Is there a solution to this functional equation?The functional equation $f(y) + fleft(frac1yright) = 0$Solution of functional equation $f(x/f(x)) = 1/f(x)$?Functional Equation $f(n) = 2 f(n / f(n) )$A late-diverging “approximating solution” for a system of functional equationsGeneral Solution to functional equationUniqueness of a solution to a functional equationChecking my understanding of Cauchy's functional equation.Finding domain where this complex logarithm identity holdsDoes Wolfram Alpha solve this equation incorrectly?solutions to this functional equation
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Is there a solution to this functional equation?
The functional equation $f(y) + fleft(frac1yright) = 0$Solution of functional equation $f(x/f(x)) = 1/f(x)$?Functional Equation $f(n) = 2 f(n / f(n) )$A late-diverging “approximating solution” for a system of functional equationsGeneral Solution to functional equationUniqueness of a solution to a functional equationChecking my understanding of Cauchy's functional equation.Finding domain where this complex logarithm identity holdsDoes Wolfram Alpha solve this equation incorrectly?solutions to this functional equation
$begingroup$
I was going through my old notebooks and I found a sheet of paper with this problem on it. I thought it would be a shame to let such an unreasonably difficult question go to waste, so I decided I would share it. The problem simply states:
Solve for $f$: $$f(x)=2log(x)^2fleft(x^frac38right)^2fleft(x^frac14right)^2$$
No other information or context is given, but I'm assuming that $f$ is a complex valued function of a single real or complex variable (since evaluating the function for negative $x$ would require $f(x)$ to be complex), and that $log$ is the natural logarithm (since no one would use it for log base-10 if they were talking about complex functions).
For curiosity's sake, I present it as a challenge to either solve for $f$ or prove that a solution does not exist there is one and only one solution (at least one solution exists, courtesy of Chrystomath). My own attempts at solving have been... unsuccessful.
Edit:
In their answer, Chrystomath provides a solution:
$$f(x)=alog(x)^-2/3quad:quad ainleftzinmathbbCmid z^9=frac6^48^9right$$
Which is quite possibly the most multivalued multivalued function I've ever seen.
I don't know whether or not the solution is unique, and I would still be interested in any other solutions.
complex-analysis recreational-mathematics functional-equations
asked Mar 21 at 17:33
R. BurtonR. Burton
732110
$endgroup$
add a comment |
$begingroup$
I was going through my old notebooks and I found a sheet of paper with this problem on it. I thought it would be a shame to let such an unreasonably difficult question go to waste, so I decided I would share it. The problem simply states:
Solve for $f$: $$f(x)=2log(x)^2fleft(x^frac38right)^2fleft(x^frac14right)^2$$
No other information or context is given, but I'm assuming that $f$ is a complex valued function of a single real or complex variable (since evaluating the function for negative $x$ would require $f(x)$ to be complex), and that $log$ is the natural logarithm (since no one would use it for log base-10 if they were talking about complex functions).
For curiosity's sake, I present it as a challenge to either solve for $f$ or prove that a solution does not exist there is one and only one solution (at least one solution exists, courtesy of Chrystomath). My own attempts at solving have been... unsuccessful.
Edit:
In their answer, Chrystomath provides a solution:
$$f(x)=alog(x)^-2/3quad:quad ainleftzinmathbbCmid z^9=frac6^48^9right$$
Which is quite possibly the most multivalued multivalued function I've ever seen.
I don't know whether or not the solution is unique, and I would still be interested in any other solutions.
complex-analysis recreational-mathematics functional-equations
asked Mar 21 at 17:33
R. BurtonR. Burton
732110
$endgroup$
1
$begingroup$
A better title would perhaps be more appreciated ;-)
$endgroup$
– tatan
Mar 21 at 17:38
1
$begingroup$
@tatan Title changed, thanks for the advice.
$endgroup$
– R. Burton
Mar 21 at 19:31
add a comment |
$begingroup$
I was going through my old notebooks and I found a sheet of paper with this problem on it. I thought it would be a shame to let such an unreasonably difficult question go to waste, so I decided I would share it. The problem simply states:
Solve for $f$: $$f(x)=2log(x)^2fleft(x^frac38right)^2fleft(x^frac14right)^2$$
No other information or context is given, but I'm assuming that $f$ is a complex valued function of a single real or complex variable (since evaluating the function for negative $x$ would require $f(x)$ to be complex), and that $log$ is the natural logarithm (since no one would use it for log base-10 if they were talking about complex functions).
For curiosity's sake, I present it as a challenge to either solve for $f$ or prove that a solution does not exist there is one and only one solution (at least one solution exists, courtesy of Chrystomath). My own attempts at solving have been... unsuccessful.
