# How can I create an equation for a Gaussian distribution based on a sum of a series?

I am trying to create an equation that will generate a Gaussian distribution such that y = the sum of f(x) in a series of integers 1->28. I have y and want to know what the value of x at each integer in the series should be, make sense?

I'll explain a different way to illustrate what I'm looking for:

If I have $100 and I want to give away that money over 28 days in increments such that it creates a nice bell curve over the 28 day period, what would that equation be? And how can I tweak it to make the curve steeper or flatter and/or adjust the points of inflection? Edit: I'll try to make this a little more clear, and please forgive me for using what I'm sure is incorrect notation. I need a function where y = f(x) produces a Gaussian distribution. I know that the sum of y across the series x = {1,2,3...,28} is 100. What is a formula that will allow me to solve for y given the values of x? - ## 2 Answers Take the bell curve and select some "window" (part of the X-axis). Divide your window into$28$equal parts, and replace each part by the integral of the bell curve on that interval. Normalize what you get so that it sums to$100$. Round everything up. Now make some small adjustment to get it to sum to exactly$100$. Edit: Here is an algorithm. Try it with$W=2$or$W=3$. void calc_array(int res[28], double W) { double sum = 0, a[28]; for (int i = 0; i < 28; i++) { double x = (i - 14) * W/14; sum += a[i] = exp(-x*x/2); } int sum2 = 0; for (int i = 0; i < 28; i++) sum2 += round(res[i] * 100/sum); // you can do better than that, but I'll leave that to you res[13] += 50 - sum2/2; res[14] += 50 - (sum2+1)/2; }  - Can you give me an example that I could use to create an algorithm in a standard programming language? – Swish May 26 '11 at 0:57 This worked great! Any idea how I would alter the equation to make the curve steeper on just left or right of center or how I could shift the peak one way or the other? – Swish May 26 '11 at 19:37 In order to shift the peak, replace the 14 in "i - 14". If you want the steepness to be different, say instead of one W have two W1/W2, then replace the entire statement by something like "if (i < 14) x = (i - 14) * W1/14; else x = (i - 14) * W2/14;" – Yuval Filmus May 27 '11 at 2:50 Great thank you! – Swish May 27 '11 at 16:25 Not sure I understand your question, but seems to me you could let $$A=50(e^{-1}+e^{-9}+e^{-25}+\cdots+e^{-729})^{-1}$$ and then use$f(x)=Ae^{-(2x-29)^2}$. The exponents in the expression for$A$are the squares of the odd numbers from$1$to$27$, which are also the exponents that will show up in the formula for$f(x)$as$x$goes$1,2,3,\dots,28$. EDIT: It occurs to me that the binomial is a good approximation to the bell curve. The 28 numbers$27\choose0$,$27\choose1$,...,$27\choose{27}$add up to$2^{27}$. Let me abbreviate the number$100/2^{27}$by$B$. Then you could give out${27\choose0}B$the first day,${27\choose1}B$the second day, etc., to${27\choose27}B$the last day. FURTHER EDIT: In response to the comment that the first gives a curve that's too steep (too concentrated at the center), there are ways to fiddle that. You could take the numbers$(100/B)e^{-(2x-29)^2/30}$for$x=1,2,\dots,28$, where$B=\sum_1^{28}e^{-(2n-29)^2/30}$. If you don't like that shape, you could replace the$30$with something bigger (to make the numbers flatter) or smaller (to make the numbers steeper). Similarly, you can fiddle with the binomials; pick some number$n$, let $$A={2n+27\choose n}+{2n+27\choose n+1}+\cdots+{2n+27\choose n+27}$$ and then let your gifts be $${100\over A}{2n+27\choose n},{100\over A}{2n+27\choose n+1},\dots,{100\over A}{2n+27\choose n+27}$$ Again, the bigger$n$you choose, the flatter the numbers you get. - Assuming that you did understand what I was after, how could I get this into a formula so I could plug in the target sum ($100 in my example above) and the day (between 1 and 28) and get the amount I should give away on that day? – Swish May 26 '11 at 0:54
The formula is $f(x)=Ae^{-(2x-29)^2}$. For, say, day 17, you put in $x=17$; the amount to give away on day 17 is $f(17)=Ae^{-25}$. Here, $A$ is the number in the displayed equation; you have to calculate that number first. – Gerry Myerson May 26 '11 at 2:42
[0]=>float(0) [1]=>float(3.4075443614396E-315) [2]=>float(5.0011241012202E-270) [3]=>float(2.4622825895526E-228) [4]=>float(4.0667951949457E-190) [5]=>float(2.2532576489861E-155) [6]=>float(4.1880663270157E-124) [7]=>float(2.611321790927E-96) [8]=>float(5.4619960211382E-72) [9]=>float(3.8325383633254E-51) [10]=>float(9.0212306531447E-34) [11]=>float(7.1234307686304E-20) [12]=>float(1.8869342761374E-9) [13]=>float(0.016767506522691) [14]=>float(49.98323249159) [15]=>float(49.98323249159) [16]=>float(0.016767506522691) [17]=>float(1.8869342761374E-9) Comment won't allow enough space for the rest – Swish May 26 '11 at 19:31
The comment above shows results I got with a calculated value for A of 135.86851260953 so something seems wrong as it produces an extremely steep curve – Swish May 26 '11 at 19:36