Taylor Series

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AP Calculus BC › Taylor Series

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$\begin{align*}&\text{Approximate }f(6) \&\text{Where }f(11)=-18;f'(11)=15;f''(11)=5;f'''(11)=-6\&\text{With a Taylor series expansion.}\end{align*}$

$\frac{189}{2}$

$-\frac{11}{2}$

$\frac{205}{12}$

Explanation

$\begin{align*}&\text{The Taylor series is a means of approximating a function value}\&\text{at given point, if values for the function and its derivatives}\&\text{are known at another point. It takes advantage of rates of change}\&\text{(i.e. function derivative values), and the distance between}\&\text{points on the x-axis to approximate changes along the y-axis.}\&\text{It is similar to using the slope of a line to find changes in}\&\text{y with regards to changes in x. The Taylor series follows the}\&\text{form:}\&f(x)\approx\sum_{i=0}^{n}\frac{f^{(i)}(a)}{i!}(x-a)^i\approx f(a)+f'(a)(x-a)+\frac{f''(a)}{2!}(x-a)^2+\frac{f^{(3)}(a)}{3!}(x-a)^3+...+\frac{f^{(n)}(a)}{n!}(x-a)^n\&\text{For the function values }f(11)=-18;f'(11)=15;f''(11)=5;f'''(11)=-6\&\text{ at }x=6\&\text{This expansion becomes:}\&f(6)\approx \frac{-18}{0!}(6-11)^{0}+\frac{15}{1!}(6-11)^{1}+\frac{5}{2!}(6-11)^{2}+\frac{-6}{3!}(6-11)^{3}\&f(6)\approx\frac{189}{2}\end{align*}$

$\begin{align*}&\text{Approximate }f(1) \&\text{Where }f(x)=- cos(5x) - sin(2x)\&\text{With a Taylor series centered at }\frac{(5\cdot \pi )}{2}\text{, using }3\text{ terms}\end{align*}$

Explanation

$\begin{align*}&\text{The Taylor series is a means of approximating a function value}\&\text{at given point, if values for the function and its derivatives}\&\text{are known at another point. It takes advantage of rates of change}\&\text{(i.e. function derivative values), and the distance between}\&\text{points on the x-axis to approximate changes along the y-axis.}\&\text{It is similar to using the slope of a line to find changes in}\&\text{y with regards to changes in x. The Taylor series follows the}\&\text{form:}\&f(x)\approx\sum_{i=0}^{n}\frac{f^{(i)}(a)}{i!}(x-a)^i\approx f(a)+f'(a)(x-a)+\frac{f''(a)}{2!}(x-a)^2+\frac{f^{(3)}(a)}{3!}(x-a)^3+...+\frac{f^{(n)}(a)}{n!}(x-a)^n\&\text{For the function }- cos(5x) - sin(2x)\text{ at }x=1\&\text{This expansion becomes:}\&f(x)\approx (- cos(5a) - sin(2a))+(5sin(5a) - 2cos(2a))(x-a)^{1}+\frac{25cos(5a) + 4sin(2a)}{2}(x-a)^{2}\&x=1\&a=\frac{(5\cdot \pi )}{2}\&f(1)\approx-47.98\end{align*}$

$\begin{align*}&\text{Approximate }f(3) \&\text{Where }f(8)=17;f'(8)=12;f''(8)=3\&\text{By using a Taylor expansion.}\end{align*}$

