Does there exist a function which satisfies follow conditions?
-1
$begingroup$
Does there exist a function which satisfies following conditions?
1) $F(x)=frac{G(x)}{H(x)}$ is a sigmoid function (or S-shaped function);
2) $G(x)$ is strictly concave;
3) $H(x)$ is not a constant.
linear-algebra
asked Nov 30 '18 at 12:35
Lee WhiteLee White
86
$endgroup$
add a comment |
-1
$begingroup$
Does there exist a function which satisfies following conditions?
1) $F(x)=frac{G(x)}{H(x)}$ is a sigmoid function (or S-shaped function);
2) $G(x)$ is strictly concave;
3) $H(x)$ is not a constant.
linear-algebra
asked Nov 30 '18 at 12:35
Lee WhiteLee White
86
$endgroup$
add a comment |
-1
-1
-1
0
$begingroup$
Does there exist a function which satisfies following conditions?
1) $F(x)=frac{G(x)}{H(x)}$ is a sigmoid function (or S-shaped function);
2) $G(x)$ is strictly concave;
3) $H(x)$ is not a constant.
linear-algebra
asked Nov 30 '18 at 12:35
Lee WhiteLee White
86
$endgroup$
Does there exist a function which satisfies following conditions?
1) $F(x)=frac{G(x)}{H(x)}$ is a sigmoid function (or S-shaped function);
2) $G(x)$ is strictly concave;
3) $H(x)$ is not a constant.
linear-algebra
linear-algebra
asked Nov 30 '18 at 12:35
Lee WhiteLee White
86
asked Nov 30 '18 at 12:35
Lee WhiteLee White
86
edited Nov 30 '18 at 13:46
Lee White
asked Nov 30 '18 at 12:35
Lee WhiteLee White
86
asked Nov 30 '18 at 12:35
Lee WhiteLee White
86
asked Nov 30 '18 at 12:35
Lee WhiteLee White
86
86
add a comment |
add a comment |
1 Answer
1
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$begingroup$
G(x) = log(x+1)
H(x) = 1+log(x+1)
F(x) = log(x+1)/(1+log(x+1))
That works grand
edit: you edited your question. Your new answer is
H = log(x+1)/(tanh(x+4)+1)
G = log(x+1)
F = tanh(x+4)+1
answered Nov 30 '18 at 12:52
D'ArcyD'Arcy
263
$endgroup$
$begingroup$
Thanks for your answers. But I can't fully understand the functions you offered. The first function, referred to as $F1$, indeed works when $x in [0,infty]$, which I can understand. But the second one, $F2=1/(tanh(x+4)+1)$, is this a sigmoid function?
$endgroup$
– Lee White
Nov 30 '18 at 13:23
$begingroup$
Yep, I had a typo in there, basically F can be any sigmoid function you like, just make G random concave function, and H the inverse of a sigmoid times G
$endgroup$
– D'Arcy
Nov 30 '18 at 13:31
1
$begingroup$
Cool, man! I get it! I will try to work as you said, it's a nice method.
$endgroup$
– Lee White
Nov 30 '18 at 13:43
add a comment |
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1 Answer
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1 Answer
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0
$begingroup$
G(x) = log(x+1)
H(x) = 1+log(x+1)
F(x) = log(x+1)/(1+log(x+1))
That works grand
edit: you edited your question. Your new answer is
H = log(x+1)/(tanh(x+4)+1)
G = log(x+1)
F = tanh(x+4)+1
answered Nov 30 '18 at 12:52
D'ArcyD'Arcy
263
$endgroup$
$begingroup$
Thanks for your answers. But I can't fully understand the functions you offered. The first function, referred to as $F1$, indeed works when $x in [0,infty]$, which I can understand. But the second one, $F2=1/(tanh(x+4)+1)$, is this a sigmoid function?
$endgroup$
– Lee White
Nov 30 '18 at 13:23
$begingroup$
Yep, I had a typo in there, basically F can be any sigmoid function you like, just make G random concave function, and H the inverse of a sigmoid times G
$endgroup$
– D'Arcy
Nov 30 '18 at 13:31
1
$begingroup$
Cool, man! I get it! I will try to work as you said, it's a nice method.
$endgroup$
– Lee White
Nov 30 '18 at 13:43
add a comment |
0
$begingroup$
G(x) = log(x+1)
H(x) = 1+log(x+1)
F(x) = log(x+1)/(1+log(x+1))
That works grand
edit: you edited your question. Your new answer is
H = log(x+1)/(tanh(x+4)+1)
G = log(x+1)
F = tanh(x+4)+1
answered Nov 30 '18 at 12:52
D'ArcyD'Arcy
263
$endgroup$
$begingroup$
Thanks for your answers. But I can't fully understand the functions you offered. The first function, referred to as $F1$, indeed works when $x in [0,infty]$, which I can understand. But the second one, $F2=1/(tanh(x+4)+1)$, is this a sigmoid function?
