sqrt(1 + x) = sum n = 0 to ∞ ((- 1) ^ n * (2n)!)/((1 - 2n) * (n!) ^ 2 * (4 ^ n)) * x ^ n =1+ 1 2 x- 1 8 x^ 2 + 1 16 x^ 3 - 5 128 x^ 4 +...

Guest Mar 1, 2023

#1**0 **

Yes, that is correct! The expression you provided is the Taylor series expansion of the square root of (1 + x) around x = 0. It is an infinite series that converges to the square root of (1 + x) for all values of x between -1 and 1.

The first few terms of the series are:

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

Each term in the series is calculated using the formula:

((-1)^n * (2n)! / ((1-2n) * (n!)^2 * (4^n))) * x^n

where n is the index of the term and x is the value at which the series is evaluated.

The series can be used to approximate the value of the square root of (1 + x) to a desired degree of accuracy, by including a sufficient number of terms in the expansion. HP Instant Ink

Amandahertzog Mar 1, 2023