Find the maclaurin series for f(x) = cos(x²) and use it to determine f⁸(0).

Solution:

We know that by using Maclaurin's series, if f(x) is continuous and differentiable over a finite interval (0,a) and if the derivatives f' ,f'' ,f''' ,... exists then

f(x) = f(0) + x f'(0) + (x²/2!) f''(0) + (x³/3!)f'''(0) + ………(xⁿ/n!)fⁿ(0)

For f(x) = cosx,

f(0) = cos 0 = 1

f'(x) = -sin x, f'(0) = 0

f''(x) = -cosx, f''(0) = -cos 0 = -1

f'''(x) = sinx = 0

f''''(x) = cosx, f''''(0) = cos 0 = 1

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

Using the definition of Maclaurin series, we can write cosx as

cosx = 1 + 0 - (x²/2) + 0 + (x⁴/ 24) - (x⁶/720) +... = \(\sum_{n=0}^{\infty}(-1)^n\dfrac{x^{2n}}{(2n)!}\) ---(1)

Replace x by x², we get Maclaurin series of cosx²

cos(x²) = 1 - ((x²)²/2 + ((x²)⁴/24 - ((x²)⁶/720) +…

cos(x²) = 1 - (x⁴/2) + (x⁸/24) - (x¹²/720) +... = \(\sum_{n=0}^{\infty}(-1)^n\dfrac{x^{4n}}{(2n)!}\) --- (2)

Now, f(x) = cos x²

f⁽⁸⁾(0) is the 8^th derivative of f(x) = cosx²evaluated at x = 0.

(x⁸/8!) f⁸(0)= x⁸/ 24

Comparing the coefficients of x⁸ from equations (1) and (2), we get

f⁸(0)/8! = 1/24

f⁸(0) = 8!/24

f⁸(0) = (8 × 7 × 6 × 5 × 4 × 3 × 2 ×1)/24

f⁸(0) = 8 × 7 × 6 × 5 = 1680