Etydhv

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Roman Numerals Equation 2

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When the match time is called, does the current turn end immediately?

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How to know the path of a particular software?

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Holding rent money for my friend which amounts to over $10k?

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Would a "ring language" be possible?

Promotion comes with unexpected 24/7/365 on-call

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Is there any deeper thematic meaning to the white horse that Arya finds in The Bells (S08E05)?

What formula to chose a nonlinear formula?

How To Create a Non-Linear Output from a Linear Input?What is the simplest formula for activation / smooth step function?Process For Building a Function?“Proportional to” - but nonlinear.Range of nonlinear functionFunctional equation - nonlinearAsymptotic functions with derivatives that are $1/2^x$Validating a function over a domain in a computer program.(nonlinear) Regression, but: What type of function is this?What is the most simple formula to achieve this pattern?

2

$begingroup$

Suppose I have a formula

$$f(x) = x,$$

where $0 leq x leq 255$

Now I want to have a formula

$$f(x) = y,$$

where $0 leq x leq 255$, where $f(0) = 0$, $f(255) = 255$ and e.g., $f(128) = 150$ (the value of $150$ might vary).

All other values should be interpolated.

So actually I want a function that is nonlinear (starts with $0$ and goes up to $255$, starting increasing fast and finishes increasing slow); opposite as a parabolic function.

What (kind of) function should I use?

functions

share|cite|improve this question

edited 1 hour ago

Pantelis Sopasakis

2,6171040

asked 1 hour ago

Michel KeijzersMichel Keijzers

1487

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  So you want a function $f(x)$ such that $f(0) = 0$ , $f(128)=150$ and $f(255) = 255$?
  $endgroup$
  – NoChance
  1 hour ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Welcome to Mathematics Stack Exchange! A quick tour will enhance your experience. Here are helpful tips to write a good question and write a good answer. For equations, please refer to this MathJax tutorial.
  $endgroup$
  – Pantelis Sopasakis
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  You may use a logistic function.
  $endgroup$
  – minori minus
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You can interpolate and get a parabolic function, I don't get what you mean by "opposite as a parabolic function".
  $endgroup$
  – Crostul
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  See wolframalpha.com/input/… . Wolfram Alpha describes an arc of a parabola.
  $endgroup$
  – Crostul
  1 hour ago
  

  
  
  
  

 | 
show 4 more comments

2

$begingroup$

Suppose I have a formula

$$f(x) = x,$$

where $0 leq x leq 255$

Now I want to have a formula

$$f(x) = y,$$

where $0 leq x leq 255$, where $f(0) = 0$, $f(255) = 255$ and e.g., $f(128) = 150$ (the value of $150$ might vary).

All other values should be interpolated.

So actually I want a function that is nonlinear (starts with $0$ and goes up to $255$, starting increasing fast and finishes increasing slow); opposite as a parabolic function.

What (kind of) function should I use?

functions

share|cite|improve this question

edited 1 hour ago

Pantelis Sopasakis

2,6171040

asked 1 hour ago

Michel KeijzersMichel Keijzers

1487

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  So you want a function $f(x)$ such that $f(0) = 0$ , $f(128)=150$ and $f(255) = 255$?
  $endgroup$
  – NoChance
  1 hour ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Welcome to Mathematics Stack Exchange! A quick tour will enhance your experience. Here are helpful tips to write a good question and write a good answer. For equations, please refer to this MathJax tutorial.
  $endgroup$
  – Pantelis Sopasakis
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  You may use a logistic function.
  $endgroup$
  – minori minus
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You can interpolate and get a parabolic function, I don't get what you mean by "opposite as a parabolic function".
  $endgroup$
  – Crostul
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  See wolframalpha.com/input/… . Wolfram Alpha describes an arc of a parabola.
  $endgroup$
  – Crostul
  1 hour ago
  

  
  
  
  

 | 
show 4 more comments

$begingroup$

Suppose I have a formula

$$f(x) = x,$$

where $0 leq x leq 255$

Now I want to have a formula

$$f(x) = y,$$

where $0 leq x leq 255$, where $f(0) = 0$, $f(255) = 255$ and e.g., $f(128) = 150$ (the value of $150$ might vary).

All other values should be interpolated.

