Mixing
Smooth transition between functions
$begingroup$
Therefor that I don't really have a mathematical background, it is kind of difficult to me, to describe what I'm looking for (but I'll give it a try):
I'm looking for a way to parameterize a function to fulfill the following constraints:
- function is a typical $y=f(x)$
- lets call the parameter $p$
- the following three coordinates shall be fixed: $f(0)=0$; $f(-100)=-100$; $f(100)=100$
$p$ shall have 100 valid values $([0-99];[1-100])$
now it comes to the tricky part:
- when 'p' has its minimum value; I want $f(x)$ to be $x^2$
- now with p increasing, I need f(x) become more and more bulgy like a $x^3$
I tried to separate the range from 1-3 in 100steps for $p$, and tried to use $y=x^p$, but that misses a lot of the above constraints (p.e. I never want the function to be $x^2$)
Maybe it becomes a little clearer, if you know what I need this for:
I want to program an configurable exponential transmission for a computergame. If you set the parameter to the minimum, the transmission is 1:1 (a movement of a joystick by one, results in a in_game_change of the value by one. If you want maximum exponential control, you need to move the joystick a lot more, before you reach a change of the in_game_value (reduced sensitivity). But in every case, 100% joystick_movement shall result in a 100% ingame_change (thats why I need the three coordinates to be fixed).
I hope I could make clear what I'm looking for, and will be very glad if someone could point me to the right idea.
Thanks in advance :)
exponentiation
edited Jan 23 at 18:31
caverac
14.8k31130
asked Jan 23 at 9:34
Tom MekkenTom Mekken
1083
$endgroup$
add a comment |
$begingroup$
Therefor that I don't really have a mathematical background, it is kind of difficult to me, to describe what I'm looking for (but I'll give it a try):
I'm looking for a way to parameterize a function to fulfill the following constraints:
- function is a typical $y=f(x)$
- lets call the parameter $p$
- the following three coordinates shall be fixed: $f(0)=0$; $f(-100)=-100$; $f(100)=100$
$p$ shall have 100 valid values $([0-99];[1-100])$
now it comes to the tricky part:
- when 'p' has its minimum value; I want $f(x)$ to be $x^2$
- now with p increasing, I need f(x) become more and more bulgy like a $x^3$
I tried to separate the range from 1-3 in 100steps for $p$, and tried to use $y=x^p$, but that misses a lot of the above constraints (p.e. I never want the function to be $x^2$)
Maybe it becomes a little clearer, if you know what I need this for:
I want to program an configurable exponential transmission for a computergame. If you set the parameter to the minimum, the transmission is 1:1 (a movement of a joystick by one, results in a in_game_change of the value by one. If you want maximum exponential control, you need to move the joystick a lot more, before you reach a change of the in_game_value (reduced sensitivity). But in every case, 100% joystick_movement shall result in a 100% ingame_change (thats why I need the three coordinates to be fixed).
I hope I could make clear what I'm looking for, and will be very glad if someone could point me to the right idea.
Thanks in advance :)
exponentiation
edited Jan 23 at 18:31
caverac
14.8k31130
asked Jan 23 at 9:34
Tom MekkenTom Mekken
1083
$endgroup$
$begingroup$
Welcome to MSE. You should choose your tags carefully. What has this to do withexponential-function?
