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This is such a basic question I'm sure, but I've been trying to find a robust solution to it for a while.

Lets say I have the numbers 0-100 in series.

As the number increases, I want another number say 1000, to decrease. so that when you got to 100, it was at 0

0 = 1000

50 = 500

100 = 0

What is this kind of series called?

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Yes it does, thanks. Do you want to add it as your answer for the rep? –  optician Sep 3 '10 at 14:32

4 Answers 4

As you mention “series” in the question I was wondering if something as simple as the term “Arithmetic Progression” is what you're after.

Your first arithmetic progression being $$a_n=n$$ for n=0,1,...,100 and your second being $$b_n=1000-10n=1000-10a_n.$$

Given your chosen name, I was wondering if your intended application has anything to do with optics? (I ask as my wife is an optician.)

Not really part of the answer but on a lighter note, here's a wonderful quote involving optics from the book “Littlewood's Miscellany”:

From time to time (since 1910) there were moves to get rid of the revolting optics and astronomy set in the Mathematical Tripos. It was discovered that over a period of years no wrangler attempted a question in either subject. An equivalent form of this proposition is that every attempt to do a question in optics or astronomy resulted in a failure to get a 1st.

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The function you want is $y=10(100-x)$. This was obtained by attempting to compute the quadratic function that passes through the three points ($n+1$ points uniquely determine an nth-degree curve), and then noting that the interpolating polynomial is actually a linear function.

Alternatively, computing the determinant


(the area of the "triangle" formed by the three points)

gives a value of 0, and thus the three points are collinear. Take any two of the three points, and you can derive the equation of the line passing through them.

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This was a comment, made an answer and expanded to include a derivation, in response to OP's comment. –  Guess who it is. Sep 3 '10 at 20:13

There is no way to uniquely answer your question without knowing more about what class of functions you permit to interpolate these points. Notice that all of your points are divisible by 50, so we may simplify a bit by scaling down by 50, yielding the points $\rm (x,y) = (0,20), (1,10), (2,0)$. Since the intermediate slopes agree $(10-20)/(1-0) = -10 = (0-10)/(2-1)\;$ it follows that a linear function interpolates these points, viz. $\rm\;y = 20-10x,\;$ or $\rm\;Y = 1000 - 10X\;$ when scaled up by 50 to $\rm (X,Y) = (50x,50y)$. But one could easily interpolate many other curves between these points. Perhaps if you tell us more about your intended application we can elaborate further.

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Here is how you can construct a linear relation for general bounds:

  • If $x$ goes from $a$ to $b$,
  • then $x-a$ goes from $0$ to $b-a$,
  • then $\frac{x-a}{b-a}$ goes from $0$ to $1$,
  • then $(x-a) \frac{d-c}{b-a}$ goes from $0$ to $d-c$,
  • then $y := c + (x-a) \frac{d-c}{b-a}$ goes from $c$ to $d$.

If you plug $a=0$, $b=100$, $c=1000$, $d=0$ as in your example, you obtain $y = 1000 + (x-0) \frac{0-1000}{100-0} = 1000 - 10x$ as in the other answers.

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