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10.11 Finding Taylor Polynomial Approximations of Functions

1 min readβ€’june 18, 2024

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.


πŸ“ˆΒ Taylor Approximations Theorem

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

βˆ‘n=0∞f(n)(a)n!β‹…(xβˆ’a)n\sum_{n=0}^\infty \frac{f^{(n)}(a)}{n!}\cdot(x-a)^n

This can be rewritten as…

f(a)+fβ€²(a)(xβˆ’a)+fβ€²β€²(a)2!(xβˆ’a)2+fβ€²β€²β€²(a)3!(xβˆ’a)3+...+f(n)(a)n!(xβˆ’a)nf(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)f^{(n)}(a) is the nthn^{\text{th}} deriviative of the function and f(0)(a)=f(x)f^{(0)}(a)=f(x). The nthn^{\text{th}}-order Taylor polynomial is the nthn^{\text{th}} partial sum of the infinite series.

Taylor series centered at x=0x=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!

nn!fn(x)fn(a)(xβˆ’a)nfn(a)n!β‹…(xβˆ’a)n01f(x)f(a)(xβˆ’a)0f(a)1β‹…(xβˆ’a)011fβ€²(x)fβ€²(a)(xβˆ’a)1fβ€²(a)1β‹…(xβˆ’a)122fβ€²β€²(x)fβ€²β€²(a)(xβˆ’a)2fβ€²β€²(a)2β‹…(xβˆ’a)236fβ€²β€²β€²(x)fβ€²β€²β€²(a)(xβˆ’a)3fβ€²β€²β€²(a)6β‹…(xβˆ’a)3..................nn!fn(x)fn(a)(xβˆ’a)nfn(a)n!β‹…(xβˆ’a)n\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 e5xe^{5x}.

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

nn!fn(x)fn(a)(xβˆ’a)nfn(a)n!β‹…(xβˆ’a)n01e5x111115e5x5x5x2225e5x25x225x2/236125e5x125x3125x3/6\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 e5xe^{5x} is:

1+5x+252x2+1256x31+5x+\frac{25}{2}x^2+\frac{125}{6}x^3

πŸ“Β Practice

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

❓Problems

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

πŸ’‘Β Solution for Question 1

Start by building your table and filling in the values:

nn!fn(x)fn(a)(xβˆ’a)nfn(a)n!β‹…(xβˆ’a)n01cos(x)11111βˆ’sin(x)0x022βˆ’cos(x)βˆ’1x2βˆ’x2/236sin(x)0x30424cos(x)1x4x4/245120βˆ’sin(x)0x50\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)=cos(x)f(x)=\text{cos}(x) is

1βˆ’x22+x241-\frac{x^2}{2}+\frac{x^2}{4}

πŸ’‘Β Solution for Question 2

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

nn!fn(x)fn(a)(xβˆ’a)nfn(a)n!β‹…(xβˆ’a)n01ln(x)010111/x1(xβˆ’1)(xβˆ’1)22βˆ’1/x2βˆ’1(xβˆ’1)2βˆ’12(xβˆ’1)2362/x32/3(xβˆ’1)319(xβˆ’1)3\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:

(xβˆ’1)βˆ’12(xβˆ’1)2+19(xβˆ’1)3(x-1)-\frac{1}{2}(x-1)^2+\frac{1}{9}(x-1)^3

πŸ’‘Β Solution for Question 3

One more table!

nn!fn(x)fn(a)(xβˆ’a)nfn(a)n!β‹…(xβˆ’a)n01x2121112x122(xβˆ’2)122β‹…(xβˆ’2)22βˆ’14x3βˆ’148(xβˆ’2)2βˆ’188β‹…(xβˆ’2)23638x53832(xβˆ’2)334832β‹…(xβˆ’2)3424βˆ’1516x7βˆ’1516128(xβˆ’2)4βˆ’15384128β‹…(xβˆ’2)4\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:

2+122β‹…(xβˆ’2)βˆ’188β‹…(xβˆ’2)2+34832β‹…(xβˆ’2)3βˆ’15384128β‹…(xβˆ’2)4\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:

2+xβˆ’222βˆ’(xβˆ’2)2162+(xβˆ’2)3642βˆ’5(xβˆ’2)410242\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=2x=2 for 2\sqrt{2}.


πŸ’«Β 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!

