Taylor Polynomial Approximation

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AP Calculus BC › Taylor Polynomial Approximation

Questions 1 - 10

$\begin{align*}&\text{Approximate }f(6) \&\text{Where }f(x)=- cos(9x) - cos(x)\&\text{With a Taylor series centered at }-\frac{(5\cdot \pi )}{2}\text{, using }3\text{ terms}\end{align*}$

Explanation

$\begin{align*}&\text{The Taylor series is a means of approximating a function value}\&\text{at given point, if values for the function and its derivatives}\&\text{are known at another point. It takes advantage of rates of change}\&\text{(i.e. function derivative values), and the distance between}\&\text{points on the x-axis to approximate changes along the y-axis.}\&\text{It is similar to using the slope of a line to find changes in}\&\text{y with regards to changes in x. The Taylor series follows the}\&\text{form:}\&f(x)\approx\sum_{i=0}^{n}\frac{f^{(i)}(a)}{i!}(x-a)^i\approx f(a)+f'(a)(x-a)+\frac{f''(a)}{2!}(x-a)^2+\frac{f^{(3)}(a)}{3!}(x-a)^3+...+\frac{f^{(n)}(a)}{n!}(x-a)^n\&\text{For the function }- cos(9x) - cos(x)\text{ at }x=6\&\text{This expansion becomes:}\&f(x)\approx (- cos(9a) - cos(a))+(9sin(9a) + sin(a))(x-a)^{1}+\frac{81cos(9a) + cos(a)}{2}(x-a)^{2}\&x=6\&a=-\frac{(5\cdot \pi )}{2}\&f(6)\approx-138.54\end{align*}$

$\begin{align*}&\text{For the function: }f(x)=cos(x) +\frac{ 1}{cos(8x)}\&\text{Approximate }f(5)\&\text{Using a Taylor expansion centered at }-4\cdot \pi\text{ to }3\text{ terms}\end{align*}$

Explanation

$\begin{align*}\&\text{form:}\&f(x)\approx\sum_{i=0}^{n}\frac{f^{(i)}(a)}{i!}(x-a)^i\approx f(a)+f'(a)(x-a)+\frac{f''(a)}{2!}(x-a)^2+\frac{f^{(3)}(a)}{3!}(x-a)^3+...+\frac{f^{(n)}(a)}{n!}(x-a)^n\&\text{For the function }cos(x) +\frac{ 1}{cos(8x)}\text{ at }x=5\&\text{This expansion becomes:}\&f(x)\approx (cos(a) +\frac{ 1}{cos(8a)})+(\frac{(8sin(8a))}{cos(8a)^{2}}- sin(a))(x-a)^{1}+\frac{\frac{64}{cos(8a)}- cos(a) +\frac{ (128sin(8a)^{2})}{cos(8a)^{3}}}{2}(x-a)^{2}\&x=5\&a=-4\cdot \pi\&f(5)\approx9722.19\end{align*}$

$\begin{align*}&\text{Approximate }f(9) \&\text{Where }f(x)=-\frac{cos(8x)}{cos(4x)}\&\text{By using a Taylor expansion to }2\text{ terms, centered at }\frac{(5\cdot \pi )}{4}\end{align*}$

Explanation

$\begin{align*}&\text{The Taylor series is a means of approximating a function value}\&\text{at given point, if values for the function and its derivatives}\&\text{are known at another point. It takes advantage of rates of change}\&\text{(i.e. function derivative values), and the distance between}\&\text{points on the x-axis to approximate changes along the y-axis.}\&\text{It is similar to using the slope of a line to find changes in}\&\text{y with regards to changes in x. The Taylor series follows the}\&\text{form:}\&f(x)\approx\sum_{i=0}^{n}\frac{f^{(i)}(a)}{i!}(x-a)^i\approx f(a)+f'(a)(x-a)+\frac{f''(a)}{2!}(x-a)^2+\frac{f^{(3)}(a)}{3!}(x-a)^3+...+\frac{f^{(n)}(a)}{n!}(x-a)^n\&\text{For the function }-\frac{cos(8x)}{cos(4x)}\text{ at }x=9\&\text{This expansion becomes:}\&f(x)\approx (-\frac{cos(8a)}{cos(4a)})+(\frac{(8sin(8a))}{cos(4a)}-\frac{ (4cos(8a)sin(4a))}{cos(4a)^{2}})(x-a)^{1}\&x=9\&a=\frac{(5\cdot \pi )}{4}\&f(9)\approx1.00\end{align*}$

