Taylor Series - AP Calculus BC

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Question

Let $P_{5}(x)$ be the fifth-degree Taylor polynomial approximation for , centered at .

What is the Lagrange error of the polynomial approximation to ?

Answer

The fifth degree Taylor polynomial approximating centered at is:

$sin(x) \approx x - \frac{x^3}{3!} + \frac{x^5}{5!}$

The Lagrange error is the absolute value of the next term in the sequence, which is equal to $\left |\frac{x^7}{7!} \right |$ .

We need only evaluate this at and thus we obtain $\frac{1^7}{7!} = \frac{1}{7!}$

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Question

Let $P_{5}(x)$ be the fifth-degree Taylor polynomial approximation for , centered at .

What is the Lagrange error of the polynomial approximation to ?

Answer

The fifth degree Taylor polynomial approximating centered at is:

$sin(x) \approx x - \frac{x^3}{3!} + \frac{x^5}{5!}$

The Lagrange error is the absolute value of the next term in the sequence, which is equal to $\left |\frac{x^7}{7!} \right |$ .

We need only evaluate this at and thus we obtain $\frac{1^7}{7!} = \frac{1}{7!}$

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Question

For which of the following functions can the Maclaurin series representation be expressed in four or fewer non-zero terms?

Answer

Recall the Maclaurin series formula:

$f(x)=\sum_{n=0}^{\infty}\frac{f^n(0)}{n!}x^n$

$=f(0)+f'(0)x+\frac{f''(0)}{2!}x^2+\frac{f'''(0)}{3!}x^3+...$

Despite being a 5th degree polynomial recall that the Maclaurin series for any polynomial is just the polynomial itself, so this function's Taylor series is identical to itself with two non-zero terms.

The only function that has four or fewer terms is $f(x)= \frac{x^{5}}{3}+2$ as its Maclaurin series is $2+\frac{1}{3}x^5$ .

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Question

Let $f(x)=(1+\frac{x}{2})^{1/3}$

Find the the first three terms of the Taylor Series for centered at .

Answer

Using the formula of a binomial series centered at 0:

$(1+x)^k =\sum_{n=0}^{\infty } \binom{k}{n}x^n$ ,

where we replace with $\frac{x}{2}$ and $k=\frac{1}{3}$ , we get:

$\sum_{n=0}^{2}\binom{\frac{1}{3}}{n}\left(\frac{x}{2}\right)^n$ for the first 3 terms.

Then, we find the terms where,

$T_2(x)=\binom{\frac{1}{3}}{0}(\frac{x}{2})^0 + \binom{\frac{1}{3}}{1}(\frac{x}{2})^1+\binom{\frac{1}{3}}{2}(\frac{x}{2})^2=1+\frac{x}{6}- \frac{x^2}{36}$

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Question

Write the first three terms of the Taylor series for the following function about :

Answer

The general form for the Taylor series (of a function f(x)) about x=a is the following:

$\sum_{n=0}^{\infty }\frac{f^{(n)}(a)}{n!}(x-a)^n$ .

Because we only want the first three terms, we simply plug in a=1, and then n=0, 1, and 2 for the first three terms (starting at n=0).

The hardest part, then, is finding the zeroth, first, and second derivative of the function given:

$f^{0}(1)=f(1)=6$

The derivative was found using the following rules:

$\frac{d}{dx}(x^n)=nx^{n-1}$

Then, simply plug in the remaining information and write out the terms:

$\frac{6(x-1)^0}{0!}+\frac{5(x-1)^1}{1!}+\frac{2(x-1)^2}{2!}=6+5(x-1)+(x-1)^2$

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Question

Write out the first four terms of the Taylor series about for the following function:

Answer

The Taylor series about x=a of any function is given by the following:

$\sum_{n=0}^{\infty }f^{(n)}(a)\frac{(x-a)^n}{n!}$

So, we must find the zeroth, first, second, and third derivatives of the function (for n=0, 1, 2, and 3 which makes the first four terms):

$f^{0}(x)=f(x)$

The derivatives were found using the following rule:

$\frac{d}{dx}(x^n)=nx^{n-1}$

Now, evaluated at x=a=1, and plugging in the correct n where appropriate, we get the following:

$\frac{4(x-1)^0}{0!}+\frac{4(x-1)^1}{1!}+\frac{2(x-1)^2}{2!}+\frac{0(x-1)^3}{3!}$

which when simplified is equal to

.

