# Find the maclaurin series for f(x) = cos(x^{2}) and use it to determine f^{8}(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) + (x^{2}/2!) f''(0) + (x^{3}/3!)f'''(0) + ………(x^{n}/n!)f^{n}(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) + (x^{2}/2!)(-1) + (x^{3}/3!)(0) + (x^{4}/ 4!)(1) + …

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

cosx = 1 + 0 - (x^{2}/2) + 0 + (x^{4}/ 24) - (x^{6}/720) +... = \(\sum_{n=0}^{\infty}(-1)^n\dfrac{x^{2n}}{(2n)!}\) ---(1)

Replace x by x^{2}, we get Maclaurin series of cosx^{2}

cos(x^{2}) = 1 - ((x^{2})^{2}/2 + ((x^{2})^{4}/24 - ((x^{2})^{6}/720) +…

cos(x^{2}) = 1 - (x^{4}/2) + (x^{8}/24) - (x^{12}/720) +... = \(\sum_{n=0}^{\infty}(-1)^n\dfrac{x^{4n}}{(2n)!}\) --- (2)

Now, f(x) = cos x^{2}

f^{(8)}(0) is the 8^{th} derivative of f(x) = cosx^{2 }evaluated at x = 0.

(x^{8}/8!) f^{8}(0)= x^{8}/ 24

Comparing the coefficients of x^{8} from equations (1) and (2), we get

f^{8}(0)/8! = 1/24

f^{8}(0) = 8!/24

f^{8}(0) = (8 × 7 × 6 × 5 × 4 × 3 × 2 ×1)/24

f^{8}(0) = 8 × 7 × 6 × 5 = 1680

## Find the maclaurin series for f(x) = cos(x^{2}) and use it to determine f^{8}(0).

**Summary:**

The value of f^{8}(0) for the maclaurin series for f(x) = cos(x^{2}) is 1680.