So actually I want a function that is nonlinear (starts with $0$ and goes up to $255$, starting increasing fast and finishes increasing slow); opposite as a parabolic function.

What (kind of) function should I use?

functions

share|cite|improve this question

edited 1 hour ago

Pantelis Sopasakis

2,6171040

asked 1 hour ago

Michel KeijzersMichel Keijzers

1487

$endgroup$

share|cite|improve this question

edited 1 hour ago

Pantelis Sopasakis

2,6171040

asked 1 hour ago

Michel KeijzersMichel Keijzers

1487

share|cite|improve this question

edited 1 hour ago

Pantelis Sopasakis

2,6171040

asked 1 hour ago

Michel KeijzersMichel Keijzers

1487

edited 1 hour ago

Pantelis Sopasakis

2,6171040

edited 1 hour ago

Pantelis Sopasakis

2,6171040

asked 1 hour ago

Michel KeijzersMichel Keijzers

1487

asked 1 hour ago

Michel KeijzersMichel Keijzers

1487

3 Answers
3

active

oldest

votes

4

$begingroup$

WA gives $$frac10933x-11x^28128$$

https://www.wolframalpha.com/input/?i=interpolate+((0,0),(128,150)(255,255))

share|cite|improve this answer

answered 1 hour ago

CrostulCrostul

28.4k22352

$endgroup$

add a comment | 

2

$begingroup$

To satisfy the requirements:
$$f(128)=150, f(255)=255$$

However, $f(0)=0$ can't be satisfied by this method. I assumed that f(1)=1, however
you can change the numbers but not use zero, otherwise, there would be no inverse.

An example of the curve looks like this Curve

I can provide more info about the derivation if you want.

share|cite|improve this answer

edited 1 hour ago

answered 1 hour ago

NoChanceNoChance

3,89021321

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks ... Crostul gave me already a satisfactory formula, but good to know it can be derived too... I'm not fully understand the derivation, but for me it's not that needed right know... I always can search on internet how to solve matrix calculations (I had them at school long time ago, but forgotten about them meanwhile).
  $endgroup$
  – Michel Keijzers
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you for your feedback. You can construct an equation of the form $f(x)=Ax^2+Bx+C$ and substitute 3 non-zero values you know x and y for. You will get a system of equations having 3 unknowns (A, B, C). You solve for theses unknowns and plug the values back into $f(x)=Ax^2+Bx+C$. Matrix calculations are optional.
  $endgroup$
  – NoChance
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks ... that makes a lot of sense indeed, thanks for that info (very useful if in future I encounter similar problems, solving such equations even I should still be able to do that :-) ).
  $endgroup$
  – Michel Keijzers
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you, I will be very happy to help.
  $endgroup$
  – NoChance
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks (although most of my questions are in the electronics/Arduino stack exchange).
  $endgroup$
  – Michel Keijzers
  1 hour ago
  

  
  
  
  

add a comment | 

1

$begingroup$

I think one classical example of nonlinearity could be the gamma for color correction.

Instead of using the interval $X=0 .. 255$ it is easier to work in $[0,1]$ by scaling $x=frac X255$.

Now, remark that any $x^gamma$ stays in $[0,1]$ since $0^gamma=0$ and $1^gamma=1$.

Remark: this is not exactly color correction, which is $x^1/gamma$, but this is just a convention.

So you have linearity for $gamma=1$ and non-linearity for any other value of the exponent.

Note that if you really want to work with bytes the result can be scaled back using $255timesleft(frac X255right)^gamma$

Depending of your choice of $gamma<1$ or $gamma>1$ you will get either a fast increase at the start of the interval or at the end of the interval.