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Question

Write out the first three terms of the Taylor series about for the following function:

Answer

The general formula for the Taylor series about x=a for a given function is

$\sum_{n=0}^{\infty }f^{(n)}(a)\frac{(x-a)^n}{n!}$

We must find the zeroth, first, and second derivative of the function (for n=0, 1, and 2). The zeroth derivative is just the function itself.

The derivatives were found using the following rule:
$\frac{\mathrm{d} }{\mathrm{d} x}x^n=nx^{n-1}$

Now, follow the above formula to write out the first three terms:

$\frac{11(x-1)^0}{0!}+\frac{21(x-1)^1}{1!}+\frac{20(x-1)^2}{2!}$

which simplified becomes

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Question

Find the Taylor series expansion of $\frac{1}{1-x^{2}}$ at .

Answer

The Taylor series is defined as

$f(x)=\sum_{k=0}^{\infty }\frac{f^{k}(x_{0})}{k!}(x-x_{0})^{k}$ where the expression within the summation is the nth term of the Taylor polynomial.

To find the Taylor series expansion of $\frac{1}{1-x}$ at , we first need to find an expression for the nth term of the Taylor polynomial.

The nth term of the Taylor polynomial is defined as

$p_{n}x=f(x_{0})+f'(x_{0})(x-x_{0})+\frac{f''(x_{0})}{2!}(x-x_{0})^{2}+\frac{f'''(x_{0})}{3!}(x-x_{0})^{3}+....+\frac{f^{n}(x_{0})}{n!}(x-x_{0})^{n}$

For, $f(x)=\frac{1}{1-x}$ and $x_{0}=0$ .

We need to find terms in the taylor polynomial until we can determine the pattern for the nth term.

$f(0)=\frac{1}{1-0}=1$

$f'(x)=\frac{d}{dx}\left [ \frac{1}{1-x} \right ]=\frac{d}{dx}\left [ \left ( 1-x\right )^{-1} ]=-1( 1-x\right )^{-2}(-1)=( 1-x)^{-2}$

$f'(0)=-1( 1-(0) )^{-2}=1=1!$

$f''(x)=\frac{d}{dx}[( 1-x)^{-2}]=-2( 1-x)^{-3}(-1)=2( 1-x)^{-3}$

$f''(0)=2( 1-0)^{-3}=2=2!$

$f'''(x)=\frac{d}{dx}[2( 1-x)^{-3}]=-6( 1-x)^{-4}(-1)=6( 1-x)^{-4}$

$f'''(0)=6( 1-0)^{-4}=6=3!$

Substituting these values into the taylor polynomial we get

$p_{n}x=1+1(x)+\frac{2!}{2!}(x)^{2}+\frac{3!}{3!}(x)^{3}+....+\frac{f^{n}(0)}{n!}(x)^{n}$

$p_{n}x=1+(x)+(x)^{2}+(x)^{3}+....+\frac{f^{n}(0)}{n!}(x)^{n}$

The Taylor polynomial simplifies to

$p_{n}x=1+(x)+(x)^{2}+(x)^{3}+....+(x)^{n}$

Now that we know the nth term of the Taylor polynomial, we can find the Taylor series.

The Taylor series is defined as

$f(x)=\sum_{k=0}^{\infty }\frac{f^{k}(x_{0})}{k!}(x-x_{0})^{k}$ where the expression within the summation is the nth term of the Taylor polynomial.

For this problem, the taylor series is

$\frac{1}{1-x}=\sum_{k=0}^{\infty }(x)^{k}=1+x+(x)^{2}+(x)^{3}+...+(x)^{n}+...$

$\frac{1}{1-x}=\sum_{k=0}^{\infty }(x)^{k}=1+x+(x)^{2}+(x)^{3}+...+(x)^{n}+...$

Use the above to calculate the taylor series expansion of $\frac{1}{1-x^{2}}$ at ,

$\frac{1}{1-x^{2}}=\sum_{k=0}^{\infty }(x^{2})^{k}=\sum_{k=0}^{\infty }(x)^{2k}=1+(x)^{2}+(x)^{4}+...+(x)^{2n}+...\textup{}$

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Question

Use the first 3 terms of the Maclaurin series to approximate $\sin (\frac{\pi}{8})$

Answer

The Maclaurin series for sine is $\sum_{n=0}^{\infty } \frac{(-1)^n}{(2n+1)!}x^{2n+1} = x - \frac{x^3}{3!} + \frac{x^5}{5! } - ...$

Plugging in $\frac{\pi }{ 8 }$ for x gives:

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Question

Use the first 4 terms of the Maclaurin series to approximate $e^{0.78}$

Answer

The Maclaurin series for is $\sum_{n=0}^{\infty} \frac{x^n}{n!} = 1 + x + \frac{x^2 }{2! } + \frac{x^3}{3!} + ...$

Plugging in 0.78 for x gives:

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Question

Use the first four terms of the Maclaurin series to approximate $e^{0.38}$ to 6 decimal places.

