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 \(y(t)\). $$ y^{\prime}(t)=\sum_{n=0}^{\infty} \frac{(t-1)^{n}}{n !}=1+(t-1)+\frac{(t-1)^{2}}{2 !}+\cdots, \quad y(1)=1 $$

Short answer

Question: Find the power series representation of the function \(y(t)\) using the given power series for \(y'(t)=1+(t-1)+\frac{(t-1)^2}{2!}+\dots\) and the initial condition \(y(1)=1\). Determine the radius of convergence for the power series representation and identify the function \(y(t)\) based on the power series representation. Answer: The power series representation of \(y(t)\) is \(t+\frac{(t-1)^2}{2}+\frac{(t-1)^3}{3!}+\dots\). The radius of convergence for the series is infinite or \(R=\infty\). Based on the power series representation, the function \(y(t)\) can be identified as \(e^{t-1}+(t-1)\).

Step-by-step solution

Find power series representation of y(t) using given y'(t)

Since we are given the power series representation of \(y'(t)\), we can find the power series representation of \(y(t)\) by integrating term by term: $$ y(t) = \int y'(t) dt = \int \left( 1+(t-1)+\frac{(t-1)^2}{2!}+\dots \right) dt $$ Now, let's integrate each term of the series.

Integrate the power series term by term

We'll integrate each term of the power series: $$ \int 1 dt = t + C_1 \\ \int (t-1)dt = \frac{1}{2}(t-1)^2 + C_2 \\ \int \frac{(t-1)^2}{2!} dt = \frac{(t-1)^3}{3!}+ C_3 \\ \dots \\ $$ After integrating each term, the power series representation of \(y(t)\) can be written as: $$ y(t) = t +\frac{(t-1)^2}{2}+\frac{(t-1)^3}{3!}+\dots + C $$

Apply the initial condition y(1) = 1

We are given the initial condition \(y(1)=1\). We can use this to find the constant C: $$ 1 = 1 + \frac{(1-1)^2}{2}+\frac{(1-1)^3}{3!}+\dots + C $$ This simplifies to: $$ 1 = 1 + C $$ So, \(C = 0\). Therefore, the power series representation of \(y(t)\) is: $$ y(t) = t +\frac{(t-1)^2}{2}+\frac{(t-1)^3}{3!}+\dots $$

Determine the radius of convergence of the series for y(t)

Since the series for \(y(t)\) is obtained by integrating the given power series representing \(y'(t)\), their radius of convergence will be the same. The given power series has a factor of \(\frac{1}{n!}\), so sum of its terms converges for all \(t\), meaning its radius of convergence is infinite or \(R=\infty\).

Identify the function y(t) if possible

The power series representation of \(y(t)\) is as follows: $$ y(t) = t +\frac{(t-1)^2}{2}+\frac{(t-1)^3}{3!}+\dots $$ This power series is similar to the Taylor series expansion for \(e^x\) with \(x=t-1\): $$ e^{t-1} = 1+(t-1)+\frac{(t-1)^2}{2!}+\frac{(t-1)^3}{3!}+\dots $$ Now, we can rewrite \(y(t)\) in terms of \(e^{t-1}\): $$ y(t) = e^{t-1} + (t-1) $$ Hence, the function \(y(t)\) is \(e^{t-1} + (t-1)\).

Key concepts

Radius of Convergence

The radius of convergence refers to the interval around a central point for which a power series converges. It is crucial in determining where the function is well-defined by its series representation. In this exercise, the power series given is \( \sum_{n=0}^{\infty} \frac{(t-1)^{n}}{n!} \), which is similar to the series for \( e^x \). Because of the presence of the factorial in the denominator \( \frac{1}{n!} \), this series converges for all real numbers \( t \).
The radius of convergence for such series is infinite, denoted as \( R = \infty \). This means no matter what value \( t \) takes, the series will converge, providing a valid and accurate representation of the function.

Integration of Power Series

Integration of power series involves integrating each coefficient term by term. When a function is represented as a power series and we need to integrate it, the integration is usually performed term-wise. This is exactly what happens in this exercise. Given the power series for \( y'(t) = \sum_{n=0}^{\infty} \frac{(t-1)^{n}}{n!} \), we integrate each term to find \( y(t) \).

• The integral of 1 is \( t \).
• The integral of \( (t-1) \) is \( \frac{1}{2}(t-1)^2 \).
• The integral of \( \frac{(t-1)^2}{2!} \) is \( \frac{(t-1)^3}{3!} \).

After integrating and applying the initial condition \( y(1)=1 \), we find \( C \), the constant of integration, to be zero. Thus, the series representation becomes\( y(t) = t +\frac{(t-1)^2}{2} + \frac{(t-1)^3}{3!} + \cdots \).

Taylor Series Expansion

A Taylor series expansion expresses a function as an infinite sum of terms calculated from the values of its derivatives at a single point. In this exercise, once the series representation of \( y(t) \) is obtained through integration, it resembles the Taylor series for \( e^x \) centered at zero. The series expression we derived from \( y'(t) \) closely matches that of \( e^{t-1} \):

• The first term matches \( 1 \).
• The next term \( (t-1) \) fits with \( x \) in \( e^x \).
• Additional terms \( \frac{(t-1)^2}{2!} \) and \( \frac{(t-1)^3}{3!} \) further fit in the expansion.

This enables us to recognize the function \( y(t) \) as \( y(t) = e^{t-1} + (t-1) \).Through understanding Taylor series, we identify how the original sequence of derivatives provides the expansion of the function.

Differential Equations Initial Conditions

Initial conditions are vital when solving differential equations, as they allow us to identify particular solutions. They help determine unknown constants after integration.
In our exercise, the differential equation given was represented by a power series for \( y'(t) \), with an initial condition \( y(1) = 1 \).After integrating term by term to get \( y(t) \), an arbitrary constant \( C \) emerged. By applying the initial condition, \( y(1) = 1 \), we plug this into our series representation:\( 1 = 1 + C \). This results in \( C = 0 \), matching the specific solution from the general solution.

• Ensure you apply each condition appropriately in order to find the exact form of the function.
• Remember that without initial conditions, multiple solutions might fit the form of differential equations.