Try it here (move the g cursor): https://www.desmos.com/calculator/cnc2diykyx

share|cite|improve this answer

answered 1 hour ago

zwimzwim

13.3k833

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Actually, this is indeed exact the reason I use it for ... but I kind of promised the acceptance of the answer to Crostul (since he exactly answered my question), but your answer is perfect for the reason I need it for. Hope you understand.
  $endgroup$
  – Michel Keijzers
  48 mins ago
  

  
  
  
  

add a comment | 

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4

$begingroup$

WA gives $$frac10933x-11x^28128$$

https://www.wolframalpha.com/input/?i=interpolate+((0,0),(128,150)(255,255))

share|cite|improve this answer

answered 1 hour ago

CrostulCrostul

28.4k22352

$endgroup$

add a comment | 

4

$begingroup$

WA gives $$frac10933x-11x^28128$$

https://www.wolframalpha.com/input/?i=interpolate+((0,0),(128,150)(255,255))

share|cite|improve this answer

answered 1 hour ago

CrostulCrostul

28.4k22352

$endgroup$

add a comment | 

$begingroup$

WA gives $$frac10933x-11x^28128$$

https://www.wolframalpha.com/input/?i=interpolate+((0,0),(128,150)(255,255))

share|cite|improve this answer

answered 1 hour ago

CrostulCrostul

28.4k22352

$endgroup$

share|cite|improve this answer

answered 1 hour ago

CrostulCrostul

28.4k22352

answered 1 hour ago

CrostulCrostul

28.4k22352

answered 1 hour ago

CrostulCrostul

28.4k22352

2

$begingroup$

To satisfy the requirements:
$$f(128)=150, f(255)=255$$

However, $f(0)=0$ can't be satisfied by this method. I assumed that f(1)=1, however
you can change the numbers but not use zero, otherwise, there would be no inverse.

An example of the curve looks like this Curve

I can provide more info about the derivation if you want.

share|cite|improve this answer

edited 1 hour ago

answered 1 hour ago

NoChanceNoChance

3,89021321

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks ... Crostul gave me already a satisfactory formula, but good to know it can be derived too... I'm not fully understand the derivation, but for me it's not that needed right know... I always can search on internet how to solve matrix calculations (I had them at school long time ago, but forgotten about them meanwhile).
  $endgroup$
  – Michel Keijzers
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you for your feedback. You can construct an equation of the form $f(x)=Ax^2+Bx+C$ and substitute 3 non-zero values you know x and y for. You will get a system of equations having 3 unknowns (A, B, C). You solve for theses unknowns and plug the values back into $f(x)=Ax^2+Bx+C$. Matrix calculations are optional.
  $endgroup$
  – NoChance
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks ... that makes a lot of sense indeed, thanks for that info (very useful if in future I encounter similar problems, solving such equations even I should still be able to do that :-) ).
  $endgroup$
  – Michel Keijzers
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you, I will be very happy to help.
  $endgroup$
  – NoChance
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks (although most of my questions are in the electronics/Arduino stack exchange).
  $endgroup$
  – Michel Keijzers
  1 hour ago
  

  
  
  
  

add a comment | 

2

$begingroup$

To satisfy the requirements:
$$f(128)=150, f(255)=255$$

However, $f(0)=0$ can't be satisfied by this method. I assumed that f(1)=1, however
you can change the numbers but not use zero, otherwise, there would be no inverse.

An example of the curve looks like this Curve

I can provide more info about the derivation if you want.

share|cite|improve this answer

edited 1 hour ago

answered 1 hour ago

NoChanceNoChance

3,89021321

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks ... Crostul gave me already a satisfactory formula, but good to know it can be derived too... I'm not fully understand the derivation, but for me it's not that needed right know... I always can search on internet how to solve matrix calculations (I had them at school long time ago, but forgotten about them meanwhile).
  $endgroup$
  – Michel Keijzers
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you for your feedback. You can construct an equation of the form $f(x)=Ax^2+Bx+C$ and substitute 3 non-zero values you know x and y for. You will get a system of equations having 3 unknowns (A, B, C). You solve for theses unknowns and plug the values back into $f(x)=Ax^2+Bx+C$. Matrix calculations are optional.
  $endgroup$
  – NoChance
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks ... that makes a lot of sense indeed, thanks for that info (very useful if in future I encounter similar problems, solving such equations even I should still be able to do that :-) ).
  $endgroup$
  – Michel Keijzers
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you, I will be very happy to help.
  $endgroup$
  – NoChance
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks (although most of my questions are in the electronics/Arduino stack exchange).
  $endgroup$
  – Michel Keijzers
  1 hour ago
  

  
  
  
  

add a comment | 

$begingroup$

To satisfy the requirements:
$$f(128)=150, f(255)=255$$

However, $f(0)=0$ can't be satisfied by this method. I assumed that f(1)=1, however
you can change the numbers but not use zero, otherwise, there would be no inverse.

