| f - 1 < x < 1 then the infinite series expansion of log(1 + x) * is

-( x + (x ^ 2)/2 + (x ^ 3)/3 +*** alpha)

(b) x - (x ^ 2)/2 + (x ^ 3)/3 -***=

(c) x - (x ^ 7)/(2!) + (x ^ 3)/(3!) -*** x

(d) x + (x ^ 2)/(2!) + (x ^ 3)/(3!) +***=

The infinite series expansion of \( \log(1 + x) \) for \( -1 < x < 1 \) is given by:

\[
\log(1 + x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \cdots
\]

This is a well-known result and matches the form of answer **(b)**:

\[
x - \frac{x^2}{2} + \frac{x^3}{3} - \cdots
\]

Thus, the correct option is **(b)**.

Would you like a detailed explanation on how this series is derived or more about Taylor expansions?

Here are 5 related questions for further exploration:
1. How is the series expansion of \( \log(1+x) \) derived using calculus?
2. What is the radius of convergence of the series for \( \log(1+x) \)?
3. How would the series change for \( \log(1-x) \)?
4. Can you find the series expansion of \( \frac{1}{1+x} \)?
5. How is this series related to the geometric series?

**Tip:** The series for functions like \( \log(1 + x) \) can be derived from the integral of simpler functions like \( \frac{1}{1 + x} \), making it a key technique in calculus!

f - 1 < x < 1 then the infinite series expansion of log(1 + x) * is -( x + (x ^ 2)/2 + (x ^ 3)/3 +*** alpha)

(b) x - (x ^ 2)/2 + (x ^ 3)/3 -***=

(c) x - (x ^ 7)/(2!) + (x ^ 3)/(3!) -*** x

(d) x + (x ^ 2)/(2!) + (x ^ 3)/(3!) +***=

Explore the infinite series expansion of the logarithmic function log(1 + x) for -1 < x < 1. Learn how to derive the series using Taylor expansion techniques and understand its convergence properties. Suitable for undergraduate calculus students or anyone studying advanced calculus.

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