# Maths Question: Distributing numbers on a curve **URL:** \ **Category:** Factual Questions\ **Created:** [December 9, 2005, 3:48pm UTC](https://boards.straightdope.com/t/maths-question-distributing-numbers-on-a-curve/334889 "2005-12-09T15:48:12Z")\ **Posts on this page:** 10\ **Page:** 1
**Author:** ![Crowbar\_of\_Irony\_3](https://avatars.discourse-cdn.com/v4/letter/c/f08c70/32.png) [@Crowbar\_of\_Irony\_3](https://boards.straightdope.com/u/Crowbar_of_Irony_3)\ **Post date:** [December 9, 2005, 3:48pm UTC](https://boards.straightdope.com/t/maths-question-distributing-numbers-on-a-curve/334889/1 "2005-12-09T15:48:12Z")
The problem situation: I need a function which when you give it a 1, it will be give you a 100, and when you give it 99, it will give you 9900. The tough part is that I need the “gap” between each output to increase as the inputs get higher. I’m not sure if I am explaining myself clearly. I could come up a function which gives an even distribution f(x) = 100 + (x-1) \* 100. So if x is 1, I will get 100 and if x is 2 then I will get 200, and when x is 99, I will get 9900. But the difference between each output is 100 each time What I like to achieve is when x gets higher, the gap between f(x) and (fx+1) grows, but in the meanwhile f(1) = 100 and f(99) = 9900. What mathematical concepts shall I use? Thanks in advance! ---
**Author:** ![David\_Simmons](https://avatars.discourse-cdn.com/v4/letter/d/9de053/32.png) [@David\_Simmons](https://boards.straightdope.com/u/David_Simmons)\ **Post date:** [December 9, 2005, 4:07pm UTC](https://boards.straightdope.com/t/maths-question-distributing-numbers-on-a-curve/334889/2 "2005-12-09T16:07:13Z")
> [@ExtraKun](#): > > The problem situation: I need a function which when you give it a 1, it will be give you a 100, and when you give it 99, it will give you 9900. The tough part is that I need the “gap” between each output to increase as the inputs get higher. > > I’m not sure if I am explaining myself clearly. I could come up a function which gives an even distribution f(x) = 100 + (x-1) \* 100. So if x is 1, I will get 100 and if x is 2 then I will get 200, and when x is 99, I will get 9900. But the difference between each output is 100 each time > > What I like to achieve is when x gets higher, the gap between f(x) and (fx+1) grows, but in the meanwhile f(1) = 100 and f(99) = 9900. > > What mathematical concepts shall I use? > > Thanks in advance! I’m not any kind of mathematician but I suspect that any function that returns 100 for 1, 200 for 2, 300 for 3, … will have a difference of 100 between successive values. And by the way, your equation 100 - (x - 1)\* 100 boils down to 100\*x. ---
**Author:** ![CJJ](https://avatars.discourse-cdn.com/v4/letter/c/ecc23a/32.png) [@CJJ](https://boards.straightdope.com/u/CJJ)\ **Post date:** [December 9, 2005, 4:21pm UTC](https://boards.straightdope.com/t/maths-question-distributing-numbers-on-a-curve/334889/3 "2005-12-09T16:21:38Z")
_What I like to achieve is when x gets higher, the gap between f(x) and (fx+1) grows, but in the meanwhile f(1) = 100 and f(99) = 9900._ Any equation with an increasing [derivative](http://mathworld.wolfram.com/Derivative.html) in the range you give will do the job; a simple choice would be a [quadratic](http://mathworld.wolfram.com/QuadraticEquation.html); one meeting the criteria above would be: f(x) = x^2 + 99 But there are other choices (cubic, exponential) which would make the “gap” between sucessive values of f(x) for x=integer increase more quickly. ---
**Author:** ![nivlac](https://avatars.discourse-cdn.com/v4/letter/n/3bc359/32.png) [@nivlac](https://boards.straightdope.com/u/nivlac)\ **Post date:** [December 9, 2005, 4:34pm UTC](https://boards.straightdope.com/t/maths-question-distributing-numbers-on-a-curve/334889/4 "2005-12-09T16:34:40Z")
This has to be an exponential function. And here it is … y = f(x) = 100e[sup]b(x-1)[/sup], where b = ln(9)/8 = 0.274653072167027. ---
**Author:** ![nivlac](https://avatars.discourse-cdn.com/v4/letter/n/3bc359/32.png) [@nivlac](https://boards.straightdope.com/u/nivlac)\ **Post date:** [December 9, 2005, 4:38pm UTC](https://boards.straightdope.com/t/maths-question-distributing-numbers-on-a-curve/334889/5 "2005-12-09T16:38:04Z")
Ooops, if you want f(9) = 9900, then b = ln(99)/8 = 0.574389981266824. So, it’s y = f(x) = 100e[sup]b(x-1)[/sup], with b = 0.574389981266824. ---
