# Proof of R = s√N

Teaching Guidance for 14-16

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 *N* strides.

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 *R* for all those walks. Because it leads to a simple result, you can find the average value of *R*^{ 2} 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 (see below) 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 *x*_{1} and *y*_{1}, stride number 2 into *x*_{2} and *y*_{2}, and so on.

Then the resultant of that walk, *R*, has an

*x*-component (*x*_{1}+*x*_{2}+...*x*_{N})*y*-component (*y*_{1}+*y*_{2}+...*y*_{N})

and

*R*^{ 2} = *( x_{1} + x_{2} + ... x_{N})*

^{ 2}+

*(*

*y*_{1}+*y*_{2}+ ...*y*_{N})^{ 2}

= *x*_{1}^{ 2} + *x*_{2}^{ 2} + etc. + 2*x*_{1}*x*_{2} + 2*x*_{1}*x*_{3} + etc.

+ *y*_{1}^{ 2} + *y*_{2}^{ 2} + etc. + 2*y*_{1}*y*_{2} + 2*y*_{1}*y*_{3} + etc.

= *s*_{1}^{ 2} + *s*_{2}^{ 2} + etc. + 0

= *s*^{ 2}*N*

The cross terms such as 2*x*_{1}*x*_{2}, 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 2*s*^{ 2}. 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 *s*cosθ and *s*sinθ. Then the cross terms in the expression for *R*^{ 2} take the form 2*s*^{ 2}cos(*θ*_{1}-*θ*_{2}), etc, and you can argue that the cosines are as often positive as negative.