What function can decrease a number with increasing proportion?
This is function can be used to reduce a number with unfixed proportion: when x equals 0, the subtracted proportion will be 0, that is x * (1-0), and when x equals infinite large, then subtracted proportion will be 1, that is x * (1-1). For some specific example, when x is rather small number such as 200, the output will be 199.9, and when x is pretty large number such as 4000, the output will be 1800, etc.
And in ideal, this function can have a coefficient to control how the decreasing trend will be like. It's kind looks like a root function, but that reduces a number too fast at beginning and too slow at the end.
In practice, it may can be used to calculate how much resistant force to apply or how much a image's pixels will be removed when compress.
If anyone can answer it, thank you in advance.
functions
asked Nov 28 '18 at 4:03
Elliott YangElliott Yang
62
add a comment |
This is function can be used to reduce a number with unfixed proportion: when x equals 0, the subtracted proportion will be 0, that is x * (1-0), and when x equals infinite large, then subtracted proportion will be 1, that is x * (1-1). For some specific example, when x is rather small number such as 200, the output will be 199.9, and when x is pretty large number such as 4000, the output will be 1800, etc.
And in ideal, this function can have a coefficient to control how the decreasing trend will be like. It's kind looks like a root function, but that reduces a number too fast at beginning and too slow at the end.
In practice, it may can be used to calculate how much resistant force to apply or how much a image's pixels will be removed when compress.
If anyone can answer it, thank you in advance.
functions
asked Nov 28 '18 at 4:03
Elliott YangElliott Yang
62
Your question is a bit confusing to be honest, but $f(x) = xleft(1 - frac{1}{800}left(frac{19x}{100} - 30right) right)$ goes through the points you have and increases up until $x = 41500/19$
– AlkaKadri
Nov 28 '18 at 4:41
add a comment |
This is function can be used to reduce a number with unfixed proportion: when x equals 0, the subtracted proportion will be 0, that is x * (1-0), and when x equals infinite large, then subtracted proportion will be 1, that is x * (1-1). For some specific example, when x is rather small number such as 200, the output will be 199.9, and when x is pretty large number such as 4000, the output will be 1800, etc.
And in ideal, this function can have a coefficient to control how the decreasing trend will be like. It's kind looks like a root function, but that reduces a number too fast at beginning and too slow at the end.
In practice, it may can be used to calculate how much resistant force to apply or how much a image's pixels will be removed when compress.
If anyone can answer it, thank you in advance.
functions
asked Nov 28 '18 at 4:03
Elliott YangElliott Yang
62
This is function can be used to reduce a number with unfixed proportion: when x equals 0, the subtracted proportion will be 0, that is x * (1-0), and when x equals infinite large, then subtracted proportion will be 1, that is x * (1-1). For some specific example, when x is rather small number such as 200, the output will be 199.9, and when x is pretty large number such as 4000, the output will be 1800, etc.
And in ideal, this function can have a coefficient to control how the decreasing trend will be like. It's kind looks like a root function, but that reduces a number too fast at beginning and too slow at the end.
In practice, it may can be used to calculate how much resistant force to apply or how much a image's pixels will be removed when compress.
If anyone can answer it, thank you in advance.
functions
functions
asked Nov 28 '18 at 4:03
Elliott YangElliott Yang
62
asked Nov 28 '18 at 4:03
Elliott YangElliott Yang
62
edited Nov 28 '18 at 6:20
Elliott Yang
asked Nov 28 '18 at 4:03
Elliott YangElliott Yang
62
asked Nov 28 '18 at 4:03
Elliott YangElliott Yang
62
asked Nov 28 '18 at 4:03
Elliott YangElliott Yang
62
62
Your question is a bit confusing to be honest, but $f(x) = xleft(1 - frac{1}{800}left(frac{19x}{100} - 30right) right)$ goes through the points you have and increases up until $x = 41500/19$
– AlkaKadri
Nov 28 '18 at 4:41
add a comment |
Your question is a bit confusing to be honest, but $f(x) = xleft(1 - frac{1}{800}left(frac{19x}{100} - 30right) right)$ goes through the points you have and increases up until $x = 41500/19$
– AlkaKadri
Nov 28 '18 at 4:41
Your question is a bit confusing to be honest, but $f(x) = xleft(1 - frac{1}{800}left(frac{19x}{100} - 30right) right)$ goes through the points you have and increases up until $x = 41500/19$
– AlkaKadri
Nov 28 '18 at 4:41
Your question is a bit confusing to be honest, but $f(x) = xleft(1 - frac{1}{800}left(frac{19x}{100} - 30right) right)$ goes through the points you have and increases up until $x = 41500/19$
– AlkaKadri
Nov 28 '18 at 4:41
add a comment |
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Your question is a bit confusing to be honest, but $f(x) = xleft(1 - frac{1}{800}left(frac{19x}{100} - 30right) right)$ goes through the points you have and increases up until $x = 41500/19$
– AlkaKadri
Nov 28 '18 at 4:41