Answer

The Maclaurin series for is $\sum_{n=0}^{\infty } \frac{x^n}{n!} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + ...$

Plugging in x = 0.38 gives:

$1 + 0.38 + .0722 + .09145 \overline{3} = 1.461345$

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Question

$\begin{align}&\text{Approximate }f(1.2) \&\text{Where }f(x)=-cos(x)\&\text{By using a Maclaurin expansion to }3\text{ terms}\end{align}$

Answer

$\begin{align}&\text{The Maclaurin series is a special case of the Taylor series.}\&\text{Like the Taylor series, it is used to approximate an unknown}\&\text{value of a function at a given point, if values of the function}\&\text{and its derivatives are known at another point. In the case}\&\text{of the Maclaurin series, this latter point has the value of}\&\text{zero (note that for many functions, such as the trigonometric}\&\text{functions, f(0) is readily known):}\&f(x)\approx\sum_{i=0}^{n}\frac{f^{(i)}(0)}{i!}x^i\approx f(0)+f'(0)x+\frac{f''(0)}{2!}x^2+\frac{f^{(3)}(0)}{3!}x^3+...+\frac{f^{(n)}(0)}{n!}x^n\&\text{For the function }-cos(x)\text{ at }x=1.2\&\text{This expansion becomes:}\&f(1.2)\approx \frac{-cos((0))}{1}(1.2)^{0}+\frac{sin((0))}{1}(1.2)^{1}+\frac{cos((0))}{2}(1.2)^{2}\&f(1.2)\approx-0.28\end{align}$

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Question

$\begin{align}&\text{Approximate }f(4) \&\text{Where }f(x)=-2^{(3x)}cos(9x)\&\text{With a Taylor series centered at zero, using }3\text{ terms}\end{align}$

Answer

$\begin{align}&\text{The Maclaurin series is a special case of the Taylor series.}\&\text{Like the Taylor series, it is used to approximate an unknown}\&\text{value of a function at a given point, if values of the function}\&\text{and its derivatives are known at another point. In the case}\&\text{of the Maclaurin series, this latter point has the value of}\&\text{zero (note that for many functions, such as the trigonometric}\&\text{functions, f(0) is readily known):}\&f(x)\approx\sum_{i=0}^{n}\frac{f^{(i)}(0)}{i!}x^i\approx f(0)+f'(0)x+\frac{f''(0)}{2!}x^2+\frac{f^{(3)}(0)}{3!}x^3+...+\frac{f^{(n)}(0)}{n!}x^n\&\text{For the function }-2^{(3x)}cos(9x)\text{ at }x=4\&\text{This expansion becomes:}\&f(4)\approx \frac{-2^{(3(0))}cos(9(0))}{1}(4)^{0}+\frac{9\cdot 2^{(3(0))}sin(9(0)) - 3\cdot 2^{(3(0))}cos(9(0))ln(2)}{1}(4)^{1}+\frac{81\cdot 2^{(3(0))}cos(9(0)) + 54\cdot 2^{(3(0))}sin(9(0))ln(2) - 9\cdot 2^{(3(0))}cos(9(0))ln(2)^{2}}{2}(4)^{2}\&f(4)\approx604.09\end{align}$

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Question

$\begin{align}&\text{Approximate }f(2.5) \&\text{Where }f(x)=-cos(6x)e^{(3x)}\&\text{With a Taylor series centered at zero, using }3\text{ terms}\end{align}$

Answer

$\begin{align}&\text{The Maclaurin series is a special case of the Taylor series.}\&\text{Like the Taylor series, it is used to approximate an unknown}\&\text{value of a function at a given point, if values of the function}\&\text{and its derivatives are known at another point. In the case}\&\text{of the Maclaurin series, this latter point has the value of}\&\text{zero (note that for many functions, such as the trigonometric}\&\text{functions, f(0) is readily known):}\&f(x)\approx\sum_{i=0}^{n}\frac{f^{(i)}(0)}{i!}x^i\approx f(0)+f'(0)x+\frac{f''(0)}{2!}x^2+\frac{f^{(3)}(0)}{3!}x^3+...+\frac{f^{(n)}(0)}{n!}x^n\&\text{For the function }-cos(6x)e^{(3x)}\text{ at }x=2.5\&\text{This expansion becomes:}\&f(2.5)\approx \frac{-cos(6(0))e^{(3(0))}}{1}(2.5)^{0}+\frac{6sin(6(0))e^{(3(0))} - 3cos(6(0))e^{(3(0))}}{1}(2.5)^{1}+\frac{27cos(6(0))e^{(3(0))} + 36sin(6(0))e^{(3(0))}}{2}(2.5)^{2}\&f(2.5)\approx75.88\end{align}$

