Determining the uniform convergence
$begingroup$
Show that the series ,whose partial sum of n terms is $S_n=frac{x}{(1+nx^2)}$, converges uniformly for all real x.
I found that the series is pointwise convergent to 0 for all x.
For showing uniform convergence, I found out that the function S attains maximum value at $x=frac{1}{sqrt{n}}$..i.e. $M= frac{1}{sqrt{n}}$.which tends to 0 as n tends to infinity.So it is proved that it is uniformly convergent.
However i am doubtful if this works for all real x or for particular closed interval.
calculus
edited Dec 6 '18 at 16:24
Alexis
2199
asked Dec 6 '18 at 15:34
KashmiraKashmira
463
$endgroup$
add a comment |
$begingroup$
Show that the series ,whose partial sum of n terms is $S_n=frac{x}{(1+nx^2)}$, converges uniformly for all real x.
I found that the series is pointwise convergent to 0 for all x.
For showing uniform convergence, I found out that the function S attains maximum value at $x=frac{1}{sqrt{n}}$..i.e. $M= frac{1}{sqrt{n}}$.which tends to 0 as n tends to infinity.So it is proved that it is uniformly convergent.
However i am doubtful if this works for all real x or for particular closed interval.
calculus
edited Dec 6 '18 at 16:24
Alexis
2199
asked Dec 6 '18 at 15:34
KashmiraKashmira
463
$endgroup$
add a comment |
$begingroup$
Show that the series ,whose partial sum of n terms is $S_n=frac{x}{(1+nx^2)}$, converges uniformly for all real x.
I found that the series is pointwise convergent to 0 for all x.
For showing uniform convergence, I found out that the function S attains maximum value at $x=frac{1}{sqrt{n}}$..i.e. $M= frac{1}{sqrt{n}}$.which tends to 0 as n tends to infinity.So it is proved that it is uniformly convergent.
However i am doubtful if this works for all real x or for particular closed interval.
calculus
edited Dec 6 '18 at 16:24
Alexis
2199
asked Dec 6 '18 at 15:34
KashmiraKashmira
463
$endgroup$
Show that the series ,whose partial sum of n terms is $S_n=frac{x}{(1+nx^2)}$, converges uniformly for all real x.
I found that the series is pointwise convergent to 0 for all x.
For showing uniform convergence, I found out that the function S attains maximum value at $x=frac{1}{sqrt{n}}$..i.e. $M= frac{1}{sqrt{n}}$.which tends to 0 as n tends to infinity.So it is proved that it is uniformly convergent.
However i am doubtful if this works for all real x or for particular closed interval.
calculus
calculus
edited Dec 6 '18 at 16:24
Alexis
2199
asked Dec 6 '18 at 15:34
KashmiraKashmira
463
edited Dec 6 '18 at 16:24
Alexis
2199
asked Dec 6 '18 at 15:34
KashmiraKashmira
463
edited Dec 6 '18 at 16:24
Alexis
2199
edited Dec 6 '18 at 16:24
Alexis
2199
edited Dec 6 '18 at 16:24
Alexis
2199
2199
asked Dec 6 '18 at 15:34
KashmiraKashmira
463
asked Dec 6 '18 at 15:34
KashmiraKashmira
463
asked Dec 6 '18 at 15:34
KashmiraKashmira
463
463
add a comment |
add a comment |
2 Answers
2
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oldest
votes
$begingroup$
Note the sum of the first $n$ terms as $S_n(x)$. We have
$$ S_n(x) = frac{x}{1+nx^2}$$
and $ lim_{n to infty} S_n(x) = 0 forall x in mathbb{R}.$ Therefore the function $S_n$(x) converges point wise to $0$ on $mathbb{R}$.
To show uniform convergence to $0$ on $D subset mathbb{R}$ we must show that
$$ forall epsilon, exists N in mathbb{N} : sup_{x in D}vert S_n(x) - 0 vert < epsilon.$$
You have shown that $$ sup_{x in mathbb{R}} vert S_n (x)vert = S_n(frac{1}{sqrt{n}}) = frac{1}{n + 1} rightarrow 0 text{ as } n to infty.$$
Remember by definition that $ lim_{n to infty } a_n = L in mathbb{R} iff forall epsilon , exists N in mathbb{N},n > N : vert a_n - L vert < epsilon$. Therefore take $a_n = sup_{x in mathbb{R}} vert S_n(x)vert$.
