Studdy AI Logo
Math  /  Calculus

QuestionUse a Maclaurin series in this table to obtain the Maclaurin series for the given function. f(x)=x2ln⁡(1+x3)∑n=1∞(□)\begin{array}{r} f(x)=x^{2} \ln \left(1+x^{3}\right) \\ \sum_{n=1}^{\infty}(\square) \end{array}

Studdy Solution

STEP 1

What is this asking? We need to find the Maclaurin series for x2ln⁡(1+x3)x^2 \ln(1 + x^3) using a given table of common Maclaurin series. Watch out! Remember that the Maclaurin series is a special case of the Taylor series centered at x=0x = 0.
Don't mix up the general Taylor series formula with the Maclaurin series!

STEP 2

1. Find the series for ln⁡(1+x)\ln(1+x).
2. Substitute and simplify.

STEP 3

We're given a table of Maclaurin series (not shown here, but assumed to be available to the student).
Let's **locate** the series for ln⁡(1+x)\ln(1+x).
It should look something like this: ln⁡(1+x)=∑n=1∞(−1)n−1nxn=x−x22+x33−x44+…\ln(1+x) = \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n} x^n = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \dots This series is **valid** for −1<x≤1-1 < x \le 1.

STEP 4

Now, we need to find the Maclaurin series for ln⁡(1+x3)\ln(1 + x^3).
We can do this by **substituting** x3x^3 for xx in the series we found for ln⁡(1+x)\ln(1+x): ln⁡(1+x3)=∑n=1∞(−1)n−1n(x3)n=∑n=1∞(−1)n−1nx3n\ln(1 + x^3) = \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n} (x^3)^n = \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n} x^{3n} This substitution is **valid** because if −1<x≤1-1 < x \le 1, then −1<x3≤1-1 < x^3 \le 1.

STEP 5

Finally, to find the Maclaurin series for x2ln⁡(1+x3)x^2 \ln(1 + x^3), we **multiply** the series for ln⁡(1+x3)\ln(1+x^3) by x2x^2: x2ln⁡(1+x3)=x2∑n=1∞(−1)n−1nx3n=∑n=1∞(−1)n−1nx3n⋅x2=∑n=1∞(−1)n−1nx3n+2x^2 \ln(1 + x^3) = x^2 \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n} x^{3n} = \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n} x^{3n} \cdot x^2 = \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n} x^{3n+2}

STEP 6

The Maclaurin series for f(x)=x2ln⁡(1+x3)f(x) = x^2 \ln(1 + x^3) is: ∑n=1∞(−1)n−1nx3n+2\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n} x^{3n+2}

Was this helpful?

Studdy solves anything!

banner

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

ParentsInfluencer programContactPolicyTerms
TwitterInstagramFacebookTikTokDiscord