$endgroup$
– José Carlos Santos
Jan 23 at 9:37
$begingroup$
isn't it? I'm sorry, but I haven't had anything to do with math for years. Please feel free to edit/correct my question to make it more clear
$endgroup$
– Tom Mekken
Jan 23 at 9:41
add a comment |
$begingroup$
Therefor that I don't really have a mathematical background, it is kind of difficult to me, to describe what I'm looking for (but I'll give it a try):
I'm looking for a way to parameterize a function to fulfill the following constraints:
- function is a typical $y=f(x)$
- lets call the parameter $p$
- the following three coordinates shall be fixed: $f(0)=0$; $f(-100)=-100$; $f(100)=100$
$p$ shall have 100 valid values $([0-99];[1-100])$
now it comes to the tricky part:
- when 'p' has its minimum value; I want $f(x)$ to be $x^2$
- now with p increasing, I need f(x) become more and more bulgy like a $x^3$
I tried to separate the range from 1-3 in 100steps for $p$, and tried to use $y=x^p$, but that misses a lot of the above constraints (p.e. I never want the function to be $x^2$)
Maybe it becomes a little clearer, if you know what I need this for:
I want to program an configurable exponential transmission for a computergame. If you set the parameter to the minimum, the transmission is 1:1 (a movement of a joystick by one, results in a in_game_change of the value by one. If you want maximum exponential control, you need to move the joystick a lot more, before you reach a change of the in_game_value (reduced sensitivity). But in every case, 100% joystick_movement shall result in a 100% ingame_change (thats why I need the three coordinates to be fixed).
I hope I could make clear what I'm looking for, and will be very glad if someone could point me to the right idea.
Thanks in advance :)
exponentiation
edited Jan 23 at 18:31
caverac
14.8k31130
asked Jan 23 at 9:34
Tom MekkenTom Mekken
1083
$endgroup$
Therefor that I don't really have a mathematical background, it is kind of difficult to me, to describe what I'm looking for (but I'll give it a try):
I'm looking for a way to parameterize a function to fulfill the following constraints:
- function is a typical $y=f(x)$
- lets call the parameter $p$
- the following three coordinates shall be fixed: $f(0)=0$; $f(-100)=-100$; $f(100)=100$
$p$ shall have 100 valid values $([0-99];[1-100])$
now it comes to the tricky part:
- when 'p' has its minimum value; I want $f(x)$ to be $x^2$
- now with p increasing, I need f(x) become more and more bulgy like a $x^3$
I tried to separate the range from 1-3 in 100steps for $p$, and tried to use $y=x^p$, but that misses a lot of the above constraints (p.e. I never want the function to be $x^2$)
Maybe it becomes a little clearer, if you know what I need this for:
I want to program an configurable exponential transmission for a computergame. If you set the parameter to the minimum, the transmission is 1:1 (a movement of a joystick by one, results in a in_game_change of the value by one. If you want maximum exponential control, you need to move the joystick a lot more, before you reach a change of the in_game_value (reduced sensitivity). But in every case, 100% joystick_movement shall result in a 100% ingame_change (thats why I need the three coordinates to be fixed).
I hope I could make clear what I'm looking for, and will be very glad if someone could point me to the right idea.
Thanks in advance :)
exponentiation
exponentiation
edited Jan 23 at 18:31
caverac
14.8k31130
asked Jan 23 at 9:34
Tom MekkenTom Mekken
1083
edited Jan 23 at 18:31
caverac
14.8k31130
asked Jan 23 at 9:34
Tom MekkenTom Mekken
1083
edited Jan 23 at 18:31
caverac
14.8k31130
edited Jan 23 at 18:31
caverac
14.8k31130
edited Jan 23 at 18:31
caverac
14.8k31130
14.8k31130
asked Jan 23 at 9:34
Tom MekkenTom Mekken
1083
asked Jan 23 at 9:34
Tom MekkenTom Mekken
1083
asked Jan 23 at 9:34
Tom MekkenTom Mekken
1083
1083
$begingroup$
Welcome to MSE. You should choose your tags carefully. What has this to do withexponential-function?
$endgroup$
– José Carlos Santos
Jan 23 at 9:37
$begingroup$
isn't it? I'm sorry, but I haven't had anything to do with math for years. Please feel free to edit/correct my question to make it more clear
$endgroup$
– Tom Mekken
Jan 23 at 9:41
add a comment |
$begingroup$
Welcome to MSE. You should choose your tags carefully. What has this to do withexponential-function?
