brendan mccloskey

how can i achieve the following, preferably with expr

but it's a bit, erm what's the word, grokky?

how can i achieve the following, preferably with expr
input: 1, output 0.1
input: 5, output 1.
input: 10, output 10.

simple-expr-math

https://cycling74.com/forums/nonlinear-scaling-of-parameter-values

not as straightforward, but same idea should get you there.

* Hint you have 3 variables and 3 equations :)

not as straightforward, but same idea should get you there.  

* Hint you have 3 variables and 3 equations :)


big_pause - I saw that thread too, but ran screaming from the equations, thanks for filling in the blanks - ps how the hell did you arrive at that function?? Give me your brain - NOW, mwah ha ha

2 excellent solutions (3 if you count mine LOL)

it's pretty much the simplest functional form that fits the criteria.

big_pause: but how did you count the coefficients? That is indeed pretty awesome

whenever you have 3 points you can go with quadratic equation (the expression big_pause estimated) by that substitution algo from school (at least in germany).

0.1 = a + b*1^c = a + b => a = 0.1 - b [1]

substituting [1] into [2] (you could sub into [3] here, doesn't matter)

take logs of both sides, to give an expression for c in terms of b

then, substituting [1] and [4] into [3], lets us find b

10 = 0.1 - b + b*10^((ln(0.9 + b) - ln(b)) / ln(5))

0 = -9.9 - b + b*10^((ln(0.9 + b) - ln(b)) / ln(5))

at this stage I just ran through a quick hacked python script to find b (life's too short)

once you have b, you can use [1] to get a, then either [2] or [3] to get c.

y = 0.1, x = 1
y = 1, x = 5
y = 10, x = 10

0.1 = a + b*1^c = a + b => a = 0.1 - b [1]
1 = a + b*5^c [2]
10 = a + b*10^c [3]

ln(0.9 + b) = ln(b*5^c)
ln(0.9 + b) = ln(b) c*ln(5)
c = (ln(0.9 + b) - ln(b)) / ln(5) [4]

10 = 0.1 - b + b*10^((ln(0.9 + b) - ln(b)) / ln(5))
0 = -9.9 - b + b*10^((ln(0.9 + b) - ln(b)) / ln(5))

so, I thought about this a little more. We can transform our linear inputs to this, and get the same problem as in the 'other' post

here's what we do. we have as our inputs (x's) that need mapping, [1, 5, 10].

we shift this down to [0, 4, 9], then normalise to [0, 4/9, 1], or algorithmically

[x1, x2, x3] -> [(x1-x1)/(x3-x1), (x2-x1)/(x3-x1), (x3-x1)/(x3-x1)] = [x1', x2', x3'] and these map = [y1, y2, y3]

we now have a similar problem as in the post above, and can get the solution to

y = a + b*x'^c : where x' = (x - x1)/(x3 - x1)

for our new mappings (both for this case, and general algorithm)

y = 0.1 + 9.9*[(x - 1)/9]^{ln(0.9/9.9)/ln(4/9)}

y = y1 + (y3 - y1)*x'^{ln((y2 - y1)/(y3 - y1))/ln(x2')}

with x' = (x - x1)/(x3 - x1) and x2' = (x2 - x1)/(x3 - x1)

and we can chuck this straight in an single expr - happy days

so, I thought about this a little more.  We can transform our linear inputs to this, and get the same problem as in the 'other' post 

here's what we do.  we have as our inputs (x's) that need mapping, [1, 5, 10].

we now have a similar problem as in the post above, and can get the solution to 

y = a + b*x'^c  : where x' = (x - x1)/(x3 - x1)

1 = 0.1 + 9.9*(4/9)^c
0.9/9.9 = (4/9)^c
c = ln(0.9/9.9)/ln(4/9)

or
y2 = y1 + (y3 - y1)*x2'^c
c = ln((y2 - y1)/(y3 - y1))/ln(x2')

y = 0.1 + 9.9*x'^{ln(0.9/9.9)/ln(4/9)}
y = 0.1 + 9.9*[(x - 1)/9]^{ln(0.9/9.9)/ln(4/9)}

and we can chuck this straight in an single expr - happy days


to find a ln or exp expression starting from 3 points - except when they are variables - you

could also just try manually which coefficient matches.

to find a ln or exp expression starting from 3 points - except when they are variables - you
could also just try manually which coefficient matches.


simple expr math