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Is the functional sequence $f_n(x)$ uniformly convergent on $[1, 2]$?
The Next CEO of Stack OverflowChecking if $f_n$ or $sum f_n$converges uniformlyDoes $(f_n)$ converge pointwise/uniformly on $I$?The sequences $f_n,f_n'$ both are uniformly convergent on $[0,1]$.If a sequence of unif. cont. Lipschitz functions $f_n$ converges unif. to an unif. cont. function f, then f is Lipschitz.Pick out the sequences which are uniformly convergent:Show that $f_n(x)=frac1x+n$ converges uniformly on $0leq x leq 1$?Counterexample for functional sequence $f_n$ that converges uniformly on $E=[0;A]$ but not on $E=[0;+infty)$Proving if $f_n$ converges uniformly.Uniform convergence of a sequence of functions on $[0, 1]$Uniform Convergence of $f_n^k$ and polynomial
$begingroup$
Consider the functional sequence
beginalign*
f_n(x) = sqrtn+1 - sqrtn+2 + sum_k=1^n frac12sqrtk+x+2 - frac12sqrtk+x+1, (1leq xleq 2, n=1,2,3,...).
endalign*
Is $f_n(x)$ uniformly convergent on $[1, 2]$?
I know that $f(x) = displaystylelim_ntoinftyf_n(x) = sum_k=1^inftyfrac12sqrtk+x+2 - frac12sqrtk+x+1 = frac-12sqrtx+2$,
but I do not know how to use the supremm test. Please guide me?
(Let $M_n = sup|f_n(x) - f(x)|$, so $f_nrightarrow f$ is uniformly convergent on $E$ if and only if $M_nrightarrow 0$ as $nrightarrow infty$).
real-analysis calculus sequences-and-series uniform-convergence
asked Mar 18 at 20:10
koohyar eslamikoohyar eslami
22727
$endgroup$
add a comment |
$begingroup$
Consider the functional sequence
beginalign*
f_n(x) = sqrtn+1 - sqrtn+2 + sum_k=1^n frac12sqrtk+x+2 - frac12sqrtk+x+1, (1leq xleq 2, n=1,2,3,...).
endalign*
Is $f_n(x)$ uniformly convergent on $[1, 2]$?
I know that $f(x) = displaystylelim_ntoinftyf_n(x) = sum_k=1^inftyfrac12sqrtk+x+2 - frac12sqrtk+x+1 = frac-12sqrtx+2$,
but I do not know how to use the supremm test. Please guide me?
(Let $M_n = sup|f_n(x) - f(x)|$, so $f_nrightarrow f$ is uniformly convergent on $E$ if and only if $M_nrightarrow 0$ as $nrightarrow infty$).
real-analysis calculus sequences-and-series uniform-convergence
asked Mar 18 at 20:10
koohyar eslamikoohyar eslami
22727
$endgroup$
add a comment |
$begingroup$
Consider the functional sequence
beginalign*
f_n(x) = sqrtn+1 - sqrtn+2 + sum_k=1^n frac12sqrtk+x+2 - frac12sqrtk+x+1, (1leq xleq 2, n=1,2,3,...).
endalign*
Is $f_n(x)$ uniformly convergent on $[1, 2]$?
I know that $f(x) = displaystylelim_ntoinftyf_n(x) = sum_k=1^inftyfrac12sqrtk+x+2 - frac12sqrtk+x+1 = frac-12sqrtx+2$,
but I do not know how to use the supremm test. Please guide me?
(Let $M_n = sup|f_n(x) - f(x)|$, so $f_nrightarrow f$ is uniformly convergent on $E$ if and only if $M_nrightarrow 0$ as $nrightarrow infty$).
real-analysis calculus sequences-and-series uniform-convergence
asked Mar 18 at 20:10
koohyar eslamikoohyar eslami
22727
$endgroup$
Consider the functional sequence
beginalign*
f_n(x) = sqrtn+1 - sqrtn+2 + sum_k=1^n frac12sqrtk+x+2 - frac12sqrtk+x+1, (1leq xleq 2, n=1,2,3,...).
endalign*
Is $f_n(x)$ uniformly convergent on $[1, 2]$?
I know that $f(x) = displaystylelim_ntoinftyf_n(x) = sum_k=1^inftyfrac12sqrtk+x+2 - frac12sqrtk+x+1 = frac-12sqrtx+2$,
but I do not know how to use the supremm test. Please guide me?
