{Use of Tech} Remainders Let 


f(x) = ∑ₖ₌₀∞ xᵏ = 1/(1−x) and Sₙ(x) = ∑ₖ₌₀ⁿ⁻¹ xᵏ


The remainder in truncating the power series after n terms is Rₙ = f(x) − Sₙ(x), which depends on x.


a. Show that Rₙ(x) = xⁿ /(1−x).
b. Graph the remainder function on the interval |x| < 1, for n=1, 2, and 3 . Discuss and interpret the graph. Where on the interval is |Rₙ(x)| largest? Smallest?
c. For fixed n, minimize |Rₙ(x)| with respect to x. Does the result agree with the observations in part (b)?
d. Let N(x) be the number of terms required to reduce |Rₙ(x)| to less than 10⁻⁶. Graph the function N(x) on the interval |x|<1.

Verified step by step guidance

1. Step 1: Understand the definitions given. The function is defined as the infinite geometric series \(f(x) = \sum_{k=0}^\infty x^k = \frac{1}{1-x}\) for \(|x| < 1\). The partial sum up to \(n\) terms is \(S_n(x) = \sum_{k=0}^{n-1} x^k\). The remainder after \(n\) terms is \(R_n(x) = f(x) - S_n(x)\).
2. Step 2: To show that \(R_n(x) = \frac{x^n}{1-x}\), start from the formula for the partial sum of a geometric series: \(S_n(x) = \frac{1 - x^n}{1 - x}\). Then express the remainder as \(R_n(x) = f(x) - S_n(x) = \frac{1}{1-x} - \frac{1 - x^n}{1 - x}\). Simplify this expression to isolate \(R_n(x)\).
3. Step 3: For the graphing part, consider the function \(R_n(x) = \frac{x^n}{1-x}\) on the interval \(|x| < 1\) for \(n=1, 2, 3\). Plot these functions to observe how the remainder behaves as \(x\) varies. Pay attention to the magnitude \(|R_n(x)|\) and note where it reaches its maximum and minimum values within the interval.
4. Step 4: To minimize \(|R_n(x)|\) for fixed \(n\), analyze the function \(|R_n(x)| = \left| \frac{x^n}{1-x} \right|\). Consider taking the derivative with respect to \(x\) and setting it to zero to find critical points. Check these points within the domain \(|x| < 1\) to determine where the minimum occurs. Compare these results with the observations from the graph.
5. Step 5: For the function \(N(x)\), which is the number of terms needed to make \(|R_n(x)| < 10^{-6}\), set up the inequality \(\left| \frac{x^n}{1-x} \right| < 10^{-6}\). Solve for \(n\) in terms of \(x\), which will give \(N(x)\). Then graph \(N(x)\) over the interval \(|x| < 1\) to visualize how many terms are required for different values of \(x\).

Key Concepts

Geometric Series and Partial Sums

A geometric series is a sum of terms where each term is a constant multiple (common ratio) of the previous one. The infinite series ∑ₖ₌₀∞ xᵏ converges to 1/(1−x) for |x|<1. The partial sum Sₙ(x) sums the first n terms and approximates the infinite sum, serving as a foundation for understanding series truncation and remainders.

Remainder (Error) of a Power Series Approximation

The remainder Rₙ(x) measures the difference between the infinite series and its partial sum after n terms. For the geometric series, Rₙ(x) = xⁿ/(1−x), quantifying the truncation error. Understanding this remainder helps analyze convergence behavior and error bounds in approximations.

Minimization and Graphical Analysis of Functions

Minimizing |Rₙ(x)| involves finding values of x that reduce the remainder's magnitude, often using calculus or inspection. Graphing Rₙ(x) over |x|<1 reveals where the error is largest or smallest, providing visual insight into convergence rates and guiding the choice of n for desired accuracy.