$endgroup$
– José Carlos Santos
Jan 23 at 9:37
$begingroup$
isn't it? I'm sorry, but I haven't had anything to do with math for years. Please feel free to edit/correct my question to make it more clear
$endgroup$
– Tom Mekken
Jan 23 at 9:41
$begingroup$
Welcome to MSE. You should choose your tags carefully. What has this to do with
exponential-function?$endgroup$
– José Carlos Santos
Jan 23 at 9:37
$begingroup$
Welcome to MSE. You should choose your tags carefully. What has this to do with
exponential-function?$endgroup$
– José Carlos Santos
Jan 23 at 9:37
$begingroup$
isn't it? I'm sorry, but I haven't had anything to do with math for years. Please feel free to edit/correct my question to make it more clear
$endgroup$
– Tom Mekken
Jan 23 at 9:41
$begingroup$
isn't it? I'm sorry, but I haven't had anything to do with math for years. Please feel free to edit/correct my question to make it more clear
$endgroup$
– Tom Mekken
Jan 23 at 9:41
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Define the functions
begin{align}
g_1(x) &= x \
g_2(x) &= 100 (x/100)^3
end{align}
Now define
$$
sigma(p) = frac{2}{1 + e^{-p/30}}-1
$$
The number 30 is kind of arbitrary, it just tells you how fast you want the transition between $g_1$ and $g_2$ to occur. Small means fast. One option is
$$
f_p(x) = g_1(x) + [g_2(x) - g_1(x)]sigma(p)
$$
Here is the result
answered Jan 23 at 10:14
caveraccaverac
14.8k31130
$endgroup$
$begingroup$
this is just perfect and exactly what I needed. Thank you :). What kind of program did you use to plot this function dynamically?
$endgroup$
– Tom Mekken
Jan 23 at 11:32
$begingroup$
@TomMekken I usedLaTeXfor this, but it can be done in a few lines withpython
$endgroup$
– caverac
Jan 23 at 12:55
add a comment |
$begingroup$
You could set $f_p(x)=mathrm{sgn}(x)|x|^p$ for $1leq pleq 3$, where $mathrm{sgn}(x)$ is the sign of $x$ and $|x|$ is the absolute value of $x$. This reduces to the desired $x^p$ when $xgeq0$, with the added benefit of being an odd function, so for example $f(-1)=-1$.
Or, you can try linear interpolation between two functions, as in $f_p(x)=(1-p)x+px^3$ for $0leq pleq 1$. This may look more elegant on paper, but it might not feel as nice in practice, since it doesn't place a dead zone around $x=0$ unless $p$ is very close to $1$.
I assume you want a dead zone to compensate for the discontinuous jump in force that is required for the player's hand to move the joystick away from neutral. In other words, we want the derivative $f'_p(0)=0$ for all $p$ greater than the minimum setting. That's what you'll get with $mathrm{sgn}(x)|x|^p$.
answered Jan 23 at 10:21
Chris CulterChris Culter
21.4k43888
$endgroup$
add a comment |
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2 Answers
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active
oldest
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2 Answers
2
active
oldest
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active
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active
oldest
votes
$begingroup$
Define the functions
begin{align}
g_1(x) &= x \
g_2(x) &= 100 (x/100)^3
end{align}
Now define
$$
sigma(p) = frac{2}{1 + e^{-p/30}}-1
$$
The number 30 is kind of arbitrary, it just tells you how fast you want the transition between $g_1$ and $g_2$ to occur. Small means fast. One option is
$$
f_p(x) = g_1(x) + [g_2(x) - g_1(x)]sigma(p)
$$
Here is the result
answered Jan 23 at 10:14
caveraccaverac
14.8k31130
$endgroup$
$begingroup$
this is just perfect and exactly what I needed. Thank you :). What kind of program did you use to plot this function dynamically?
$endgroup$
– Tom Mekken
Jan 23 at 11:32
$begingroup$
@TomMekken I usedLaTeXfor this, but it can be done in a few lines withpython
$endgroup$
– caverac
Jan 23 at 12:55
add a comment |
$begingroup$
Define the functions
begin{align}
g_1(x) &= x \
g_2(x) &= 100 (x/100)^3
end{align}
Now define
$$
sigma(p) = frac{2}{1 + e^{-p/30}}-1
$$
The number 30 is kind of arbitrary, it just tells you how fast you want the transition between $g_1$ and $g_2$ to occur. Small means fast. One option is
$$
f_p(x) = g_1(x) + [g_2(x) - g_1(x)]sigma(p)
$$
Here is the result
answered Jan 23 at 10:14
caveraccaverac
14.8k31130
$endgroup$
$begingroup$
this is just perfect and exactly what I needed. Thank you :). What kind of program did you use to plot this function dynamically?
