Uniform convergence of Binomial series on $Bbb{R}$
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Consider the Binomial series
begin{align} (1+x)^alpha =sum^{infty}_{k=0}{alphachoose k}x^k,;text{for };xin Bbb{R};text{and};alphainBbb{R} end{align}
I want to show that it converges uniformly on $Bbb{R}$.
However, it can be shown by D'Alembert's Ratio test that the following series converges absolutely begin{align} F(alpha) =sum^{infty}_{k=0}{alphachoose k}x^k,;text{for fixed};|x|<1;text{and};alphainBbb{R} end{align}
since begin{align} limlimits_{ktoinfty}left|dfrac{^alpha C_{k+1}}{^alpha C_{k}}right|=limlimits_{ktoinfty}left|dfrac{alpha-k}{k+1}right|=|x|<1,;text{for fixed};|x|<1;text{and};alphainBbb{R} end{align}
QUESTION: How do I get uniform convergence of $F$ on $Bbb{R}$ from there? Alternatively, if there's any other way of showing that it converges uniformly on $Bbb{R}$, I will appreciate.
real-analysis sequences-and-series analysis convergence power-series
asked Jan 8 at 6:58
Omojola MichealOmojola Micheal
1,787324
$endgroup$
add a comment |
$begingroup$
Consider the Binomial series
begin{align} (1+x)^alpha =sum^{infty}_{k=0}{alphachoose k}x^k,;text{for };xin Bbb{R};text{and};alphainBbb{R} end{align}
I want to show that it converges uniformly on $Bbb{R}$.
However, it can be shown by D'Alembert's Ratio test that the following series converges absolutely begin{align} F(alpha) =sum^{infty}_{k=0}{alphachoose k}x^k,;text{for fixed};|x|<1;text{and};alphainBbb{R} end{align}
since begin{align} limlimits_{ktoinfty}left|dfrac{^alpha C_{k+1}}{^alpha C_{k}}right|=limlimits_{ktoinfty}left|dfrac{alpha-k}{k+1}right|=|x|<1,;text{for fixed};|x|<1;text{and};alphainBbb{R} end{align}
QUESTION: How do I get uniform convergence of $F$ on $Bbb{R}$ from there? Alternatively, if there's any other way of showing that it converges uniformly on $Bbb{R}$, I will appreciate.
real-analysis sequences-and-series analysis convergence power-series
asked Jan 8 at 6:58
Omojola MichealOmojola Micheal
1,787324
$endgroup$
$begingroup$
It won't converge uniformly on $Bbb R$ (unless $alpha=0$). It may or may not converge uniformly on $(-1,1)$, but that will depend on the value of $alpha$.
$endgroup$
– Lord Shark the Unknown
Jan 8 at 7:00
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@Lord Shark the Unknown: But William R. Wade-Introduction to Analysis-Pearson tells me it converges uniformly on $Bbb{R}$
$endgroup$
– Omojola Micheal
Jan 8 at 7:03
$begingroup$
In line 2 you say 'for $|x| <1$' and in line 3 you ask for uniform convergence of $mathbb R$
$endgroup$
– Kavi Rama Murthy
Jan 8 at 7:21
$begingroup$
@Kavi Rama Murthy: Sorry, I corrected the error.
$endgroup$
– Omojola Micheal
Jan 8 at 7:24
add a comment |
$begingroup$
Consider the Binomial series
begin{align} (1+x)^alpha =sum^{infty}_{k=0}{alphachoose k}x^k,;text{for };xin Bbb{R};text{and};alphainBbb{R} end{align}
I want to show that it converges uniformly on $Bbb{R}$.
However, it can be shown by D'Alembert's Ratio test that the following series converges absolutely begin{align} F(alpha) =sum^{infty}_{k=0}{alphachoose k}x^k,;text{for fixed};|x|<1;text{and};alphainBbb{R} end{align}
since begin{align} limlimits_{ktoinfty}left|dfrac{^alpha C_{k+1}}{^alpha C_{k}}right|=limlimits_{ktoinfty}left|dfrac{alpha-k}{k+1}right|=|x|<1,;text{for fixed};|x|<1;text{and};alphainBbb{R} end{align}
QUESTION: How do I get uniform convergence of $F$ on $Bbb{R}$ from there? Alternatively, if there's any other way of showing that it converges uniformly on $Bbb{R}$, I will appreciate.
real-analysis sequences-and-series analysis convergence power-series
asked Jan 8 at 6:58
Omojola MichealOmojola Micheal
1,787324
$endgroup$
Consider the Binomial series
begin{align} (1+x)^alpha =sum^{infty}_{k=0}{alphachoose k}x^k,;text{for };xin Bbb{R};text{and};alphainBbb{R} end{align}
I want to show that it converges uniformly on $Bbb{R}$.
