The problem you want help with is Exercise 6:

$T_n[\log(1+x)] = \sum_{k=1}^{n} \frac{(-1)^{k+1}x^k}{k}$.

This expression is the Taylor polynomial for the function $\log(1 + x)$ around $x = 0$ (Maclaurin series), and the task asks you to find the polynomial of degree $n$.

Step-by-step approach:

1. Recognize the Taylor series expansion of $\log(1 + x)$:

The Taylor series for $\log(1 + x)$ at $x = 0$ is given by: $\log(1 + x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \cdots$ This is an alternating series with the general term: $(-1)^{k+1} \frac{x^k}{k}, \quad \text{for} \ k = 1, 2, 3, \dots$

2. The general term:

The general form of the Taylor polynomial of degree $n$ for $\log(1 + x)$ is the sum of the first $n$ terms: $T_n[\log(1 + x)] = \sum_{k=1}^{n} \frac{(-1)^{k+1}x^k}{k}$ This matches the series you provided.

3. Example with a specific $n$:

Let's compute this for $n = 4$: $T_4[\log(1 + x)] = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4}$ For $n = 5$, it becomes: $T_5[\log(1 + x)] = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \frac{x^5}{5}$

4. General Procedure:

To compute $T_n[\log(1 + x)]$, simply follow the general formula and substitute the degree $n$, computing each term up to $x^n$.

Summary:

• The Taylor polynomial $T_n[\log(1 + x)]$ is given by the formula $\sum_{k=1}^{n} \frac{(-1)^{k+1}x^k}{k}$.
• Plug in any value of $n$ to obtain the specific Taylor polynomial of degree $n$.

Let me know if you'd like to explore any particular part of this further!

Questions:

1. How is the general term for the Taylor series derived?
2. What is the radius of convergence for the Taylor series of $\log(1 + x)$?
3. Can you find $T_6[\log(1 + x)]$?
4. How does the Taylor polynomial approximate $\log(1 + x)$ for values of $x$ near zero?
5. Can you visualize how the series converges as $n$ increases?

Tip: The Taylor series for $\log(1 + x)$ converges for $-1 < x \leq 1$, so keep this in mind when using it for approximation!