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10.11 Finding Taylor Polynomial Approximations of Functions

1 min readβ€’june 18, 2024

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.


πŸ“ˆΒ Taylor Approximations Theorem

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

βˆ‘n=0∞f(n)(a)n!β‹…(xβˆ’a)n\sum_{n=0}^\infty \frac{f^{(n)}(a)}{n!}\cdot(x-a)^n

This can be rewritten as…

f(a)+fβ€²(a)(xβˆ’a)+fβ€²β€²(a)2!(xβˆ’a)2+fβ€²β€²β€²(a)3!(xβˆ’a)3+...+f(n)(a)n!(xβˆ’a)nf(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)f^{(n)}(a) is the nthn^{\text{th}} deriviative of the function and f(0)(a)=f(x)f^{(0)}(a)=f(x). The nthn^{\text{th}}-order Taylor polynomial is the nthn^{\text{th}} partial sum of the infinite series.

Taylor series centered at x=0x=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!

nn!fn(x)fn(a)(xβˆ’a)nfn(a)n!β‹…(xβˆ’a)n01f(x)f(a)(xβˆ’a)0f(a)1β‹…(xβˆ’a)011fβ€²(x)fβ€²(a)(xβˆ’a)1fβ€²(a)1β‹…(xβˆ’a)122fβ€²β€²(x)fβ€²β€²(a)(xβˆ’a)2fβ€²β€²(a)2β‹…(xβˆ’a)236fβ€²β€²β€²(x)fβ€²β€²β€²(a)(xβˆ’a)3fβ€²β€²β€²(a)6β‹…(xβˆ’a)3..................nn!fn(x)fn(a)(xβˆ’a)nfn(a)n!β‹…(xβˆ’a)n\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 e5xe^{5x}.

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

nn!fn(x)fn(a)(xβˆ’a)nfn(a)n!β‹…(xβˆ’a)n01e5x111115e5x5x5x2225e5x25x225x2/236125e5x125x3125x3/6\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 e5xe^{5x} is:

1+5x+252x2+1256x31+5x+\frac{25}{2}x^2+\frac{125}{6}x^3

πŸ“Β Practice

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

❓Problems

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

πŸ’‘Β Solution for Question 1

Start by building your table and filling in the values:

nn!fn(x)fn(a)(xβˆ’a)nfn(a)n!β‹…(xβˆ’a)n01cos(x)11111βˆ’sin(x)0x022βˆ’cos(x)βˆ’1x2βˆ’x2/236sin(x)0x30424cos(x)1x4x4/245120βˆ’sin(x)0x50\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)=cos(x)f(x)=\text{cos}(x) is

1βˆ’x22+x241-\frac{x^2}{2}+\frac{x^2}{4}

πŸ’‘Β Solution for Question 2

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

nn!fn(x)fn(a)(xβˆ’a)nfn(a)n!β‹…(xβˆ’a)n01ln(x)010111/x1(xβˆ’1)(xβˆ’1)22βˆ’1/x2βˆ’1(xβˆ’1)2βˆ’12(xβˆ’1)2362/x32/3(xβˆ’1)319(xβˆ’1)3\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:

(xβˆ’1)βˆ’12(xβˆ’1)2+19(xβˆ’1)3(x-1)-\frac{1}{2}(x-1)^2+\frac{1}{9}(x-1)^3

πŸ’‘Β Solution for Question 3

One more table!

nn!fn(x)fn(a)(xβˆ’a)nfn(a)n!β‹…(xβˆ’a)n01x2121112x122(xβˆ’2)122β‹…(xβˆ’2)22βˆ’14x3βˆ’148(xβˆ’2)2βˆ’188β‹…(xβˆ’2)23638x53832(xβˆ’2)334832β‹…(xβˆ’2)3424βˆ’1516x7βˆ’1516128(xβˆ’2)4βˆ’15384128β‹…(xβˆ’2)4\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:

2+122β‹…(xβˆ’2)βˆ’188β‹…(xβˆ’2)2+34832β‹…(xβˆ’2)3βˆ’15384128β‹…(xβˆ’2)4\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:

2+xβˆ’222βˆ’(xβˆ’2)2162+(xβˆ’2)3642βˆ’5(xβˆ’2)410242\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=2x=2 for 2\sqrt{2}.


πŸ’«Β 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!