$\begin{align*}&\text{Approximate }f(5) \&\text{Where }f(x)=sin(6x) - cos(2x)\&\text{By using a Taylor expansion to }3\text{ terms, centered at }\pi\end{align*}$

Explanation

$\begin{align*}&\text{The Taylor series is a means of approximating a function value}\&\text{at given point, if values for the function and its derivatives}\&\text{are known at another point. It takes advantage of rates of change}\&\text{(i.e. function derivative values), and the distance between}\&\text{points on the x-axis to approximate changes along the y-axis.}\&\text{It is similar to using the slope of a line to find changes in}\&\text{y with regards to changes in x. The Taylor series follows the}\&\text{form:}\&f(x)\approx\sum_{i=0}^{n}\frac{f^{(i)}(a)}{i!}(x-a)^i\approx f(a)+f'(a)(x-a)+\frac{f''(a)}{2!}(x-a)^2+\frac{f^{(3)}(a)}{3!}(x-a)^3+...+\frac{f^{(n)}(a)}{n!}(x-a)^n\&\text{For the function }sin(6x) - cos(2x)\text{ at }x=5\&\text{This expansion becomes:}\&f(x)\approx (sin(6a) - cos(2a))+(6cos(6a) + 2sin(2a))(x-a)^{1}+\frac{4cos(2a) - 36sin(6a)}{2}(x-a)^{2}\&x=5\&a=\pi\&f(5)\approx17.06\end{align*}$

$\begin{align*}&\text{For the function: }f(x)=-cos(4x)^{2}\&\text{Approximate }f(3)\&\text{Using a Taylor expansion centered at }6\cdot \pi\text{ to }2\text{ terms}\end{align*}$

Explanation

$\begin{align*}&\text{The Taylor series is a means of approximating a function value}\&\text{at given point, if values for the function and its derivatives}\&\text{are known at another point. It takes advantage of rates of change}\&\text{(i.e. function derivative values), and the distance between}\&\text{points on the x-axis to approximate changes along the y-axis.}\&\text{It is similar to using the slope of a line to find changes in}\&\text{y with regards to changes in x. The Taylor series follows the}\&\text{form:}\&f(x)\approx\sum_{i=0}^{n}\frac{f^{(i)}(a)}{i!}(x-a)^i\approx f(a)+f'(a)(x-a)+\frac{f''(a)}{2!}(x-a)^2+\frac{f^{(3)}(a)}{3!}(x-a)^3+...+\frac{f^{(n)}(a)}{n!}(x-a)^n\&\text{For the function }-cos(4x)^{2}\text{ at }x=3\&\text{This expansion becomes:}\&f(x)\approx (-cos(4a)^{2})+(8cos(4a)sin(4a))(x-a)^{1}\&x=3\&a=6\cdot \pi\&f(3)\approx-1.00\end{align*}$

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

Explanation

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

$\begin{align*}&\text{For the function: }f(x)=-cos(6x)sin(2x)\&\text{Approximate }f(9)\&\text{Using a Taylor expansion centered at }-\frac{(3\cdot \pi )}{2}\text{ to }2\text{ terms}\end{align*}$

