Problem 387 - Advanced Calculus

Question

a) Find an expansion in powers of \(\mathrm{x}\) of the function $$ \mathrm{f}(\mathrm{x})={ }^{1} \int_{0}\left[\left(1-\mathrm{e}^{-\mathrm{tx}}\right) / \mathrm{t}\right] \mathrm{dt} \text { . } $$ b) Use the results from part (a) to find \(\mathrm{f}(1 / 2)\) approximately.

Step-by-Step Solution

Answer

Using a power series expansion, we find the function \(f(x) = x - \frac{x^2}{2\cdot 2!} + \frac{x^3}{3\cdot 3!} - \cdots\). To approximate \(f(1/2)\), we can use the first few terms of this power series. Plugging in \(x = 1/2\), we get \(f(1/2) \approx \frac{1}{2} - \frac{1}{4} + \frac{1}{18} = \frac{9}{36}\), so the approximate value of \(f(1/2)\) is \(\frac{1}{4}\).

Step 1: Write down the given function

The given function is: \[ f(x) = \int_{0}^{1}\frac{1 - e^{-tx}}{t} dt \]

Step 2: Expand the exponential function using Taylor series

Recall the Taylor series for an exponential function: \[ e^{-tx} = 1 - tx + \frac{(-tx)^2}{2!} - \frac{(-tx)^3}{3!} + \cdots \] Now substitute this expansion back into the expression inside the integral: \[ f(x) = \int_{0}^{1}\frac{1 - (1 - tx + (-tx)^2/2! - (-tx)^3/3! + \cdots)}{t} dt \]

Step 3: Simplify the integrand

After substituting the expansion, the integrand can be simplified as: \[ f(x) = \int_{0}^{1}(x - \frac{(-x)^2t}{2!} + \frac{(-x)^3t^2}{3!} - \cdots) dt \]

Step 4: Integrate term by term

Now, we integrate each term of the series function between the limits 0 to 1: \[ f(x) = \left[x\int_{0}^{1} dt - \frac{x^2}{2!}\int_{0}^{1}t dt + \frac{x^3}{3!}\int_{0}^{1}t^2 dt \cdots\right] \] Perform the integration for each term: \[ f(x) = \left[x - \frac{x^2}{2\cdot 2!} + \frac{x^3}{3\cdot 3!} - \cdots\right] \] This results in a power series representation of the function f(x). Now, we need to use the newly found expansion to estimate the value of the function at x = 1/2.

Step 5: Compute f(1/2) using the expansion found in step 4

Now we will plug x = 1/2 into the power series we obtained in the previous step to find the approximation of f(1/2). Take the first few terms of the expansion to get an approximate value: \[ f(1/2) \approx \left[\frac{1}{2} - \frac{1}{2\cdot 2!} + \frac{1}{3\cdot 3!} - \cdots\right] \] Evaluate the first few terms for a good approximation: \[ f(1/2) \approx \frac{1}{2} - \frac{1}{4} + \frac{1}{18} = \frac{9}{36} \] Hence, the approximate value of f(1/2) is 9/36 or 1/4.

Key Concepts

Exponential Function Expansion

The exponential function, typically denoted as \(e^x\), is pivotal in mathematics and appears frequently in calculus, particularly in solving differential equations. Its power series, often referred to as the Taylor series expansion, can be used to approximate the function around a specific point, usually near \(x = 0\) (known as the Maclaurin series).

For any real number \(x\), the expansion is given by:
\[e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots\]
This elegant formula represents an infinite sum, where \(n!\) denotes the factorial of \(n\). Understanding this expansion is critical because it allows us to approximate \(e^x\) with a polynomial of finite degree, providing a pragmatic approach in scenarios where an exact solution is out of reach or unnecessary. The function \(f(x)\) from the exercise involves an exponential term \(e^{-tx}\), and using this expansion we can transform complex integrals into a series of simpler ones.

Power Series Integration

Power series integration is a technique whereby each term of a power series is integrated individually. This type of integration can help to solve integrals that would otherwise be challenging or impossible to evaluate using standard methods.

When integrating a power series term by term such as:
\[\int (a_0 + a_1x + a_2x^2 + a_3x^3 + \cdots) dx\]
we simply integrate each term, usually with respect to \(x\), yielding:
\[A_0x + \frac{a_1x^2}{2} + \frac{a_2x^3}{3} + \frac{a_3x^4}{4} + \cdots\]
provided the series converges. In our exercise, this method is used to integrate the terms obtained after expanding \(e^{-tx}\) through Taylor series. It's important to note that integrating each term individually is valid only if the integral and sum are both convergent over the interval of integration.

Calculating Integrals of Series

Calculating the integral of a series involves summing up the integrals of the individual terms in the series. This approach is based on the principle that, under proper convergence conditions, integration and summation operations can be interchanged.

For a given series \(\sum_{n=0}^{\infty} a_nx^n\), the integrated series can be expressed as:
\[\int\sum_{n=0}^{\infty} a_nx^n dx = \sum_{n=0}^{\infty} \int a_nx^n dx\]
The result is another series where each term is the integral of the corresponding term in the original series:
\[\sum_{n=0}^{\infty} \frac{a_nx^{n+1}}{n+1}\]
In our example, each term of the power series obtained from the exponential function expansion is integrated with respect to \(t\). Care must be taken to ensure the series converges; otherwise, the interchange of the integral and summation may lead to an incorrect result. The ease of executing this process paves the way for approximations like the one we find in the exercise, where \(f(1/2)\) is estimated using the initial terms of the integrated series.