Looking for infinite series resembling an exponential
$begingroup$
I'm looking for some $f(x)$ that has the following property:
$sum_{x=1}^infty f(kx) = r^k$
for some real $0 < r < 1$, and at least for strictly positive integer $k$.
Does such an $f(x)$ exist?
This could also be thought of in terms of some sequence of real numbers $f[n]$.
I posted this at MSE but got no answer, so thought I would try MO.
real-analysis sequences-and-series power-series
asked Jan 22 at 0:26
Mike BattagliaMike Battaglia
1,153623
$endgroup$
add a comment |
$begingroup$
I'm looking for some $f(x)$ that has the following property:
$sum_{x=1}^infty f(kx) = r^k$
for some real $0 < r < 1$, and at least for strictly positive integer $k$.
Does such an $f(x)$ exist?
This could also be thought of in terms of some sequence of real numbers $f[n]$.
I posted this at MSE but got no answer, so thought I would try MO.
real-analysis sequences-and-series power-series
asked Jan 22 at 0:26
Mike BattagliaMike Battaglia
1,153623
$endgroup$
$begingroup$
The m.se post is math.stackexchange.com/questions/3078889/…
$endgroup$
– Gerry Myerson
Jan 23 at 19:59
add a comment |
$begingroup$
I'm looking for some $f(x)$ that has the following property:
$sum_{x=1}^infty f(kx) = r^k$
for some real $0 < r < 1$, and at least for strictly positive integer $k$.
Does such an $f(x)$ exist?
This could also be thought of in terms of some sequence of real numbers $f[n]$.
I posted this at MSE but got no answer, so thought I would try MO.
real-analysis sequences-and-series power-series
asked Jan 22 at 0:26
Mike BattagliaMike Battaglia
1,153623
$endgroup$
I'm looking for some $f(x)$ that has the following property:
$sum_{x=1}^infty f(kx) = r^k$
for some real $0 < r < 1$, and at least for strictly positive integer $k$.
Does such an $f(x)$ exist?
This could also be thought of in terms of some sequence of real numbers $f[n]$.
I posted this at MSE but got no answer, so thought I would try MO.
real-analysis sequences-and-series power-series
real-analysis sequences-and-series power-series
asked Jan 22 at 0:26
Mike BattagliaMike Battaglia
1,153623
asked Jan 22 at 0:26
Mike BattagliaMike Battaglia
1,153623
asked Jan 22 at 0:26
Mike BattagliaMike Battaglia
1,153623
asked Jan 22 at 0:26
Mike BattagliaMike Battaglia
1,153623
asked Jan 22 at 0:26
Mike BattagliaMike Battaglia
1,153623
1,153623
$begingroup$
The m.se post is math.stackexchange.com/questions/3078889/…
$endgroup$
– Gerry Myerson
Jan 23 at 19:59
add a comment |
$begingroup$
The m.se post is math.stackexchange.com/questions/3078889/…
$endgroup$
– Gerry Myerson
Jan 23 at 19:59
$begingroup$
The m.se post is math.stackexchange.com/questions/3078889/…
$endgroup$
– Gerry Myerson
Jan 23 at 19:59
$begingroup$
The m.se post is math.stackexchange.com/questions/3078889/…
$endgroup$
– Gerry Myerson
Jan 23 at 19:59
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You can solve this system of equations explicitly in terms of the function
$$F(z)=sum_{m=1}^{infty} mu(m)z^m=z-z^2-z^3-z^5+z^6+cdots$$
where $mu(m)$ is the Möbius function.
The function $F(z)$ can be shown to be transcendental (see the paper "Transcendence of Power Series for Some Number Theoretic Functions", by Coons and Borwein). With this notation your sequence can be succinctly expressed as
$$f(n)=F(r^n).$$
Indeed, one can check that
$$sum_{n=1}^{infty} f(kn)=sum_{n=1}^{infty} F(r^{kn})=sum_{m,n=1}^{infty} mu(m)r^{nmk}=sum_{m=1}^{infty}sum_{d|m}mu(d)r^{mk}=r^k$$
as desired.
answered Jan 22 at 1:00
Gjergji ZaimiGjergji Zaimi
63.2k4165313
$endgroup$
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You can solve this system of equations explicitly in terms of the function
$$F(z)=sum_{m=1}^{infty} mu(m)z^m=z-z^2-z^3-z^5+z^6+cdots$$
where $mu(m)$ is the Möbius function.
