The Maclaurin series is the expansion of a function $f(x)$ about $x = 0$. Let's derive the Maclaurin series of

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

to $n$-th power.

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Step 1: Simplify the function

For ease of handling, rewrite $f(x)$ as: $f(x) = \frac{1 - \sqrt{1+x^2}}{1 + \sqrt{1+x^2}}.$

Define $u = \sqrt{1 + x^2}$, so the function becomes: $f(x) = \frac{1 - u}{1 + u}.$

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Step 2: Expand $u = \sqrt{1+x^2}$ around $x=0$

Using the binomial expansion for $\sqrt{1+x^2}$, we get: $\sqrt{1+x^2} = 1 + \frac{x^2}{2} - \frac{x^4}{8} + \frac{x^6}{16} - \cdots.$

Thus: $u = \sqrt{1+x^2} = 1 + \frac{x^2}{2} - \frac{x^4}{8} + \frac{x^6}{16} - \cdots.$

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Step 3: Expand $f(x)$

Substitute $u$ into $f(x) = \frac{1 - u}{1 + u}$: $f(x) = \frac{1 - (1 + \frac{x^2}{2} - \frac{x^4}{8} + \cdots)}{1 + (1 + \frac{x^2}{2} - \frac{x^4}{8} + \cdots)}.$

Simplify the numerator and denominator: $1 - u = -\frac{x^2}{2} + \frac{x^4}{8} - \cdots,$ $1 + u = 2 + \frac{x^2}{2} - \frac{x^4}{8} + \cdots.$

Thus: $f(x) = \frac{-\frac{x^2}{2} + \frac{x^4}{8} - \cdots}{2 + \frac{x^2}{2} - \frac{x^4}{8} + \cdots}.$

Divide term by term, using $\frac{1}{2 + \cdots} \approx \frac{1}{2} - \frac{1}{4}(\text{higher terms})$. To second order: $f(x) \approx -\frac{x^2}{4} + \frac{3x^4}{32} - \cdots.$

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Step 4: Raise $f(x)$ to the $n$-th power

For $f(x)^n$, use the series expansion: $(1 + a + b + \cdots)^n \approx 1 + n(a + b + \cdots) + \frac{n(n-1)}{2}(a + b + \cdots)^2 + \cdots.$

Thus: $f(x)^n = \left(-\frac{x^2}{4} + \frac{3x^4}{32} - \cdots\right)^n.$

For small $x$, the dominant term is: $f(x)^n \approx \left(-\frac{x^2}{4}\right)^n = (-1)^n \frac{x^{2n}}{4^n}.$

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Final Answer

The Maclaurin series for $f(x)^n$ up to the $x^{2n}$-term is: $f(x)^n \approx (-1)^n \frac{x^{2n}}{4^n} + \text{higher-order terms}.$

Would you like a detailed step-by-step expansion for specific values of $n$?