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I have three variables:

  • a = 1
  • b = 4
  • c = 5

I wanted a linear function which give me a percentage based on that the bigger the value is, the bigger the percentage will be : I's easy, it's the percent of the total : $$y = {x\over (a+b+c)}$$ (With x one of the values above).

And so, for a : 10%, b : 40% and c : 50%.

But now, I would like a linear function for the reverse : The smaller the value, the bigger the percentage. I've tried many things and the "best" function I've find so far is : $$y = {(a+b+c) - x \over 2(a+b+c)}$$ (With x one of the values above).

Which gives me : 45% for a, 30% for b and 25% for c.

But, it's clearly not a linear fonction.

Edit: I was wrong as it's indeed linear, and it was what I wanted. Sorry for bothering you.

So, I'm asking you :

  • Can a linear fonction for what I'm asking is even possible ?
  • If yes, can you help me in finding a function that suit what I would like ?

Thanks.

Pierre Demessence
  • 111
  • Why do you keep saying "with x on of the values above"? You're saying that the domain of your function is ${a, b, c}$? – Jack M Sep 14 '13 at 23:29
  • What do you mean by linear? Any function $y=ax+b$ will have a line as its graph, and the function you give is of that type. Do you mean linear as used in Linear Algebra? Then $y=-x/(a+b+c)$ will do, but it gives only negative answers. – Gerry Myerson Sep 14 '13 at 23:37
  • @JackM I mean that the function will only by used for these parameters. – Pierre Demessence Sep 14 '13 at 23:55
  • @GerryMyerson By "linear", I mean "proportionnal", like in the first function. – Pierre Demessence Sep 15 '13 at 00:00
  • Well, that's what I called the Linear Algebra usage, and in that case the only thing that can work is a function of the form $Qx$, where $Q$ is a negative constant (such as $Q=-1/(a+b+c)$). – Gerry Myerson Sep 15 '13 at 00:02
  • How is that the best function you have found so far? That is just Gerry Myerson's suggestion divided by 2. Also it is by definition linear. You need to more precisely describe what you want, and you are not allowed to make up your own definition of "linear" here. – J. W. Perry Sep 15 '13 at 00:31

1 Answers1

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How about $$1-{x\over a+b+c}$$ Does that work for you?

Gerry Myerson
  • 179,216