The expression you provided is the logarithmic series expansion of $\ln(1 + 3x + 2x^2)$. To understand how this series relates to the logarithm, let's break it down.

Taylor Series Expansion of $\ln(1 + f(x))$:

The Taylor series expansion of $\ln(1 + u)$ around $u = 0$ is given by:

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

Here, $u$ is a function of $x$. If we set $u = 3x + 2x^2$, we can expand the logarithm as:

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

Given Series:

The given series is:

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

This suggests that the coefficients of each term in the expansion are derived from the Taylor series, applied specifically to the function $3x + 2x^2$.

Matching Terms:

To match the terms of the series expansion, each coefficient is derived by substituting $3x + 2x^2$ into the general formula for the logarithm’s Taylor series:

• Linear Term: $u = 3x + 2x^2$ $3x + 2x^2$

• Quadratic Term: $-\frac{(3x + 2x^2)^2}{2} = -\frac{(9x^2 + 12x^3 + 4x^4)}{2}$

• Cubic Term: $\frac{(3x + 2x^2)^3}{3} = \frac{(27x^3 + 54x^4 + \ldots)}{3}$

• And so on for higher-order terms.

Conclusion:

The series expansion provided matches the expected result of expanding $\ln(1 + 3x + 2x^2)$ using a Taylor series. Each coefficient in the expansion corresponds to the series generated by substituting $3x + 2x^2$ into the Taylor series formula.

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Related Questions:

1. How do you derive the Taylor series for common functions?
2. What is the radius of convergence for a Taylor series?
3. How can you use the Taylor series to approximate functions near a point?
4. What are some applications of logarithmic series expansions in engineering?
5. How does the error in a Taylor series approximation change with the order of the expansion?

Tip:

When expanding functions as a series, ensure that the series converges for the values of $x$ you are interested in; otherwise, the expansion may not be valid for those values.