sufficiently small. Now $|f(x))=|f(x)-f(0)|=|x||f'zeta)|$ for some $zeta$ between $0$ and $x$. Is it clear now that $|f(x)| leq Mx^{2}$ for some $M$, for $|x|$ is sufficiently small?
$endgroup$
– Kavi Rama Murthy
Feb 1 at 23:18
add a comment |
$begingroup$
As $f''(0)$ exists, $f'(0)$ is can be linearly approximated:
$$ f'(h) = mcdot h+ r(h)$$
With some function $r$ with $frac {r(h)} hto 0$ for $hto 0 $. Therefore we can find an interval $U:=[-a,a]$ so that $ r(h) le mcdot h$, and by that:
$$xin Uimplies f'(x) le 2 mcdot x
$$
From this follows $f(x) le mx^2$ per integration for $xin U$, and thus the sum converges
(as for some $N$ the tail of the sum, $sum_{i=N}^infty f(1/i)$, is completely in $U$ for all $i$ it assumes, and we know that $sum_{i=0}^infty frac 1 {x^2}$ converges).
answered Feb 1 at 7:12
SudixSudix
9601316
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
We can use comparison test based on the following fact
$$
lim_{xto 0} frac{f(x)}{x^2}=lim_{xto 0}frac{f'(x)}{2x}=frac{f''(0)}{2}
$$ where L'Hopital's rule is used in the first equation. Thus for all sufficiently large $nge N$,
$$
n^2|a_n| = n^2left|fleft(frac1nright)right|le |f''(0)|
$$ hence
$$
sum_{nge 1}|a_n| le sum_{n=1}^{N-1} |a_n|+sum_{nge N} frac{|f''(0)|}{n^2}<infty.
$$ The series converges absolutely, which implies that the series converges.
answered Feb 1 at 7:20
SongSong
18.6k21651
$endgroup$
$begingroup$
Why did you compare it to $x^2$?
$endgroup$
– davidllerenav
Feb 1 at 16:33
$begingroup$
@davidllerenav It is a natural choice in view of the Taylor-Mclaurin expansion $f(x) = f(0)+f'(0)x +frac{f''(0)}2 x^2 +cdots cong frac{f''(0)}2 x^2$. Also, $2$ is the first exponent $rin Bbb N$ such that $sum_{nge 1} frac1{n^r}<infty$. (Thus we find that $sum_{nge 1}f(1/n)cong sum_{nge 1} frac{f'(0)}{n}$ does not converge if $f'(0)ne 0$.)
$endgroup$
– Song
Feb 1 at 16:59
$begingroup$
Ok, I think I got it. Can't I also do it by the limit comparison? Because I already proved that if $f'(0)$ exists and $sum_{n=1}^infty a_n$ converges then $f'(0)=0$. I did it by using the derivative definition and supposing that $f'(0)$ isn't 0, ending up with $lim_ntoinfty nf(frac 1 n)=t$. Then by the limit comparison test, I reached a contradiction, thus $f'(0)=0$. Can't I do it that way instead?
$endgroup$
– davidllerenav
Feb 1 at 17:08
$begingroup$
@davidllerenav Yes, you can also do it by limit comparison test. Indeed, what I wrote in my answer can be viewed as a logic behind applying limit comparison test.
$endgroup$
– Song
Feb 1 at 17:10
$begingroup$
So if I just write the same thing but with $f''(0)$ and show that $f''(0)=0$, then $sum_{n=1}^infty a_n$ must converge?
$endgroup$
– davidllerenav
Feb 1 at 17:13
|
show 2 more comments
$begingroup$
We can use comparison test based on the following fact
$$
lim_{xto 0} frac{f(x)}{x^2}=lim_{xto 0}frac{f'(x)}{2x}=frac{f''(0)}{2}
$$ where L'Hopital's rule is used in the first equation. Thus for all sufficiently large $nge N$,
$$
n^2|a_n| = n^2left|fleft(frac1nright)right|le |f''(0)|
$$ hence
$$
sum_{nge 1}|a_n| le sum_{n=1}^{N-1} |a_n|+sum_{nge N} frac{|f''(0)|}{n^2}<infty.
$$ The series converges absolutely, which implies that the series converges.
answered Feb 1 at 7:20
SongSong
18.6k21651
$endgroup$
$begingroup$
Why did you compare it to $x^2$?
