Mixing
power series inequality
$begingroup$
I seem to be stuck on showing the following inequality.
Let $$sum a_nx^n$$ be a power series with a radius of convergence of 2. Then there exists a M such that
$leftlvert sum_{n=1}^infty a_nx^nrightrvert le Mleftlvert x rightrvert$
where $leftlvert x rightrvert le 1$
Note the constant term is 0.
Im not sure on where to exactly to start. I've tried showing the series are bounded and attempted to rearrange the expression into the given form, but that was not successful.
analysis power-series
asked Feb 2 at 12:38
J.DoeJ.Doe
132
$endgroup$
add a comment |
$begingroup$
I seem to be stuck on showing the following inequality.
Let $$sum a_nx^n$$ be a power series with a radius of convergence of 2. Then there exists a M such that
$leftlvert sum_{n=1}^infty a_nx^nrightrvert le Mleftlvert x rightrvert$
where $leftlvert x rightrvert le 1$
Note the constant term is 0.
Im not sure on where to exactly to start. I've tried showing the series are bounded and attempted to rearrange the expression into the given form, but that was not successful.
analysis power-series
asked Feb 2 at 12:38
J.DoeJ.Doe
132
$endgroup$
add a comment |
$begingroup$
I seem to be stuck on showing the following inequality.
Let $$sum a_nx^n$$ be a power series with a radius of convergence of 2. Then there exists a M such that
$leftlvert sum_{n=1}^infty a_nx^nrightrvert le Mleftlvert x rightrvert$
where $leftlvert x rightrvert le 1$
Note the constant term is 0.
Im not sure on where to exactly to start. I've tried showing the series are bounded and attempted to rearrange the expression into the given form, but that was not successful.
analysis power-series
asked Feb 2 at 12:38
J.DoeJ.Doe
132
$endgroup$
I seem to be stuck on showing the following inequality.
Let $$sum a_nx^n$$ be a power series with a radius of convergence of 2. Then there exists a M such that
$leftlvert sum_{n=1}^infty a_nx^nrightrvert le Mleftlvert x rightrvert$
where $leftlvert x rightrvert le 1$
Note the constant term is 0.
Im not sure on where to exactly to start. I've tried showing the series are bounded and attempted to rearrange the expression into the given form, but that was not successful.
analysis power-series
analysis power-series
asked Feb 2 at 12:38
J.DoeJ.Doe
132
asked Feb 2 at 12:38
J.DoeJ.Doe
132
asked Feb 2 at 12:38
J.DoeJ.Doe
132
asked Feb 2 at 12:38
J.DoeJ.Doe
132
asked Feb 2 at 12:38
J.DoeJ.Doe
132
132
add a comment |
add a comment |
1 Answer
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$begingroup$
$g(x)=sumlimits_{n=1}^{infty}a_nx^{n-1}$ also has radius of convergence $2$. Hence $g$ is analytic, in particular continuous, for $|x| leq 1$. Since continuous functions are bounded on compact sets there exists $M <infty$ such that $|g(x)| leq M$ for $|x| leq 1$. Hence $|f(x)|=|xg(x)| leq M|x|$ for $|x| leq 1$.
answered Feb 2 at 12:50
Kavi Rama MurthyKavi Rama Murthy
74.5k53270
$endgroup$
$begingroup$
I see, thank you!
$endgroup$
– J.Doe
Feb 2 at 12:56
add a comment |
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1 Answer
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1 Answer
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active
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$begingroup$
$g(x)=sumlimits_{n=1}^{infty}a_nx^{n-1}$ also has radius of convergence $2$. Hence $g$ is analytic, in particular continuous, for $|x| leq 1$. Since continuous functions are bounded on compact sets there exists $M <infty$ such that $|g(x)| leq M$ for $|x| leq 1$. Hence $|f(x)|=|xg(x)| leq M|x|$ for $|x| leq 1$.
answered Feb 2 at 12:50
Kavi Rama MurthyKavi Rama Murthy
74.5k53270
$endgroup$
$begingroup$
I see, thank you!
$endgroup$
– J.Doe
Feb 2 at 12:56
add a comment |
$begingroup$
$g(x)=sumlimits_{n=1}^{infty}a_nx^{n-1}$ also has radius of convergence $2$. Hence $g$ is analytic, in particular continuous, for $|x| leq 1$. Since continuous functions are bounded on compact sets there exists $M <infty$ such that $|g(x)| leq M$ for $|x| leq 1$. Hence $|f(x)|=|xg(x)| leq M|x|$ for $|x| leq 1$.
answered Feb 2 at 12:50
Kavi Rama MurthyKavi Rama Murthy
74.5k53270
$endgroup$
$begingroup$
I see, thank you!
$endgroup$
– J.Doe
Feb 2 at 12:56
add a comment |
$begingroup$
$g(x)=sumlimits_{n=1}^{infty}a_nx^{n-1}$ also has radius of convergence $2$. Hence $g$ is analytic, in particular continuous, for $|x| leq 1$. Since continuous functions are bounded on compact sets there exists $M <infty$ such that $|g(x)| leq M$ for $|x| leq 1$. Hence $|f(x)|=|xg(x)| leq M|x|$ for $|x| leq 1$.
answered Feb 2 at 12:50
Kavi Rama MurthyKavi Rama Murthy
74.5k53270
$endgroup$
$g(x)=sumlimits_{n=1}^{infty}a_nx^{n-1}$ also has radius of convergence $2$. Hence $g$ is analytic, in particular continuous, for $|x| leq 1$. Since continuous functions are bounded on compact sets there exists $M <infty$ such that $|g(x)| leq M$ for $|x| leq 1$. Hence $|f(x)|=|xg(x)| leq M|x|$ for $|x| leq 1$.
answered Feb 2 at 12:50
Kavi Rama MurthyKavi Rama Murthy
74.5k53270
answered Feb 2 at 12:50
Kavi Rama MurthyKavi Rama Murthy
74.5k53270
answered Feb 2 at 12:50
Kavi Rama MurthyKavi Rama Murthy
74.5k53270
answered Feb 2 at 12:50
Kavi Rama MurthyKavi Rama Murthy
74.5k53270
74.5k53270
$begingroup$
I see, thank you!
$endgroup$
– J.Doe
Feb 2 at 12:56
add a comment |
$begingroup$
I see, thank you!
$endgroup$
– J.Doe
Feb 2 at 12:56
$begingroup$
I see, thank you!
$endgroup$
– J.Doe
Feb 2 at 12:56
$begingroup$
I see, thank you!
$endgroup$
– J.Doe
Feb 2 at 12:56
add a comment |
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