Given the equation:

$\ln(1 + 3x + 2x^2) = 3x - \frac{5x^2}{2} + \frac{9x^3}{3} - \frac{17x^4}{4} + \ldots$

This is a power series expansion for the logarithm of a function of $x$. To verify and derive the coefficients in the power series on the right-hand side, we can expand the logarithm as a Taylor series around $x = 0$.

The general form of the series expansion for $\ln(1 + f(x))$ is:

$\ln(1 + f(x)) = f(x) - \frac{f(x)^2}{2} + \frac{f(x)^3}{3} - \frac{f(x)^4}{4} + \cdots$

Let $f(x) = 3x + 2x^2$. We now expand $\ln(1 + 3x + 2x^2)$ using the Taylor series formula and compare it term by term with the given series.

Step-by-step Expansion:

1. The first term is $f(x) = 3x + 2x^2$.
2. The second term is $-\frac{1}{2} \cdot (3x + 2x^2)^2$.
3. The third term is $\frac{1}{3} \cdot (3x + 2x^2)^3$, and so on.

I will now compute and expand the first few terms to see how they compare with the provided series.from sympy import symbols, ln, series, simplify

Define the variable and function

x = symbols('x') f_x = 3x + 2x**2

Series expansion of ln(1 + f(x)) around x = 0

expansion = series(ln(1 + f_x), x, 0, 5)

Simplify the series to get the result

simplified_expansion = simplify(expansion) simplified_expansionThe series expansion of $\ln(1 + 3x + 2x^2)$ around $x = 0$ is:

$\ln(1 + 3x + 2x^2) = 3x - \frac{5x^2}{2} + 3x^3 - \frac{17x^4}{4} + \mathcal{O}(x^5)$

Comparing this with the given series:

$3x - \frac{5x^2}{2} + \frac{9x^3}{3} - \frac{17x^4}{4} + \dots$

We can see that the coefficients for the first four terms match exactly. Therefore, the provided series accurately represents the expansion of $\ln(1 + 3x + 2x^2)$.

Would you like further details on any part of this solution?

Here are 5 related questions you might explore:

1. How do you derive the Taylor series of a function step by step?
2. What are the conditions under which a power series converges?
3. How does the logarithmic series compare to other common series expansions?
4. How would you expand more complicated functions, such as trigonometric or exponential functions?
5. Can you generalize this technique to approximate functions to higher degrees?

Tip: When working with power series, remember that truncating the series provides an approximation, which becomes more accurate as you include higher-order terms.