Question

Using Taylor’s Theorem (and taking x₀= 0), show that (for |x| << 1) (1+ x)ⁿ ≈ 1+ nx This can be especially useful for approximating the values of square roots, for which n = ½. (The full expansion of (1+x) n is sometimes called a binomial series, and the first order approximation a “binomial approximation.”)

Answer 1

"Taylor's theorem If the function of and its first n + 1 derivatives are continuous on an internal containing a and x, then the value of function at x is given by f(x) = f(a) + f'(a) (x - a) + f''(a)/2! (x - a)^2 + ... + f^(n) (a)/n! (x - a)^n + R_n where remainder R_n is defined as R_n = integrals^x_a (x - t)^n/n_1 f^(n + 1) (t) dt there, f(x) = (1 + x)^n }x} < < 1 a = 0 Rightarrow f(x) = f(0) + f'(0)(x - 0) + f''(0)/2! (x - 0)^2 + - - - = 1^n + nx (1 + 0)^n + 1 (x - 0) + n(n - 1)/2! (1 + 0) (x - 0)^2 = 1 + nx + n(n - 1) x^2 + ... (1 + x)^2 almostequalto 1 + hx (because