Problem 20 - Calculus

Question

using known Taylor series, find the first four nonzero terms of the Taylor series about 0 for the function. $$\sqrt{(1+t)} \sin t$$

Step-by-Step Solution

Answer

The first four nonzero terms are: \(t + \frac{t^2}{2} - \frac{7t^3}{24}\).

Step 1: Identify Taylor series for individual functions

We begin by identifying the Taylor series for each of the two functions, \(\sqrt{1+t}\) and \(\sin t\), about 0. The Taylor series for \(\sqrt{1+t}\) is:\[1 + \frac{t}{2} - \frac{t^2}{8} + \frac{t^3}{16} - \cdots\]The Taylor series for \(\sin t\) is:\[t - \frac{t^3}{6} + \frac{t^5}{120} - \cdots\]We will use these series to find the Taylor series for the product \(\sqrt{1+t} \sin t\).

Step 2: Multiply the two series

We perform multiplication of the Taylor series for \(\sqrt{1+t}\) and \(\sin t\). We are interested in finding four nonzero terms. To achieve this, we calculate up to the third degree terms involved in the multiplication.Begin by writing down the initial terms for the multiplication:\[(1 + \frac{t}{2} - \frac{t^2}{8} + \cdots)(t - \frac{t^3}{6} + \cdots)\]Multiply them term by term and collect like terms up to \(t^3\):\[1 \cdot t = t\]\[\frac{t}{2} \cdot t = \frac{t^2}{2}\]\[1 \cdot \left(-\frac{t^3}{6}\right) = -\frac{t^3}{6}\]\[\frac{t}{2} \cdot \left(-\frac{t^3}{6}\right) + \left(-\frac{t^2}{8}\right) \cdot t = -\frac{t^4}{12} - \frac{t^3}{8}\]Ignoring the higher power terms \(t^4\) for now, we proceed to collecting like terms.

Step 3: Simplify and collect terms

Collect the like terms from the multiplication results and stop where we get four nonzero terms:\[t + \frac{t^2}{2} - \frac{t^3}{6} - \frac{t^3}{8} + \cdots\]Combine the \(t^3\) terms:\[-\frac{t^3}{6} - \frac{t^3}{8} = -\frac{4t^3}{24} - \frac{3t^3}{24} = -\frac{7t^3}{24}\]Thus, the first four nonzero terms are:\[t + \frac{t^2}{2} - \frac{7t^3}{24}\]

Step 4: State the result

The Taylor series for \(\sqrt{1+t} \sin t\) about 0, with the first four nonzero terms is:\[t + \frac{t^2}{2} - \frac{7t^3}{24}\]These are the first four nonzero terms calculated by considering up to the third degree of terms in the multiplication of the individual Taylor series for \(\sqrt{1+t}\) and \(\sin t\).

Key Concepts

Maclaurin Series

The Maclaurin Series is a special case of the Taylor Series that is expanded about 0. When we talk about the Taylor Series, we are often focusing on a function being approximated around a point. However, with the Maclaurin Series, this point is always 0.
The general formula for the Maclaurin Series of a function \( f(t) \) is given by:

• \( f(t) = f(0) + f'(0)t + \frac{f''(0)}{2!}t^2 + \frac{f'''(0)}{3!}t^3 + \ldots \)

Maclaurin Series are particularly useful because they make calculations simpler and help in understanding how a function behaves very close to the point 0.
For example, in our exercise, each function (\( \sqrt{1+t} \) and \( \sin t \)) is expressed as a Maclaurin Series to approximate the function \( \sqrt{1+t} \sin t \) around 0.

Function Approximation

Function Approximation is all about finding a simpler expression that approximates the behavior of a function near a certain point. In mathematical analysis, it's a powerful tool that allows us to analyze complex functions.
Using series like the Taylor or Maclaurin Series, we can create polynomials that approximate the function. These polynomials are easier to manage, compute, and visualize.

• In our exercise, approximating \( \sqrt{1+t} \sin t \) involves using these series to reveal an expression that is accurate near \( t = 0 \).
• This gives us a clearer understanding of the function without having to deal with its more complicated form directly.

Function approximation is crucial in fields like physics and engineering, where working with the exact form of functions can be practical only around certain points.

Series Expansion

A Series Expansion, such as a Taylor or Maclaurin Series, expresses a function as an infinite sum of terms calculated from the derivatives of a function at a single point.
This process breaks down complex functions into an infinite series of simpler polynomial terms that approximates the original function's values.

• The initial terms provide a close approximation of the function, especially near the chosen point (like 0 in a Maclaurin Series).
• By considering more terms in the series, the approximation becomes more accurate, hence why higher-order derivatives contribute to better approximations.
• In our example, this idea is used to decompose both \( \sqrt{1+t} \) and \( \sin t \) into series and to approximate their product by examining only the first few terms.

Series expansions are extensively used when directly evaluating a function is complex or when computational ease is desired.

Mathematical Analysis

Mathematical Analysis refers to the rigorous study of mathematics where concepts like limits, continuity, derivatives, integrals, and infinite series are carefully defined and studied.
By employing the Maclaurin and Taylor series, mathematical analysis can explore the behavior of functions around specific points.

• In the context of our exercise, analysis involves understanding how \( \sqrt{1+t} \sin t \) can be approximated around \( t = 0 \).
• This process involves combining smaller, well-understood series expansions to yield an approximation that is mathematically sound.
• Through analysis, we gain insights into not just the functional outputs but also their behavior, continuity, and even more complex characteristics such as convergence and differentiability.

Mathematical analysis is foundational for proving results and ensuring that approximations hold true under anticipated conditions.