$endgroup$
– Tom Mekken
Jan 23 at 11:32
$begingroup$
@TomMekken I usedLaTeXfor this, but it can be done in a few lines withpython
$endgroup$
– caverac
Jan 23 at 12:55
add a comment |
$begingroup$
Define the functions
begin{align}
g_1(x) &= x \
g_2(x) &= 100 (x/100)^3
end{align}
Now define
$$
sigma(p) = frac{2}{1 + e^{-p/30}}-1
$$
The number 30 is kind of arbitrary, it just tells you how fast you want the transition between $g_1$ and $g_2$ to occur. Small means fast. One option is
$$
f_p(x) = g_1(x) + [g_2(x) - g_1(x)]sigma(p)
$$
Here is the result
answered Jan 23 at 10:14
caveraccaverac
14.8k31130
$endgroup$
Define the functions
begin{align}
g_1(x) &= x \
g_2(x) &= 100 (x/100)^3
end{align}
Now define
$$
sigma(p) = frac{2}{1 + e^{-p/30}}-1
$$
The number 30 is kind of arbitrary, it just tells you how fast you want the transition between $g_1$ and $g_2$ to occur. Small means fast. One option is
$$
f_p(x) = g_1(x) + [g_2(x) - g_1(x)]sigma(p)
$$
Here is the result
answered Jan 23 at 10:14
caveraccaverac
14.8k31130
answered Jan 23 at 10:14
caveraccaverac
14.8k31130
answered Jan 23 at 10:14
caveraccaverac
14.8k31130
answered Jan 23 at 10:14
caveraccaverac
14.8k31130
14.8k31130
$begingroup$
this is just perfect and exactly what I needed. Thank you :). What kind of program did you use to plot this function dynamically?
$endgroup$
– Tom Mekken
Jan 23 at 11:32
$begingroup$
@TomMekken I usedLaTeXfor this, but it can be done in a few lines withpython
$endgroup$
– caverac
Jan 23 at 12:55
add a comment |
$begingroup$
this is just perfect and exactly what I needed. Thank you :). What kind of program did you use to plot this function dynamically?
$endgroup$
– Tom Mekken
Jan 23 at 11:32
$begingroup$
@TomMekken I usedLaTeXfor this, but it can be done in a few lines withpython
$endgroup$
– caverac
Jan 23 at 12:55
$begingroup$
this is just perfect and exactly what I needed. Thank you :). What kind of program did you use to plot this function dynamically?
$endgroup$
– Tom Mekken
Jan 23 at 11:32
$begingroup$
this is just perfect and exactly what I needed. Thank you :). What kind of program did you use to plot this function dynamically?
$endgroup$
– Tom Mekken
Jan 23 at 11:32
$begingroup$
@TomMekken I used
LaTeX for this, but it can be done in a few lines with python$endgroup$
– caverac
Jan 23 at 12:55
$begingroup$
@TomMekken I used
LaTeX for this, but it can be done in a few lines with python$endgroup$
– caverac
Jan 23 at 12:55
add a comment |
$begingroup$
You could set $f_p(x)=mathrm{sgn}(x)|x|^p$ for $1leq pleq 3$, where $mathrm{sgn}(x)$ is the sign of $x$ and $|x|$ is the absolute value of $x$. This reduces to the desired $x^p$ when $xgeq0$, with the added benefit of being an odd function, so for example $f(-1)=-1$.
Or, you can try linear interpolation between two functions, as in $f_p(x)=(1-p)x+px^3$ for $0leq pleq 1$. This may look more elegant on paper, but it might not feel as nice in practice, since it doesn't place a dead zone around $x=0$ unless $p$ is very close to $1$.