The function $F(z)$ can be shown to be transcendental (see the paper "Transcendence of Power Series for Some Number Theoretic Functions", by Coons and Borwein). With this notation your sequence can be succinctly expressed as
$$f(n)=F(r^n).$$
Indeed, one can check that
$$sum_{n=1}^{infty} f(kn)=sum_{n=1}^{infty} F(r^{kn})=sum_{m,n=1}^{infty} mu(m)r^{nmk}=sum_{m=1}^{infty}sum_{d|m}mu(d)r^{mk}=r^k$$
as desired.
answered Jan 22 at 1:00
Gjergji ZaimiGjergji Zaimi
63.2k4165313
$endgroup$
add a comment |
$begingroup$
You can solve this system of equations explicitly in terms of the function
$$F(z)=sum_{m=1}^{infty} mu(m)z^m=z-z^2-z^3-z^5+z^6+cdots$$
where $mu(m)$ is the Möbius function.
The function $F(z)$ can be shown to be transcendental (see the paper "Transcendence of Power Series for Some Number Theoretic Functions", by Coons and Borwein). With this notation your sequence can be succinctly expressed as
$$f(n)=F(r^n).$$
Indeed, one can check that
$$sum_{n=1}^{infty} f(kn)=sum_{n=1}^{infty} F(r^{kn})=sum_{m,n=1}^{infty} mu(m)r^{nmk}=sum_{m=1}^{infty}sum_{d|m}mu(d)r^{mk}=r^k$$
as desired.
answered Jan 22 at 1:00
Gjergji ZaimiGjergji Zaimi
63.2k4165313
$endgroup$
add a comment |
$begingroup$
You can solve this system of equations explicitly in terms of the function
$$F(z)=sum_{m=1}^{infty} mu(m)z^m=z-z^2-z^3-z^5+z^6+cdots$$
where $mu(m)$ is the Möbius function.
The function $F(z)$ can be shown to be transcendental (see the paper "Transcendence of Power Series for Some Number Theoretic Functions", by Coons and Borwein). With this notation your sequence can be succinctly expressed as
$$f(n)=F(r^n).$$
Indeed, one can check that
$$sum_{n=1}^{infty} f(kn)=sum_{n=1}^{infty} F(r^{kn})=sum_{m,n=1}^{infty} mu(m)r^{nmk}=sum_{m=1}^{infty}sum_{d|m}mu(d)r^{mk}=r^k$$
as desired.
answered Jan 22 at 1:00
Gjergji ZaimiGjergji Zaimi
63.2k4165313
$endgroup$
You can solve this system of equations explicitly in terms of the function
$$F(z)=sum_{m=1}^{infty} mu(m)z^m=z-z^2-z^3-z^5+z^6+cdots$$
where $mu(m)$ is the Möbius function.
The function $F(z)$ can be shown to be transcendental (see the paper "Transcendence of Power Series for Some Number Theoretic Functions", by Coons and Borwein). With this notation your sequence can be succinctly expressed as
$$f(n)=F(r^n).$$
Indeed, one can check that
$$sum_{n=1}^{infty} f(kn)=sum_{n=1}^{infty} F(r^{kn})=sum_{m,n=1}^{infty} mu(m)r^{nmk}=sum_{m=1}^{infty}sum_{d|m}mu(d)r^{mk}=r^k$$
as desired.
answered Jan 22 at 1:00
Gjergji ZaimiGjergji Zaimi
63.2k4165313
edited Jan 23 at 17:58
answered Jan 22 at 1:00
Gjergji ZaimiGjergji Zaimi
63.2k4165313
answered Jan 22 at 1:00
Gjergji ZaimiGjergji Zaimi
63.2k4165313
answered Jan 22 at 1:00
Gjergji ZaimiGjergji Zaimi
63.2k4165313
63.2k4165313
add a comment |
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$begingroup$
The m.se post is math.stackexchange.com/questions/3078889/…
$endgroup$
– Gerry Myerson
Jan 23 at 19:59