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Question

$\begin{align}&\text{Approximate }f(2.9) \&\text{Where }f(x)=-3^{(4x)}cos(9x)\&\text{With a Taylor series centered at zero, using }3\text{ terms}\end{align}$

Answer

$\begin{align}&\text{The Maclaurin series is a special case of the Taylor series.}\&\text{Similarly, it is used to approximate an unknown function value,}\&\text{if values of the function and its derivatives are known at a}\&\text{particular point. In the case of the Maclaurin series, this}\&\text{ point has the value of zero (for many functions f(0) is readily}\&\text{known):}\&f(x)\approx\sum_{i=0}^{n}\frac{f^{(i)}(0)}{i!}x^i\approx f(0)+f'(0)x+\frac{f''(0)}{2!}x^2+\frac{f^{(3)}(0)}{3!}x^3+...+\frac{f^{(n)}(0)}{n!}x^n\&\text{For the function }-3^{(4x)}cos(9x)\text{ at }x=2.9\&\text{This expansion becomes:}\&f(2.9)\approx (-3^{(4(0))}cos(9(0)))(2.9)^{0}+(9\cdot 3^{(4(0))}sin(9(0)) - 4\cdot 3^{(4(0))}cos(9(0))ln(3))(2.9)^{1}+\frac{81\cdot 3^{(4(0))}cos(9(0)) + 72\cdot 3^{(4(0))}sin(9(0))ln(3) - 16\cdot 3^{(4(0))}cos(9(0))ln(3)^{2}}{2}(2.9)^{2}\&f(2.9)\approx245.66\end{align}$

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Question

$\begin{align}&\text{For the function: }f(x)=cos(4x)sin(x)\&\text{Approximate }f(3.1)\&\text{Using a Maclaurin expansion to }3\text{ terms}\end{align}$

Answer

$\begin{align}&\text{The Maclaurin series is a special case of the Taylor series.}\&\text{Similarly, it is used to approximate an unknown function value,}\&\text{if values of the function and its derivatives are known at a}\&\text{particular point. In the case of the Maclaurin series, this}\&\text{ point has the value of zero (for many functions f(0) is readily}\&\text{known):}\&f(x)\approx\sum_{i=0}^{n}\frac{f^{(i)}(0)}{i!}x^i\approx f(0)+f'(0)x+\frac{f''(0)}{2!}x^2+\frac{f^{(3)}(0)}{3!}x^3+...+\frac{f^{(n)}(0)}{n!}x^n\&\text{For the function }cos(4x)sin(x)\text{ at }x=3.1\&\text{This expansion becomes:}\&f(3.1)\approx (cos(4(0))sin((0)))(3.1)^{0}+(cos(4(0))cos((0)) - 4sin(4(0))sin((0)))(3.1)^{1}+\frac{- 17cos(4(0))sin((0)) - 8sin(4(0))cos((0))}{2}(3.1)^{2}\&f(3.1)\approx3.10\end{align}$

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Question

$\begin{align}&\text{Approximate }f(1.8) \&\text{Where }f(x)=-cos(5x)e^{(-3x)}\&\text{By using a Maclaurin expansion to }3\text{ terms}\end{align}$

Answer

$\begin{align}&\text{The Maclaurin series is a special case of the Taylor series.}\&\text{Similarly, it is used to approximate an unknown function value,}\&\text{if values of the function and its derivatives are known at a}\&\text{particular point. In the case of the Maclaurin series, this}\&\text{ point has the value of zero (for many functions f(0) is readily}\&\text{known):}\&f(x)\approx\sum_{i=0}^{n}\frac{f^{(i)}(0)}{i!}x^i\approx f(0)+f'(0)x+\frac{f''(0)}{2!}x^2+\frac{f^{(3)}(0)}{3!}x^3+...+\frac{f^{(n)}(0)}{n!}x^n\&\text{For the function }-cos(5x)e^{(-3x)}\text{ at }x=1.8\&\text{This expansion becomes:}\&f(1.8)\approx (-cos(5(0))e^{(-3(0))})(1.8)^{0}+(3cos(5(0))e^{(-3(0))} + 5sin(5(0))e^{(-3(0))})(1.8)^{1}+\frac{16cos(5(0))e^{(-3(0))} - 30sin(5(0))e^{(-3(0))}}{2}(1.8)^{2}\&f(1.8)\approx30.32\end{align}$