**Author:** ![Dead\_Cat](https://avatars.discourse-cdn.com/v4/letter/d/9fc29f/32.png) [@Dead\_Cat](https://boards.straightdope.com/u/Dead_Cat)\ **Post date:** [December 9, 2005, 5:08pm UTC](https://boards.straightdope.com/t/maths-question-distributing-numbers-on-a-curve/334889/6 "2005-12-09T17:08:49Z")
I’m some kind of ex-mathematician, and I agree with the first reply to the OP: if you want f(x) = 100x, the difference between each successive value of f(x) has to be 100 - your question makes no sense. Or do you mean that after x = 99 you want the derivative to increase? But it would be easy to come up with an arbitrary function that does that. In any case, neither of the other two replies seem to give you what you want; **CJJ** \*'s equation gives f(x) = 103 when x = 2 (you want f(x) = 200), and both of **nivlac** ’s provide infinite non-recurring decimals except at certain values of x. But until we know what the OP is trying to ask, I can’t say for certain whether they’re completely wrong :). ---
**Author:** ![Crowbar\_of\_Irony\_3](https://avatars.discourse-cdn.com/v4/letter/c/f08c70/32.png) [@Crowbar\_of\_Irony\_3](https://boards.straightdope.com/u/Crowbar_of_Irony_3)\ **Post date:** [December 9, 2005, 5:28pm UTC](https://boards.straightdope.com/t/maths-question-distributing-numbers-on-a-curve/334889/7 "2005-12-09T17:28:19Z")
> [@Dead Cat](#): > > But until we know what the OP is trying to ask, I can’t say for certain whether they’re completely wrong :). I’m sorry I didn’t clarify (I tried, but I don’t know enough Math jargon). I need to have a function, when you plug in f(1), it gives you 100 and when you plug in f(99) it gives you 9900. But I want the difference between f(1) and f(2) to be different. **If** I take f(x) = 100 + (x - 1) \* 100, I will get a function in which f(x+1) - f(x), for any value of x between 1 and 99, will be 100, _which is not what I want_. What I wish is as x gets higher, the difference between f(x) and f(x+1) grows. So I guess I shall take a stab at **niviac’s** and _CJJ’s_\* answers. ---
**Author:** ![Crowbar\_of\_Irony\_3](https://avatars.discourse-cdn.com/v4/letter/c/f08c70/32.png) [@Crowbar\_of\_Irony\_3](https://boards.straightdope.com/u/Crowbar_of_Irony_3)\ **Post date:** [December 9, 2005, 5:30pm UTC](https://boards.straightdope.com/t/maths-question-distributing-numbers-on-a-curve/334889/8 "2005-12-09T17:30:56Z")
> [@ExtraKun](#): > > I need to have a function, when you plug in f(1), it gives you 100 and when you plug in f(99) it gives you 9900. [del]But I want the difference between f(1) and f(2) to be different.[/del] What am I talking about :smack: Ignore that line. What I meant is I want the different between f(1) and f(2) to be different from that of f(3) and f(4) and so on… ---
**Author:** ![Dead\_Cat](https://avatars.discourse-cdn.com/v4/letter/d/9fc29f/32.png) [@Dead\_Cat](https://boards.straightdope.com/u/Dead_Cat)\ **Post date:** [December 9, 2005, 5:57pm UTC](https://boards.straightdope.com/t/maths-question-distributing-numbers-on-a-curve/334889/9 "2005-12-09T17:57:41Z")
OK, so we’re getting to the bottom of this now! **CJJ** \*'s suggestion seems to give you what you need - looks like he understood it better than me without needing further explanation! **nivlac** ’s second attempt does do what he says it does, but you wanted f(99) = 9900, not f(9) = 9900. It would seem the equation he posted, but with b = ln(99)/98 would work. ---
**Author:** ![nivlac](https://avatars.discourse-cdn.com/v4/letter/n/3bc359/32.png) [@nivlac](https://boards.straightdope.com/u/nivlac)\ **Post date:** [December 9, 2005, 6:18pm UTC](https://boards.straightdope.com/t/maths-question-distributing-numbers-on-a-curve/334889/10 "2005-12-09T18:18:25Z")
> [@Dead Cat](#): > > OK, so we’re getting to the bottom of this now! **CJJ** \*'s suggestion seems to give you what you need - looks like he understood it better than me without needing further explanation! **nivlac** ’s second attempt does do what he says it does, but you wanted f(99) = 9900, not f(9) = 9900. It would seem the equation he posted, but with b = ln(99)/98 would work. Thanks for the correction! Yes, I read f(9) = 9900. With b = ln(99)/98, f(99) = 9900. So the solution is y = f(x) = 100e[sup]b(x-1)[/sup], with b = 0.046888978062598.