How do you find the Maclaurin series of the function #f(x)=5cos(10x^2)#?

Question

Evelyn Agard · Accepted Answer

The Maclaurin series of #f_{(x)}# is #f_{(x)}=f(0)+f'(0)x+{f''(0)}/{2!}  x^2+{f'''(0)}/{3!}  x^3+ . . .# Using this formula, get that the Maclaurin series of #cos(z)# is  #cos(z)=1-z^2/2+z^4/{4!}+....# #5cos(z)=5-5/2z^2+5/{4!}z^4+....# Substitute #z=10x^2#, The Maclaurin series of #5cos(10x^2)# is then #5cos(10x^2)=5-5/2(10x^2)^2+5/{4!}(10x^2)^4+ . . .#

William Abdalla · Answer

undefined To find the Maclaurin series of the function $ f(x) = 5\cos(10x^2) $, you follow these steps:

1. Start with the Maclaurin series for $ \cos(x) $, which is:
$$ \cos(x) = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!}x^{2n} $$

2. Replace $ x $ with $ 10x^2 $ in the Maclaurin series for $ \cos(x) $ to get the Maclaurin series for $ \cos(10x^2) $:
$$ \cos(10x^2) = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!}(10x^2)^{2n} $$

3. Multiply each term by $ 5 $ to get the Maclaurin series for $ 5\cos(10x^2) $:
$$ 5\cos(10x^2) = \sum_{n=0}^{\infty} 5 \cdot \frac{(-1)^n}{(2n)!}(10x^2)^{2n} $$

4. Simplify each term to get the final Maclaurin series for $ f(x) = 5\cos(10x^2) $.

The Maclaurin series for $ f(x) = 5\cos(10x^2) $ is:
$$ \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!} \cdot 5 \cdot (10x^2)^{2n} $$