$endgroup$
– Lee White
Nov 30 '18 at 13:23
$begingroup$
Yep, I had a typo in there, basically F can be any sigmoid function you like, just make G random concave function, and H the inverse of a sigmoid times G
$endgroup$
– D'Arcy
Nov 30 '18 at 13:31
1
$begingroup$
Cool, man! I get it! I will try to work as you said, it's a nice method.
$endgroup$
– Lee White
Nov 30 '18 at 13:43
add a comment |
0
0
0
$begingroup$
G(x) = log(x+1)
H(x) = 1+log(x+1)
F(x) = log(x+1)/(1+log(x+1))
That works grand
edit: you edited your question. Your new answer is
H = log(x+1)/(tanh(x+4)+1)
G = log(x+1)
F = tanh(x+4)+1
answered Nov 30 '18 at 12:52
D'ArcyD'Arcy
263
$endgroup$
G(x) = log(x+1)
H(x) = 1+log(x+1)
F(x) = log(x+1)/(1+log(x+1))
That works grand
edit: you edited your question. Your new answer is
H = log(x+1)/(tanh(x+4)+1)
G = log(x+1)
F = tanh(x+4)+1
answered Nov 30 '18 at 12:52
D'ArcyD'Arcy
263
edited Nov 30 '18 at 13:30
answered Nov 30 '18 at 12:52
D'ArcyD'Arcy
263
answered Nov 30 '18 at 12:52
D'ArcyD'Arcy
263
answered Nov 30 '18 at 12:52
D'ArcyD'Arcy
263
263
$begingroup$
Thanks for your answers. But I can't fully understand the functions you offered. The first function, referred to as $F1$, indeed works when $x in [0,infty]$, which I can understand. But the second one, $F2=1/(tanh(x+4)+1)$, is this a sigmoid function?
$endgroup$
– Lee White
Nov 30 '18 at 13:23
$begingroup$
Yep, I had a typo in there, basically F can be any sigmoid function you like, just make G random concave function, and H the inverse of a sigmoid times G
$endgroup$
– D'Arcy
Nov 30 '18 at 13:31
1
$begingroup$
Cool, man! I get it! I will try to work as you said, it's a nice method.
$endgroup$
– Lee White
Nov 30 '18 at 13:43
add a comment |
$begingroup$
Thanks for your answers. But I can't fully understand the functions you offered. The first function, referred to as $F1$, indeed works when $x in [0,infty]$, which I can understand. But the second one, $F2=1/(tanh(x+4)+1)$, is this a sigmoid function?
$endgroup$
– Lee White
Nov 30 '18 at 13:23
$begingroup$
Yep, I had a typo in there, basically F can be any sigmoid function you like, just make G random concave function, and H the inverse of a sigmoid times G
$endgroup$
– D'Arcy
Nov 30 '18 at 13:31
1
$begingroup$
Cool, man! I get it! I will try to work as you said, it's a nice method.
$endgroup$
– Lee White
Nov 30 '18 at 13:43
$begingroup$
Thanks for your answers. But I can't fully understand the functions you offered. The first function, referred to as $F1$, indeed works when $x in [0,infty]$, which I can understand. But the second one, $F2=1/(tanh(x+4)+1)$, is this a sigmoid function?
$endgroup$
– Lee White
Nov 30 '18 at 13:23
$begingroup$
Thanks for your answers. But I can't fully understand the functions you offered. The first function, referred to as $F1$, indeed works when $x in [0,infty]$, which I can understand. But the second one, $F2=1/(tanh(x+4)+1)$, is this a sigmoid function?
$endgroup$
– Lee White
Nov 30 '18 at 13:23
$begingroup$
Yep, I had a typo in there, basically F can be any sigmoid function you like, just make G random concave function, and H the inverse of a sigmoid times G
$endgroup$
– D'Arcy
Nov 30 '18 at 13:31
$begingroup$
Yep, I had a typo in there, basically F can be any sigmoid function you like, just make G random concave function, and H the inverse of a sigmoid times G
$endgroup$
– D'Arcy
Nov 30 '18 at 13:31
1
1
$begingroup$
Cool, man! I get it! I will try to work as you said, it's a nice method.
$endgroup$
– Lee White
Nov 30 '18 at 13:43
$begingroup$
Cool, man! I get it! I will try to work as you said, it's a nice method.
$endgroup$
– Lee White
Nov 30 '18 at 13:43
add a comment |
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