If the second derivative test reveals that a critical point corresponds to a local minimum value, what can you infer about the sign on the coefficient at $(x - c)^n$ in its nth-degree Taylor polynomial expansion if n is even?

2. It could be either positive or negative, depending on other terms in the expansion.

For question number five, what part of functions is used to determine coefficients in Taylor series?

The function and its derivatives

Just the initial function value

First few terms in the power series expression

What is the fourth-degree Taylor polynomial centered at $x = 0$ for the function $f(x) = \frac{1}{1-x}$?

$-x - \frac{x^2}{2} - \frac{x^3}{3} - \frac{x^4}{4}

$\ln(1+x) + \frac{x}{2}+\frac{x^2}{3}+\frac{x^3}{4}

Using L'Hôpital's Rule, what limit comparison could you use to determine whether a function has a finite non-zero radius convergence when expanded as a power series around $x=0$?

The limit of $\frac{a_{n+1}}{a_n}$ as $n$ approaches infinity.

The sum from $k=1$ to infinity $a_kx^{k-1}$

The integral from $0$ to $1$ $f(t)dt$ where $f$ is our function.

The limit as $x$ approaches infinity of $\frac{f(x)}{g(x)}$, where $g$ is another function.

To determine if the graph of a given polar equation intersects itself how should you proceed?

By checking whether different values of theta lead to the same set of radial distances r

By converting each point into rectangular coordinates first

Assume it never crosses because that's rare in polar coordinates

Simplifying the equation so that it does not depend on theta

Which represents a possible fifth-degree Taylor polynomial approximation for $sin(x)$ centered at zero?

$x - \frac{x^3}{6} + \frac{x^5}{120}

1. $x - \frac{x^4 }{24}+ \frac {x ^5 } {120 } $

2. $\sin (0)+\cos (0)x-\sin (0)\cdot\!\!\left(\!\dfrac { x } { 2 }\right)^{\!2}\,+o(x^{5}) $

3. $\sin(0) + \cos(0)x - \sin(0)\cdot\dfrac { x ^5 } {120 } $

When approximating an oscillating function like $\cos(x)$ near $x = \frac{\pi}{4}$ with a fourth degree Maclaurin polynomial, which strategy can reduce error significantly?

Use more terms in the Maclaurin series approximation beyond fourth degree to capture more oscillations accurately.

Omit all odd-powered terms as they do not contribute to an oscillating function’s behavior near $x=\frac{\pi}{4}$.

Which of these represents a fourth-degree Maclaurin polynomial for $\sin(x)$?

$\sum_{n=0}^{4}\dfrac{(-1)^nx^{2n+1}}{(2n+1)!}$

$\sin(x) - \dfrac{\sin''(0)x^2}{2!}+\dfrac{\sin''''(0)x^4}{4!}$

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10.11 Finding Taylor Polynomial Approximations of Functions

1 min read•december 14, 2020

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

📈 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.

10.11 Finding Taylor Polynomial Approximations of Functions

1 min read•december 14, 2020

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

📈 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

p-Series

Power Series

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

Tangent Line Approximation

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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