$endgroup$
– davidllerenav
Feb 1 at 16:33
$begingroup$
@davidllerenav It is a natural choice in view of the Taylor-Mclaurin expansion $f(x) = f(0)+f'(0)x +frac{f''(0)}2 x^2 +cdots cong frac{f''(0)}2 x^2$. Also, $2$ is the first exponent $rin Bbb N$ such that $sum_{nge 1} frac1{n^r}<infty$. (Thus we find that $sum_{nge 1}f(1/n)cong sum_{nge 1} frac{f'(0)}{n}$ does not converge if $f'(0)ne 0$.)
$endgroup$
– Song
Feb 1 at 16:59
$begingroup$
Ok, I think I got it. Can't I also do it by the limit comparison? Because I already proved that if $f'(0)$ exists and $sum_{n=1}^infty a_n$ converges then $f'(0)=0$. I did it by using the derivative definition and supposing that $f'(0)$ isn't 0, ending up with $lim_ntoinfty nf(frac 1 n)=t$. Then by the limit comparison test, I reached a contradiction, thus $f'(0)=0$. Can't I do it that way instead?
$endgroup$
– davidllerenav
Feb 1 at 17:08
$begingroup$
@davidllerenav Yes, you can also do it by limit comparison test. Indeed, what I wrote in my answer can be viewed as a logic behind applying limit comparison test.
$endgroup$
– Song
Feb 1 at 17:10
$begingroup$
So if I just write the same thing but with $f''(0)$ and show that $f''(0)=0$, then $sum_{n=1}^infty a_n$ must converge?
$endgroup$
– davidllerenav
Feb 1 at 17:13
|
show 2 more comments
$begingroup$
We can use comparison test based on the following fact
$$
lim_{xto 0} frac{f(x)}{x^2}=lim_{xto 0}frac{f'(x)}{2x}=frac{f''(0)}{2}
$$ where L'Hopital's rule is used in the first equation. Thus for all sufficiently large $nge N$,
$$
n^2|a_n| = n^2left|fleft(frac1nright)right|le |f''(0)|
$$ hence
$$
sum_{nge 1}|a_n| le sum_{n=1}^{N-1} |a_n|+sum_{nge N} frac{|f''(0)|}{n^2}<infty.
$$ The series converges absolutely, which implies that the series converges.
answered Feb 1 at 7:20
SongSong
18.6k21651
$endgroup$
We can use comparison test based on the following fact
$$
lim_{xto 0} frac{f(x)}{x^2}=lim_{xto 0}frac{f'(x)}{2x}=frac{f''(0)}{2}
$$ where L'Hopital's rule is used in the first equation. Thus for all sufficiently large $nge N$,
$$
n^2|a_n| = n^2left|fleft(frac1nright)right|le |f''(0)|
$$ hence
$$
sum_{nge 1}|a_n| le sum_{n=1}^{N-1} |a_n|+sum_{nge N} frac{|f''(0)|}{n^2}<infty.
$$ The series converges absolutely, which implies that the series converges.
answered Feb 1 at 7:20
SongSong
18.6k21651
answered Feb 1 at 7:20
SongSong
18.6k21651
answered Feb 1 at 7:20
SongSong
18.6k21651
answered Feb 1 at 7:20
SongSong
18.6k21651
18.6k21651
$begingroup$
Why did you compare it to $x^2$?
$endgroup$
– davidllerenav
Feb 1 at 16:33
$begingroup$
@davidllerenav It is a natural choice in view of the Taylor-Mclaurin expansion $f(x) = f(0)+f'(0)x +frac{f''(0)}2 x^2 +cdots cong frac{f''(0)}2 x^2$. Also, $2$ is the first exponent $rin Bbb N$ such that $sum_{nge 1} frac1{n^r}<infty$. (Thus we find that $sum_{nge 1}f(1/n)cong sum_{nge 1} frac{f'(0)}{n}$ does not converge if $f'(0)ne 0$.)
