10.11 Finding Taylor Polynomial Approximations of Functions

10.11 Finding Taylor Polynomial Approximations of Functions

Welcome to AP Calc 10.11! In this lesson, you’ll learn how to approximate a function over at a point.

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📈 Taylor Approximations Theorem

This theorem states that for a function $f(x)$, it’s Taylor series approximation at $x=a$ is…

$$\sum_{n=0}^\infty \frac{f^{(n)}(a)}{n!}\cdot(x-a)^n$$

This can be rewritten as…

$$f(a)+f'(a)(x-a)+\frac{f''(a)}{2!}(x-a)^2+\frac{f'''(a)}{3!}(x-a)^3+...+\frac{f^{(n)}(a)}{n!}(x-a)^n$$

where $f^{(n)}(a)$ is the $n^{\text{th}}$ deriviative of the function and $f^{(0)}(a)=f(x)$. The $n^{\text{th}}$-order Taylor polynomial is the $n^{\text{th}}$ partial sum of the infinite series.

Taylor series centered at $x=0$ are common and are called Maclaurin series.

🧱 Breaking Down the Theorem

Taylor series look very daunting when you first approach them. Let’s define each portion and build a table that will help you tackle problems of this type!

$$\begin{array}{ |c|c|c|c|c|c| } \hline n & n! & f^n(x) & f^n(a)&(x-a)^n & \frac{f^n(a)}{n!}\cdot(x-a)^n \\ 0 & 1 & f(x) & f(a)&(x-a)^0 & \frac{f(a)}{1}\cdot(x-a)^0 \\ 1 & 1 & f'(x) & f'(a)&(x-a)^1 & \frac{f'(a)}{1}\cdot(x-a)^1 \\ 2 & 2 & f''(x) & f''(a)&(x-a)^2 & \frac{f''(a)}{2}\cdot(x-a)^2 \\ 3 & 6 & f'''(x) & f'''(a)&(x-a)^3 & \frac{f'''(a)}{6}\cdot(x-a)^3 \\ ... & ... & ... & ... & ... & ... \\ n & n! & f^n(x) & f^n(a)&(x-a)^n & \frac{f^n(a)}{n!}\cdot(x-a)^n \\ \hline \end{array}$$

Now, let’s try a practice problem using this table to walk through it step by step.

✏️ Applying the Theorem

Find the third-degree Maclaurin polynomial for $e^{5x}$.

Solution: First, let’s build our table. Remember that a Maclaurin series is just a Taylor series where $a=0$!

$$\begin{array}{ |c|c|c|c|c|c| } \hline n & n! & f^n(x) & f^n(a)&(x-a)^n & \frac{f^n(a)}{n!}\cdot(x-a)^n \\ 0 & 1 & e^{5x} & 1&1 & 1 \\ 1 & 1 & 5e^{5x} & 5&x & 5x \\ 2 & 2 & 25e^{5x} & 25&x^2 & 25x^2/2 \\ 3 & 6 & 125e^{5x} & 125&x^3 & 125x^3/6 \\ \hline \end{array}$$

Now, we just put the terms in our final column together as a full formula. The third-degree Maclaurin polynomial for $e^{5x}$ is:

$$1+5x+\frac{25}{2}x^2+\frac{125}{6}x^3$$

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📝 Practice

Now it’s your turn to apply what you’ve learned!

❓Problems

1. Find the fifth-degree Maclaurin polynomial for $f(x)=\text{cos}(x)$.
2. Find the third-degree Taylor polynomial for $f(x)=\text{ln}(x)$ about $x=1$.
3. Find the fourth-degree Taylor polynomial about $x=2$ for $f(x)=\sqrt{x}$.

💡 Solution for Question 1

Start by building your table and filling in the values:

$$\begin{array}{ |c|c|c|c|c|c| } \hline n & n! & f^n(x) & f^n(a)&(x-a)^n & \frac{f^n(a)}{n!}\cdot(x-a)^n \\ 0 & 1 & \text{cos}(x) & 1&1 & 1 \\ 1 & 1 & -\text{sin}(x) & 0&x & 0 \\ 2 & 2 & -\text{cos}(x) & -1&x^2 & -x^2/2 \\ 3 & 6 & \text{sin}(x) & 0&x^3 & 0 \\ 4 & 24 & \text{cos}(x) & 1&x^4 & x^4/24 \\ 5 & 120 & -\text{sin}(x) & 0&x^5 & 0 \\ \hline \end{array}$$

Putting it all together, we get that the fifth-degree Maclaurin polynomial for $f(x)=\text{cos}(x)$ is

$$1-\frac{x^2}{2}+\frac{x^2}{4}$$

💡 Solution for Question 2

Keep on building your tables! This time, our $(x-a)^n$ column will be a bit more complicated.

$$\begin{array}{ |c|c|c|c|c|c| } \hline n & n! & f^n(x) & f^n(a)&(x-a)^n & \frac{f^n(a)}{n!}\cdot(x-a)^n \\ 0 & 1 & \text{ln}(x) & 0&1 & 0 \\ 1 & 1 & 1/x & 1&(x-1) & (x-1) \\ 2 & 2 & -1/x^2 & -1&(x-1)^2 & -\frac{1}{2}(x-1)^2 \\ 3 & 6 & 2/x^3 & 2/3&(x-1)^3 & \frac{1}{9}(x-1)^3 \\ \hline \end{array}$$

We then find the polynomial to be equal to:

$$(x-1)-\frac{1}{2}(x-1)^2+\frac{1}{9}(x-1)^3$$

💡 Solution for Question 3

One more table!

$$\begin{array}{ |c|c|c|c|c|c| } \hline n & n! & f^n(x) & f^n(a)&(x-a)^n & \frac{f^n(a)}{n!}\cdot(x-a)^n \\ 0 & 1 & \sqrt{x} & \sqrt{2}&1 & \sqrt{2} \\ 1 & 1 & \frac{1}{2\sqrt{x}} & \frac{1}{2\sqrt{2}}&(x-2) & \frac{1}{2\sqrt{2}}\cdot(x-2)\\ 2 & 2 & -\frac{1}{4\sqrt{x^3}} & -\frac{1}{4\sqrt{8}}&(x-2)^2 & -\frac{1}{8\sqrt{8}}\cdot(x-2)^2 \\ 3 & 6 & \frac{3}{8\sqrt{x^5}} & \frac{3}{8\sqrt{32}}&(x-2)^3 & \frac{3}{48\sqrt{32}}\cdot (x-2)^3 \\ 4 & 24 & -\frac{15}{16\sqrt{x^7}} & -\frac{15}{16\sqrt{128}}&(x-2)^4 & -\frac{15}{384\sqrt{128}}\cdot(x-2)^4 \\ \hline \end{array}$$

If we put this all together, we get:

$$\sqrt{2}+\frac{1}{2\sqrt{2}}\cdot(x-2)-\frac{1}{8\sqrt{8}}\cdot(x-2)^2+\frac{3}{48\sqrt{32}}\cdot (x-2)^3 -\frac{15}{384\sqrt{128}}\cdot(x-2)^4$$

We can simplify a few of these terms a bit more using exponent rules to get:

$$\sqrt{2}+\frac{x-2}{2\sqrt{2}}-\frac{(x-2)^2}{16\sqrt{2}}+\frac{(x-2)^3}{64\sqrt{2}} -\frac{5(x-2)^4}{1024\sqrt{2}}$$

This is the fourth-degree Taylor polynomial centered at $x=2$ for $\sqrt{2}$.

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💫 Closing

Great work! Taylor polynomials may seem daunting at first, but when in doubt, break it down with a table and you’ll be sure to master them!