To prove the Maclaurin series expansion for $\ln(1 + x)$, we derive it step-by-step and verify its validity.

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1. Definition of the Maclaurin Series

A Maclaurin series is a Taylor series expanded about $x = 0$: $f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f^{(3)}(0)}{3!}x^3 + \dots$

For $\ln(1 + x)$, the general term is: $\frac{f^{(n)}(0)}{n!}x^n$

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2. Function and Derivatives

The function is $f(x) = \ln(1 + x)$.

First derivative:

$f'(x) = \frac{1}{1 + x}$

Second derivative:

$f''(x) = -\frac{1}{(1 + x)^2}$

Third derivative:

$f^{(3)}(x) = \frac{2}{(1 + x)^3}$

General $n$-th derivative:

$f^{(n)}(x) = (-1)^{n-1}(n-1)!\frac{1}{(1 + x)^n}$

Evaluate these derivatives at $x = 0$: $f^{(n)}(0) = (-1)^{n-1}(n-1)!$

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3. Substitute into Maclaurin Series

Using the general term of the Maclaurin series: $\frac{f^{(n)}(0)}{n!}x^n$

Substitute $f^{(n)}(0) = (-1)^{n-1}(n-1)!$: $\frac{(-1)^{n-1}(n-1)!}{n!}x^n$

Simplify: $\frac{(-1)^{n-1}}{n}x^n$

Thus, the series expansion becomes: $\ln(1 + x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \dots + (-1)^{n-1}\frac{x^n}{n} + \dots$

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4. Verification

The series converges for $|x| < 1$ and represents the natural logarithm function. By differentiating term-by-term, we can verify that the derivative of this series equals $\frac{1}{1 + x}$, confirming it is the expansion of $\ln(1 + x)$.

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Do you want further details or have questions?
Here are 5 related questions to consider:

1. How do you test the radius of convergence for this series?
2. What happens when $x > 1$ or $x < -1$?
3. How is the series used in numerical approximations?
4. Can you derive a similar series for $\ln(1 - x)$?
5. How do the alternating terms influence the speed of convergence?

Tip: Remember, the convergence of series is faster for smaller $|x|$, so truncating at fewer terms gives accurate results for $|x| \ll 1$.