What exactly is the “hyperbolic” tanh function used in the context of activation functions?
$begingroup$
I know the plot of $tanh$ activation function looks like. I also know that its output has a range of $[-1, 1]$. Furthermore, I also know the it is defined as follows
$$
tanh(x) = frac{sinh(x)}{cosh(x)} = frac{e^x - e^{-x}}{e^x + e^{-x}}
$$
- What exactly is the meaning of term "hyperbolic"?
- What is the interpretation of the word hyperbolic in this context? How is the equation above derived?
machine-learning neural-network deep-learning activation-function linear-algebra
asked Mar 6 '18 at 15:35
GeorgeOfTheRFGeorgeOfTheRF
4682817
$endgroup$
add a comment |
$begingroup$
I know the plot of $tanh$ activation function looks like. I also know that its output has a range of $[-1, 1]$. Furthermore, I also know the it is defined as follows
$$
tanh(x) = frac{sinh(x)}{cosh(x)} = frac{e^x - e^{-x}}{e^x + e^{-x}}
$$
- What exactly is the meaning of term "hyperbolic"?
- What is the interpretation of the word hyperbolic in this context? How is the equation above derived?
machine-learning neural-network deep-learning activation-function linear-algebra
asked Mar 6 '18 at 15:35
GeorgeOfTheRFGeorgeOfTheRF
4682817
$endgroup$
$begingroup$
The answer by Tophat is a good one, but I'll add a fun fact. Rotations trade a bit of the component in one dimension to another, turning "x"ness into "y"ness. In the theory of special relativity, you can also trade a bit of the component of a spacial dimension with the time dimension, turning length into duration. This is why "clocks slow down" and "objects contract" when moving relative to an observer. The difference, though, is that the latter uses hyperbolic trig functions, while the former uses standard (circular) trig functions.
$endgroup$
– Tac-Tics
Mar 6 '18 at 17:25
add a comment |
$begingroup$
I know the plot of $tanh$ activation function looks like. I also know that its output has a range of $[-1, 1]$. Furthermore, I also know the it is defined as follows
$$
tanh(x) = frac{sinh(x)}{cosh(x)} = frac{e^x - e^{-x}}{e^x + e^{-x}}
$$
- What exactly is the meaning of term "hyperbolic"?
- What is the interpretation of the word hyperbolic in this context? How is the equation above derived?
machine-learning neural-network deep-learning activation-function linear-algebra
asked Mar 6 '18 at 15:35
GeorgeOfTheRFGeorgeOfTheRF
4682817
$endgroup$
I know the plot of $tanh$ activation function looks like. I also know that its output has a range of $[-1, 1]$. Furthermore, I also know the it is defined as follows
$$
tanh(x) = frac{sinh(x)}{cosh(x)} = frac{e^x - e^{-x}}{e^x + e^{-x}}
$$
- What exactly is the meaning of term "hyperbolic"?
- What is the interpretation of the word hyperbolic in this context? How is the equation above derived?
machine-learning neural-network deep-learning activation-function linear-algebra
machine-learning neural-network deep-learning activation-function linear-algebra
asked Mar 6 '18 at 15:35
GeorgeOfTheRFGeorgeOfTheRF
4682817
asked Mar 6 '18 at 15:35
GeorgeOfTheRFGeorgeOfTheRF
4682817
edited 9 mins ago
GeorgeOfTheRF
asked Mar 6 '18 at 15:35
GeorgeOfTheRFGeorgeOfTheRF
4682817
asked Mar 6 '18 at 15:35
GeorgeOfTheRFGeorgeOfTheRF
4682817
asked Mar 6 '18 at 15:35
GeorgeOfTheRFGeorgeOfTheRF
4682817
4682817
$begingroup$
The answer by Tophat is a good one, but I'll add a fun fact. Rotations trade a bit of the component in one dimension to another, turning "x"ness into "y"ness. In the theory of special relativity, you can also trade a bit of the component of a spacial dimension with the time dimension, turning length into duration. This is why "clocks slow down" and "objects contract" when moving relative to an observer. The difference, though, is that the latter uses hyperbolic trig functions, while the former uses standard (circular) trig functions.
$endgroup$
– Tac-Tics
Mar 6 '18 at 17:25
add a comment |
$begingroup$
The answer by Tophat is a good one, but I'll add a fun fact. Rotations trade a bit of the component in one dimension to another, turning "x"ness into "y"ness. In the theory of special relativity, you can also trade a bit of the component of a spacial dimension with the time dimension, turning length into duration. This is why "clocks slow down" and "objects contract" when moving relative to an observer. The difference, though, is that the latter uses hyperbolic trig functions, while the former uses standard (circular) trig functions.