I assume you want a dead zone to compensate for the discontinuous jump in force that is required for the player's hand to move the joystick away from neutral. In other words, we want the derivative $f'_p(0)=0$ for all $p$ greater than the minimum setting. That's what you'll get with $mathrm{sgn}(x)|x|^p$.
answered Jan 23 at 10:21
Chris CulterChris Culter
21.4k43888
$endgroup$
add a comment |
$begingroup$
You could set $f_p(x)=mathrm{sgn}(x)|x|^p$ for $1leq pleq 3$, where $mathrm{sgn}(x)$ is the sign of $x$ and $|x|$ is the absolute value of $x$. This reduces to the desired $x^p$ when $xgeq0$, with the added benefit of being an odd function, so for example $f(-1)=-1$.
Or, you can try linear interpolation between two functions, as in $f_p(x)=(1-p)x+px^3$ for $0leq pleq 1$. This may look more elegant on paper, but it might not feel as nice in practice, since it doesn't place a dead zone around $x=0$ unless $p$ is very close to $1$.
I assume you want a dead zone to compensate for the discontinuous jump in force that is required for the player's hand to move the joystick away from neutral. In other words, we want the derivative $f'_p(0)=0$ for all $p$ greater than the minimum setting. That's what you'll get with $mathrm{sgn}(x)|x|^p$.
answered Jan 23 at 10:21
Chris CulterChris Culter
21.4k43888
$endgroup$
add a comment |
$begingroup$
You could set $f_p(x)=mathrm{sgn}(x)|x|^p$ for $1leq pleq 3$, where $mathrm{sgn}(x)$ is the sign of $x$ and $|x|$ is the absolute value of $x$. This reduces to the desired $x^p$ when $xgeq0$, with the added benefit of being an odd function, so for example $f(-1)=-1$.
Or, you can try linear interpolation between two functions, as in $f_p(x)=(1-p)x+px^3$ for $0leq pleq 1$. This may look more elegant on paper, but it might not feel as nice in practice, since it doesn't place a dead zone around $x=0$ unless $p$ is very close to $1$.
I assume you want a dead zone to compensate for the discontinuous jump in force that is required for the player's hand to move the joystick away from neutral. In other words, we want the derivative $f'_p(0)=0$ for all $p$ greater than the minimum setting. That's what you'll get with $mathrm{sgn}(x)|x|^p$.
answered Jan 23 at 10:21
Chris CulterChris Culter
21.4k43888
$endgroup$
You could set $f_p(x)=mathrm{sgn}(x)|x|^p$ for $1leq pleq 3$, where $mathrm{sgn}(x)$ is the sign of $x$ and $|x|$ is the absolute value of $x$. This reduces to the desired $x^p$ when $xgeq0$, with the added benefit of being an odd function, so for example $f(-1)=-1$.
Or, you can try linear interpolation between two functions, as in $f_p(x)=(1-p)x+px^3$ for $0leq pleq 1$. This may look more elegant on paper, but it might not feel as nice in practice, since it doesn't place a dead zone around $x=0$ unless $p$ is very close to $1$.
I assume you want a dead zone to compensate for the discontinuous jump in force that is required for the player's hand to move the joystick away from neutral. In other words, we want the derivative $f'_p(0)=0$ for all $p$ greater than the minimum setting. That's what you'll get with $mathrm{sgn}(x)|x|^p$.
answered Jan 23 at 10:21
Chris CulterChris Culter
21.4k43888
answered Jan 23 at 10:21
Chris CulterChris Culter
21.4k43888
answered Jan 23 at 10:21
Chris CulterChris Culter
21.4k43888
answered Jan 23 at 10:21
Chris CulterChris Culter
21.4k43888
21.4k43888
add a comment |
add a comment |
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$begingroup$
Welcome to MSE. You should choose your tags carefully. What has this to do with
exponential-function?$endgroup$
– José Carlos Santos
Jan 23 at 9:37
$begingroup$
isn't it? I'm sorry, but I haven't had anything to do with math for years. Please feel free to edit/correct my question to make it more clear
$endgroup$
– Tom Mekken
Jan 23 at 9:41