How do you find the maclaurin series expansion of #f(x) = x^2 arctan (x^3)#?

Question

Emilia Aguilar · Accepted Answer

#sum_{i=0}^infty (-1)^{n} x^{6n+5}/{2n+1}#

Substitute #y=x^3# (note that when #x\to 0#, also #x^3# and thus #y# tend to 0, so it's a good substitution).

Write down the McLaurin series for #arctan(y)#, which is known: it's the sum, with alternate sign, of the odd powers of #y# divided by the power itself:

#y-y^3/3+y^5/5-y^7/7+y^9/9+...#,

Alternatively, compactly

#sum_{i=0}^infty (-1)^{n} y^{2n+1}/{2n+1}#

Now substitute back #y=x^3#:

#sum_{i=0}^infty (-1)^{n} (x^3)^{2n+1}/{2n+1}= sum_{i=0}^infty (-1)^{n} x^{6n+3}/{2n+1}#

Multiply the whole expression by #x^2#:

#sum_{i=0}^infty (-1)^{n} x^2*x^{6n+3}/{2n+1}= sum_{i=0}^infty (-1)^{n} x^{6n+5}/{2n+1}#

Elijah Abram · Answer

undefined To find the Maclaurin series expansion of $ f(x) = x^2 \arctan(x^3) $, you would follow these steps:

1. Express the function as a series using known Maclaurin series expansions.
2. Find the derivatives of the function.
3. Evaluate the derivatives at $ x = 0 $.
4. Write the Maclaurin series expansion using the coefficients obtained.

The Maclaurin series expansion of $ f(x) = x^2 \arctan(x^3) $ is:

$$ f(x) = x^2 \arctan(x^3) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}x^n $$

where $ f^{(n)}(0) $ denotes the $ n $-th derivative of $ f(x) $ evaluated at $ x = 0 $.