Find the Taylor Series of f ( x ) = cos ⁡ x f\left(x\right)=\cos x centered x = π x=\pi . Then, write the power series using summation notation.

∑ n = 0 ∞ ( − 1 ) n + 1 ( 2 n ) ! ( x − π ) 2 n \sum_{n=0}^{\infty}\frac{\left(-1\right)^{n+1}}{\left(2n\right)!}\left(x-\pi\right)^{2n}

∑ n = 0 ∞ ( 1 ) n + 1 ( 2 n ) ! ( x − π ) 2 n \sum_{n=0}^{\infty}\frac{\left(1\right)^{n+1}}{\left(2n\right)!}\left(x-\pi\right)^{2n}

∑ n = 0 ∞ ( π ) n ( 2 n ) ! ( x − π ) 2 n \sum_{n=0}^{\infty}\frac{\left(\pi\right)^{n}}{\left(2n\right)!}\left(x-\pi\right)^{2n}

∑ n = 0 ∞ ( − 1 ) n ( 2 n ) ! ( x + π ) 2 n \sum_{n=0}^{\infty}\frac{\left(-1\right)^{n}}{\left(2n\right)!}\left(x+\pi\right)^{2n}

Find the interval of convergence for the Taylor series for f ( x ) = sin ⁡ x f\left(x\right)=\sin x centered at x = π 2 x=\frac{\pi}{2} .

( − ∞ , ∞ ) \left(-\infty,\infty\right)

( − π 2 , 3 π 2 ) \left(-\frac{\pi}{2},\frac{3\pi}{2}\right)

[ − π 2 , 3 π 2 ] \left\lbrack-\frac{\pi}{2},\frac{3\pi}{2}\right\rbrack

( − π 2 , π 2 ) \left(-\frac{\pi}{2},\frac{\pi}{2}\right)

Find the Taylor polynomials of order 0 , 1 , 2 0,1,2 , and 3 3 for f ( x ) = ln ⁡ ( x ) f\left(x\right)=\ln\left(x\right) centered at x = 1 x=1 .

<math alttext="p sub 0 of x equals 0 comma p sub 1 of x equals x minus 1 comma p sub 2 of x equals open paren x minus 1 close paren minus one half times open paren x minus 1 close paren squared comma p sub 3 of x equals open paren x minus 1 close paren minus one half times open paren x minus 1 close paren squared plus one third times open paren x minus 1 close paren cubed" data-value="p_0\left(x\right)=0,p_1\left(x\right)=x-1,p_2\left(x\right)=\left(x-1\right)-\frac12\left(x-1\right)^2,p_3\left(x\right)=\left(x-1\right)-\frac12\left(x-1\right)^2+\frac13\left(x-1\right)^3"> p 0 ( x ) = 0 , p 1 ( x ) = x − 1 , p 2 ( x ) = ( x − 1 ) − 1 2 ( x − 1 ) 2 , p 3 ( x ) = ( x − 1 ) − 1 2 ( x − 1 ) 2 + 1 3 ( x − 1 ) 3 p_0\left(x\right)=0,p_1\left(x\right)=x-1,p_2\left(x\right)=\left(x-1\right)-\frac12\left(x-1\right)^2,p_3\left(x\right)=\left(x-1\right)-\frac12\left(x-1\right)^2+\frac13\left(x-1\right)^3

<math alttext="p sub 0 of x equals 0 comma p sub 1 of x equals x minus 1 comma p sub 2 of x equals open paren x minus 1 close paren minus open paren x minus 1 close paren squared comma p sub 3 of x equals open paren x minus 1 close paren minus one half times open paren x minus 1 close paren squared plus one third times open paren x minus 1 close paren cubed" data-value="p_0\left(x\right)=0,p_1\left(x\right)=x-1,p_2\left(x\right)=\left(x-1\right)-\left(x-1\right)^2,p_3\left(x\right)=\left(x-1\right)-\frac12\left(x-1\right)^2+\frac13\left(x-1\right)^3"> p 0 ( x ) = 0 , p 1 ( x ) = x − 1 , p 2 ( x ) = ( x − 1 ) − ( x − 1 ) 2 , p 3 ( x ) = ( x − 1 ) − 1 2 ( x − 1 ) 2 + 1 3 ( x − 1 ) 3 p_0\left(x\right)=0,p_1\left(x\right)=x-1,p_2\left(x\right)=\left(x-1\right)-\left(x-1\right)^2,p_3\left(x\right)=\left(x-1\right)-\frac12\left(x-1\right)^2+\frac13\left(x-1\right)^3

