Find the McLaurin series expansion of f(x) = cos3x.

Solution:

Given, y = f(x) = cos3x

McLaurin’s series is given as;

“If f(x) can be expanded as an infinite series then, 

f(x) = f(0) + x.f'(0) + x² / 2!. f''(0) + x³ /3!.f'''(0) + x⁴ /4!.f''''(0) +... ” ……….(1)

f(x) = cos 3x ; f(0) = 1

f'(x) = -3sin 3x ; f'(0) = 0

f''(x) = -9cos 3x ; f''(0) = -9

f'''(x) = 27 sin 3x ; f'''(0) = 0 

f''''(x) = -81 cos 3x ; f''''(0) = -81

∴ f(x) = cos 3x = 1 + 0 + x² / 2!.(-9) + 0 + x⁴ /4!.(-81) +...

⇒cos 3x = 1 + x² / 2!.(-9) + x⁴ /4!.(-81) +...

The expansion of f(x) = cos 3x = 1 -9 x² / 2! - 81 x⁴ /4!+...

Find the McLaurin series expansion of f(x) = cos3x.

Summary:

The maclaurin series expansion of f(x)=cos 3x is 1 -9 x² / 2! - 81 x⁴ /4!+.....