$-\frac{11}{2}$

$\frac{229}{2}$

$\frac{5}{2}$

Explanation

$\begin{align*}&\text{The Taylor series is a means of approximating a function value}\&\text{at given point, if values for the function and its derivatives}\&\text{are known at another point. It takes advantage of rates of change}\&\text{(i.e. function derivative values), and the distance between}\&\text{points on the x-axis to approximate changes along the y-axis.}\&\text{It is similar to using the slope of a line to find changes in}\&\text{y with regards to changes in x. The Taylor series follows the}\&\text{form:}\&f(x)\approx\sum_{i=0}^{n}\frac{f^{(i)}(a)}{i!}(x-a)^i\approx f(a)+f'(a)(x-a)+\frac{f''(a)}{2!}(x-a)^2+\frac{f^{(3)}(a)}{3!}(x-a)^3+...+\frac{f^{(n)}(a)}{n!}(x-a)^n\&\text{For the function values }f(8)=17;f'(8)=12;f''(8)=3\&\text{ at }x=3\&\text{This expansion becomes:}\&f(3)\approx \frac{17}{0!}(3-8)^{0}+\frac{12}{1!}(3-8)^{1}+\frac{3}{2!}(3-8)^{2}\&f(3)\approx-\frac{11}{2}\end{align*}$

$\begin{align*}&\text{For the function values: }\&f(9)=10;f'(9)=-16;f''(9)=7;f'''(9)=-2;f''''(9)=-12\&\text{Approximate }f(6)\text{ using a Taylor expansion.}\end{align*}$

$-\frac{2319}{20}$

Explanation

$\begin{align*}&\text{The Taylor series is a means of approximating a function value}\&\text{at given point, if values for the function and its derivatives}\&\text{are known at another point. It takes advantage of rates of change}\&\text{(i.e. function derivative values), and the distance between}\&\text{points on the x-axis to approximate changes along the y-axis.}\&\text{It is similar to using the slope of a line to find changes in}\&\text{y with regards to changes in x. The Taylor series follows the}\&\text{form:}\&f(x)\approx\sum_{i=0}^{n}\frac{f^{(i)}(a)}{i!}(x-a)^i\approx f(a)+f'(a)(x-a)+\frac{f''(a)}{2!}(x-a)^2+\frac{f^{(3)}(a)}{3!}(x-a)^3+...+\frac{f^{(n)}(a)}{n!}(x-a)^n\&\text{For the function values }f(9)=10;f'(9)=-16;f''(9)=7;f'''(9)=-2;f''''(9)=-12\&\text{ at }x=6\&\text{This expansion becomes:}\&f(6)\approx \frac{10}{0!}(6-9)^{0}+\frac{-16}{1!}(6-9)^{1}+\frac{7}{2!}(6-9)^{2}+\frac{-2}{3!}(6-9)^{3}+\frac{-12}{4!}(6-9)^{4}\&f(6)\approx58\end{align*}$

$\begin{align*}&\text{Approximate }f(3) \&\text{Where }f(8)=17;f'(8)=12;f''(8)=3\&\text{By using a Taylor expansion.}\end{align*}$

$-\frac{11}{2}$

$\frac{229}{2}$

$\frac{5}{2}$

Explanation

$\begin{align*}&\text{The Taylor series is a means of approximating a function value}\&\text{at given point, if values for the function and its derivatives}\&\text{are known at another point. It takes advantage of rates of change}\&\text{(i.e. function derivative values), and the distance between}\&\text{points on the x-axis to approximate changes along the y-axis.}\&\text{It is similar to using the slope of a line to find changes in}\&\text{y with regards to changes in x. The Taylor series follows the}\&\text{form:}\&f(x)\approx\sum_{i=0}^{n}\frac{f^{(i)}(a)}{i!}(x-a)^i\approx f(a)+f'(a)(x-a)+\frac{f''(a)}{2!}(x-a)^2+\frac{f^{(3)}(a)}{3!}(x-a)^3+...+\frac{f^{(n)}(a)}{n!}(x-a)^n\&\text{For the function values }f(8)=17;f'(8)=12;f''(8)=3\&\text{ at }x=3\&\text{This expansion becomes:}\&f(3)\approx \frac{17}{0!}(3-8)^{0}+\frac{12}{1!}(3-8)^{1}+\frac{3}{2!}(3-8)^{2}\&f(3)\approx-\frac{11}{2}\end{align*}$

