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Proof of R = s√N

This proof of R = s√N describes an algebraic approach to the random walk problem, sometimes known as the drunkard’s walk. 

If the drunk takes a large number of strides (N) each of the same length (s) in succession, but in random directions, what is the resultant distance of travel (R)?

Obviously this will vary from one batch of N strides to another and may often be zero or as large as s x N. We want the average distance from start to finish, averaged over many batches of Nstrides. Observe a walk of N strides and find the resultant travel distance (R) from start to finish. Observe a large number of such walks starting afresh each time and find the average value of Rfor all those walks. Because it leads to a simple result, you can find the average value of R2 and take the square root, obtaining a root mean square (R.M.S.) average. You can show that this average should approach the value s √N. The two-dimensional proof is shown below. The three-dimensional one is similar. 
 
The diagram shows a few strides of a random walk. Choose a set of perpendicular co-ordinates, such as x and y, then resolve stride number 1 into components x1 and y1, stride number 2 into x2and y2, and so on. Then the resultant of that walk, R, has an 
 

x-component (x1 x2 +.... xN

and a y-component (y1 + y2 + .... yN
 
and 
 
R2 = (x1 + x2 + .... xN)2        + (y1 + y2 + .... yN2) 
 
= x1 2 + x2 2 + etc                  +2x1x2 + 2x1x3 + etc              
+ y1 2 + y2 2 + etc                  + 2y1y2 + 2y1y3 + etc                  
 
= s12 + s2 2 + etc                   + ZERO 
 
= s2N 
 
The cross terms such as 2x1x2, add up to zero in averaging over many walks, because those terms are as often negative as positive, and they range similarly from 0 to 2s2. Similarly for the y cross terms. 
 
Then the average value of R = s√N. 
 
The proof is better if you use trigonometry and resolve each stride, s, into horizontal and vertical components scos θ and s sin θ. Then the cross terms in the expression for R2 take the form 2s2 cos (θ 1 - θ 2), etc, and you can argue that the cosines are as often positive as negative. 

Using trigonometry to resolve each stride