Problem 18 - Essential Calculus

Question

Use the binomial series to find the Maclaurin series for the function. $$ f(x)=\sqrt[4]{1+x} $$

Step-by-Step Solution

Answer

The Maclaurin series for the function \( f(x)=\sqrt[4]{1+x}\) is \( f(x) = 1 + \frac{1}{4}x - \frac{3}{32}x^2 + \frac{5}{256}x^3 - ... \).

Step 1: Write the function in suitable form for binomial series

The function \( f(x)=\sqrt[4]{1+x}\) in the format applicable for binomial series can be written as \( (1+x)^{1/4}\).

Step 2: Apply binomial series

The general form of binomial series (binomial theorem) is \( (1+x)^n = 1+nx+\frac{n(n-1)x^2}{2!}+\frac{n(n-1)(n-2)x^3}{3!}+..\). Now, for \( n=1/4 \), the function can be expanded as: \( (1+x)^{1/4} = 1 + \frac{1}{4}x - \frac{3}{32}x^2 + \frac{5}{256}x^3 - ... \).

Step 3: Write down the Maclaurin series for the function

The Maclaurin series for the function is the same as the expanded binomial series. Therefore, the Maclaurin series for the function \( f(x)=\sqrt[4]{1+x}\) is \( f(x) = 1 + \frac{1}{4}x - \frac{3}{32}x^2 + \frac{5}{256}x^3 - ... \).

Key Concepts

Understanding the Binomial Series

The binomial series is a powerful mathematical tool that helps us expand expressions of the form \((1+x)^n\) into infinite series. It is especially useful when \(|x|<1\). This series provides a way to express these complicated functions as sums of simpler terms. The formula for the binomial series is given by: \[(1+x)^n = 1 + nx + \frac{n(n-1)x^2}{2!} + \frac{n(n-1)(n-2)x^3}{3!} + \dots\] Each term in the series can be calculated using successive powers of \(x\) and decreasing factorial products of \(n\). When \(n\) is a fraction, as with our exercise's \(n=1/4\), the series provides a way to calculate values that aren't immediately intuitive. It's like peeling back the layers of a complex function to reveal its simpler components, making calculations and approximations much easier.

Applying the Binomial Theorem

The binomial theorem is a cornerstone of algebra and is often introduced early in one's mathematical journey. It provides a straightforward way to expand expressions raised to any power. The theorem states that for any real number \(n\), the expansion of \((1 + x)^n\) is: \[(1 + x)^n = \sum_{k=0}^{\infty} \binom{n}{k}x^k\] where \(\binom{n}{k}\) represents the binomial coefficients. These coefficients are calculated as \(\frac{n(n-1)(n-2)\cdots(n-k+1)}{k!}\), matching the formula you see in a typical binomial series expansion. This general form can handle not just whole numbers, but fractions and negatives. It's particularly helpful in calculus and probability, offering solutions for derivatives, growth models, and statistical calculations. In our solution, using \(n=1/4\) allows us to expand \((1+x)^{1/4}\) and derive a Maclaurin series that approximates the square root function.

Exploring Function Expansion with Maclaurin Series

Function expansion allows us to represent functions that might initially appear difficult to handle. It transforms them into infinite series that are often easier to work with numerically and analytically. The Maclaurin series is particularly popular for this purpose. It's a Taylor series expanded about 0, giving it a central position in the realm of calculus. The Maclaurin series of a function \(f(x)\) is given by: \[f(x) = f(0) + f'(0)x + \frac{f''(0)x^2}{2!} + \frac{f'''(0)x^3}{3!} + \dots\] Each term in this series represents higher-order derivatives of the function at 0, expanded progressively. For our problem, the Maclaurin series for \(f(x)=\sqrt[4]{1+x}\) is derived from the binomial series for \((1+x)^{1/4}\). The resulting series provides an effective approximation, especially for small values of \(x\), and allows us to analyze the behavior of the function near 0.