Explanation

$\begin{align*}&\text{The Taylor series is a means of approximating a function value}\&\text{at given point, if values for the function and its derivatives}\&\text{are known at another point. It takes advantage of rates of change}\&\text{(i.e. function derivative values), and the distance between}\&\text{points on the x-axis to approximate changes along the y-axis.}\&\text{It is similar to using the slope of a line to find changes in}\&\text{y with regards to changes in x. The Taylor series follows the}\&\text{form:}\&f(x)\approx\sum_{i=0}^{n}\frac{f^{(i)}(a)}{i!}(x-a)^i\approx f(a)+f'(a)(x-a)+\frac{f''(a)}{2!}(x-a)^2+\frac{f^{(3)}(a)}{3!}(x-a)^3+...+\frac{f^{(n)}(a)}{n!}(x-a)^n\&\text{For the function }-cos(6x)sin(2x)\text{ at }x=9\&\text{This expansion becomes:}\&f(x)\approx (-cos(6a)sin(2a))+(6sin(2a)sin(6a) - 2cos(2a)cos(6a))(x-a)^{1}\&x=9\&a=-\frac{(3\cdot \pi )}{2}\&f(9)\approx-27.42\end{align*}$

$\begin{align*}&\text{Approximate }f(-7) \&\text{Where }f(-2)=7;f'(-2)=-19;f''(-2)=13;f'''(-2)=10;f''''(-2)=-16\&\text{By using a Taylor expansion.}\end{align*}$

$-\frac{721}{2}$

$-\frac{803}{6}$

$\frac{535}{4}$

Explanation

$\begin{align*}\&\text{y with regards to changes in x. The Taylor series follows the}\&\text{form:}\&f(x)\approx\sum_{i=0}^{n}\frac{f^{(i)}(a)}{i!}(x-a)^i\approx f(a)+f'(a)(x-a)+\frac{f''(a)}{2!}(x-a)^2+\frac{f^{(3)}(a)}{3!}(x-a)^3+...+\frac{f^{(n)}(a)}{n!}(x-a)^n\&\text{For the function values }f(-2)=7;f'(-2)=-19;f''(-2)=13;f'''(-2)=10;f''''(-2)=-16\&\text{ at }x=-7\&\text{This expansion becomes:}\&f(-7)\approx \frac{7}{0!}(-7-(-2))^{0}+\frac{-19}{1!}(-7-(-2))^{1}+\frac{13}{2!}(-7-(-2))^{2}+\frac{10}{3!}(-7-(-2))^{3}+\frac{-16}{4!}(-7-(-2))^{4}\&f(-7)\approx-\frac{721}{2}\end{align*}$

$\begin{align*}&\text{Approximate }f(7) \&\text{Where }f(3)=7;f'(3)=16;f''(3)=1\&\text{With a Taylor series expansion.}\end{align*}$

$\frac{500}{3}$

Explanation

$\begin{align*}&\text{The Taylor series is a means of approximating a function value}\&\text{at given point, if values for the function and its derivatives}\&\text{are known at another point. It takes advantage of rates of change}\&\text{(i.e. function derivative values), and the distance between}\&\text{points on the x-axis to approximate changes along the y-axis.}\&\text{It is similar to using the slope of a line to find changes in}\&\text{y with regards to changes in x. The Taylor series follows the}\&\text{form:}\&f(x)\approx\sum_{i=0}^{n}\frac{f^{(i)}(a)}{i!}(x-a)^i\approx f(a)+f'(a)(x-a)+\frac{f''(a)}{2!}(x-a)^2+\frac{f^{(3)}(a)}{3!}(x-a)^3+...+\frac{f^{(n)}(a)}{n!}(x-a)^n\&\text{For the function values }f(3)=7;f'(3)=16;f''(3)=1\&\text{ at }x=7\&\text{This expansion becomes:}\&f(7)\approx \frac{7}{0!}(7-3)^{0}+\frac{16}{1!}(7-3)^{1}+\frac{1}{2!}(7-3)^{2}\&f(7)\approx79\end{align*}$

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

$\frac{189}{2}$

$-\frac{11}{2}$

$\frac{205}{12}$

Explanation

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

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