<math alttext="p sub 0 of x equals 1 comma p sub 1 of x equals x minus 1 comma p sub 2 of x equals open paren x minus 1 close paren minus one half times open paren x minus 1 close paren squared comma p sub 3 of x equals open paren x minus 1 close paren minus one half times open paren x minus 1 close paren squared plus one third times open paren x minus 1 close paren cubed" data-value="p_0\left(x\right)=1,p_1\left(x\right)=x-1,p_2\left(x\right)=\left(x-1\right)-\frac12\left(x-1\right)^2,p_3\left(x\right)=\left(x-1\right)-\frac12\left(x-1\right)^2+\frac13\left(x-1\right)^3"> p 0 ( x ) = 1 , p 1 ( x ) = x − 1 , p 2 ( x ) = ( x − 1 ) − 1 2 ( x − 1 ) 2 , p 3 ( x ) = ( x − 1 ) − 1 2 ( x − 1 ) 2 + 1 3 ( x − 1 ) 3 p_0\left(x\right)=1,p_1\left(x\right)=x-1,p_2\left(x\right)=\left(x-1\right)-\frac12\left(x-1\right)^2,p_3\left(x\right)=\left(x-1\right)-\frac12\left(x-1\right)^2+\frac13\left(x-1\right)^3

<math alttext="p sub 0 of x equals 0 comma p sub 1 of x equals 1 comma p sub 2 of x equals open paren x minus 1 close paren comma p sub 3 of x equals open paren x minus 1 close paren squared plus one third times open paren x minus 1 close paren cubed" data-value="p_0\left(x\right)=0,p_1\left(x\right)=1,p_2\left(x\right)=\left(x-1\right),p_3\left(x\right)=\left(x-1\right)^2+\frac13\left(x-1\right)^3"> p 0 ( x ) = 0 , p 1 ( x ) = 1 , p 2 ( x ) = ( x − 1 ) , p 3 ( x ) = ( x − 1 ) 2 + 1 3 ( x − 1 ) 3 p_0\left(x\right)=0,p_1\left(x\right)=1,p_2\left(x\right)=\left(x-1\right),p_3\left(x\right)=\left(x-1\right)^2+\frac13\left(x-1\right)^3

Find the Maclaurin polynomials of order 0 , 1 , 2 0,1,2 , and 3 3 for f ( x ) = sin ⁡ x f\left(x\right)=\sin x

p o ( x ) = 0 , p 1 ( x ) = x , p 2 ( x ) = x , p 3 ( x ) = x − 1 6 x 3 p_{o}(x)=0,p_1(x)=x,p_2(x)=x,p_3(x)=x-\frac16x^3

p o ( x ) = 0 , p 1 ( x ) = 0 , p 2 ( x ) = x , p 3 ( x ) = x − 1 6 x 3 p_{o}(x)=0,p_1(x)=0,p_2(x)=x,p_3(x)=x-\frac16x^3

p o ( x ) = 0 , p 1 ( x ) = x , p 2 ( x ) = x , p 3 ( x ) = x − 1 6 x 2 p_{o}(x)=0,p_1(x)=x,p_2(x)=x,p_3(x)=x-\frac16x^2

p o ( x ) = 0 , p 1 ( x ) = x , p 2 ( x ) = x , p 3 ( x ) = x − 1 3 x 2 p_{o}(x)=0,p_1(x)=x,p_2(x)=x,p_3(x)=x-\frac13x^2

What is the Taylor series, and how is it derived?

The Taylor series is a representation of a function as an infinite sum of terms, calculated from the values of its derivatives at a single point. It is derived using the formula: f ( x ) = ∑ n = 0 ∞ f ( n ) ( a ) n ! ( x − a ) n Here, f ( n ) ( a ) is the nth derivative of the function evaluated at a , and n ! is the factorial of n . The series approximates the function near a .

What is the difference between a Taylor series and a Maclaurin series?

The Taylor series is a general representation of a function as an infinite sum of terms based on its derivatives at a specific point a . The formula is: f ( x ) = ∑ n = 0 ∞ f ( n ) ( a ) n ! ( x − a ) n The Maclaurin series is a special case of the Taylor series where a = 0 . Its formula is: f ( x ) = ∑ n = 0 ∞ f ( n ) ( 0 ) n ! ( x ) n Thus, the Maclaurin series is centered at x = 0 , while the Taylor series can be centered at any point a .

How do you find the nth-degree Taylor polynomial for a function?

The nth-degree Taylor polynomial for a function f ( x ) centered at a is a finite approximation of the Taylor series. The formula is: P n ( x ) = ∑ k = 0 n f ( k ) ( a ) k ! ( x − a ) k To find it: Compute the derivatives of f up to the nth order. Evaluate each derivative at a . Substitute these values into the formula. This polynomial approximates f near a .

