Mixing
About $ f(x^2) = f(x) + f(x/2) $
$begingroup$
I was thinking about the equation
$$ f(x^2) = f(x) + f(x/2) $$
This should be consistant for $x>1$.
And probably for all reals. But I focus on $x>1$.
This equation implies that $f$ grows slower than a logarithm or any power of a logarithm. But still faster than a double logarithm.
[*] Maybe like $exp( sqrt ln (ln(x)) ) $ ??
It is a simple Equation, but I have not seen it before.
The estimate [*] comes from The similar equation
$$ g(n+1) = g(n) + g(n/2) $$
Or
$$ h’(x) = h(x/2) $$
Which I discussed here before and is strongly related to the so-called binary partitions function.
$h$ can be given by an infinite sum.
Can $f$ be given by an infinite sum ?
See If $f(x) = sum limits_{n=0}^{infty} frac{x^n}{2^{n(n-1)/2} n!}$ then $f^{-1}(f(x)-f(x-1))-frac{x}{2}$ is bounded
sequences-and-series asymptotics functional-equations closed-form
asked Jan 24 at 23:46
mickmick
5,19332164
$endgroup$
add a comment |
$begingroup$
I was thinking about the equation
$$ f(x^2) = f(x) + f(x/2) $$
This should be consistant for $x>1$.
And probably for all reals. But I focus on $x>1$.
This equation implies that $f$ grows slower than a logarithm or any power of a logarithm. But still faster than a double logarithm.
[*] Maybe like $exp( sqrt ln (ln(x)) ) $ ??
It is a simple Equation, but I have not seen it before.
The estimate [*] comes from The similar equation
$$ g(n+1) = g(n) + g(n/2) $$
Or
$$ h’(x) = h(x/2) $$
Which I discussed here before and is strongly related to the so-called binary partitions function.
$h$ can be given by an infinite sum.
Can $f$ be given by an infinite sum ?
See If $f(x) = sum limits_{n=0}^{infty} frac{x^n}{2^{n(n-1)/2} n!}$ then $f^{-1}(f(x)-f(x-1))-frac{x}{2}$ is bounded
sequences-and-series asymptotics functional-equations closed-form
asked Jan 24 at 23:46
mickmick
5,19332164
$endgroup$
$begingroup$
You may try to write it as a formal series near 0 and find coefficients, so that you can guess what $f$ should be.
$endgroup$
– Seewoo Lee
Jan 24 at 23:56
1
$begingroup$
You can show that the zero function is the only such function which is analytic on a neighbourhood of zero.
$endgroup$
– nathan.j.mcdougall
Jan 25 at 0:01
add a comment |
$begingroup$
I was thinking about the equation
$$ f(x^2) = f(x) + f(x/2) $$
This should be consistant for $x>1$.
And probably for all reals. But I focus on $x>1$.
This equation implies that $f$ grows slower than a logarithm or any power of a logarithm. But still faster than a double logarithm.
[*] Maybe like $exp( sqrt ln (ln(x)) ) $ ??
It is a simple Equation, but I have not seen it before.
The estimate [*] comes from The similar equation
$$ g(n+1) = g(n) + g(n/2) $$
Or
$$ h’(x) = h(x/2) $$
Which I discussed here before and is strongly related to the so-called binary partitions function.
$h$ can be given by an infinite sum.
Can $f$ be given by an infinite sum ?
See If $f(x) = sum limits_{n=0}^{infty} frac{x^n}{2^{n(n-1)/2} n!}$ then $f^{-1}(f(x)-f(x-1))-frac{x}{2}$ is bounded
sequences-and-series asymptotics functional-equations closed-form
asked Jan 24 at 23:46
mickmick
5,19332164
$endgroup$
I was thinking about the equation
$$ f(x^2) = f(x) + f(x/2) $$
This should be consistant for $x>1$.
And probably for all reals. But I focus on $x>1$.
This equation implies that $f$ grows slower than a logarithm or any power of a logarithm. But still faster than a double logarithm.
[*] Maybe like $exp( sqrt ln (ln(x)) ) $ ??
It is a simple Equation, but I have not seen it before.
The estimate [*] comes from The similar equation
$$ g(n+1) = g(n) + g(n/2) $$
Or
$$ h’(x) = h(x/2) $$
Which I discussed here before and is strongly related to the so-called binary partitions function.
$h$ can be given by an infinite sum.
Can $f$ be given by an infinite sum ?
See If $f(x) = sum limits_{n=0}^{infty} frac{x^n}{2^{n(n-1)/2} n!}$ then $f^{-1}(f(x)-f(x-1))-frac{x}{2}$ is bounded
sequences-and-series asymptotics functional-equations closed-form
sequences-and-series asymptotics functional-equations closed-form
asked Jan 24 at 23:46
mickmick
5,19332164
asked Jan 24 at 23:46
mickmick
5,19332164
edited Jan 25 at 0:03
mick
asked Jan 24 at 23:46
mickmick
5,19332164
asked Jan 24 at 23:46
mickmick
5,19332164
asked Jan 24 at 23:46
mickmick
5,19332164
5,19332164
$begingroup$
You may try to write it as a formal series near 0 and find coefficients, so that you can guess what $f$ should be.
$endgroup$
– Seewoo Lee
Jan 24 at 23:56
1
$begingroup$
You can show that the zero function is the only such function which is analytic on a neighbourhood of zero.
$endgroup$
– nathan.j.mcdougall
Jan 25 at 0:01
add a comment |
$begingroup$
You may try to write it as a formal series near 0 and find coefficients, so that you can guess what $f$ should be.
