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

Question

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

Luke Alvarez · Accepted Answer

#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 #

Since it is very hard to find power expansions for any case, we are unable to expand this in a general case; however, we can estimate it in the limit of one being much larger than the other, and then discuss that case.

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

Now, we know that #h/x# is a very small value, so we can expand this using Taylor expansion. We need to find the #n#th derivative at #u=0# of #f(u) = sqrt(1+u)#. This is simple to find #f(u) = (1+u)^(1/2) implies f(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.

Adam Adkins · Answer

undefined The expansion of sqrt(x+h) in powers of x and h is given by:

sqrt(x+h) = sqrt(x) + (1/2)*(h/x)*sqrt(x) - (1/8)*(h^2/x^2)*sqrt(x) + (1/16)*(h^3/x^3)*sqrt(x) - (5/128)*(h^4/x^4)*sqrt(x) + ...

This expansion can be continued indefinitely, with each term involving higher powers of h and lower powers of x.