🚧 This is an AP Calculus BC topic only! If you are taking Calculus AB, you can skip this material. If you’re taking AP Calculus BC, here you go! ⬇️

📈 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

📝 Practice

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

❓Problems

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

💫 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!

Key Terms to Review (5)

Function Approximations: Function approximations are mathematical techniques used to estimate the value of a function at a particular point or within a certain range. They involve using simpler functions, such as polynomials, to closely mimic the behavior of the original function.

P-Series: A p-series is a series of the form Σ(1/n^p), where n starts from 1 and goes to infinity, and p is a positive constant. It converges if the value of p is greater than 1, and diverges if the value of p is less than or equal to 1.

Power Series: A power series is an infinite series that represents a function as an infinite polynomial expression.

Tangent Line Approximation: Tangent line approximation, also known as linear approximation or tangent line estimation, is a method that uses the equation of a tangent line at a specific point on a curve to approximate the value of the function near that point. It provides a close estimate when dealing with small intervals.

Taylor Polynomial: A Taylor polynomial is a polynomial approximation of a function centered around a specific point. It is used to estimate the value of the function at nearby points.

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♾️ap calculus ab/bc review

10.11 Finding Taylor Polynomial Approximations of Functions