Edit:
In their answer, Chrystomath provides a solution:
$$f(x)=alog(x)^-2/3quad:quad ainleftzinmathbbCmid z^9=frac6^48^9right$$
Which is quite possibly the most multivalued multivalued function I've ever seen.
I don't know whether or not the solution is unique, and I would still be interested in any other solutions.
complex-analysis recreational-mathematics functional-equations
asked Mar 21 at 17:33
R. BurtonR. Burton
732110
$endgroup$
I was going through my old notebooks and I found a sheet of paper with this problem on it. I thought it would be a shame to let such an unreasonably difficult question go to waste, so I decided I would share it. The problem simply states:
Solve for $f$: $$f(x)=2log(x)^2fleft(x^frac38right)^2fleft(x^frac14right)^2$$
No other information or context is given, but I'm assuming that $f$ is a complex valued function of a single real or complex variable (since evaluating the function for negative $x$ would require $f(x)$ to be complex), and that $log$ is the natural logarithm (since no one would use it for log base-10 if they were talking about complex functions).
For curiosity's sake, I present it as a challenge to either solve for $f$ or prove that a solution does not exist there is one and only one solution (at least one solution exists, courtesy of Chrystomath). My own attempts at solving have been... unsuccessful.
Edit:
In their answer, Chrystomath provides a solution:
$$f(x)=alog(x)^-2/3quad:quad ainleftzinmathbbCmid z^9=frac6^48^9right$$
Which is quite possibly the most multivalued multivalued function I've ever seen.
I don't know whether or not the solution is unique, and I would still be interested in any other solutions.
complex-analysis recreational-mathematics functional-equations
complex-analysis recreational-mathematics functional-equations
asked Mar 21 at 17:33
R. BurtonR. Burton
732110
asked Mar 21 at 17:33
R. BurtonR. Burton
732110
edited Mar 21 at 20:57
R. Burton
asked Mar 21 at 17:33
R. BurtonR. Burton
732110
asked Mar 21 at 17:33
R. BurtonR. Burton
732110
asked Mar 21 at 17:33
R. BurtonR. Burton
732110
732110
1
$begingroup$
A better title would perhaps be more appreciated ;-)
$endgroup$
– tatan
Mar 21 at 17:38
1
$begingroup$
@tatan Title changed, thanks for the advice.
$endgroup$
– R. Burton
Mar 21 at 19:31
add a comment |
1
$begingroup$
A better title would perhaps be more appreciated ;-)
$endgroup$
– tatan
Mar 21 at 17:38
1
$begingroup$
@tatan Title changed, thanks for the advice.
$endgroup$
– R. Burton
Mar 21 at 19:31
1
1
$begingroup$
A better title would perhaps be more appreciated ;-)
$endgroup$
– tatan
Mar 21 at 17:38
$begingroup$
A better title would perhaps be more appreciated ;-)
$endgroup$
– tatan
Mar 21 at 17:38
1
1
$begingroup$
@tatan Title changed, thanks for the advice.
$endgroup$
– R. Burton
Mar 21 at 19:31
$begingroup$
@tatan Title changed, thanks for the advice.
$endgroup$
– R. Burton
Mar 21 at 19:31
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
One possible solution. Try $f(x)=a(ln x)^k$, then $$a(ln x)^k=2(ln x)^2a^2(frac38)^2k(ln x)^2ka^2(frac14)^2k(ln x)^2k=2a^4(frac332)^2k(ln x)^2(1+2k)$$ from which $a$ and $k=-2/3$ can be found.
answered Mar 21 at 17:57
ChrystomathChrystomath
1,978513
$endgroup$
$begingroup$
I can't believe I didn't think of that... Well, it's certainly a solution. It also appears that $a$ and $k$ are unique, with $k=-2/3$ and $ainleftzinmathbbCmid z^9=6^4/8^9right$, the only real $a$ being $6^4/9/8$ (verified numerically by finding the zeros of $h(x,y)=y(ln x)^k-2y^4(frac332)^2k(ln x)^2(1+2k)biggvert_k=-2/3$).
$endgroup$
– R. Burton
Mar 21 at 20:18
add a comment |
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$begingroup$
One possible solution. Try $f(x)=a(ln x)^k$, then $$a(ln x)^k=2(ln x)^2a^2(frac38)^2k(ln x)^2ka^2(frac14)^2k(ln x)^2k=2a^4(frac332)^2k(ln x)^2(1+2k)$$ from which $a$ and $k=-2/3$ can be found.
answered Mar 21 at 17:57
ChrystomathChrystomath
1,978513
$endgroup$
$begingroup$
I can't believe I didn't think of that... Well, it's certainly a solution. It also appears that $a$ and $k$ are unique, with $k=-2/3$ and $ainleftzinmathbbCmid z^9=6^4/8^9right$, the only real $a$ being $6^4/9/8$ (verified numerically by finding the zeros of $h(x,y)=y(ln x)^k-2y^4(frac332)^2k(ln x)^2(1+2k)biggvert_k=-2/3$).
