Finding $a$ such that $x^a cdot sin(1/x)$ is uniformly continuous.
$begingroup$
Assuming that $sin x$ is continuous on $mathbb R$, find all real $alpha$ such that $x^alphasin (1/x)$ is uniformly continuous on the open interval (0,1).
I'm guessing that I need to show that $x^alphasin x$ is continuously extendable to [0,1]. Doing that for $x=1$ is pretty trivial, but I am having trouble doing that for $x=0$. I believe that the $lim_{xto 0}x^alphasin (1/x)=0$, but how can I find what $f(0)$ equals?
I would appreciate any guidance! Thanks for your help in advance.
analysis continuity
edited Dec 20 '18 at 21:14
Jam
5,01921432
asked Oct 28 '12 at 7:59
JessJess
491716
$endgroup$
add a comment |
$begingroup$
Assuming that $sin x$ is continuous on $mathbb R$, find all real $alpha$ such that $x^alphasin (1/x)$ is uniformly continuous on the open interval (0,1).
I'm guessing that I need to show that $x^alphasin x$ is continuously extendable to [0,1]. Doing that for $x=1$ is pretty trivial, but I am having trouble doing that for $x=0$. I believe that the $lim_{xto 0}x^alphasin (1/x)=0$, but how can I find what $f(0)$ equals?
I would appreciate any guidance! Thanks for your help in advance.
analysis continuity
edited Dec 20 '18 at 21:14
Jam
5,01921432
asked Oct 28 '12 at 7:59
JessJess
491716
$endgroup$
add a comment |
$begingroup$
Assuming that $sin x$ is continuous on $mathbb R$, find all real $alpha$ such that $x^alphasin (1/x)$ is uniformly continuous on the open interval (0,1).
I'm guessing that I need to show that $x^alphasin x$ is continuously extendable to [0,1]. Doing that for $x=1$ is pretty trivial, but I am having trouble doing that for $x=0$. I believe that the $lim_{xto 0}x^alphasin (1/x)=0$, but how can I find what $f(0)$ equals?
I would appreciate any guidance! Thanks for your help in advance.
analysis continuity
edited Dec 20 '18 at 21:14
Jam
5,01921432
asked Oct 28 '12 at 7:59
JessJess
491716
$endgroup$
Assuming that $sin x$ is continuous on $mathbb R$, find all real $alpha$ such that $x^alphasin (1/x)$ is uniformly continuous on the open interval (0,1).
I'm guessing that I need to show that $x^alphasin x$ is continuously extendable to [0,1]. Doing that for $x=1$ is pretty trivial, but I am having trouble doing that for $x=0$. I believe that the $lim_{xto 0}x^alphasin (1/x)=0$, but how can I find what $f(0)$ equals?
I would appreciate any guidance! Thanks for your help in advance.
analysis continuity
analysis continuity
edited Dec 20 '18 at 21:14
Jam
5,01921432
asked Oct 28 '12 at 7:59
JessJess
491716
edited Dec 20 '18 at 21:14
Jam
5,01921432
asked Oct 28 '12 at 7:59
JessJess
491716
edited Dec 20 '18 at 21:14
Jam
5,01921432
edited Dec 20 '18 at 21:14
Jam
5,01921432
edited Dec 20 '18 at 21:14
Jam
5,01921432
5,01921432
asked Oct 28 '12 at 7:59
JessJess
491716
asked Oct 28 '12 at 7:59
JessJess
491716
asked Oct 28 '12 at 7:59
JessJess
491716
491716
add a comment |
add a comment |
1 Answer
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oldest
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$begingroup$
You're on the right track. If you can show that $f_alpha(x) = x^alpha sin(1/x)$ can be continuously extended to $[0,1]$, you are done. We will show that $lim_{x to 0} f_alpha(x)$ exists exactly for $alpha > 0$ (as is 0 in this case). That means that for $alpha > 0$, the definition $f_alpha(0) := 0$ makes $f_alphacolon [0,1]to mathbb R$ continuous, hence uniformly (as $[0,1]$ is compact). If $alpha > 0$, we have
$$ |f_alpha(x)| le x^alpha cdot |sin(1/x)|le x^alpha to 0, quad x to 0 $$
If $alpha le 0$, consider $x_n = 1/(pi/2 + npi)$, then $x_n to 0$, but
$$ f_alpha(x_n) = (pi/2+ npi)^{|alpha|} (-1)^n $$
and this doesn't converge for $n to infty$.
