Show $f(x) = sum_{n = 1}^{infty} frac{sin(nx)}{2^{n}}$ is infinitely differentiable
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I want to show that the function $f(x) = sum_{n = 1}^{infty} frac{sin(nx)}{2^{n}}$ is infinitely differentiable on $mathbb{R}$. I have no idea how to do this. I think it's pretty obvious that the function converges since $2^{n}$ increases really quickly. I tried doing something with analyticity which implies infinitely differentiable; however, I got nowhere.
real-analysis sequences-and-series
asked Dec 14 '18 at 16:35
stackofhay42stackofhay42
2267
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add a comment |
$begingroup$
I want to show that the function $f(x) = sum_{n = 1}^{infty} frac{sin(nx)}{2^{n}}$ is infinitely differentiable on $mathbb{R}$. I have no idea how to do this. I think it's pretty obvious that the function converges since $2^{n}$ increases really quickly. I tried doing something with analyticity which implies infinitely differentiable; however, I got nowhere.
real-analysis sequences-and-series
asked Dec 14 '18 at 16:35
stackofhay42stackofhay42
2267
$endgroup$
2
$begingroup$
This series is uniformly convergent on $Bbb R$, and the series of the derivatives converges uniformly on $Bbb R$. This is enough to conclude that the derivative of $f(x)$ is simply the series of the derivatives.
$endgroup$
– Crostul
Dec 14 '18 at 16:38
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how do we know the series is uniformly convergent?
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– stackofhay42
Dec 14 '18 at 17:08
$begingroup$
Do you know Weierstrass' M-test? It's so simple to use here.
$endgroup$
– Crostul
Dec 14 '18 at 17:14
$begingroup$
i am allowed to use weierstrass m test, yes
$endgroup$
– stackofhay42
Dec 14 '18 at 17:17
add a comment |
$begingroup$
I want to show that the function $f(x) = sum_{n = 1}^{infty} frac{sin(nx)}{2^{n}}$ is infinitely differentiable on $mathbb{R}$. I have no idea how to do this. I think it's pretty obvious that the function converges since $2^{n}$ increases really quickly. I tried doing something with analyticity which implies infinitely differentiable; however, I got nowhere.
real-analysis sequences-and-series
asked Dec 14 '18 at 16:35
stackofhay42stackofhay42
2267
$endgroup$
I want to show that the function $f(x) = sum_{n = 1}^{infty} frac{sin(nx)}{2^{n}}$ is infinitely differentiable on $mathbb{R}$. I have no idea how to do this. I think it's pretty obvious that the function converges since $2^{n}$ increases really quickly. I tried doing something with analyticity which implies infinitely differentiable; however, I got nowhere.
real-analysis sequences-and-series
real-analysis sequences-and-series
asked Dec 14 '18 at 16:35
stackofhay42stackofhay42
2267
asked Dec 14 '18 at 16:35
stackofhay42stackofhay42
2267
asked Dec 14 '18 at 16:35
stackofhay42stackofhay42
2267
asked Dec 14 '18 at 16:35
stackofhay42stackofhay42
2267
asked Dec 14 '18 at 16:35
stackofhay42stackofhay42
2267
2267
2
$begingroup$
This series is uniformly convergent on $Bbb R$, and the series of the derivatives converges uniformly on $Bbb R$. This is enough to conclude that the derivative of $f(x)$ is simply the series of the derivatives.
$endgroup$
– Crostul
Dec 14 '18 at 16:38
$begingroup$
how do we know the series is uniformly convergent?
$endgroup$
– stackofhay42
Dec 14 '18 at 17:08
$begingroup$
Do you know Weierstrass' M-test? It's so simple to use here.
$endgroup$
– Crostul
Dec 14 '18 at 17:14
$begingroup$
i am allowed to use weierstrass m test, yes
$endgroup$
– stackofhay42
Dec 14 '18 at 17:17
add a comment |
2
$begingroup$
This series is uniformly convergent on $Bbb R$, and the series of the derivatives converges uniformly on $Bbb R$. This is enough to conclude that the derivative of $f(x)$ is simply the series of the derivatives.
$endgroup$
– Crostul
Dec 14 '18 at 16:38
$begingroup$
how do we know the series is uniformly convergent?
$endgroup$
– stackofhay42
Dec 14 '18 at 17:08
$begingroup$
Do you know Weierstrass' M-test? It's so simple to use here.
$endgroup$
– Crostul
Dec 14 '18 at 17:14
$begingroup$
i am allowed to use weierstrass m test, yes
$endgroup$
– stackofhay42
Dec 14 '18 at 17:17
2
2
$begingroup$
This series is uniformly convergent on $Bbb R$, and the series of the derivatives converges uniformly on $Bbb R$. This is enough to conclude that the derivative of $f(x)$ is simply the series of the derivatives.
