Mixing
Can this clamp function (designed for calculators without ABS) be optimized?
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I realized one day that I could evaluate whether a value is negative or not with one simple function:
$$frac{x}{sqrt{x^2}}$$
This would return -1 if x is negative, and 1 is x is positive. I then used it to "re-straighten" functions that were reflected off of the x-axis, for example
$$y=|x|*frac{x}{sqrt{x^2}}$$
I refined these techniques until I came up with a function that removes values from a function that are under supplied value "c".
$$y=(frac{1}{(frac{x-c}{sqrt{(x-c)^2}})+1})(2f(x))$$
This function is quite messy though, and I was hoping there may be a better way. If this is the best way, I would be pretty happy that I nailed this on the first try! I know that this community likes to pick apart inefficiencies in equations, so please do tell if there is a better way! Here is a desmos demonstration...
functions graphing-functions
asked Jan 14 at 16:36
Wasabi ThumbsWasabi Thumbs
11
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add a comment |
$begingroup$
I realized one day that I could evaluate whether a value is negative or not with one simple function:
$$frac{x}{sqrt{x^2}}$$
This would return -1 if x is negative, and 1 is x is positive. I then used it to "re-straighten" functions that were reflected off of the x-axis, for example
$$y=|x|*frac{x}{sqrt{x^2}}$$
I refined these techniques until I came up with a function that removes values from a function that are under supplied value "c".
$$y=(frac{1}{(frac{x-c}{sqrt{(x-c)^2}})+1})(2f(x))$$
This function is quite messy though, and I was hoping there may be a better way. If this is the best way, I would be pretty happy that I nailed this on the first try! I know that this community likes to pick apart inefficiencies in equations, so please do tell if there is a better way! Here is a desmos demonstration...
functions graphing-functions
asked Jan 14 at 16:36
Wasabi ThumbsWasabi Thumbs
11
$endgroup$
$begingroup$
Since $|x|=sqrt{x^2}$ then that is how one would have to calculate $|x|$ on calculators with a square root function but without an absolute value function.
$endgroup$
– John Wayland Bales
Jan 14 at 21:32
add a comment |
$begingroup$
I realized one day that I could evaluate whether a value is negative or not with one simple function:
$$frac{x}{sqrt{x^2}}$$
This would return -1 if x is negative, and 1 is x is positive. I then used it to "re-straighten" functions that were reflected off of the x-axis, for example
$$y=|x|*frac{x}{sqrt{x^2}}$$
I refined these techniques until I came up with a function that removes values from a function that are under supplied value "c".
$$y=(frac{1}{(frac{x-c}{sqrt{(x-c)^2}})+1})(2f(x))$$
This function is quite messy though, and I was hoping there may be a better way. If this is the best way, I would be pretty happy that I nailed this on the first try! I know that this community likes to pick apart inefficiencies in equations, so please do tell if there is a better way! Here is a desmos demonstration...
functions graphing-functions
asked Jan 14 at 16:36
Wasabi ThumbsWasabi Thumbs
11
$endgroup$
I realized one day that I could evaluate whether a value is negative or not with one simple function:
$$frac{x}{sqrt{x^2}}$$
This would return -1 if x is negative, and 1 is x is positive. I then used it to "re-straighten" functions that were reflected off of the x-axis, for example
$$y=|x|*frac{x}{sqrt{x^2}}$$
I refined these techniques until I came up with a function that removes values from a function that are under supplied value "c".
$$y=(frac{1}{(frac{x-c}{sqrt{(x-c)^2}})+1})(2f(x))$$
This function is quite messy though, and I was hoping there may be a better way. If this is the best way, I would be pretty happy that I nailed this on the first try! I know that this community likes to pick apart inefficiencies in equations, so please do tell if there is a better way! Here is a desmos demonstration...
functions graphing-functions
functions graphing-functions
asked Jan 14 at 16:36
Wasabi ThumbsWasabi Thumbs
11
asked Jan 14 at 16:36
Wasabi ThumbsWasabi Thumbs
11
edited Jan 14 at 16:42
Wasabi Thumbs
asked Jan 14 at 16:36
Wasabi ThumbsWasabi Thumbs
11
asked Jan 14 at 16:36
Wasabi ThumbsWasabi Thumbs
11
asked Jan 14 at 16:36
Wasabi ThumbsWasabi Thumbs
11
11
$begingroup$
Since $|x|=sqrt{x^2}$ then that is how one would have to calculate $|x|$ on calculators with a square root function but without an absolute value function.
$endgroup$
– John Wayland Bales
Jan 14 at 21:32
add a comment |
$begingroup$
Since $|x|=sqrt{x^2}$ then that is how one would have to calculate $|x|$ on calculators with a square root function but without an absolute value function.
$endgroup$
– John Wayland Bales
Jan 14 at 21:32
$begingroup$
Since $|x|=sqrt{x^2}$ then that is how one would have to calculate $|x|$ on calculators with a square root function but without an absolute value function.
$endgroup$
– John Wayland Bales
Jan 14 at 21:32
$begingroup$
Since $|x|=sqrt{x^2}$ then that is how one would have to calculate $|x|$ on calculators with a square root function but without an absolute value function.
$endgroup$
– John Wayland Bales
Jan 14 at 21:32
add a comment |
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$begingroup$
Since $|x|=sqrt{x^2}$ then that is how one would have to calculate $|x|$ on calculators with a square root function but without an absolute value function.
$endgroup$
– John Wayland Bales
Jan 14 at 21:32