$endgroup$
– Tac-Tics
Mar 6 '18 at 17:25
$begingroup$
The answer by Tophat is a good one, but I'll add a fun fact. Rotations trade a bit of the component in one dimension to another, turning "x"ness into "y"ness. In the theory of special relativity, you can also trade a bit of the component of a spacial dimension with the time dimension, turning length into duration. This is why "clocks slow down" and "objects contract" when moving relative to an observer. The difference, though, is that the latter uses hyperbolic trig functions, while the former uses standard (circular) trig functions.
$endgroup$
– Tac-Tics
Mar 6 '18 at 17:25
$begingroup$
The answer by Tophat is a good one, but I'll add a fun fact. Rotations trade a bit of the component in one dimension to another, turning "x"ness into "y"ness. In the theory of special relativity, you can also trade a bit of the component of a spacial dimension with the time dimension, turning length into duration. This is why "clocks slow down" and "objects contract" when moving relative to an observer. The difference, though, is that the latter uses hyperbolic trig functions, while the former uses standard (circular) trig functions.
$endgroup$
– Tac-Tics
Mar 6 '18 at 17:25
add a comment |
1 Answer
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oldest
votes
$begingroup$
The hyperbolic trig functions follow the equation for a Rectangular hyperbola, which is something you should be familiar from analytical geometry. The recentgular hyperbola is defined by $x^2 - y^2 = 1$ and if you let $x = cosh(t)$ and $y = sinh(t)$ and plug it into the rectangular hyperbola equation you can verify this fact quite easily.
From this idea you can derive the rest of the hyperbolic trig functions. (Also they reason they are given trig names is because they resemble the properties trig functions have).
answered Mar 6 '18 at 16:04
TophatTophat
1,342212
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add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
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active
oldest
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active
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votes
$begingroup$
The hyperbolic trig functions follow the equation for a Rectangular hyperbola, which is something you should be familiar from analytical geometry. The recentgular hyperbola is defined by $x^2 - y^2 = 1$ and if you let $x = cosh(t)$ and $y = sinh(t)$ and plug it into the rectangular hyperbola equation you can verify this fact quite easily.
From this idea you can derive the rest of the hyperbolic trig functions. (Also they reason they are given trig names is because they resemble the properties trig functions have).
answered Mar 6 '18 at 16:04
TophatTophat
1,342212
$endgroup$
add a comment |
$begingroup$
The hyperbolic trig functions follow the equation for a Rectangular hyperbola, which is something you should be familiar from analytical geometry. The recentgular hyperbola is defined by $x^2 - y^2 = 1$ and if you let $x = cosh(t)$ and $y = sinh(t)$ and plug it into the rectangular hyperbola equation you can verify this fact quite easily.
From this idea you can derive the rest of the hyperbolic trig functions. (Also they reason they are given trig names is because they resemble the properties trig functions have).
answered Mar 6 '18 at 16:04
TophatTophat
1,342212
$endgroup$
add a comment |
$begingroup$
The hyperbolic trig functions follow the equation for a Rectangular hyperbola, which is something you should be familiar from analytical geometry. The recentgular hyperbola is defined by $x^2 - y^2 = 1$ and if you let $x = cosh(t)$ and $y = sinh(t)$ and plug it into the rectangular hyperbola equation you can verify this fact quite easily.
From this idea you can derive the rest of the hyperbolic trig functions. (Also they reason they are given trig names is because they resemble the properties trig functions have).
answered Mar 6 '18 at 16:04
TophatTophat
1,342212
$endgroup$
The hyperbolic trig functions follow the equation for a Rectangular hyperbola, which is something you should be familiar from analytical geometry. The recentgular hyperbola is defined by $x^2 - y^2 = 1$ and if you let $x = cosh(t)$ and $y = sinh(t)$ and plug it into the rectangular hyperbola equation you can verify this fact quite easily.
From this idea you can derive the rest of the hyperbolic trig functions. (Also they reason they are given trig names is because they resemble the properties trig functions have).
answered Mar 6 '18 at 16:04
TophatTophat
1,342212
answered Mar 6 '18 at 16:04
TophatTophat
1,342212
answered Mar 6 '18 at 16:04
TophatTophat
1,342212
answered Mar 6 '18 at 16:04
TophatTophat
1,342212
1,342212
add a comment |
add a comment |
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$begingroup$
The answer by Tophat is a good one, but I'll add a fun fact. Rotations trade a bit of the component in one dimension to another, turning "x"ness into "y"ness. In the theory of special relativity, you can also trade a bit of the component of a spacial dimension with the time dimension, turning length into duration. This is why "clocks slow down" and "objects contract" when moving relative to an observer. The difference, though, is that the latter uses hyperbolic trig functions, while the former uses standard (circular) trig functions.
$endgroup$
– Tac-Tics
Mar 6 '18 at 17:25