You left out two special solutions to the Cauchy functional equation g(xy) = g(x)g(y); where g(x) is identically equal to zero or to one. Since g(x) =ln(f(x)) this leads to the special solutions f(x) = 1 always and f(x) = e always. It could be argued that you covered the latter: let n = 0 in f(x) = e^x^n and you get f(x) = 1 for non-zero x. I'll leave the case x = 0 which leads to 0^0 which is controversial. But you definitely left out the solution f(x) = 1. And of course you left out non-continuous solutions but you basically mentioned that.
Take ln on both sides and put x=0 . this gives f(y) is e .. now subtituting in initial eqn. gives f(xy)=f(×) which implies f(x) is constant function whose value is equal to e
Is there ever a way for stopping something without a mirroring effect a simpler equation to get there faster?😂 could you ever tell within that equation whether or not the number would be positive or negative I’m sorry I’m just a chimpanzee traveling in your world.😂
You left out two special solutions to the Cauchy functional equation g(xy) = g(x)g(y); where g(x) is identically equal to zero or to one. Since g(x) =ln(f(x)) this leads to the special solutions f(x) = 1 always and f(x) = e always. It could be argued that you covered the latter: let n = 0 in f(x) = e^x^n and you get f(x) = 1 for non-zero x. I'll leave the case x = 0 which leads to 0^0 which is controversial. But you definitely left out the solution f(x) = 1. And of course you left out non-continuous solutions but you basically mentioned that.
Take ln on both sides and put x=0 . this gives f(y) is e .. now subtituting in initial eqn. gives f(xy)=f(×) which implies f(x) is constant function whose value is equal to e
Is there ever a way for stopping something without a mirroring effect a simpler equation to get there faster?😂
Nice!
Is there ever a way for stopping something without a mirroring effect a simpler equation to get there faster?😂 could you ever tell within that equation whether or not the number would be positive or negative I’m sorry I’m just a chimpanzee traveling in your world.😂
f(xy)=f(x)^(ln(f(y))
y=1 → f(x)=f(x)^(ln(f(1)) →f(1)=e
x=1 → f(y)=f(1)^(ln(f(y))=e^(ln(f(y))
y=x → f(x²)=f(x)^(ln(f(x))
ln(f(x²))=(ln(f(x))²
ln(f(x))=x →f(x)=e^x
f(xy)=e^(xy)=(e^x)^y
=(e^x)^(ln(e^y))=f(x)^(ln(f(y))