$\begin{align*}&\text{Approximate }f(1) \&\text{Where }f(x)=- sin(6x) - sin(7x)\&\text{With a Taylor series centered at }\frac{(15\cdot \pi )}{4}\text{, using }3\text{ terms}\end{align*}$

Explanation

$\begin{align*}&\text{The Taylor series is a means of approximating a function value}\&\text{at given point, if values for the function and its derivatives}\&\text{are known at another point. It takes advantage of rates of change}\&\text{(i.e. function derivative values), and the distance between}\&\text{points on the x-axis to approximate changes along the y-axis.}\&\text{It is similar to using the slope of a line to find changes in}\&\text{y with regards to changes in x. The Taylor series follows the}\&\text{form:}\&f(x)\approx\sum_{i=0}^{n}\frac{f^{(i)}(a)}{i!}(x-a)^i\approx f(a)+f'(a)(x-a)+\frac{f''(a)}{2!}(x-a)^2+\frac{f^{(3)}(a)}{3!}(x-a)^3+...+\frac{f^{(n)}(a)}{n!}(x-a)^n\&\text{For the function }- sin(6x) - sin(7x)\text{ at }x=1\&\text{This expansion becomes:}\&f(x)\approx (- sin(6a) - sin(7a))+(- 6cos(6a) - 7cos(7a))(x-a)^{1}+\frac{36sin(6a) + 49sin(7a)}{2}(x-a)^{2}\&x=1\&a=\frac{(15\cdot \pi )}{4}\&f(1)\approx4157.36\end{align*}$

$\begin{align*}&\text{Approximate }f(-5) \&\text{Where }f(-2)=-18;f'(-2)=11;f''(-2)=7\&\text{With a Taylor series expansion.}\end{align*}$

$-\frac{39}{2}$

$\frac{93}{2}$

Explanation

$\begin{align*}&\text{The Taylor series is a means of approximating a function value}\&\text{at given point, if values for the function and its derivatives}\&\text{are known at another point. It takes advantage of rates of change}\&\text{(i.e. function derivative values), and the distance between}\&\text{points on the x-axis to approximate changes along the y-axis.}\&\text{It is similar to using the slope of a line to find changes in}\&\text{y with regards to changes in x. The Taylor series follows the}\&\text{form:}\&f(x)\approx\sum_{i=0}^{n}\frac{f^{(i)}(a)}{i!}(x-a)^i\approx f(a)+f'(a)(x-a)+\frac{f''(a)}{2!}(x-a)^2+\frac{f^{(3)}(a)}{3!}(x-a)^3+...+\frac{f^{(n)}(a)}{n!}(x-a)^n\&\text{For the function values }f(-2)=-18;f'(-2)=11;f''(-2)=7\&\text{ at }x=-5\&\text{This expansion becomes:}\&f(-5)\approx \frac{-18}{0!}(-5-(-2))^{0}+\frac{11}{1!}(-5-(-2))^{1}+\frac{7}{2!}(-5-(-2))^{2}\&f(-5)\approx-\frac{39}{2}\end{align*}$

$\begin{align*}&\text{For the function values: }\&f(-12)=6;f'(-12)=19;f''(-12)=18;f'''(-12)=-2\&\text{Approximate }f(-10)\text{ using a Taylor expansion.}\end{align*}$