By definition $S_n$ converges uniformly to $0$ on $mathbb{R}$.
answered Dec 6 '18 at 16:51
DigitalisDigitalis
528216
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$begingroup$
Note that the function finally tends to $0$ any where. Therefore for uniform continuity we must show that$$forallepsilon>0quad exists Nquadforall xquad n>Nto|{xover 1+nx^2}|<epsilon$$where $N=N(epsilon)ne N(epsilon,x)$. Also $$|{xover 1+nx^2}|<epsiloniff 1+nx^2>{|x|over epsilon}iff n>{1over epsilon |x|}-{1over x^2}$$therefore by choosing $N(epsilon)=max_{xne 0}{1over epsilon |x|}-{1over x^2}={1over 4epsilon^2}$ we obtain $$n>N(epsilon)to n>{1over epsilon |x|}-{1over x^2}to |S_n(x)|<epsilon$$ and the proof is complete.
answered Dec 6 '18 at 20:35
Mostafa AyazMostafa Ayaz
15.3k3939
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add a comment |
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2 Answers
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2 Answers
2
active
oldest
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active
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votes
$begingroup$
Note the sum of the first $n$ terms as $S_n(x)$. We have
$$ S_n(x) = frac{x}{1+nx^2}$$
and $ lim_{n to infty} S_n(x) = 0 forall x in mathbb{R}.$ Therefore the function $S_n$(x) converges point wise to $0$ on $mathbb{R}$.
To show uniform convergence to $0$ on $D subset mathbb{R}$ we must show that
$$ forall epsilon, exists N in mathbb{N} : sup_{x in D}vert S_n(x) - 0 vert < epsilon.$$
You have shown that $$ sup_{x in mathbb{R}} vert S_n (x)vert = S_n(frac{1}{sqrt{n}}) = frac{1}{n + 1} rightarrow 0 text{ as } n to infty.$$
Remember by definition that $ lim_{n to infty } a_n = L in mathbb{R} iff forall epsilon , exists N in mathbb{N},n > N : vert a_n - L vert < epsilon$. Therefore take $a_n = sup_{x in mathbb{R}} vert S_n(x)vert$.
By definition $S_n$ converges uniformly to $0$ on $mathbb{R}$.
answered Dec 6 '18 at 16:51
DigitalisDigitalis
528216
$endgroup$
add a comment |
$begingroup$
Note the sum of the first $n$ terms as $S_n(x)$. We have
$$ S_n(x) = frac{x}{1+nx^2}$$
and $ lim_{n to infty} S_n(x) = 0 forall x in mathbb{R}.$ Therefore the function $S_n$(x) converges point wise to $0$ on $mathbb{R}$.
To show uniform convergence to $0$ on $D subset mathbb{R}$ we must show that
$$ forall epsilon, exists N in mathbb{N} : sup_{x in D}vert S_n(x) - 0 vert < epsilon.$$
You have shown that $$ sup_{x in mathbb{R}} vert S_n (x)vert = S_n(frac{1}{sqrt{n}}) = frac{1}{n + 1} rightarrow 0 text{ as } n to infty.$$
Remember by definition that $ lim_{n to infty } a_n = L in mathbb{R} iff forall epsilon , exists N in mathbb{N},n > N : vert a_n - L vert < epsilon$. Therefore take $a_n = sup_{x in mathbb{R}} vert S_n(x)vert$.
By definition $S_n$ converges uniformly to $0$ on $mathbb{R}$.
answered Dec 6 '18 at 16:51
DigitalisDigitalis
528216
$endgroup$
add a comment |
$begingroup$
Note the sum of the first $n$ terms as $S_n(x)$. We have
$$ S_n(x) = frac{x}{1+nx^2}$$
and $ lim_{n to infty} S_n(x) = 0 forall x in mathbb{R}.$ Therefore the function $S_n$(x) converges point wise to $0$ on $mathbb{R}$.
To show uniform convergence to $0$ on $D subset mathbb{R}$ we must show that
$$ forall epsilon, exists N in mathbb{N} : sup_{x in D}vert S_n(x) - 0 vert < epsilon.$$
You have shown that $$ sup_{x in mathbb{R}} vert S_n (x)vert = S_n(frac{1}{sqrt{n}}) = frac{1}{n + 1} rightarrow 0 text{ as } n to infty.$$
Remember by definition that $ lim_{n to infty } a_n = L in mathbb{R} iff forall epsilon , exists N in mathbb{N},n > N : vert a_n - L vert < epsilon$. Therefore take $a_n = sup_{x in mathbb{R}} vert S_n(x)vert$.
By definition $S_n$ converges uniformly to $0$ on $mathbb{R}$.
answered Dec 6 '18 at 16:51
DigitalisDigitalis
528216
$endgroup$
Note the sum of the first $n$ terms as $S_n(x)$. We have
$$ S_n(x) = frac{x}{1+nx^2}$$
and $ lim_{n to infty} S_n(x) = 0 forall x in mathbb{R}.$ Therefore the function $S_n$(x) converges point wise to $0$ on $mathbb{R}$.