An example of the curve looks like this Curve

I can provide more info about the derivation if you want.

share|cite|improve this answer

edited 1 hour ago

answered 1 hour ago

NoChanceNoChance

3,89021321

$endgroup$

share|cite|improve this answer

edited 1 hour ago

answered 1 hour ago

NoChanceNoChance

3,89021321

answered 1 hour ago

NoChanceNoChance

3,89021321

answered 1 hour ago

NoChanceNoChance

3,89021321

1

$begingroup$

I think one classical example of nonlinearity could be the gamma for color correction.

Instead of using the interval $X=0 .. 255$ it is easier to work in $[0,1]$ by scaling $x=frac X255$.

Now, remark that any $x^gamma$ stays in $[0,1]$ since $0^gamma=0$ and $1^gamma=1$.

Remark: this is not exactly color correction, which is $x^1/gamma$, but this is just a convention.

So you have linearity for $gamma=1$ and non-linearity for any other value of the exponent.

Note that if you really want to work with bytes the result can be scaled back using $255timesleft(frac X255right)^gamma$

Depending of your choice of $gamma<1$ or $gamma>1$ you will get either a fast increase at the start of the interval or at the end of the interval.

Try it here (move the g cursor): https://www.desmos.com/calculator/cnc2diykyx

share|cite|improve this answer

answered 1 hour ago

zwimzwim

13.3k833

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Actually, this is indeed exact the reason I use it for ... but I kind of promised the acceptance of the answer to Crostul (since he exactly answered my question), but your answer is perfect for the reason I need it for. Hope you understand.
  $endgroup$
  – Michel Keijzers
  48 mins ago
  

  
  
  
  

add a comment | 

1

$begingroup$

I think one classical example of nonlinearity could be the gamma for color correction.

Instead of using the interval $X=0 .. 255$ it is easier to work in $[0,1]$ by scaling $x=frac X255$.

Now, remark that any $x^gamma$ stays in $[0,1]$ since $0^gamma=0$ and $1^gamma=1$.

Remark: this is not exactly color correction, which is $x^1/gamma$, but this is just a convention.

So you have linearity for $gamma=1$ and non-linearity for any other value of the exponent.

Note that if you really want to work with bytes the result can be scaled back using $255timesleft(frac X255right)^gamma$

Depending of your choice of $gamma<1$ or $gamma>1$ you will get either a fast increase at the start of the interval or at the end of the interval.

Try it here (move the g cursor): https://www.desmos.com/calculator/cnc2diykyx

share|cite|improve this answer

answered 1 hour ago

zwimzwim

13.3k833

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Actually, this is indeed exact the reason I use it for ... but I kind of promised the acceptance of the answer to Crostul (since he exactly answered my question), but your answer is perfect for the reason I need it for. Hope you understand.
  $endgroup$
  – Michel Keijzers
  48 mins ago
  

  
  
  
  

add a comment | 

$begingroup$

I think one classical example of nonlinearity could be the gamma for color correction.

Instead of using the interval $X=0 .. 255$ it is easier to work in $[0,1]$ by scaling $x=frac X255$.

Now, remark that any $x^gamma$ stays in $[0,1]$ since $0^gamma=0$ and $1^gamma=1$.

Remark: this is not exactly color correction, which is $x^1/gamma$, but this is just a convention.

So you have linearity for $gamma=1$ and non-linearity for any other value of the exponent.

Note that if you really want to work with bytes the result can be scaled back using $255timesleft(frac X255right)^gamma$

Depending of your choice of $gamma<1$ or $gamma>1$ you will get either a fast increase at the start of the interval or at the end of the interval.

Try it here (move the g cursor): https://www.desmos.com/calculator/cnc2diykyx

share|cite|improve this answer

answered 1 hour ago

zwimzwim

13.3k833

$endgroup$

share|cite|improve this answer

answered 1 hour ago

zwimzwim

13.3k833

answered 1 hour ago

zwimzwim

13.3k833

answered 1 hour ago

zwimzwim

13.3k833