However, it can be shown by D'Alembert's Ratio test that the following series converges absolutely begin{align} F(alpha) =sum^{infty}_{k=0}{alphachoose k}x^k,;text{for fixed};|x|<1;text{and};alphainBbb{R} end{align}
since begin{align} limlimits_{ktoinfty}left|dfrac{^alpha C_{k+1}}{^alpha C_{k}}right|=limlimits_{ktoinfty}left|dfrac{alpha-k}{k+1}right|=|x|<1,;text{for fixed};|x|<1;text{and};alphainBbb{R} end{align}
QUESTION: How do I get uniform convergence of $F$ on $Bbb{R}$ from there? Alternatively, if there's any other way of showing that it converges uniformly on $Bbb{R}$, I will appreciate.
real-analysis sequences-and-series analysis convergence power-series
real-analysis sequences-and-series analysis convergence power-series
asked Jan 8 at 6:58
Omojola MichealOmojola Micheal
1,787324
asked Jan 8 at 6:58
Omojola MichealOmojola Micheal
1,787324
edited Jan 8 at 7:24
Omojola Micheal
asked Jan 8 at 6:58
Omojola MichealOmojola Micheal
1,787324
asked Jan 8 at 6:58
Omojola MichealOmojola Micheal
1,787324
asked Jan 8 at 6:58
Omojola MichealOmojola Micheal
1,787324
1,787324
$begingroup$
It won't converge uniformly on $Bbb R$ (unless $alpha=0$). It may or may not converge uniformly on $(-1,1)$, but that will depend on the value of $alpha$.
$endgroup$
– Lord Shark the Unknown
Jan 8 at 7:00
$begingroup$
@Lord Shark the Unknown: But William R. Wade-Introduction to Analysis-Pearson tells me it converges uniformly on $Bbb{R}$
$endgroup$
– Omojola Micheal
Jan 8 at 7:03
$begingroup$
In line 2 you say 'for $|x| <1$' and in line 3 you ask for uniform convergence of $mathbb R$
$endgroup$
– Kavi Rama Murthy
Jan 8 at 7:21
$begingroup$
@Kavi Rama Murthy: Sorry, I corrected the error.
$endgroup$
– Omojola Micheal
Jan 8 at 7:24
add a comment |
$begingroup$
It won't converge uniformly on $Bbb R$ (unless $alpha=0$). It may or may not converge uniformly on $(-1,1)$, but that will depend on the value of $alpha$.
$endgroup$
– Lord Shark the Unknown
Jan 8 at 7:00
$begingroup$
@Lord Shark the Unknown: But William R. Wade-Introduction to Analysis-Pearson tells me it converges uniformly on $Bbb{R}$
$endgroup$
– Omojola Micheal
Jan 8 at 7:03
$begingroup$
In line 2 you say 'for $|x| <1$' and in line 3 you ask for uniform convergence of $mathbb R$
$endgroup$
– Kavi Rama Murthy
Jan 8 at 7:21
$begingroup$
@Kavi Rama Murthy: Sorry, I corrected the error.
$endgroup$
– Omojola Micheal
Jan 8 at 7:24
$begingroup$
It won't converge uniformly on $Bbb R$ (unless $alpha=0$). It may or may not converge uniformly on $(-1,1)$, but that will depend on the value of $alpha$.
$endgroup$
– Lord Shark the Unknown
Jan 8 at 7:00
$begingroup$
It won't converge uniformly on $Bbb R$ (unless $alpha=0$). It may or may not converge uniformly on $(-1,1)$, but that will depend on the value of $alpha$.
