Welcome to AP Calc 10.11! In this lesson, you’ll learn how to approximate a function over at a point.
This theorem states that for a function f ( x ) f(x) f ( x ) x = a x=a x = a
∑ n = 0 ∞ f ( n ) ( a ) n ! ⋅ ( x − a ) n \sum_{n=0}^\infty \frac{f^{(n)}(a)}{n!}\cdot(x-a)^n n = 0 ∑ ∞ n ! f ( n ) ( a ) ⋅ ( x − a ) n This can be rewritten as…
f ( a ) + f ′ ( a ) ( x − a ) + f ′ ′ ( a ) 2 ! ( x − a ) 2 + f ′ ′ ′ ( a ) 3 ! ( x − a ) 3 + . . . + f ( n ) ( a ) n ! ( x − a ) n 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 f ( a ) + f ′ ( a ) ( x − a ) + 2 ! f ′′ ( a ) ( x − a ) 2 + 3 ! f ′′′ ( a ) ( x − a ) 3 + ... + n ! f ( n ) ( a ) ( x − a ) n where f ( n ) ( a ) f^{(n)}(a) f ( n ) ( a ) n th n^{\text{th}} n th f ( 0 ) ( a ) = f ( x ) f^{(0)}(a)=f(x) f ( 0 ) ( a ) = f ( x ) n th n^{\text{th}} n th n th n^{\text{th}} n th
Taylor series centered at x = 0 x=0 x = 0 Maclaurin series .
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!
n n ! f n ( x ) f n ( a ) ( x − a ) n f n ( a ) n ! ⋅ ( x − a ) n 0 1 f ( x ) f ( a ) ( x − a ) 0 f ( a ) 1 ⋅ ( x − a ) 0 1 1 f ′ ( x ) f ′ ( a ) ( x − a ) 1 f ′ ( a ) 1 ⋅ ( x − a ) 1 2 2 f ′ ′ ( x ) f ′ ′ ( a ) ( x − a ) 2 f ′ ′ ( a ) 2 ⋅ ( x − a ) 2 3 6 f ′ ′ ′ ( x ) f ′ ′ ′ ( a ) ( x − a ) 3 f ′ ′ ′ ( a ) 6 ⋅ ( x − a ) 3 . . . . . . . . . . . . . . . . . . n n ! f n ( x ) f n ( a ) ( x − a ) n f n ( a ) n ! ⋅ ( x − 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} n 0 1 2 3 ... n n ! 1 1 2 6 ... n ! f n ( x ) f ( x ) f ′ ( x ) f ′′ ( x ) f ′′′ ( x ) ... f n ( x ) f n ( a ) f ( a ) f ′ ( a ) f ′′ ( a ) f ′′′ ( a ) ... f n ( a ) ( x − a ) n ( x − a ) 0 ( x − a ) 1 ( x − a ) 2 ( x − a ) 3 ... ( x − a ) n n ! f n ( a ) ⋅ ( x − a ) n 1 f ( a ) ⋅ ( x − a ) 0 1 f ′ ( a ) ⋅ ( x − a ) 1 2 f ′′ ( a ) ⋅ ( x − a ) 2 6 f ′′′ ( a ) ⋅ ( x − a ) 3 ... n ! f n ( a ) ⋅ ( x − a ) n Now, let’s try a practice problem using this table to walk through it step by step.
Find the third-degree Maclaurin polynomial for e 5 x e^{5x} e 5 x
Solution: First, let’s build our table. Remember that a Maclaurin series is just a Taylor series where a = 0 a=0 a = 0
n n ! f n ( x ) f n ( a ) ( x − a ) n f n ( a ) n ! ⋅ ( x − a ) n 0 1 e 5 x 1 1 1 1 1 5 e 5 x 5 x 5 x 2 2 25 e 5 x 25 x 2 25 x 2 / 2 3 6 125 e 5 x 125 x 3 125 x 3 / 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} n 0 1 2 3 n ! 1 1 2 6 f n ( x ) e 5 x 5 e 5 x 25 e 5 x 125 e 5 x f n ( a ) 1 5 25 125 ( x − a ) n 1 x x 2 x 3 n ! f n ( a ) ⋅ ( x − a ) n 1 5 x 25 x 2 /2 125 x 3 /6 Now, we just put the terms in our final column together as a full formula. The third-degree Maclaurin polynomial for e 5 x e^{5x} e 5 x
1 + 5 x + 25 2 x 2 + 125 6 x 3 1+5x+\frac{25}{2}x^2+\frac{125}{6}x^3 1 + 5 x + 2 25 x 2 + 6 125 x 3 Now it’s your turn to apply what you’ve learned!
