What will be the expansion of sqrt(x+h)x+h in powers of x and hxandh?

1 Answer
Jul 18, 2018

sqrt(x+h) approx x^(1/2) + 1/2 hx^(-1/2) - 1/8 h^2 x^(-3/2) + 1/16 h^3x^(-5/2) + O(h/x)^4 x+hx12+12hx1218h2x32+116h3x52+O(hx)4

Explanation:

We can't expand this in a general case since it's very difficult to find power expansions for any case. However, we can estimate it in the limit of one being far larger than the other and then discuss that case.

We will assume that xx is much greater than hh. Therefore,
sqrt(x+h) approx sqrtx * sqrt(1 + (h/x)) x+hx1+(hx)

Now, we know that h/xhx is a very small value, so we can expand this using Taylor expansion. We need to find the nnth derivative at u=0u=0 of f(u) = sqrt(1+u)f(u)=1+u. This is simple to find
f(u) = (1+u)^(1/2) implies f(0) = 1 f(u)=(1+u)12f(0)=1
f'(u) = 1/2 * (1+u)^(-1/2) implies f'(0) = 1/2 * 1
f''(u) = -1/2 * 1/2 * (1+u)^(-3/2) implies f''(0) = -1/2 * 1/2 * 1 = -1/4

Imagining continuing this, it is clear that we have
f^(n) (u) = 1/2 * (-1)/2 * ... (3-2n)/2 (1+u)^(1/2-n)
= -(-1/2)^n * (1 * 3 * 5 * ... (2n-3)) * (1+u)^(1/2-n)
This middle value takes on the special character of a double factorial, hence for n > 2,

f^(n) (0) = (-1)^(n+1)2^-n (2n-3)!!

This gives us the expansion via Taylor's formula:
sqrt(1 + u) = 1 + 1/2 u- 1/8 u^2 - sum_(n=3)^(infty) (-1/2)^n ((2n-3)!!)/(n!) u^n

From this, we can recover the answer to the original from this:
sqrt(x+h) approx sqrt(x) * sqrt(1 + (h/x)) = sqrt(x) * f(h/x)
approx sqrt(x) * [ 1 + 1/2 (h/x)- 1/8 (h/x)^2 - sum_(n=3)^(infty) (-1/2)^n ((2n-3)!!)/(n!) (h/x)^n ]

Using just the first four terms, we get
sqrt(x+h) approx x^(1/2) + 1/2 hx^(-1/2) - 1/8 h^2 x^(-3/2) + 1/16 h^3x^(-5/2)

To see how accurate this is, let's say x = 225 and h = 64. Therefore sqrt(x+h) = 17. Plugging in these numbers to the above formula,
sqrt(x+h) approx 15 + 1/2 * 64 / 15 - 1/8 * 64^2 / (15^3) + 1/16 * 64^3 / (15^5) = 17.003
which is quite good already.