$endgroup$
– Seewoo Lee
Jan 24 at 23:56
1
$begingroup$
You can show that the zero function is the only such function which is analytic on a neighbourhood of zero.
$endgroup$
– nathan.j.mcdougall
Jan 25 at 0:01
$begingroup$
You may try to write it as a formal series near 0 and find coefficients, so that you can guess what $f$ should be.
$endgroup$
– Seewoo Lee
Jan 24 at 23:56
$begingroup$
You may try to write it as a formal series near 0 and find coefficients, so that you can guess what $f$ should be.
$endgroup$
– Seewoo Lee
Jan 24 at 23:56
1
1
$begingroup$
You can show that the zero function is the only such function which is analytic on a neighbourhood of zero.
$endgroup$
– nathan.j.mcdougall
Jan 25 at 0:01
$begingroup$
You can show that the zero function is the only such function which is analytic on a neighbourhood of zero.
$endgroup$
– nathan.j.mcdougall
Jan 25 at 0:01
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
$f(x)=1+log_2(x)$ solves the equation for $x > 0$:
$f(x^2)=1+log_2(x^2)= 1+2log_2(x)$
$f(x)+f(x/2)=1+log_2(x) + 1+log_2(x/2)=1+log_2(x) + 1+log_2(x)-1=1+2log_2(x)$.
answered Jan 24 at 23:58
IngixIngix
4,912159
$endgroup$
$begingroup$
Oh yes. Wow. Nice +1. Are there other solutions ?
$endgroup$
– mick
Jan 25 at 0:02
$begingroup$
And every constant multiple of that solution ofcourse.
$endgroup$
– mick
Jan 25 at 0:25
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$f(x)=1+log_2(x)$ solves the equation for $x > 0$:
$f(x^2)=1+log_2(x^2)= 1+2log_2(x)$
$f(x)+f(x/2)=1+log_2(x) + 1+log_2(x/2)=1+log_2(x) + 1+log_2(x)-1=1+2log_2(x)$.
answered Jan 24 at 23:58
IngixIngix
4,912159
$endgroup$
$begingroup$
Oh yes. Wow. Nice +1. Are there other solutions ?
$endgroup$
– mick
Jan 25 at 0:02
$begingroup$
And every constant multiple of that solution ofcourse.
$endgroup$
– mick
Jan 25 at 0:25
add a comment |
$begingroup$
$f(x)=1+log_2(x)$ solves the equation for $x > 0$:
$f(x^2)=1+log_2(x^2)= 1+2log_2(x)$
$f(x)+f(x/2)=1+log_2(x) + 1+log_2(x/2)=1+log_2(x) + 1+log_2(x)-1=1+2log_2(x)$.
answered Jan 24 at 23:58
IngixIngix
4,912159
$endgroup$
$begingroup$
Oh yes. Wow. Nice +1. Are there other solutions ?
$endgroup$
– mick
Jan 25 at 0:02
$begingroup$
And every constant multiple of that solution ofcourse.
$endgroup$
– mick
Jan 25 at 0:25
add a comment |
$begingroup$
$f(x)=1+log_2(x)$ solves the equation for $x > 0$:
$f(x^2)=1+log_2(x^2)= 1+2log_2(x)$
$f(x)+f(x/2)=1+log_2(x) + 1+log_2(x/2)=1+log_2(x) + 1+log_2(x)-1=1+2log_2(x)$.
answered Jan 24 at 23:58
IngixIngix
4,912159
$endgroup$
$f(x)=1+log_2(x)$ solves the equation for $x > 0$:
$f(x^2)=1+log_2(x^2)= 1+2log_2(x)$
$f(x)+f(x/2)=1+log_2(x) + 1+log_2(x/2)=1+log_2(x) + 1+log_2(x)-1=1+2log_2(x)$.
answered Jan 24 at 23:58
IngixIngix
4,912159
answered Jan 24 at 23:58
IngixIngix
4,912159
answered Jan 24 at 23:58
IngixIngix
4,912159
answered Jan 24 at 23:58
IngixIngix
4,912159
4,912159
$begingroup$
Oh yes. Wow. Nice +1. Are there other solutions ?
$endgroup$
– mick
Jan 25 at 0:02
$begingroup$
And every constant multiple of that solution ofcourse.
$endgroup$
– mick
Jan 25 at 0:25
add a comment |
$begingroup$
Oh yes. Wow. Nice +1. Are there other solutions ?
$endgroup$
– mick
Jan 25 at 0:02
$begingroup$
And every constant multiple of that solution ofcourse.
$endgroup$
– mick
Jan 25 at 0:25
$begingroup$
Oh yes. Wow. Nice +1. Are there other solutions ?
$endgroup$
– mick
Jan 25 at 0:02
$begingroup$
Oh yes. Wow. Nice +1. Are there other solutions ?
$endgroup$
– mick
Jan 25 at 0:02
$begingroup$
And every constant multiple of that solution ofcourse.
$endgroup$
– mick
Jan 25 at 0:25
$begingroup$
And every constant multiple of that solution ofcourse.
$endgroup$
– mick
Jan 25 at 0:25
add a comment |
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$begingroup$
You may try to write it as a formal series near 0 and find coefficients, so that you can guess what $f$ should be.
$endgroup$
– Seewoo Lee
Jan 24 at 23:56
1
$begingroup$
You can show that the zero function is the only such function which is analytic on a neighbourhood of zero.
$endgroup$
– nathan.j.mcdougall
Jan 25 at 0:01