How is the Taylor series used to approximate functions in real-world applications?

The Taylor series is widely used to approximate functions in real-world applications, especially when exact computation is difficult. For example: Physics: Approximating complex functions like sine, cosine, or exponential functions in mechanics and wave analysis. Engineering: Simplifying control systems and signal processing equations. Computer Science: Calculating values of transcendental functions in numerical algorithms. Economics: Modeling nonlinear relationships in financial data. By using a finite number of terms, the Taylor series provides a manageable approximation that balances accuracy and computational efficiency. The accuracy depends on the number of terms used and the proximity to the expansion point.

15. Power Series

Taylor Series & Taylor Polynomials

15. Power Series

Taylor Series & Taylor Polynomials: Videos & Practice Problems

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Topic summary

To find the first four terms of the sequence defined by $a_{n} = n^2 n_{- 1}!$ , substitute values of n from 1 to 4. The calculations yield: a₁ = 1, a₂ = 4, a₃ = 18, and a₄ = 96. This sequence illustrates the interaction of polynomial growth and factorial functions, showcasing how factorials can influence the growth rate of sequences.

concept

Taylor Series

Video duration:

Taylor Series Video Summary

To find the first four terms of the sequence defined by $ a_n = n^2 \times (n-1)! $, we will evaluate the expression for $ n = 1 $ through $ n = 4 $.

Starting with $ n = 1 $:

We have:

$ a_1 = 1^2 \times (1-1)! $

Calculating this gives:

$ a_1 = 1 \times 0! $

Since $ 0! = 1 $, we find:

$ a_1 = 1 \times 1 = 1 $

Next, for $ n = 2 $:

We calculate:

$ a_2 = 2^2 \times (2-1)! $

This simplifies to:

$ a_2 = 4 \times 1! $

And since $ 1! = 1 $, we have:

$ a_2 = 4 \times 1 = 4 $

Now, for $ n = 3 $:

We find:

$ a_3 = 3^2 \times (3-1)! $

This becomes:

$ a_3 = 9 \times 2! $

Calculating $ 2! $ gives $ 2 \times 1 = 2 $, so:

$ a_3 = 9 \times 2 = 18 $

Finally, for $ n = 4 $:

We compute:

$ a_4 = 4^2 \times (4-1)! $

This simplifies to:

$ a_4 = 16 \times 3! $

Calculating $ 3! $ gives $ 3 \times 2 \times 1 = 6 $, thus:

$ a_4 = 16 \times 6 = 96 $

In summary, the first four terms of the sequence are:

1, 4, 18, and 96.

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Problem

Find the Taylor Series of $f (x) = \cos x$ centered $x = π$ . Then, write the power series using summation notation.

$\sum_{n = 0}^{\infty} \frac{{(1)}^{n + 1}}{(2 n)!} {(x - π)}^{2 n}$

$\sum_{n = 0}^{\infty} \frac{{(π)}^{n}}{(2 n)!} {(x - π)}^{2 n}$

$\sum_{n = 0}^{\infty} \frac{{(- 1)}^{n + 1}}{(2 n)!} {(x - π)}^{2 n}$

$\sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{(2 n)!} {(x + π)}^{2 n}$

example

Taylor Series Example 1

Video duration:

Taylor Series Example 1 Video Summary

To find the first four terms of the sequence defined by $ a_n = n^2 \times (n-1)! $, we will evaluate the expression for $ n = 1 $ through $ n = 4 $.

Starting with $ n = 1 $:

We have:

$ a_1 = 1^2 \times (1-1)! $

Calculating this gives:

$ a_1 = 1 \times 0! $

Since $ 0! = 1 $, we find:

$ a_1 = 1 \times 1 = 1 $

Next, for $ n = 2 $:

We calculate:

$ a_2 = 2^2 \times (2-1)! $

This simplifies to:

$ a_2 = 4 \times 1! $

And since $ 1! = 1 $, we have:

$ a_2 = 4 \times 1 = 4 $

Now, for $ n = 3 $:

We find:

$ a_3 = 3^2 \times (3-1)! $

This becomes:

$ a_3 = 9 \times 2! $

Calculating $ 2! $ gives $ 2 \times 1 = 2 $, so:

$ a_3 = 9 \times 2 = 18 $

Finally, for $ n = 4 $:

We compute:

$ a_4 = 4^2 \times (4-1)! $

This simplifies to:

$ a_4 = 16 \times 3! $

Calculating $ 3! $ gives $ 3 \times 2 \times 1 = 6 $, thus:

$ a_4 = 16 \times 6 = 96 $

In summary, the first four terms of the sequence are:

1, 4, 18, and 96.

concept

Convergence of Taylor & Maclaurin Series

Video duration:

Convergence of Taylor & Maclaurin Series Video Summary

To find the first four terms of the sequence defined by $ a_n = n^2 \times (n-1)! $, we will evaluate the expression for $ n = 1 $ through $ n = 4 $.