To show uniform convergence to $0$ on $D subset mathbb{R}$ we must show that
$$ forall epsilon, exists N in mathbb{N} : sup_{x in D}vert S_n(x) - 0 vert < epsilon.$$
You have shown that $$ sup_{x in mathbb{R}} vert S_n (x)vert = S_n(frac{1}{sqrt{n}}) = frac{1}{n + 1} rightarrow 0 text{ as } n to infty.$$
Remember by definition that $ lim_{n to infty } a_n = L in mathbb{R} iff forall epsilon , exists N in mathbb{N},n > N : vert a_n - L vert < epsilon$. Therefore take $a_n = sup_{x in mathbb{R}} vert S_n(x)vert$.
By definition $S_n$ converges uniformly to $0$ on $mathbb{R}$.
answered Dec 6 '18 at 16:51
DigitalisDigitalis
528216
answered Dec 6 '18 at 16:51
DigitalisDigitalis
528216
answered Dec 6 '18 at 16:51
DigitalisDigitalis
528216
answered Dec 6 '18 at 16:51
DigitalisDigitalis
528216
528216
add a comment |
add a comment |
$begingroup$
Note that the function finally tends to $0$ any where. Therefore for uniform continuity we must show that$$forallepsilon>0quad exists Nquadforall xquad n>Nto|{xover 1+nx^2}|<epsilon$$where $N=N(epsilon)ne N(epsilon,x)$. Also $$|{xover 1+nx^2}|<epsiloniff 1+nx^2>{|x|over epsilon}iff n>{1over epsilon |x|}-{1over x^2}$$therefore by choosing $N(epsilon)=max_{xne 0}{1over epsilon |x|}-{1over x^2}={1over 4epsilon^2}$ we obtain $$n>N(epsilon)to n>{1over epsilon |x|}-{1over x^2}to |S_n(x)|<epsilon$$ and the proof is complete.
answered Dec 6 '18 at 20:35
Mostafa AyazMostafa Ayaz
15.3k3939
$endgroup$
add a comment |
$begingroup$
Note that the function finally tends to $0$ any where. Therefore for uniform continuity we must show that$$forallepsilon>0quad exists Nquadforall xquad n>Nto|{xover 1+nx^2}|<epsilon$$where $N=N(epsilon)ne N(epsilon,x)$. Also $$|{xover 1+nx^2}|<epsiloniff 1+nx^2>{|x|over epsilon}iff n>{1over epsilon |x|}-{1over x^2}$$therefore by choosing $N(epsilon)=max_{xne 0}{1over epsilon |x|}-{1over x^2}={1over 4epsilon^2}$ we obtain $$n>N(epsilon)to n>{1over epsilon |x|}-{1over x^2}to |S_n(x)|<epsilon$$ and the proof is complete.
answered Dec 6 '18 at 20:35
Mostafa AyazMostafa Ayaz
15.3k3939
$endgroup$
add a comment |
$begingroup$
Note that the function finally tends to $0$ any where. Therefore for uniform continuity we must show that$$forallepsilon>0quad exists Nquadforall xquad n>Nto|{xover 1+nx^2}|<epsilon$$where $N=N(epsilon)ne N(epsilon,x)$. Also $$|{xover 1+nx^2}|<epsiloniff 1+nx^2>{|x|over epsilon}iff n>{1over epsilon |x|}-{1over x^2}$$therefore by choosing $N(epsilon)=max_{xne 0}{1over epsilon |x|}-{1over x^2}={1over 4epsilon^2}$ we obtain $$n>N(epsilon)to n>{1over epsilon |x|}-{1over x^2}to |S_n(x)|<epsilon$$ and the proof is complete.
answered Dec 6 '18 at 20:35
Mostafa AyazMostafa Ayaz
15.3k3939
$endgroup$
Note that the function finally tends to $0$ any where. Therefore for uniform continuity we must show that$$forallepsilon>0quad exists Nquadforall xquad n>Nto|{xover 1+nx^2}|<epsilon$$where $N=N(epsilon)ne N(epsilon,x)$. Also $$|{xover 1+nx^2}|<epsiloniff 1+nx^2>{|x|over epsilon}iff n>{1over epsilon |x|}-{1over x^2}$$therefore by choosing $N(epsilon)=max_{xne 0}{1over epsilon |x|}-{1over x^2}={1over 4epsilon^2}$ we obtain $$n>N(epsilon)to n>{1over epsilon |x|}-{1over x^2}to |S_n(x)|<epsilon$$ and the proof is complete.
answered Dec 6 '18 at 20:35
Mostafa AyazMostafa Ayaz
15.3k3939
answered Dec 6 '18 at 20:35
Mostafa AyazMostafa Ayaz
15.3k3939
answered Dec 6 '18 at 20:35
Mostafa AyazMostafa Ayaz
15.3k3939
answered Dec 6 '18 at 20:35
Mostafa AyazMostafa Ayaz
15.3k3939
15.3k3939
add a comment |
add a comment |
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