How do you find the Maclaurin Series for #f(x)=sin(x^4)#?

Question

Isaiah Allen · Accepted Answer

#x^4-x^12/(3!)+x^20/(5!)-...+(-1)^(n-1)((x^4)^(2n+1))/((2n+1)!)+...#

Let #u=x^4# Then, fx)=g(u)=sin u#

Maclaurin series for #g(u)=u-u^3/(3!)+u^5/(5!)-..=f(x#)

Going back to the function of x,

#f(x)=x^4-x^12/(3!)+x^20/(5!)-...+(-1)^(n-1)(x^4)^(2n=1)/((2n+1)!)+...#

Maclaurin series is valid for any #x in (-oo, oo)#. Here, the mapping is #xtox^4#, for a subset of #(-oo. oo)# ...

Axel Alvarez · Answer

undefined To find the Maclaurin series for $ f(x) = \sin(x^4) $, follow these steps:

1. Find the derivatives of $ f(x) = \sin(x^4) $ up to the desired order.
2. Evaluate each derivative at $ x = 0 $ to find the coefficients of the Maclaurin series.
3. Write out the Maclaurin series using the coefficients found in step 2.

The Maclaurin series for $ f(x) = \sin(x^4) $ is:

$$ \sin(x^4) = x^4 - \frac{x^{12}}{3!} + \frac{x^{20}}{5!} - \frac{x^{28}}{7!} + \cdots $$

The coefficients of the series come from the derivatives of $ \sin(x^4) $ evaluated at $ x = 0 $. The $ n $th term of the series has the form $ \frac{f^{(n)}(0)}{n!}x^n $, where $ f^{(n)}(0) $ is the $ n $th derivative of $ f(x) = \sin(x^4) $ evaluated at $ x = 0 $.