$endgroup$
– Song
Feb 1 at 16:59
$begingroup$
Ok, I think I got it. Can't I also do it by the limit comparison? Because I already proved that if $f'(0)$ exists and $sum_{n=1}^infty a_n$ converges then $f'(0)=0$. I did it by using the derivative definition and supposing that $f'(0)$ isn't 0, ending up with $lim_ntoinfty nf(frac 1 n)=t$. Then by the limit comparison test, I reached a contradiction, thus $f'(0)=0$. Can't I do it that way instead?
$endgroup$
– davidllerenav
Feb 1 at 17:08
$begingroup$
@davidllerenav Yes, you can also do it by limit comparison test. Indeed, what I wrote in my answer can be viewed as a logic behind applying limit comparison test.
$endgroup$
– Song
Feb 1 at 17:10
$begingroup$
So if I just write the same thing but with $f''(0)$ and show that $f''(0)=0$, then $sum_{n=1}^infty a_n$ must converge?
$endgroup$
– davidllerenav
Feb 1 at 17:13
|
show 2 more comments
$begingroup$
Why did you compare it to $x^2$?
$endgroup$
– davidllerenav
Feb 1 at 16:33
$begingroup$
@davidllerenav It is a natural choice in view of the Taylor-Mclaurin expansion $f(x) = f(0)+f'(0)x +frac{f''(0)}2 x^2 +cdots cong frac{f''(0)}2 x^2$. Also, $2$ is the first exponent $rin Bbb N$ such that $sum_{nge 1} frac1{n^r}<infty$. (Thus we find that $sum_{nge 1}f(1/n)cong sum_{nge 1} frac{f'(0)}{n}$ does not converge if $f'(0)ne 0$.)
$endgroup$
– Song
Feb 1 at 16:59
$begingroup$
Ok, I think I got it. Can't I also do it by the limit comparison? Because I already proved that if $f'(0)$ exists and $sum_{n=1}^infty a_n$ converges then $f'(0)=0$. I did it by using the derivative definition and supposing that $f'(0)$ isn't 0, ending up with $lim_ntoinfty nf(frac 1 n)=t$. Then by the limit comparison test, I reached a contradiction, thus $f'(0)=0$. Can't I do it that way instead?
$endgroup$
– davidllerenav
Feb 1 at 17:08
$begingroup$
@davidllerenav Yes, you can also do it by limit comparison test. Indeed, what I wrote in my answer can be viewed as a logic behind applying limit comparison test.
$endgroup$
– Song
Feb 1 at 17:10
$begingroup$
So if I just write the same thing but with $f''(0)$ and show that $f''(0)=0$, then $sum_{n=1}^infty a_n$ must converge?
$endgroup$
– davidllerenav
Feb 1 at 17:13
$begingroup$
Why did you compare it to $x^2$?
$endgroup$
– davidllerenav
Feb 1 at 16:33
$begingroup$
Why did you compare it to $x^2$?
$endgroup$
– davidllerenav
Feb 1 at 16:33
$begingroup$
@davidllerenav It is a natural choice in view of the Taylor-Mclaurin expansion $f(x) = f(0)+f'(0)x +frac{f''(0)}2 x^2 +cdots cong frac{f''(0)}2 x^2$. Also, $2$ is the first exponent $rin Bbb N$ such that $sum_{nge 1} frac1{n^r}<infty$. (Thus we find that $sum_{nge 1}f(1/n)cong sum_{nge 1} frac{f'(0)}{n}$ does not converge if $f'(0)ne 0$.)
