## Wednesday, December 14, 2011

### Getting a little help from Newton

I've posted here on a number of topics and derivations in fundamental calculus. This is about finding the derivative for powers of x with negative or fractional exponents. It seems a trivial proof, until I got stopped in my tracks by the cofactors of the binomial distribution.

The idea is to find the slope of the curve at x by adding a little bit to x, (x + h), calculating f(x + h), then subtract f(x), and finally divide by h and finding the limit as h gets very small. It works great for y = x2. But consider

 `y = x1/2`

We want to do:

 `1/h [(x + h)1/2 - x1/2]`

So thinking about the binomial expansion of the x + h term, I'm trying to figure what is the expansion for the 1/2 power? I mean, what can "n choose k" mean when n is 1/2?

The other thing that is a good approach here is to bring x out of the sum as follows

 ```(x + h)1/2 = x1/2(1 + h/x)1/2```

To expand this, we just remember Newton's super-duper binomial expansion from here. If the expression is

 `(1 + Q)m/n`

the expansion is

 `1 + (m/n)Q + (m/n)((m-n)/2n)Q2 + (m/n)((m-n)/2n)((m-2n)/3n)Q3 + .. `

We only need the first two terms because Q2 is (h/x)2 which we will ignore. So we have for the expansion

 ```(1 + h/x)1/2 = 1 + 1/2(h/x)```

and the derivative is the limit as h => 0 of

 ```1/h [x1/2(1 + 1/2(h/x)) - x1/2] = 1/h (1/2 h) 1/x1/2 = 1/2 x-1/2```

Similarly for m = -2 and n = 1 the first two terms are 1 and -2 and the expansion we need will be

 ```(1 + (h/x))-2 = = 1 - 2(h/x)```

The derivative is the limit as h => 0 of

 ```1/h [x-2 (1 - 2(h/x)) - x-2] = = 1/h x-2 (-2)(h/x) = -2 x-3```

Easy when you know how.