$\frac{232}{3}$

$\frac{20}{3}$

$\frac{218}{3}$

Explanation

$\begin{align*}&\text{The Taylor series is a means of approximating a function value}\&\text{at given point, if values for the function and its derivatives}\&\text{are known at another point. It takes advantage of rates of change}\&\text{(i.e. function derivative values), and the distance between}\&\text{points on the x-axis to approximate changes along the y-axis.}\&\text{It is similar to using the slope of a line to find changes in}\&\text{y with regards to changes in x. The Taylor series follows the}\&\text{form:}\&f(x)\approx\sum_{i=0}^{n}\frac{f^{(i)}(a)}{i!}(x-a)^i\approx f(a)+f'(a)(x-a)+\frac{f''(a)}{2!}(x-a)^2+\frac{f^{(3)}(a)}{3!}(x-a)^3+...+\frac{f^{(n)}(a)}{n!}(x-a)^n\&\text{For the function values }f(-12)=6;f'(-12)=19;f''(-12)=18;f'''(-12)=-2\&\text{ at }x=-10\&\text{This expansion becomes:}\&f(-10)\approx \frac{6}{0!}(-10-(-12))^{0}+\frac{19}{1!}(-10-(-12))^{1}+\frac{18}{2!}(-10-(-12))^{2}+\frac{-2}{3!}(-10-(-12))^{3}\&f(-10)\approx\frac{232}{3}\end{align*}$

$\begin{align*}&\text{Approximate }f(3) \&\text{Where }f(2)=13;f'(2)=-7;f''(2)=13;f'''(2)=12\&\text{With a Taylor series expansion.}\end{align*}$

$\frac{29}{2}$

$\frac{49}{2}$

$\frac{73}{6}$

Explanation

$\begin{align*}&\text{The Taylor series is a means of approximating a function value}\&\text{at given point, if values for the function and its derivatives}\&\text{are known at another point. It takes advantage of rates of change}\&\text{(i.e. function derivative values), and the distance between}\&\text{points on the x-axis to approximate changes along the y-axis.}\&\text{It is similar to using the slope of a line to find changes in}\&\text{y with regards to changes in x. The Taylor series follows the}\&\text{form:}\&f(x)\approx\sum_{i=0}^{n}\frac{f^{(i)}(a)}{i!}(x-a)^i\approx f(a)+f'(a)(x-a)+\frac{f''(a)}{2!}(x-a)^2+\frac{f^{(3)}(a)}{3!}(x-a)^3+...+\frac{f^{(n)}(a)}{n!}(x-a)^n\&\text{For the function values }f(2)=13;f'(2)=-7;f''(2)=13;f'''(2)=12\&\text{ at }x=3\&\text{This expansion becomes:}\&f(3)\approx \frac{13}{0!}(3-2)^{0}+\frac{-7}{1!}(3-2)^{1}+\frac{13}{2!}(3-2)^{2}+\frac{12}{3!}(3-2)^{3}\&f(3)\approx\frac{29}{2}\end{align*}$

$\begin{align*}&\text{Approximate }f(1) \&\text{Where }f(x)=- sin(6x) - sin(7x)\&\text{With a Taylor series centered at }\frac{(15\cdot \pi )}{4}\text{, using }3\text{ terms}\end{align*}$

Explanation

$\begin{align*}&\text{The Taylor series is a means of approximating a function value}\&\text{at given point, if values for the function and its derivatives}\&\text{are known at another point. It takes advantage of rates of change}\&\text{(i.e. function derivative values), and the distance between}\&\text{points on the x-axis to approximate changes along the y-axis.}\&\text{It is similar to using the slope of a line to find changes in}\&\text{y with regards to changes in x. The Taylor series follows the}\&\text{form:}\&f(x)\approx\sum_{i=0}^{n}\frac{f^{(i)}(a)}{i!}(x-a)^i\approx f(a)+f'(a)(x-a)+\frac{f''(a)}{2!}(x-a)^2+\frac{f^{(3)}(a)}{3!}(x-a)^3+...+\frac{f^{(n)}(a)}{n!}(x-a)^n\&\text{For the function }- sin(6x) - sin(7x)\text{ at }x=1\&\text{This expansion becomes:}\&f(x)\approx (- sin(6a) - sin(7a))+(- 6cos(6a) - 7cos(7a))(x-a)^{1}+\frac{36sin(6a) + 49sin(7a)}{2}(x-a)^{2}\&x=1\&a=\frac{(15\cdot \pi )}{4}\&f(1)\approx4157.36\end{align*}$

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