$endgroup$
– Crostul
Dec 14 '18 at 16:38
$begingroup$
This series is uniformly convergent on $Bbb R$, and the series of the derivatives converges uniformly on $Bbb R$. This is enough to conclude that the derivative of $f(x)$ is simply the series of the derivatives.
$endgroup$
– Crostul
Dec 14 '18 at 16:38
$begingroup$
how do we know the series is uniformly convergent?
$endgroup$
– stackofhay42
Dec 14 '18 at 17:08
$begingroup$
how do we know the series is uniformly convergent?
$endgroup$
– stackofhay42
Dec 14 '18 at 17:08
$begingroup$
Do you know Weierstrass' M-test? It's so simple to use here.
$endgroup$
– Crostul
Dec 14 '18 at 17:14
$begingroup$
Do you know Weierstrass' M-test? It's so simple to use here.
$endgroup$
– Crostul
Dec 14 '18 at 17:14
$begingroup$
i am allowed to use weierstrass m test, yes
$endgroup$
– stackofhay42
Dec 14 '18 at 17:17
$begingroup$
i am allowed to use weierstrass m test, yes
$endgroup$
– stackofhay42
Dec 14 '18 at 17:17
add a comment |
1 Answer
1
active
oldest
votes
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Write $sin(nx)={1over 2i}bigl(e^{inx}-e^{-inx}bigr)$, so that your $f$ appears as $$f(x)={1over 2i}left(sum_{ngeq1} p^n-sum_{ngeq1} q^nright) .$$
Simplifying in the end will give you a simple expression for $f$, namely
$$f(x)={2sin xover5-4cos x} .$$
answered Dec 14 '18 at 17:02
Christian BlatterChristian Blatter
174k8115327
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add a comment |
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
votes
active
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active
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votes
$begingroup$
Write $sin(nx)={1over 2i}bigl(e^{inx}-e^{-inx}bigr)$, so that your $f$ appears as $$f(x)={1over 2i}left(sum_{ngeq1} p^n-sum_{ngeq1} q^nright) .$$
Simplifying in the end will give you a simple expression for $f$, namely
$$f(x)={2sin xover5-4cos x} .$$
answered Dec 14 '18 at 17:02
Christian BlatterChristian Blatter
174k8115327
$endgroup$
add a comment |
$begingroup$
Write $sin(nx)={1over 2i}bigl(e^{inx}-e^{-inx}bigr)$, so that your $f$ appears as $$f(x)={1over 2i}left(sum_{ngeq1} p^n-sum_{ngeq1} q^nright) .$$
Simplifying in the end will give you a simple expression for $f$, namely
$$f(x)={2sin xover5-4cos x} .$$
answered Dec 14 '18 at 17:02
Christian BlatterChristian Blatter
174k8115327
$endgroup$
add a comment |
$begingroup$
Write $sin(nx)={1over 2i}bigl(e^{inx}-e^{-inx}bigr)$, so that your $f$ appears as $$f(x)={1over 2i}left(sum_{ngeq1} p^n-sum_{ngeq1} q^nright) .$$
Simplifying in the end will give you a simple expression for $f$, namely
$$f(x)={2sin xover5-4cos x} .$$
answered Dec 14 '18 at 17:02
Christian BlatterChristian Blatter
174k8115327
$endgroup$
Write $sin(nx)={1over 2i}bigl(e^{inx}-e^{-inx}bigr)$, so that your $f$ appears as $$f(x)={1over 2i}left(sum_{ngeq1} p^n-sum_{ngeq1} q^nright) .$$
Simplifying in the end will give you a simple expression for $f$, namely
$$f(x)={2sin xover5-4cos x} .$$
answered Dec 14 '18 at 17:02
Christian BlatterChristian Blatter
174k8115327
edited Dec 16 '18 at 16:37
answered Dec 14 '18 at 17:02
Christian BlatterChristian Blatter
174k8115327
answered Dec 14 '18 at 17:02
Christian BlatterChristian Blatter
174k8115327
answered Dec 14 '18 at 17:02
Christian BlatterChristian Blatter
174k8115327
174k8115327
add a comment |
add a comment |
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This series is uniformly convergent on $Bbb R$, and the series of the derivatives converges uniformly on $Bbb R$. This is enough to conclude that the derivative of $f(x)$ is simply the series of the derivatives.
$endgroup$
– Crostul
Dec 14 '18 at 16:38
$begingroup$
how do we know the series is uniformly convergent?
$endgroup$
– stackofhay42
Dec 14 '18 at 17:08
$begingroup$
Do you know Weierstrass' M-test? It's so simple to use here.
$endgroup$
– Crostul
Dec 14 '18 at 17:14
$begingroup$
i am allowed to use weierstrass m test, yes
$endgroup$
– stackofhay42
Dec 14 '18 at 17:17