Active Calculus

1. Finding coefficients in a power series expansion of a rational function.

2. Finding coefficients in a power series expansion of a function involving \(\arctan(x)\).

3.

We can use power series to approximate definite integrals to which known techniques of integration do not apply. We will illustrate this in this exercise with the definite integral \(\int_0^1 \sin(x^2) \,ds\text{.}\)

1. Use the Taylor series for \(\sin(x)\) to find the Taylor series for \(\sin(x^2)\text{.}\) What is the interval of convergence for the Taylor series for \(\sin(x^2)\text{?}\) Explain.

2. Integrate the Taylor series for \(\sin(x^2)\) term by term to obtain a power series expansion for \(\int \sin(x^2)\,dx\text{.}\)

3. Use the result from part (b) to explain how to evaluate \(\int_0^1 \sin(x^2) \ dx\text{.}\) Determine the number of terms you will need to approximate \(\int_0^1 \sin(x^2) \,dx\) to 3 decimal places.

4.

There is an important connection between power series and Taylor series. Suppose \(f\) is defined by a power series centered at 0 so that

1. Determine the first 4 derivatives of \(f\) evaluated at 0 in terms of the coefficients \(a_k\text{.}\)

2. Show that \(f^{(n)}(0) = n!a_n\) for each positive integer \(n\text{.}\)

3. Explain how the result of (b) tells us the following:

   On its interval of convergence, a power series is the Taylor series of its sum.

5.

In this exercise we will begin with a strange power series and then find its sum. The Fibonacci sequence \(\{f_n\}\) is a famous sequence whose first few terms are

where each term in the sequence after the first two is the sum of the preceding two terms. That is, \(f_0 = 0\text{,}\) \(f_1 = 1\) and for \(n \geq 2\) we have

Now consider the power series

We will determine the sum of this power series in this exercise.

1. Explain why each of the following is true.

   1. \(\displaystyle xF(x) = \sum_{k=1}^{\infty} f_{k-1}x^k\)

   2. \(\displaystyle x^2F(x) = \sum_{k=2}^{\infty} f_{k-2}x^k\)

2. Show that

   \begin{equation*} F(x) - xF(x) - x^2F(x) = x\text{.} \end{equation*}

3. Now use the equation

   \begin{equation*} F(x) - xF(x) - x^2F(x) = x \end{equation*}

   to find a simple form for \(F(x)\) that doesn't involve a sum.

4. Use a computer algebra system or some other method to calculate the first 8 derivatives of \(\frac{x}{1-x-x^2}\) evaluated at 0. Why shouldn't the results surprise you?

6.

Airy's equation²

can be used to model an undamped vibrating spring with spring constant \(x\) (note that \(y\) is an unknown function of \(x\)). So the solution to this differential equation will tell us the behavior of a spring-mass system as the spring ages (like an automobile shock absorber). Assume that a solution \(y=f(x)\) has a Taylor series that can be written in the form

where the coefficients are undetermined. Our job is to find the coefficients.

1. Differentiate the series for \(y\) term by term to find the series for \(y'\text{.}\) Then repeat to find the series for \(y''\text{.}\)

2. Substitute your results from part (a) into the Airy equation and show that we can write Equation (8.6.4) in the form

   \begin{equation} \sum_{k=2}^{\infty} (k-1)ka_kx^{k-2} - \sum_{k=0}^{\infty} a_kx^{k+1} = 0\text{.}\tag{8.6.5} \end{equation}

3. At this point, it would be convenient if we could combine the series on the left in (8.6.5), but one written with terms of the form \(x^{k-2}\) and the other with terms in the form \(x^{k+1}\text{.}\) Explain why

   \begin{equation} \sum_{k=2}^{\infty} (k-1)ka_kx^{k-2} = \sum_{k=0}^{\infty} (k+1)(k+2)a_{k+2}x^{k}\text{.}\tag{8.6.6} \end{equation}

4. Now show that

   \begin{equation} \sum_{k=0}^{\infty} a_kx^{k+1} = \sum_{k=1}^{\infty} a_{k-1}x^k\text{.}\tag{8.6.7} \end{equation}

5. We can now substitute (8.6.6) and (8.6.7) into (8.6.5) to obtain

   \begin{equation} \sum_{n=0}^{\infty} (n+1)(n+2)a_{n+2}x^{n} - \sum_{n=1}^{\infty} a_{n-1}x^{n} = 0\text{.}\tag{8.6.8} \end{equation}

   Combine the like powers of \(x\) in the two series to show that our solution must satisfy

   \begin{equation} 2a_2 + \sum_{k=1}^{\infty} \left[(k+1)(k+2)a_{k+2}-a_{k-1} \right] x^{k} = 0\text{.}\tag{8.6.9} \end{equation}

6. Use equation (8.6.9) to show the following:

   1. \(a_{3k+2} = 0\) for every positive integer \(k\text{,}\)

   2. \(a_{3k} = \frac{1}{(2)(3)(5)(6) \cdots (3k-1)(3k)} a_0 \text{ for } k \geq 1\text{,}\)

   3. \(a_{3k+1} = \frac{1}{(3)(4)(6)(7) \cdots (3k)(3k+1)} a_1 \text{ for } k \geq 1\text{.}\)

7. Use the previous part to conclude that the general solution to the Airy equation (8.6.4) is

   \begin{align*} y \amp= a_0\left( 1+\sum_{k=1}^{\infty} \frac{x^{3k}}{(2)(3)(5)(6) \cdots (3k-1)(3k)} \right)\\ \amp\phantom{={}}+ a_1 \left( x + \sum_{k=1}^{\infty} \frac{x^{3k+1}}{(3)(4)(6)(7) \cdots (3k)(3k+1)} \right)\text{.} \end{align*}

   Any values for \(a_0\) and \(a_1\) then determine a specific solution that we can approximate as closely as we like using this series solution.