$endgroup$
– R. Burton
Mar 21 at 20:18
add a comment |
$begingroup$
One possible solution. Try $f(x)=a(ln x)^k$, then $$a(ln x)^k=2(ln x)^2a^2(frac38)^2k(ln x)^2ka^2(frac14)^2k(ln x)^2k=2a^4(frac332)^2k(ln x)^2(1+2k)$$ from which $a$ and $k=-2/3$ can be found.
answered Mar 21 at 17:57
ChrystomathChrystomath
1,978513
$endgroup$
$begingroup$
I can't believe I didn't think of that... Well, it's certainly a solution. It also appears that $a$ and $k$ are unique, with $k=-2/3$ and $ainleftzinmathbbCmid z^9=6^4/8^9right$, the only real $a$ being $6^4/9/8$ (verified numerically by finding the zeros of $h(x,y)=y(ln x)^k-2y^4(frac332)^2k(ln x)^2(1+2k)biggvert_k=-2/3$).
$endgroup$
– R. Burton
Mar 21 at 20:18
add a comment |
$begingroup$
One possible solution. Try $f(x)=a(ln x)^k$, then $$a(ln x)^k=2(ln x)^2a^2(frac38)^2k(ln x)^2ka^2(frac14)^2k(ln x)^2k=2a^4(frac332)^2k(ln x)^2(1+2k)$$ from which $a$ and $k=-2/3$ can be found.
answered Mar 21 at 17:57
ChrystomathChrystomath
1,978513
$endgroup$
One possible solution. Try $f(x)=a(ln x)^k$, then $$a(ln x)^k=2(ln x)^2a^2(frac38)^2k(ln x)^2ka^2(frac14)^2k(ln x)^2k=2a^4(frac332)^2k(ln x)^2(1+2k)$$ from which $a$ and $k=-2/3$ can be found.
answered Mar 21 at 17:57
ChrystomathChrystomath
1,978513
answered Mar 21 at 17:57
ChrystomathChrystomath
1,978513
answered Mar 21 at 17:57
ChrystomathChrystomath
1,978513
answered Mar 21 at 17:57
ChrystomathChrystomath
1,978513
1,978513
$begingroup$
I can't believe I didn't think of that... Well, it's certainly a solution. It also appears that $a$ and $k$ are unique, with $k=-2/3$ and $ainleftzinmathbbCmid z^9=6^4/8^9right$, the only real $a$ being $6^4/9/8$ (verified numerically by finding the zeros of $h(x,y)=y(ln x)^k-2y^4(frac332)^2k(ln x)^2(1+2k)biggvert_k=-2/3$).
$endgroup$
– R. Burton
Mar 21 at 20:18
add a comment |
$begingroup$
I can't believe I didn't think of that... Well, it's certainly a solution. It also appears that $a$ and $k$ are unique, with $k=-2/3$ and $ainleftzinmathbbCmid z^9=6^4/8^9right$, the only real $a$ being $6^4/9/8$ (verified numerically by finding the zeros of $h(x,y)=y(ln x)^k-2y^4(frac332)^2k(ln x)^2(1+2k)biggvert_k=-2/3$).
$endgroup$
– R. Burton
Mar 21 at 20:18
$begingroup$
I can't believe I didn't think of that... Well, it's certainly a solution. It also appears that $a$ and $k$ are unique, with $k=-2/3$ and $ainleftzinmathbbCmid z^9=6^4/8^9right$, the only real $a$ being $6^4/9/8$ (verified numerically by finding the zeros of $h(x,y)=y(ln x)^k-2y^4(frac332)^2k(ln x)^2(1+2k)biggvert_k=-2/3$).
$endgroup$
– R. Burton
Mar 21 at 20:18
$begingroup$
I can't believe I didn't think of that... Well, it's certainly a solution. It also appears that $a$ and $k$ are unique, with $k=-2/3$ and $ainleftzinmathbbCmid z^9=6^4/8^9right$, the only real $a$ being $6^4/9/8$ (verified numerically by finding the zeros of $h(x,y)=y(ln x)^k-2y^4(frac332)^2k(ln x)^2(1+2k)biggvert_k=-2/3$).
$endgroup$
– R. Burton
Mar 21 at 20:18
add a comment |
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$begingroup$
A better title would perhaps be more appreciated ;-)
$endgroup$
– tatan
Mar 21 at 17:38
1
$begingroup$
@tatan Title changed, thanks for the advice.
$endgroup$
– R. Burton
Mar 21 at 19:31