$endgroup$
– Lord Shark the Unknown
Jan 8 at 7:00
$begingroup$
@Lord Shark the Unknown: But William R. Wade-Introduction to Analysis-Pearson tells me it converges uniformly on $Bbb{R}$
$endgroup$
– Omojola Micheal
Jan 8 at 7:03
$begingroup$
@Lord Shark the Unknown: But William R. Wade-Introduction to Analysis-Pearson tells me it converges uniformly on $Bbb{R}$
$endgroup$
– Omojola Micheal
Jan 8 at 7:03
$begingroup$
In line 2 you say 'for $|x| <1$' and in line 3 you ask for uniform convergence of $mathbb R$
$endgroup$
– Kavi Rama Murthy
Jan 8 at 7:21
$begingroup$
In line 2 you say 'for $|x| <1$' and in line 3 you ask for uniform convergence of $mathbb R$
$endgroup$
– Kavi Rama Murthy
Jan 8 at 7:21
$begingroup$
@Kavi Rama Murthy: Sorry, I corrected the error.
$endgroup$
– Omojola Micheal
Jan 8 at 7:24
$begingroup$
@Kavi Rama Murthy: Sorry, I corrected the error.
$endgroup$
– Omojola Micheal
Jan 8 at 7:24
add a comment |
1 Answer
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A series of the type $sum a_k x^{k}$ cannot converge uniformly on $mathbb R$ unless $a_k=0$ for all but finite number of $k$'s. (This is because the general term does not tend to $0$ uniformly).
answered Jan 8 at 7:23
Kavi Rama MurthyKavi Rama Murthy
53.2k32055
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add a comment |
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$begingroup$
A series of the type $sum a_k x^{k}$ cannot converge uniformly on $mathbb R$ unless $a_k=0$ for all but finite number of $k$'s. (This is because the general term does not tend to $0$ uniformly).
answered Jan 8 at 7:23
Kavi Rama MurthyKavi Rama Murthy
53.2k32055
$endgroup$
add a comment |
$begingroup$
A series of the type $sum a_k x^{k}$ cannot converge uniformly on $mathbb R$ unless $a_k=0$ for all but finite number of $k$'s. (This is because the general term does not tend to $0$ uniformly).
answered Jan 8 at 7:23
Kavi Rama MurthyKavi Rama Murthy
53.2k32055
$endgroup$
add a comment |
$begingroup$
A series of the type $sum a_k x^{k}$ cannot converge uniformly on $mathbb R$ unless $a_k=0$ for all but finite number of $k$'s. (This is because the general term does not tend to $0$ uniformly).
answered Jan 8 at 7:23
Kavi Rama MurthyKavi Rama Murthy
53.2k32055
$endgroup$
A series of the type $sum a_k x^{k}$ cannot converge uniformly on $mathbb R$ unless $a_k=0$ for all but finite number of $k$'s. (This is because the general term does not tend to $0$ uniformly).
answered Jan 8 at 7:23
Kavi Rama MurthyKavi Rama Murthy
53.2k32055
answered Jan 8 at 7:23
Kavi Rama MurthyKavi Rama Murthy
53.2k32055
answered Jan 8 at 7:23
Kavi Rama MurthyKavi Rama Murthy
53.2k32055
answered Jan 8 at 7:23
Kavi Rama MurthyKavi Rama Murthy
53.2k32055
53.2k32055
add a comment |
add a comment |
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$begingroup$
It won't converge uniformly on $Bbb R$ (unless $alpha=0$). It may or may not converge uniformly on $(-1,1)$, but that will depend on the value of $alpha$.
$endgroup$
– Lord Shark the Unknown
Jan 8 at 7:00
$begingroup$
@Lord Shark the Unknown: But William R. Wade-Introduction to Analysis-Pearson tells me it converges uniformly on $Bbb{R}$
$endgroup$
– Omojola Micheal
Jan 8 at 7:03
$begingroup$
In line 2 you say 'for $|x| <1$' and in line 3 you ask for uniform convergence of $mathbb R$
$endgroup$
– Kavi Rama Murthy
Jan 8 at 7:21
$begingroup$
@Kavi Rama Murthy: Sorry, I corrected the error.
$endgroup$
– Omojola Micheal
Jan 8 at 7:24