❓Problems
Find the fifth-degree Maclaurin polynomial for f ( x ) = cos ( x ) f(x)=\text{cos}(x) f ( x ) = cos ( x )
Find the third-degree Taylor polynomial for f ( x ) = ln ( x ) f(x)=\text{ln}(x) f ( x ) = ln ( x ) x = 1 x=1 x = 1
Find the fourth-degree Taylor polynomial about x = 2 x=2 x = 2 f ( x ) = x f(x)=\sqrt{x} f ( x ) = x
💡 Solution for Question 1
Start by building your table and filling in the values:
n n ! f n ( x ) f n ( a ) ( x − a ) n f n ( a ) n ! ⋅ ( x − a ) n 0 1 cos ( x ) 1 1 1 1 1 − sin ( x ) 0 x 0 2 2 − cos ( x ) − 1 x 2 − x 2 / 2 3 6 sin ( x ) 0 x 3 0 4 24 cos ( x ) 1 x 4 x 4 / 24 5 120 − sin ( x ) 0 x 5 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 & \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} n 0 1 2 3 4 5 n ! 1 1 2 6 24 120 f n ( x ) cos ( x ) − sin ( x ) − cos ( x ) sin ( x ) cos ( x ) − sin ( x ) f n ( a ) 1 0 − 1 0 1 0 ( x − a ) n 1 x x 2 x 3 x 4 x 5 n ! f n ( a ) ⋅ ( x − a ) n 1 0 − x 2 /2 0 x 4 /24 0 Putting it all together, we get that the fifth-degree Maclaurin polynomial for f ( x ) = cos ( x ) f(x)=\text{cos}(x) f ( x ) = cos ( x )
1 − x 2 2 + x 2 4 1-\frac{x^2}{2}+\frac{x^2}{4} 1 − 2 x 2 + 4 x 2 💡 Solution for Question 2
Keep on building your tables! This time, our ( x − a ) n (x-a)^n ( x − a ) n
n n ! f n ( x ) f n ( a ) ( x − a ) n f n ( a ) n ! ⋅ ( x − a ) n 0 1 ln ( x ) 0 1 0 1 1 1 / x 1 ( x − 1 ) ( x − 1 ) 2 2 − 1 / x 2 − 1 ( x − 1 ) 2 − 1 2 ( x − 1 ) 2 3 6 2 / x 3 2 / 3 ( x − 1 ) 3 1 9 ( x − 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} n 0 1 2 3 n ! 1 1 2 6 f n ( x ) ln ( x ) 1/ x − 1/ x 2 2/ x 3 f n ( a ) 0 1 − 1 2/3 ( x − a ) n 1 ( x − 1 ) ( x − 1 ) 2 ( x − 1 ) 3 n ! f n ( a ) ⋅ ( x − a ) n 0 ( x − 1 ) − 2 1 ( x − 1 ) 2 9 1 ( x − 1 ) 3 We then find the polynomial to be equal to:
( x − 1 ) − 1 2 ( x − 1 ) 2 + 1 9 ( x − 1 ) 3 (x-1)-\frac{1}{2}(x-1)^2+\frac{1}{9}(x-1)^3 ( x − 1 ) − 2 1 ( x − 1 ) 2 + 9 1 ( x − 1 ) 3 💡 Solution for Question 3
One more table!
n n ! f n ( x ) f n ( a ) ( x − a ) n f n ( a ) n ! ⋅ ( x − a ) n 0 1 x 2 1 2 1 1 1 2 x 1 2 2 ( x − 2 ) 1 2 2 ⋅ ( x − 2 ) 2 2 − 1 4 x 3 − 1 4 8 ( x − 2 ) 2 − 1 8 8 ⋅ ( x − 2 ) 2 3 6 3 8 x 5 3 8 32 ( x − 2 ) 3 3 48 32 ⋅ ( x − 2 ) 3 4 24 − 15 16 x 7 − 15 16 128 ( x − 2 ) 4 − 15 384 128 ⋅ ( x − 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} n 0 1 2 3 4 n ! 1 1 2 6 24 f n ( x ) x 2 x 1 − 4 x 3 1 8 x 5 3 − 16 x 7 15 f n ( a ) 2 2 2 1 − 4 8 1 8 32 3 − 16 128 15 ( x − a ) n 1 ( x − 2 ) ( x − 2 ) 2 ( x − 2 ) 3 ( x − 2 ) 4 n ! f n ( a ) ⋅ ( x − a ) n 2 2 2 1 ⋅ ( x − 2 ) − 8 8 1 ⋅ ( x − 2 ) 2 48 32 3 ⋅ ( x − 2 ) 3 − 384 128 15 ⋅ ( x − 2 ) 4 If we put this all together, we get:
2 + 1 2 2 ⋅ ( x − 2 ) − 1 8 8 ⋅ ( x − 2 ) 2 + 3 48 32 ⋅ ( x − 2 ) 3 − 15 384 128 ⋅ ( x − 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 2 + 2 2 1 ⋅ ( x − 2 ) − 8 8 1 ⋅ ( x − 2 ) 2 + 48 32 3 ⋅ ( x − 2 ) 3 − 384 128 15 ⋅ ( x − 2 ) 4 We can simplify a few of these terms a bit more using exponent rules to get:
2 + x − 2 2 2 − ( x − 2 ) 2 16 2 + ( x − 2 ) 3 64 2 − 5 ( x − 2 ) 4 1024 2 \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}} 2 + 2 2 x − 2 − 16 2 ( x − 2 ) 2 + 64 2 ( x − 2 ) 3 − 1024 2 5 ( x − 2 ) 4 This is the fourth-degree Taylor polynomial centered at x = 2 x=2 x = 2 2 \sqrt{2} 2
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!