(Let $M_n = sup|f_n(x) - f(x)|$, so $f_nrightarrow f$ is uniformly convergent on $E$ if and only if $M_nrightarrow 0$ as $nrightarrow infty$).
real-analysis calculus sequences-and-series uniform-convergence
real-analysis calculus sequences-and-series uniform-convergence
asked Mar 18 at 20:10
koohyar eslamikoohyar eslami
22727
asked Mar 18 at 20:10
koohyar eslamikoohyar eslami
22727
asked Mar 18 at 20:10
koohyar eslamikoohyar eslami
22727
asked Mar 18 at 20:10
koohyar eslamikoohyar eslami
22727
asked Mar 18 at 20:10
koohyar eslamikoohyar eslami
22727
22727
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
For $xin[1,2]$,
$$
beginalign
f_n(x)
&=sqrtn+1-sqrtn+2+sum_k=1^nleft(frac12sqrtk+x+2-frac12sqrtk+x+1right)\
&=-underbracefrac1sqrtn+1+sqrtn+2_lefrac12sqrtn+1+underbracefrac12sqrtx+n+2_lefrac12sqrtn+3-frac12sqrtx+2\
endalign
$$
Can you finish from here?
answered Mar 18 at 20:20
robjohn♦robjohn
270k27312640
$endgroup$
$begingroup$
robjohn. Thanks. $M_n = sup|f_n(x) - f(x)| = sup|frac-1sqrtn+1 + sqrtn+2 + frac12sqrtn+x+2 - frac12sqrtx+2 + frac12sqrtx+2|leq frac12sqrtn+1+frac1sqrtn+3$, since $displaystylelim_ntoinfty(frac12sqrtn+1+frac1sqrtn+3) = 0$ as $nrightarrow infty$, so $M_nrightarrow 0$. Is it right?
$endgroup$
– koohyar eslami
Mar 19 at 4:43
1
$begingroup$
@koohyareslami: yes. The important point is that there is no dependence on $x$ in $sup|f_n-f|$.
$endgroup$
– robjohn♦
Mar 19 at 8:06
add a comment |
Your Answer
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1 Answer
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1 Answer
1
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oldest
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active
oldest
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oldest
votes
$begingroup$
For $xin[1,2]$,
$$
beginalign
f_n(x)
&=sqrtn+1-sqrtn+2+sum_k=1^nleft(frac12sqrtk+x+2-frac12sqrtk+x+1right)\
&=-underbracefrac1sqrtn+1+sqrtn+2_lefrac12sqrtn+1+underbracefrac12sqrtx+n+2_lefrac12sqrtn+3-frac12sqrtx+2\
endalign
$$
Can you finish from here?
answered Mar 18 at 20:20
robjohn♦robjohn
270k27312640
$endgroup$
$begingroup$
robjohn. Thanks. $M_n = sup|f_n(x) - f(x)| = sup|frac-1sqrtn+1 + sqrtn+2 + frac12sqrtn+x+2 - frac12sqrtx+2 + frac12sqrtx+2|leq frac12sqrtn+1+frac1sqrtn+3$, since $displaystylelim_ntoinfty(frac12sqrtn+1+frac1sqrtn+3) = 0$ as $nrightarrow infty$, so $M_nrightarrow 0$. Is it right?
$endgroup$
– koohyar eslami
Mar 19 at 4:43
1
$begingroup$
@koohyareslami: yes. The important point is that there is no dependence on $x$ in $sup|f_n-f|$.
$endgroup$
– robjohn♦
Mar 19 at 8:06
add a comment |
$begingroup$
For $xin[1,2]$,
$$
beginalign
f_n(x)
&=sqrtn+1-sqrtn+2+sum_k=1^nleft(frac12sqrtk+x+2-frac12sqrtk+x+1right)\
&=-underbracefrac1sqrtn+1+sqrtn+2_lefrac12sqrtn+1+underbracefrac12sqrtx+n+2_lefrac12sqrtn+3-frac12sqrtx+2\
endalign
$$
Can you finish from here?
answered Mar 18 at 20:20
robjohn♦robjohn
270k27312640
$endgroup$
$begingroup$
robjohn. Thanks. $M_n = sup|f_n(x) - f(x)| = sup|frac-1sqrtn+1 + sqrtn+2 + frac12sqrtn+x+2 - frac12sqrtx+2 + frac12sqrtx+2|leq frac12sqrtn+1+frac1sqrtn+3$, since $displaystylelim_ntoinfty(frac12sqrtn+1+frac1sqrtn+3) = 0$ as $nrightarrow infty$, so $M_nrightarrow 0$. Is it right?
$endgroup$
– koohyar eslami
Mar 19 at 4:43
1
$begingroup$
@koohyareslami: yes. The important point is that there is no dependence on $x$ in $sup|f_n-f|$.