$endgroup$
– Song
Feb 1 at 16:59
$begingroup$
@davidllerenav It is a natural choice in view of the Taylor-Mclaurin expansion $f(x) = f(0)+f'(0)x +frac{f''(0)}2 x^2 +cdots cong frac{f''(0)}2 x^2$. Also, $2$ is the first exponent $rin Bbb N$ such that $sum_{nge 1} frac1{n^r}<infty$. (Thus we find that $sum_{nge 1}f(1/n)cong sum_{nge 1} frac{f'(0)}{n}$ does not converge if $f'(0)ne 0$.)
$endgroup$
– Song
Feb 1 at 16:59
$begingroup$
Ok, I think I got it. Can't I also do it by the limit comparison? Because I already proved that if $f'(0)$ exists and $sum_{n=1}^infty a_n$ converges then $f'(0)=0$. I did it by using the derivative definition and supposing that $f'(0)$ isn't 0, ending up with $lim_ntoinfty nf(frac 1 n)=t$. Then by the limit comparison test, I reached a contradiction, thus $f'(0)=0$. Can't I do it that way instead?
$endgroup$
– davidllerenav
Feb 1 at 17:08
$begingroup$
Ok, I think I got it. Can't I also do it by the limit comparison? Because I already proved that if $f'(0)$ exists and $sum_{n=1}^infty a_n$ converges then $f'(0)=0$. I did it by using the derivative definition and supposing that $f'(0)$ isn't 0, ending up with $lim_ntoinfty nf(frac 1 n)=t$. Then by the limit comparison test, I reached a contradiction, thus $f'(0)=0$. Can't I do it that way instead?
$endgroup$
– davidllerenav
Feb 1 at 17:08
$begingroup$
@davidllerenav Yes, you can also do it by limit comparison test. Indeed, what I wrote in my answer can be viewed as a logic behind applying limit comparison test.
$endgroup$
– Song
Feb 1 at 17:10
$begingroup$
@davidllerenav Yes, you can also do it by limit comparison test. Indeed, what I wrote in my answer can be viewed as a logic behind applying limit comparison test.
$endgroup$
– Song
Feb 1 at 17:10
$begingroup$
So if I just write the same thing but with $f''(0)$ and show that $f''(0)=0$, then $sum_{n=1}^infty a_n$ must converge?
$endgroup$
– davidllerenav
Feb 1 at 17:13
$begingroup$
So if I just write the same thing but with $f''(0)$ and show that $f''(0)=0$, then $sum_{n=1}^infty a_n$ must converge?
$endgroup$
– davidllerenav
Feb 1 at 17:13
|
show 2 more comments
$begingroup$
By Taylor's Formula with remainder $|f(x)| leq cx^{2}$ for some constant $C$ for $x$ in a neighborhoodod $0$. Hence $sum f(frac 1 n)$ is dominated by a constant times $sum frac 1 {n^{2}}$.
answered Feb 1 at 6:00
Kavi Rama MurthyKavi Rama Murthy
73.7k53170
$endgroup$
$begingroup$
Why is the remainder $cx^2$? Is there another way of doing it?
$endgroup$
– davidllerenav
Feb 1 at 16:31
$begingroup$
$frac {f'(x)|} {|x|}=frac {|f'(x)-f'(0)|} {|x|} to f''(0)$. From this it follows that $|f'(x)| <(1+|f''(0)|) |x|$ if $|x|$ is sufficiently small. Now $|f(x))=|f(x)-f(0)|=|x||f'zeta)|$ for some $zeta$ between $0$ and $x$. Is it clear now that $|f(x)| leq Mx^{2}$ for some $M$, for $|x|$ is sufficiently small?
$endgroup$
– Kavi Rama Murthy
Feb 1 at 23:18
add a comment |
$begingroup$
By Taylor's Formula with remainder $|f(x)| leq cx^{2}$ for some constant $C$ for $x$ in a neighborhoodod $0$. Hence $sum f(frac 1 n)$ is dominated by a constant times $sum frac 1 {n^{2}}$.
answered Feb 1 at 6:00
Kavi Rama MurthyKavi Rama Murthy
73.7k53170
$endgroup$
$begingroup$
Why is the remainder $cx^2$? Is there another way of doing it?
