Taylor Series - AP Calculus BC

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Question

Let 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

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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$$

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.

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Question

Let

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

Answer

Using the formula of a binomial series centered at 0:

,

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:

The derivative was found using the following rules:

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

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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):

The derivatives were found using the following rule:

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

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:

which simplified becomes

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Question

Find the Taylor series expansion of 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

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

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 at ,

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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

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 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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Question

$\begin{align}&$\text{Approximate }$f(1.6) \&$\text{Where }$f(x)=-cos(3x)cos(7x)\&$\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 }$-cos(3x)cos(7x)$\text{ at }$x=1.6\&$\text{This expansion becomes:}$\&f(1.6)\approx $(-cos(3(0))cos(7(0)))(1.6)^{0}$+(7cos(3(0))sin(7(0)) + $3cos(7(0))sin(3(0)))(1.6)^{1}$+\&$\frac{58cos(3(0))cos(7(0)) - $42sin(3(0))sin(7(0))}{2}$(1.6)^{2}$\&f(1.6)\approx73.24\end{align}$

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