$endgroup$
– robjohn♦
Mar 19 at 8:06
add a comment |
$begingroup$
For $xin[1,2]$,
$$
beginalign
f_n(x)
&=sqrtn+1-sqrtn+2+sum_k=1^nleft(frac12sqrtk+x+2-frac12sqrtk+x+1right)\
&=-underbracefrac1sqrtn+1+sqrtn+2_lefrac12sqrtn+1+underbracefrac12sqrtx+n+2_lefrac12sqrtn+3-frac12sqrtx+2\
endalign
$$
Can you finish from here?
answered Mar 18 at 20:20
robjohn♦robjohn
270k27312640
$endgroup$
For $xin[1,2]$,
$$
beginalign
f_n(x)
&=sqrtn+1-sqrtn+2+sum_k=1^nleft(frac12sqrtk+x+2-frac12sqrtk+x+1right)\
&=-underbracefrac1sqrtn+1+sqrtn+2_lefrac12sqrtn+1+underbracefrac12sqrtx+n+2_lefrac12sqrtn+3-frac12sqrtx+2\
endalign
$$
Can you finish from here?
answered Mar 18 at 20:20
robjohn♦robjohn
270k27312640
answered Mar 18 at 20:20
robjohn♦robjohn
270k27312640
answered Mar 18 at 20:20
robjohn♦robjohn
270k27312640
answered Mar 18 at 20:20
robjohn♦robjohn
270k27312640
270k27312640
$begingroup$
robjohn. Thanks. $M_n = sup|f_n(x) - f(x)| = sup|frac-1sqrtn+1 + sqrtn+2 + frac12sqrtn+x+2 - frac12sqrtx+2 + frac12sqrtx+2|leq frac12sqrtn+1+frac1sqrtn+3$, since $displaystylelim_ntoinfty(frac12sqrtn+1+frac1sqrtn+3) = 0$ as $nrightarrow infty$, so $M_nrightarrow 0$. Is it right?
$endgroup$
– koohyar eslami
Mar 19 at 4:43
1
$begingroup$
@koohyareslami: yes. The important point is that there is no dependence on $x$ in $sup|f_n-f|$.
$endgroup$
– robjohn♦
Mar 19 at 8:06
add a comment |
$begingroup$
robjohn. Thanks. $M_n = sup|f_n(x) - f(x)| = sup|frac-1sqrtn+1 + sqrtn+2 + frac12sqrtn+x+2 - frac12sqrtx+2 + frac12sqrtx+2|leq frac12sqrtn+1+frac1sqrtn+3$, since $displaystylelim_ntoinfty(frac12sqrtn+1+frac1sqrtn+3) = 0$ as $nrightarrow infty$, so $M_nrightarrow 0$. Is it right?
$endgroup$
– koohyar eslami
Mar 19 at 4:43
1
$begingroup$
@koohyareslami: yes. The important point is that there is no dependence on $x$ in $sup|f_n-f|$.
$endgroup$
– robjohn♦
Mar 19 at 8:06
$begingroup$
robjohn. Thanks. $M_n = sup|f_n(x) - f(x)| = sup|frac-1sqrtn+1 + sqrtn+2 + frac12sqrtn+x+2 - frac12sqrtx+2 + frac12sqrtx+2|leq frac12sqrtn+1+frac1sqrtn+3$, since $displaystylelim_ntoinfty(frac12sqrtn+1+frac1sqrtn+3) = 0$ as $nrightarrow infty$, so $M_nrightarrow 0$. Is it right?
$endgroup$
– koohyar eslami
Mar 19 at 4:43
$begingroup$
robjohn. Thanks. $M_n = sup|f_n(x) - f(x)| = sup|frac-1sqrtn+1 + sqrtn+2 + frac12sqrtn+x+2 - frac12sqrtx+2 + frac12sqrtx+2|leq frac12sqrtn+1+frac1sqrtn+3$, since $displaystylelim_ntoinfty(frac12sqrtn+1+frac1sqrtn+3) = 0$ as $nrightarrow infty$, so $M_nrightarrow 0$. Is it right?
$endgroup$
– koohyar eslami
Mar 19 at 4:43
1
1
$begingroup$
@koohyareslami: yes. The important point is that there is no dependence on $x$ in $sup|f_n-f|$.
$endgroup$
– robjohn♦
Mar 19 at 8:06
$begingroup$
@koohyareslami: yes. The important point is that there is no dependence on $x$ in $sup|f_n-f|$.
$endgroup$
– robjohn♦
Mar 19 at 8:06
add a comment |
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