Question

Write the Maclaurin series expansion of the following functions:

tan^–1 (x); – 1 ≤ x ≤ 1

Answer 1

f(x) = tan^–1x, f(0) = 0

f'(x) = `1/(1 + x^2)` f'(0) = 1

= 1 – x² + x⁴ – x⁶ + …..

f”(x) = – 2x + 4x³ – 6x⁵ + ….. f”(0) = 0

f”'(x) = – 2 + 12x² – 30x⁴ + ….. f”(0) = – 2

f^IV(x) = 24x – 120x³ + …… f^IV(0) = 0

f^V(x) = 24 – 360x² + ….. f^V(0) = 24 .

f^VI(x) = – 720x + ….. f^VI(0) = 0

f^VII(x) = – 720 + … f^VII(0) = – 720

Maclaurin ‘s expansion is

f(x) = `sum_("n" = 0)^oo ("f"^(("n"))(0)"n"^"n")/("n"!)`

= `"f"(0) + ("f'"(0)x)/(1!) + ("f''"(0) x^2)/(2!) + ("f"^(("n"))(0)x^"n")/("n"!) + ...`

tan^–1x = `0 + x/(1!) + 0 + 0 ((-2))/31 x^3 + 24/(5!) x^5 + 0 + ((-720))/(7!) x^7 + ...`

tan^–1x = `x - x^3/3 + x^5/5 - x^7/7 + ...`