$endgroup$
– davidllerenav
Feb 1 at 16:31
$begingroup$
$frac {f'(x)|} {|x|}=frac {|f'(x)-f'(0)|} {|x|} to f''(0)$. From this it follows that $|f'(x)| <(1+|f''(0)|) |x|$ if $|x|$ is sufficiently small. Now $|f(x))=|f(x)-f(0)|=|x||f'zeta)|$ for some $zeta$ between $0$ and $x$. Is it clear now that $|f(x)| leq Mx^{2}$ for some $M$, for $|x|$ is sufficiently small?
$endgroup$
– Kavi Rama Murthy
Feb 1 at 23:18
add a comment |
$begingroup$
By Taylor's Formula with remainder $|f(x)| leq cx^{2}$ for some constant $C$ for $x$ in a neighborhoodod $0$. Hence $sum f(frac 1 n)$ is dominated by a constant times $sum frac 1 {n^{2}}$.
answered Feb 1 at 6:00
Kavi Rama MurthyKavi Rama Murthy
73.7k53170
$endgroup$
By Taylor's Formula with remainder $|f(x)| leq cx^{2}$ for some constant $C$ for $x$ in a neighborhoodod $0$. Hence $sum f(frac 1 n)$ is dominated by a constant times $sum frac 1 {n^{2}}$.
answered Feb 1 at 6:00
Kavi Rama MurthyKavi Rama Murthy
73.7k53170
answered Feb 1 at 6:00
Kavi Rama MurthyKavi Rama Murthy
73.7k53170
answered Feb 1 at 6:00
Kavi Rama MurthyKavi Rama Murthy
73.7k53170
answered Feb 1 at 6:00
Kavi Rama MurthyKavi Rama Murthy
73.7k53170
73.7k53170
$begingroup$
Why is the remainder $cx^2$? Is there another way of doing it?
$endgroup$
– davidllerenav
Feb 1 at 16:31
$begingroup$
$frac {f'(x)|} {|x|}=frac {|f'(x)-f'(0)|} {|x|} to f''(0)$. From this it follows that $|f'(x)| <(1+|f''(0)|) |x|$ if $|x|$ is sufficiently small. Now $|f(x))=|f(x)-f(0)|=|x||f'zeta)|$ for some $zeta$ between $0$ and $x$. Is it clear now that $|f(x)| leq Mx^{2}$ for some $M$, for $|x|$ is sufficiently small?
$endgroup$
– Kavi Rama Murthy
Feb 1 at 23:18
add a comment |
$begingroup$
Why is the remainder $cx^2$? Is there another way of doing it?
$endgroup$
– davidllerenav
Feb 1 at 16:31
$begingroup$
$frac {f'(x)|} {|x|}=frac {|f'(x)-f'(0)|} {|x|} to f''(0)$. From this it follows that $|f'(x)| <(1+|f''(0)|) |x|$ if $|x|$ is sufficiently small. Now $|f(x))=|f(x)-f(0)|=|x||f'zeta)|$ for some $zeta$ between $0$ and $x$. Is it clear now that $|f(x)| leq Mx^{2}$ for some $M$, for $|x|$ is sufficiently small?
$endgroup$
– Kavi Rama Murthy
Feb 1 at 23:18
$begingroup$
Why is the remainder $cx^2$? Is there another way of doing it?
$endgroup$
– davidllerenav
Feb 1 at 16:31
$begingroup$
Why is the remainder $cx^2$? Is there another way of doing it?
$endgroup$
– davidllerenav
Feb 1 at 16:31
$begingroup$
$frac {f'(x)|} {|x|}=frac {|f'(x)-f'(0)|} {|x|} to f''(0)$. From this it follows that $|f'(x)| <(1+|f''(0)|) |x|$ if $|x|$ is sufficiently small. Now $|f(x))=|f(x)-f(0)|=|x||f'zeta)|$ for some $zeta$ between $0$ and $x$. Is it clear now that $|f(x)| leq Mx^{2}$ for some $M$, for $|x|$ is sufficiently small?
