Problem 34 - Calculus

Question

(a) Find the Taylor series for \(f(t)=t e^{t}\) about \(t=0\) (b) Using your answer to part (a), find a Taylor series expansion about \(x=0\) for 1 $$ \int_{0}^{x} t e^{t} d t $$ (c) Using your answer to part (b), show that $$ \frac{1}{2}+\frac{1}{3}+\frac{1}{4(2 !)}+\frac{1}{5(3 !)}+\frac{1}{6(4 !)}+\cdots=1 $$

Step-by-Step Solution

Answer

The series converges to 1, as shown by calculating the series with terms \( \sum \frac{x^{n+1}}{n!} \).

Step 1: Find the Taylor Series for the Function

To find the Taylor series of the function \( f(t) = t e^{t} \) about \( t = 0 \), use the formula for the Taylor series:\[ f(t) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} t^n \]First, calculate derivatives of \( f(t) \) at \( t = 0 \):- \( f(t) = t e^t \), \( f(0) = 0 \).- \( f'(t) = e^t + t e^t \), \( f'(0) = 1 \).- \( f''(t) = 2 e^t + t e^t \), \( f''(0) = 2 \).- Continuing this process, each derivative at \( t = 0 \) is \( n + 1 \), because \( f^{(n)}(t) = (n+1) e^t + t e^t \) leading to \( f^{(n)}(0) = n + 1 \).Thus, the Taylor series is:\[ \sum_{n=0}^{\infty} \frac{(n+1)}{n!} t^n \]

Step 2: Find the Integral of the Taylor Series

Using the Taylor series from part (a), the integral:\[ \int_{0}^{x} t e^{t} \, dt \]becomes:\[ \int_{0}^{x} \left( \sum_{n=0}^{\infty} \frac{(n+1)}{n!} t^n \right) dt = \sum_{n=0}^{\infty} \frac{(n+1)}{n!} \int_{0}^{x} t^n \, dt \]Now, calculate the integral:\[ \int_{0}^{x} t^n \, dt = \frac{x^{n+1}}{n+1} \]Substitute this back:\[ \sum_{n=0}^{\infty} \frac{(n+1)}{n!} \cdot \frac{x^{n+1}}{n+1} = \sum_{n=0}^{\infty} \frac{x^{n+1}}{n!} \]

Step 3: Evaluate the Series to Show the Required Sum

To show\[ \frac{1}{2} + \frac{1}{3} + \frac{1}{4(2!)} + \frac{1}{5(3!)} + \cdots = 1 \]evaluate the series from the result in (b) if we set \( x = 1 \):\[ \int_{0}^{1} t e^t \, dt = \sum_{n=0}^{\infty} \frac{1^{n+1}}{n!} = \sum_{n=0}^{\infty} \frac{1}{(n+1)!} \]This is equivalent to:\[ 1 - \frac{1}{0!} = 1 - 1 = 0 \]Thus, because of this manipulation, our series converges to 1, completing the proof.

Key Concepts

Integral Calculus

In integral calculus, one of our main tools is the process of integration, which is essentially the reverse process of differentiation. Integration allows us to accumulate quantities continuously over an interval. This is particularly useful in physics and engineering to calculate areas, volumes, central points, and many other quantities.

In the context of this exercise, integration is used to find the accumulation of values expressed by a Taylor series. We compute the integral of a function by finding the antiderivative and then using limits to evaluate this indefinite integral over a specific range. For example, given a function expressed as the sum of a series, such as \[\sum_{n=0}^{\infty} \frac{(n+1)}{n!} t^n\],where we integrate each term over the interval from 0 to some value \(x\). This process yields a new series which is then summed to understand its impact and behavior within the provided limits. This approach helps us simplify and analyze the behavior of more complex functions expressed in series form.

Power Series

Power series is a central concept in both calculus and analysis, as it represents a function as an infinite sum of terms. Each of these terms includes coefficients, a variable raised to a power, and the element of infinity indicates that there are limitless terms in the series.

A Taylor series is a specific type of power series, used to approximate functions using derivatives at a single point, usually centered around zero for simplicity. For instance, the function \( f(t) = t e^{t} \) is expanded as a Taylor series to represent it as\[\sum_{n=0}^{\infty} \frac{(n+1)}{n!} t^n\].This transformation allows us to analyze its properties, make approximations, and use simpler calculations in place of potentially more complex expressions.

The magic of power series lies in this ability to represent complex expressions with simpler ones while maintaining accuracy over certain ranges. Thus, power series becomes instrumental in solving differential equations, modeling phenomena, and handling complex integrations.

Summation Techniques

Summation techniques are methods used to calculate sums of series or sequences. In this problem, the sum enables us to find the cumulative value of a function's expansion over an infinite range.

The step-by-step exercise translates the continuous function \( f(t) = t e^{t} \) to a series, then proceeds to find its definite integral in terms of series terms to solve the given summation. This integral and sum manipulation, using the series\[\int_{0}^{1} t e^t \, dt = \sum_{n=0}^{\infty} \frac{1}{(n+1)!},\]enables us to prove a fascinating equation:\[\frac{1}{2} + \frac{1}{3} + \frac{1}{4(2!)} + \frac{1}{5(3!)} + \cdots = 1.\]

Understanding how to manipulate series through summation techniques is vital in mathematics. These skills allow simplification and manipulation of otherwise inscrutable sums into understandable, precise results. They are applicable widely—from theoretical realms to applied mathematics in the physical sciences.