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Question

$\begin{align}&\text{Approximate }f(3.4) \&\text{Where }f(x)=-sin(5x)sin(6x)\&\text{By using a Maclaurin expansion to }3\text{ terms}\end{align}$

Answer

$\begin{align}&\text{The Maclaurin series is a special case of the Taylor series.}\&\text{Similarly, it is used to approximate an unknown function value,}\&\text{if values of the function and its derivatives are known at a}\&\text{particular point. In the case of the Maclaurin series, this}\&\text{ point has the value of zero (for many functions f(0) is readily}\&\text{known):}\&f(x)\approx\sum_{i=0}^{n}\frac{f^{(i)}(0)}{i!}x^i\approx f(0)+f'(0)x+\frac{f''(0)}{2!}x^2+\frac{f^{(3)}(0)}{3!}x^3+...+\frac{f^{(n)}(0)}{n!}x^n\&\text{For the function }-sin(5x)sin(6x)\text{ at }x=3.4\&\text{This expansion becomes:}\&f(3.4)\approx (-sin(5(0))sin(6(0)))(3.4)^{0}+(- 5cos(5(0))sin(6(0)) - 6cos(6(0))sin(5(0)))(3.4)^{1}+\frac{61sin(5(0))sin(6(0)) - 60cos(5(0))cos(6(0))}{2}(3.4)^{2}\&f(3.4)\approx-346.80\end{align}$

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Question

$\begin{align}&\text{Approximate }f(3) \&\text{Where }f(x)=-cos(5x)e^{(-3x)}\&\text{With a Taylor series centered at zero, using }3\text{ terms}\end{align}$

Answer

$\begin{align}&\text{The Maclaurin series is a special case of the Taylor series.}\&\text{Similarly, it is used to approximate an unknown function value,}\&\text{if values of the function and its derivatives are known at a}\&\text{particular point. In the case of the Maclaurin series, this}\&\text{ point has the value of zero (for many functions f(0) is readily}\&\text{known):}\&f(x)\approx\sum_{i=0}^{n}\frac{f^{(i)}(0)}{i!}x^i\approx f(0)+f'(0)x+\frac{f''(0)}{2!}x^2+\frac{f^{(3)}(0)}{3!}x^3+...+\frac{f^{(n)}(0)}{n!}x^n\&\text{For the function }-cos(5x)e^{(-3x)}\text{ at }x=3\&\text{This expansion becomes:}\&f(3)\approx (-cos(5(0))e^{(-3(0))})(3)^{0}+(3cos(5(0))e^{(-3(0))} + 5sin(5(0))e^{(-3(0))})(3)^{1}+\&\frac{16cos(5(0))e^{(-3(0))} - 30sin(5(0))e^{(-3(0))}}{2}(3)^{2}\&f(3)\approx80.00\end{align}$

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Question

$\begin{align}&\text{Approximate }f(2.1) \&\text{Where }f(x)=-sin(3x)sin(7x)\&\text{By using a Maclaurin expansion to }3\text{ terms}\end{align}$

Answer

$\begin{align}&\text{The Maclaurin series is a special case of the Taylor series.}\&\text{Similarly, it is used to approximate an unknown function value,}\&\text{if values of the function and its derivatives are known at a}\&\text{particular point. In the case of the Maclaurin series, this}\&\text{ point has the value of zero (for many functions f(0) is readily}\&\text{known):}\&f(x)\approx\sum_{i=0}^{n}\frac{f^{(i)}(0)}{i!}x^i\approx f(0)+f'(0)x+\frac{f''(0)}{2!}x^2+\frac{f^{(3)}(0)}{3!}x^3+...+\frac{f^{(n)}(0)}{n!}x^n\&\text{For the function }-sin(3x)sin(7x)\text{ at }x=2.1\&\text{This expansion becomes:}\&f(2.1)\approx (-sin(3(0))sin(7(0)))(2.1)^{0}+(- 3cos(3(0))sin(7(0)) - 7cos(7(0))sin(3(0)))(2.1)^{1}+\&\frac{58sin(3(0))sin(7(0)) - 42cos(3(0))cos(7(0))}{2}(2.1)^{2}\&f(2.1)\approx-92.61\end{align}$

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