Problem 1 - Calculus

Question

Find the first four nonzero terms of the Taylor series for the function about 0. $$(1+x)^{3 / 2}$$

Step-by-Step Solution

Answer

The first four nonzero terms are: 1, \(\frac{3}{2}x\), \(\frac{3}{8}x^2\), \(-\frac{1}{16}x^3\).

Step 1: Recognize the Function and Taylor Series Formula

To find the Taylor series of a function about 0, use the formula: \[T(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \ldots\]Here, the given function is \((1+x)^{3/2}\). Identify and calculate the necessary derivatives at \(x = 0\).

Step 2: Calculate the Zeroth Derivative

The zeroth derivative is simply the function value at \(x = 0\).\[f(0) = (1+0)^{3/2} = 1\]This means the first term of the Taylor series is 1.

Step 3: Calculate the First Derivative

Use the power rule to compute the first derivative of \((1+x)^{3/2}\):\[f'(x) = \frac{3}{2}(1+x)^{1/2}\]Evaluate at \(x = 0\):\[f'(0) = \frac{3}{2}(1+0)^{1/2} = \frac{3}{2}\]This means the second term of the series is \(\frac{3}{2}x\).

Step 4: Calculate the Second Derivative

Differentiate \(f'(x) = \frac{3}{2}(1+x)^{1/2}\) one more time:\[f''(x) = \frac{3}{4}(1+x)^{-1/2}\]Evaluate at \(x = 0\):\[f''(0) = \frac{3}{4}(1+0)^{-1/2} = \frac{3}{4}\]The third term of the series is \(\frac{3}{4 \times 2!}x^2 = \frac{3}{8}x^2\).

Step 5: Calculate the Third Derivative

Differentiate \(f''(x) = \frac{3}{4}(1+x)^{-1/2}\):\[f'''(x) = -\frac{3}{8}(1+x)^{-3/2}\]Evaluate at \(x = 0\):\[f'''(0) = -\frac{3}{8}(1+0)^{-3/2} = -\frac{3}{8}\]The fourth term of the series is \(-\frac{3}{8 \times 3!}x^3 = -\frac{1}{16}x^3\).

Step 6: Combine the First Four Nonzero Terms

Combine all the terms calculated in each step to write the Taylor series up to four terms:\[(1+x)^{3/2} \approx 1 + \frac{3}{2}x + \frac{3}{8}x^2 - \frac{1}{16}x^3\]These are the first four nonzero terms of the Taylor series expansion about \(x = 0\).

Key Concepts

Power Rule

The Power Rule is a fundamental concept in calculus used to find derivatives of power functions. It's stated quite simply:

• If you have a function of the form \(f(x) = x^n\), then its derivative \(f'(x)\) is \(nx^{n-1}\).

In the context of our exercise, the function \(f(x) = (1+x)^{3/2}\) needs to be differentiated to create a Taylor series.
Applying the Power Rule helps compute each subsequent derivative, starting from the first. It unveils how each higher-order term affects the function's approximation around a specific point, typically \(x=0\).
Understanding the Power Rule is essential, as it becomes the stepping stone to work on more complex functions and their series expansions.

Function Derivative

The concept of Function Derivative revolves around calculating how a function changes at any point. Derivatives offer critical insights into a function's slope, rate of change, and behavior near specific points.

• A derivative, noted as \(f'(x)\), tells you how the function \(f(x)\) changes as \(x\) changes.
• Higher derivatives, such as \(f''(x)\) or \(f'''(x)\), reveal more about the function's concavity and the behavior of its slope.

For the series expansion of \((1+x)^{3/2}\), each derivative calculated—up to the third—is used to articulate how the function behaves around \(x = 0\).
With each derivative, a new term in the Taylor series emerges, showcasing finer details of the function’s behavior in its approximation.

Maclaurin Series

A Maclaurin Series is a special type of Taylor Series centered about \(x = 0\). It approximates functions using derivatives evaluated at this point.

• It is represented as \[f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \ldots\]
• Given that it starts around zero, the Maclaurin Series is particularly useful for functions smooth at the origin.

In examining \((1+x)^{3/2}\), we use the first four terms of the Maclaurin Series to gain a coherent and manageable approximation of the function around zero.
The calculated terms—\(1 + \frac{3}{2}x + \frac{3}{8}x^2 - \frac{1}{16}x^3\)—offer a simplified expression capturing the essential behavior of the function near \(x = 0\).

Series Expansion

Series Expansion is the mathematical way of representing functions as an infinite sum of terms. It's a powerful technique as it transforms complex functions into simpler polynomial terms.

• This method bridges the gap between simple polynomial functions and more intricate ones.
• The resulting series offers insights into functions over an interval where direct calculations may be challenging.

In the case of our function \((1+x)^{3/2}\), the Series Expansion approximates this otherwise nontrivial expression into accessible polynomial terms.
By evaluating derivatives at \(x=0\) and calculating terms up to \(x^3\), we develop a series representation that is locally accurate and computationally more straightforward than the original function.
Such expansions are vital not only for simplification but also for deepening our understanding of mathematical models and phenomena.