To find the Maclaurin series for the function $f(x) = \frac{1}{1-x}$, we start by recalling the general definition of a Maclaurin series:

$f(x) = f(0) + f'(0)x + \frac{f''(0)x^2}{2!} + \frac{f^{(3)}(0)x^3}{3!} + \dots$

Step-by-step Solution:

1. Start with the function: $f(x) = \frac{1}{1-x}.$

2. Find the derivatives of $f(x)$:

   • First derivative: $f'(x) = \frac{d}{dx}\left(\frac{1}{1-x}\right) = \frac{1}{(1-x)^2}.$
   • Second derivative: $f''(x) = \frac{d}{dx}\left(\frac{1}{(1-x)^2}\right) = \frac{2}{(1-x)^3}.$
   • Third derivative: $f^{(3)}(x) = \frac{d}{dx}\left(\frac{2}{(1-x)^3}\right) = \frac{6}{(1-x)^4}.$
   • For the $n$-th derivative: $f^{(n)}(x) = \frac{n!}{(1-x)^{n+1}}.$

3. Evaluate each derivative at $x = 0$:

   • $f(0) = 1$,
   • $f'(0) = 1$,
   • $f''(0) = 2$,
   • $f^{(3)}(0) = 6$,
   • $f^{(n)}(0) = n!$.

4. Substitute these values into the Maclaurin series formula: $f(x) = f(0) + f'(0)x + \frac{f''(0)x^2}{2!} + \frac{f^{(3)}(0)x^3}{3!} + \dots$ $f(x) = 1 + x + \frac{2x^2}{2!} + \frac{6x^3}{3!} + \dots$

5. Simplify the series: Each term simplifies as: $\frac{n!x^n}{n!} = x^n.$ Thus: $f(x) = 1 + x + x^2 + x^3 + \dots$

6. Write the Maclaurin series in summation form: $f(x) = \sum_{n=0}^\infty x^n, \quad \text{for } |x| < 1.$

Final Result:

The Maclaurin series for $f(x) = \frac{1}{1-x}$ is: $f(x) = \sum_{n=0}^\infty x^n, \quad |x| < 1.$

Would you like further details or a graphical representation?
Here are 5 related questions and a tip for deeper understanding:

1. How does the radius of convergence $|x| < 1$ arise in this series?
2. What happens if $x$ is outside the interval of convergence?
3. Can you derive the Maclaurin series for $\frac{1}{(1-x)^2}$ using a similar method?
4. How does the Maclaurin series for $\ln(1+x)$ compare to this one?
5. How would the series change if $f(x) = \frac{1}{1+x}$?

Tip: The geometric series sum $\sum_{n=0}^\infty x^n = \frac{1}{1-x}$ is foundational in calculus—master it, and many series expansions become easier!