$endgroup$
– Kavi Rama Murthy
Feb 1 at 23:18
$begingroup$
$frac {f'(x)|} {|x|}=frac {|f'(x)-f'(0)|} {|x|} to f''(0)$. From this it follows that $|f'(x)| <(1+|f''(0)|) |x|$ if $|x|$ is sufficiently small. Now $|f(x))=|f(x)-f(0)|=|x||f'zeta)|$ for some $zeta$ between $0$ and $x$. Is it clear now that $|f(x)| leq Mx^{2}$ for some $M$, for $|x|$ is sufficiently small?
$endgroup$
– Kavi Rama Murthy
Feb 1 at 23:18
add a comment |
$begingroup$
As $f''(0)$ exists, $f'(0)$ is can be linearly approximated:
$$ f'(h) = mcdot h+ r(h)$$
With some function $r$ with $frac {r(h)} hto 0$ for $hto 0 $. Therefore we can find an interval $U:=[-a,a]$ so that $ r(h) le mcdot h$, and by that:
$$xin Uimplies f'(x) le 2 mcdot x
$$
From this follows $f(x) le mx^2$ per integration for $xin U$, and thus the sum converges
(as for some $N$ the tail of the sum, $sum_{i=N}^infty f(1/i)$, is completely in $U$ for all $i$ it assumes, and we know that $sum_{i=0}^infty frac 1 {x^2}$ converges).
answered Feb 1 at 7:12
SudixSudix
9601316
$endgroup$
add a comment |
$begingroup$
As $f''(0)$ exists, $f'(0)$ is can be linearly approximated:
$$ f'(h) = mcdot h+ r(h)$$
With some function $r$ with $frac {r(h)} hto 0$ for $hto 0 $. Therefore we can find an interval $U:=[-a,a]$ so that $ r(h) le mcdot h$, and by that:
$$xin Uimplies f'(x) le 2 mcdot x
$$
From this follows $f(x) le mx^2$ per integration for $xin U$, and thus the sum converges
(as for some $N$ the tail of the sum, $sum_{i=N}^infty f(1/i)$, is completely in $U$ for all $i$ it assumes, and we know that $sum_{i=0}^infty frac 1 {x^2}$ converges).
answered Feb 1 at 7:12
SudixSudix
9601316
$endgroup$
add a comment |
$begingroup$
As $f''(0)$ exists, $f'(0)$ is can be linearly approximated:
$$ f'(h) = mcdot h+ r(h)$$
With some function $r$ with $frac {r(h)} hto 0$ for $hto 0 $. Therefore we can find an interval $U:=[-a,a]$ so that $ r(h) le mcdot h$, and by that:
$$xin Uimplies f'(x) le 2 mcdot x
$$
From this follows $f(x) le mx^2$ per integration for $xin U$, and thus the sum converges
(as for some $N$ the tail of the sum, $sum_{i=N}^infty f(1/i)$, is completely in $U$ for all $i$ it assumes, and we know that $sum_{i=0}^infty frac 1 {x^2}$ converges).
answered Feb 1 at 7:12
SudixSudix
9601316
$endgroup$
As $f''(0)$ exists, $f'(0)$ is can be linearly approximated:
$$ f'(h) = mcdot h+ r(h)$$
With some function $r$ with $frac {r(h)} hto 0$ for $hto 0 $. Therefore we can find an interval $U:=[-a,a]$ so that $ r(h) le mcdot h$, and by that:
$$xin Uimplies f'(x) le 2 mcdot x
$$
From this follows $f(x) le mx^2$ per integration for $xin U$, and thus the sum converges
(as for some $N$ the tail of the sum, $sum_{i=N}^infty f(1/i)$, is completely in $U$ for all $i$ it assumes, and we know that $sum_{i=0}^infty frac 1 {x^2}$ converges).
answered Feb 1 at 7:12
SudixSudix
9601316
answered Feb 1 at 7:12
SudixSudix
9601316
answered Feb 1 at 7:12
SudixSudix
9601316
answered Feb 1 at 7:12
SudixSudix
9601316
9601316
add a comment |
add a comment |
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