Question

In each exercise, (a) Use the given information to determine a power series representation of the function \(y(t)\). (b) Determine the radius of convergence of the series found in part (a). (c) Where possible, use (7) to identify the function \(y(t)\). $$ y^{\prime}(t)=\sum_{n=1}^{\infty} n t^{n-1}=1+2 t+3 t^{2}+\cdots, \quad y(0)=1 $$

Short answer

2. What is the radius of convergence for the power series of \(y(t)\)? 3. What is the function \(y(t)\) identified as? Answer: 1. The power series representation for the function \(y(t)\) is: $$y(t)=e^t = 1+t+\frac{t^2}{2}+\frac{t^3}{3!}+\cdots.$$ 2. The radius of convergence for the power series of \(y(t)\) is infinite. 3. The function \(y(t)\) is identified as \(e^t\).

Step-by-step solution

Integrate the power series

We are given the power series of the derivative of \(y(t)\) as $$y^{\prime}(t)=\sum_{n=1}^{\infty} n t^{n-1}=1+2 t+3 t^{2}+\cdots$$ To find the function \(y(t)\) itself, we need to integrate the power series term-by-term: $$y(t) = \int\left(\sum_{n=1}^{\infty} n t^{n-1}\right) dt = \int (1+2t+3t^2+\cdots)dt.$$ By integrating each term, we get: $$y(t) = \sum_{n=1}^{\infty} \int nt^{n-1} dt = t+C+\frac{2t^2}{2}+\frac{3t^3}{3}+\cdots,$$ where \(C\) is the constant of integration.

Find the constant term for the power series

We are given the initial condition \(y(0) = 1\). Plug in \(t=0\) into the power series representation of \(y(t)\), and solve for the constant \(C\). The power series representation of \(y(t)\) now becomes: $$y(t) = \sum_{n=0}^{\infty} \frac{nt^n}{n} = 1+t+\frac{2t^2}{2}+\frac{3t^3}{3}+\cdots.$$

Determine the radius of convergence

Now we need to find the radius of convergence of the power series. If the given power series of the derivative has a non-zero radius of convergence, the radius of convergence for a finite number of differentiations or integrations will be the same. The radius of convergence for the power series representation of the derivative is infinite, so the radius of convergence for \(y(t)\) is also infinite.

Identify the function (if possible)

Next, we will try to identify the power series representation of \(y(t)\) as a known function. Notice that the power series representation of \(y(t)\) is the same as the Maclaurin series for the exponential function with base \(e\): $$y(t) = 1+t+\frac{t^2}{2}+\frac{t^3}{3!}+\cdots = e^t.$$ So, the function is: $$y(t) = e^t.$$ In conclusion, the power series representation of the function \(y(t)\) is: $$y(t)=e^t = 1+t+\frac{t^2}{2}+\frac{t^3}{3!}+\cdots.$$ The radius of convergence is infinite, and the function is \(e^t\).

Key concepts

Radius of Convergence

The radius of convergence is an essential concept when working with power series. It shows us the interval over which a power series converges to the function it represents.
For any given power series, knowing its radius of convergence helps us determine whether the series would provide accurate results for all values of interest in its domain. In mathematical terms, a power series \( \sum_{n=0}^{\infty} a_n(t-c)^n \) converges if the distance \(|t-c|\) is less than the radius of convergence, denoted by \(R\).
In this exercise, the power series of the derivative \(y'(t)\) and hence of \(y(t)\) itself were noted to have an infinite radius of convergence. This means the series converges for all values of \(t\). Having an infinite radius is a special property, often seen in power series representing entire functions, like exponential functions.

Integration in Power Series

Integration is one of the primary processes applied to power series in order to discover the original function from its derivative. The magic of power series and polynomials is that you can integrate term-by-term to find results quite easily. The process seems daunting, but let's break it down.
When integrating each term of a power series like \(y'(t) = \sum_{n=1}^{\infty} n t^{n-1} \), each term is approached as a simple power of \(t\). For example, integrating \(nt^{n-1}\) gives \(\frac{nt^n}{n} = t^n\).
The constant of integration \(C\) is crucial because it adjusts the result to fit initial conditions. In this case, we have \(y(0) = 1\), helping us to find \(C\) and yield a precise function. This method provides a clear path from derivative back to the function itself through integration.

Maclaurin Series

Maclaurin series is a helpful tool for depicting functions as infinite sums of their derivatives at zero. It is often used to find approximations or exact expressions for well-known functions. The standard formula for a Maclaurin series representation is:\[f(t) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}t^n\]
In this exercise, we discovered that \(y(t)\) can be expressed via a power series identical to a Maclaurin series for \(e^t\). This series is: \[1 + t + \frac{t^2}{2!} + \frac{t^3}{3!} + \cdots = e^t\]
Using Maclaurin series not only allows clear representation but also helps in simplifying or understanding how functions behave, especially around \(t = 0\). It’s an elegant way to represent many smooth functions, particularly exponential functions.

Exponential Functions

Exponential functions are a fascinating class of mathematical functions characterized by their growth patterns and ubiquity in natural phenomena. Written commonly as \(e^t\), where \(e\) is the mathematical constant approximately equal to 2.718, these functions are expressive powerhouses.
These functions hold special properties such as their rate of growth and continual doubling. The power series representation of an exponential function \[ e^t = 1 + t + \frac{t^2}{2!} + \frac{t^3}{3!} + \cdots \] is infinitely convergent across all \(t\) values.
In our example, we demonstrated that the function \(y(t)\) found by integrating its derivative falls into the exponential function category, with an infinite radius of convergence. This means \(y(t)\) adheres completely to the form, growth, and behavior of \(e^t\), which ties neatly into both mathematical theory and practical applications.