Lesson 19 - Taylor Polynomial, Power Series and Shifting Index

Let the function $y (x)$ have derivatives of all orders in an interval centered at the point $x_{0}$ . The $n$ th degree Taylor Polynomial centered at $x_{0}$ is

P_{n} (x) = y (x_{0}) + y^{'} (x_{0}) (x - x_{0}) + \frac{y^{″} (x_{0})}{2!} (x - x_{0})^{2} + \frac{y^{‴} (x_{0})}{3!} (x - x_{0})^{3} + \dots + \frac{y^{(n)} (x_{0})}{n!} (x - x_{0})^{n}

= \sum_{k = 0}^{n} \frac{y^{(k)} (x_{0})}{k!} (x - x_{0})^{k}

There's a lot going on here. We are essentially beginning by using the initial conditions and finding each subsequent term in the polynomial by utilizing the derivatives of the given differential equation.

Example 1

Find the third degree polynomial approximating the solution of the differential equation.

$y^{'} = \sin y + e^{x}$ , $y (0) = 0$

First, notice that our DE contains $x$ and $y$ and we are given the initial conditions. Let's solve for $y^{'} (0)$ .

$y^{'} (0) = \sin (0) + e^{(0)}$
$y^{'} (0) = 1$

Next, we will find the second and third derivative of the equation, recalling chain rule from fundamental calculus.

$y^{″} (x) = \cos (y) y^{'} + e^{x}$

Using our initial conditions and the condition we found above, let's solve for $y^{″} (0)$ .

$y^{″} (0) = \cos (0) (1) + e^{(0)}$
$y^{″} (0) = 2$

Lastly, let's do the same with the third derivative.

$y^{‴} (x) = - \sin (y) (y^{'})^{2} + \cos (y) y^{″} + e^{x}$
$y^{‴} (0) = - \sin (0) (1)^{2} + \cos (0) (2) + e^{(0)}$
$y^{‴} (0) = 3$

Now we have the first $3$ terms for the solution approximation, using the method above.

$P_{3} (x) = 0 + x + \frac{2 x^{2}}{2!} + \frac{3 x^{3}}{3!}$

The Taylor polynomial $P_{3} (x)$ gives a way to approximate the solution $y (x)$ of the differential equation near $x = 0$ . It uses the initial condition and the first three derivatives of $y (x)$ to describe how the solution behaves close to that point. Each term in the polynomial adds more detail about how $y (x)$ changes, but the further you move from $x = 0$ , the less accurate the approximation becomes. This method is especially helpful when finding an exact solution is too difficult, providing a simple way to estimate $y (x)$ around the starting point.

Try for Yourself

Determine the first three non-zero terms of the solution of $y^{'} = x^{2} + y^{2}$ , $y (0) = 1$ .
Find the first three non-zero terms approximating the solution of the differential equation: $y^{″} + \sin y = 0$ , $y (0) = 1$ , $y^{'} (0) = 0$ .
Find the fourth degree polynomial approximating the solution of the differential equation: $y^{″} + 3 y^{'} + x^{2} y = 0$ , $y (0) = 10$ , $y^{'} (0) = 5$ .

Power Series

A power series has the general form $\sum_{n = 0}^{\infty} a_{n} (x - x_{0})^{n}$ . In the general form, the index of the series begins at $n = 0$ . For instances in which the starting index does not start at 0, we must implement a shift of index.

Shifting the index of a power series means changing where the summation starts to make the terms easier to work with. This is done by replacing the index $n$ with something like $k + c$ , where $c$ is a constant. This adjustment reorganizes the terms so that the powers of $x$ or the coefficients match up, which is helpful when combining series or solving equations.

Example

Express the following series as a series where the generic terms is $x^{k}$ .

$\sum_{n = 2}^{\infty} n (n - 1) a_{n} x^{n - 2}$

We want $x^{n - 2}$ to be equivalent to $x^{k}$ . So, we will substitute $n - 2 = k$ . Now, we can replace $x^{n - 2}$ with $x^{k}$ . We also have a relation between $n$ and $k$ . Since $n - 2 = k$ , we can say that $n = k + 2$ . Now, let's substitute all values of $n$ .

$\sum_{n = 2}^{\infty} n (n - 1) a_{n} x^{n - 2}$ $\to$ $\sum_{k + 2 = 2}^{\infty} (k + 2) ((k + 2) - 1) a_{k + 2} x^{(k)}$ $\to$ $\sum_{k = 0}^{\infty} (k + 2) (k + 1) a_{k + 2} x^{k}$

Notice the index shifted from $n = 2$ to $k = 0$ .

Try for Yourself

$\sum_{n = 1}^{\infty} \frac{a_{n}}{n + 3} x^{n + 3}$
$\sum_{n = 1}^{\infty} n a_{n} x^{n - 1}$

We can also show that two summations are the same.

Example

Show that $x^{2} \sum_{n = 0}^{\infty} n (n + 1) a_{n} x^{n} = \sum_{n = 2}^{\infty} (n - 2) (n - 1) a_{n - 2} x^{n}$

Make $n = n - 2$ $\to$ $n + 2 = n$

$x^{2} \sum_{(n - 2) = 0}^{\infty} (n - 2) ((n - 2) + 1) a_{n + 2} x^{(n - 2)}$ $\to$ $x^{2} \sum_{n = 2}^{\infty} (n - 2) ((n - 1) a_{n + 2} x^{(n - 2)}$

Now, distribute $x^{2}$ .

$\sum_{n = 2}^{\infty} (n - 2) ((n - 1) a_{n + 2} x^{n}$ $✓$

Try for Yourself

Show that $\sum_{n = 1}^{\infty} n a_{n} x^{n - 1} + \sum_{n = 0}^{\infty} a_{n} x^{n + 1}$ $=$ $a_{1} + \sum_{n = 0}^{\infty} [(n + 2) a_{n + 2} + a_{n} x^{n + 1}]$ .

Next Lesson: Lesson 20 - Power Series Solutions

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