Hence, $f_alpha$ is uniformly continuous on $(0,1)$ iff $alpha > 0$.
answered Oct 28 '12 at 8:33
martinimartini
70.9k45991
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add a comment |
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1 Answer
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1 Answer
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oldest
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$begingroup$
You're on the right track. If you can show that $f_alpha(x) = x^alpha sin(1/x)$ can be continuously extended to $[0,1]$, you are done. We will show that $lim_{x to 0} f_alpha(x)$ exists exactly for $alpha > 0$ (as is 0 in this case). That means that for $alpha > 0$, the definition $f_alpha(0) := 0$ makes $f_alphacolon [0,1]to mathbb R$ continuous, hence uniformly (as $[0,1]$ is compact). If $alpha > 0$, we have
$$ |f_alpha(x)| le x^alpha cdot |sin(1/x)|le x^alpha to 0, quad x to 0 $$
If $alpha le 0$, consider $x_n = 1/(pi/2 + npi)$, then $x_n to 0$, but
$$ f_alpha(x_n) = (pi/2+ npi)^{|alpha|} (-1)^n $$
and this doesn't converge for $n to infty$.
Hence, $f_alpha$ is uniformly continuous on $(0,1)$ iff $alpha > 0$.
answered Oct 28 '12 at 8:33
martinimartini
70.9k45991
$endgroup$
add a comment |
$begingroup$
You're on the right track. If you can show that $f_alpha(x) = x^alpha sin(1/x)$ can be continuously extended to $[0,1]$, you are done. We will show that $lim_{x to 0} f_alpha(x)$ exists exactly for $alpha > 0$ (as is 0 in this case). That means that for $alpha > 0$, the definition $f_alpha(0) := 0$ makes $f_alphacolon [0,1]to mathbb R$ continuous, hence uniformly (as $[0,1]$ is compact). If $alpha > 0$, we have
$$ |f_alpha(x)| le x^alpha cdot |sin(1/x)|le x^alpha to 0, quad x to 0 $$
If $alpha le 0$, consider $x_n = 1/(pi/2 + npi)$, then $x_n to 0$, but
$$ f_alpha(x_n) = (pi/2+ npi)^{|alpha|} (-1)^n $$
and this doesn't converge for $n to infty$.
Hence, $f_alpha$ is uniformly continuous on $(0,1)$ iff $alpha > 0$.
answered Oct 28 '12 at 8:33
martinimartini
70.9k45991
$endgroup$
add a comment |
$begingroup$
You're on the right track. If you can show that $f_alpha(x) = x^alpha sin(1/x)$ can be continuously extended to $[0,1]$, you are done. We will show that $lim_{x to 0} f_alpha(x)$ exists exactly for $alpha > 0$ (as is 0 in this case). That means that for $alpha > 0$, the definition $f_alpha(0) := 0$ makes $f_alphacolon [0,1]to mathbb R$ continuous, hence uniformly (as $[0,1]$ is compact). If $alpha > 0$, we have
$$ |f_alpha(x)| le x^alpha cdot |sin(1/x)|le x^alpha to 0, quad x to 0 $$
If $alpha le 0$, consider $x_n = 1/(pi/2 + npi)$, then $x_n to 0$, but
$$ f_alpha(x_n) = (pi/2+ npi)^{|alpha|} (-1)^n $$
and this doesn't converge for $n to infty$.
Hence, $f_alpha$ is uniformly continuous on $(0,1)$ iff $alpha > 0$.
answered Oct 28 '12 at 8:33
martinimartini
70.9k45991
$endgroup$
You're on the right track. If you can show that $f_alpha(x) = x^alpha sin(1/x)$ can be continuously extended to $[0,1]$, you are done. We will show that $lim_{x to 0} f_alpha(x)$ exists exactly for $alpha > 0$ (as is 0 in this case). That means that for $alpha > 0$, the definition $f_alpha(0) := 0$ makes $f_alphacolon [0,1]to mathbb R$ continuous, hence uniformly (as $[0,1]$ is compact). If $alpha > 0$, we have
$$ |f_alpha(x)| le x^alpha cdot |sin(1/x)|le x^alpha to 0, quad x to 0 $$
If $alpha le 0$, consider $x_n = 1/(pi/2 + npi)$, then $x_n to 0$, but
$$ f_alpha(x_n) = (pi/2+ npi)^{|alpha|} (-1)^n $$
and this doesn't converge for $n to infty$.
Hence, $f_alpha$ is uniformly continuous on $(0,1)$ iff $alpha > 0$.
answered Oct 28 '12 at 8:33
martinimartini
70.9k45991
answered Oct 28 '12 at 8:33
martinimartini
70.9k45991
answered Oct 28 '12 at 8:33
martinimartini
70.9k45991
answered Oct 28 '12 at 8:33
martinimartini
70.9k45991
70.9k45991
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add a comment |
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