Starting with $ n = 1 $:

We have:

$ a_1 = 1^2 \times (1-1)! $

Calculating this gives:

$ a_1 = 1 \times 0! $

Since $ 0! = 1 $, we find:

$ a_1 = 1 \times 1 = 1 $

Next, for $ n = 2 $:

We calculate:

$ a_2 = 2^2 \times (2-1)! $

This simplifies to:

$ a_2 = 4 \times 1! $

And since $ 1! = 1 $, we have:

$ a_2 = 4 \times 1 = 4 $

Now, for $ n = 3 $:

We find:

$ a_3 = 3^2 \times (3-1)! $

This becomes:

$ a_3 = 9 \times 2! $

Calculating $ 2! = 2 \times 1 = 2 $, we get:

$ a_3 = 9 \times 2 = 18 $

Finally, for $ n = 4 $:

We compute:

$ a_4 = 4^2 \times (4-1)! $

This simplifies to:

$ a_4 = 16 \times 3! $

Since $ 3! = 3 \times 2 \times 1 = 6 $, we find:

$ a_4 = 16 \times 6 = 96 $

Thus, the first four terms of the sequence are $ 1, 4, 18, $ and $ 96 $. This sequence can be extended further by continuing to apply the same formula for higher values of $ n $.

Problem

Find the interval of convergence for the Taylor series for $f (x) = \sin x$ centered at $x = \frac{π}{2}$ .

$(- \frac{π}{2}, \frac{3 π}{2})$

$[- \frac{π}{2}, \frac{3 π}{2}]$

$(- \frac{π}{2}, \frac{π}{2})$

$(- \infty, \infty)$

Problem

Find the interval of convergence for the Maclaurin series for $f (x) = \tan^{- 1} x$

$open bracket negative 1 comma 1 close bracket$

$open paren negative 1 comma 1 close bracket$

$(- \sqrt{3}, \sqrt{3})$

$[- \sqrt{3}, \sqrt{3})$

concept

Taylor Polynomials

Video duration:

Taylor Polynomials Video Summary

To find the first four terms of the sequence defined by $ a_n = n^2 \times (n-1)! $, we will evaluate the expression for $ n = 1 $ through $ n = 4 $.

Starting with $ n = 1 $:

We have:

$ a_1 = 1^2 \times (1-1)! $

Calculating this gives:

$ a_1 = 1 \times 0! $

Since $ 0! = 1 $, we find:

$ a_1 = 1 \times 1 = 1 $

Next, for $ n = 2 $:

We calculate:

$ a_2 = 2^2 \times (2-1)! $

This simplifies to:

$ a_2 = 4 \times 1! $

And since $ 1! = 1 $, we have:

$ a_2 = 4 \times 1 = 4 $

Now, for $ n = 3 $:

We find:

$ a_3 = 3^2 \times (3-1)! $

This becomes:

$ a_3 = 9 \times 2! $

Calculating $ 2! = 2 \times 1 = 2 $, we get:

$ a_3 = 9 \times 2 = 18 $

Finally, for $ n = 4 $:

We compute:

$ a_4 = 4^2 \times (4-1)! $

This simplifies to:

$ a_4 = 16 \times 3! $

Since $ 3! = 3 \times 2 \times 1 = 6 $, we find:

$ a_4 = 16 \times 6 = 96 $

Thus, the first four terms of the sequence are $ 1, 4, 18, $ and $ 96 $. This sequence can be extended further by continuing to apply the same formula for higher values of $ n $.

Problem

Find the Taylor polynomials of order $0 comma 1 comma 2$ , and $3$ for $f (x) = \ln (x)$ centered at $x equals 1$ .

$p_{0} (x) = 0, p_{1} (x) = x - 1, p_{2} (x) = (x - 1) - {(x - 1)}^{2}, p_{3} (x) = (x - 1) - \frac{1}{2} {(x - 1)}^{2} + \frac{1}{3} {(x - 1)}^{3}$

$p_{0} (x) = 1, p_{1} (x) = x - 1, p_{2} (x) = (x - 1) - \frac{1}{2} {(x - 1)}^{2}, p_{3} (x) = (x - 1) - \frac{1}{2} {(x - 1)}^{2} + \frac{1}{3} {(x - 1)}^{3}$

$p_{0} (x) = 0, p_{1} (x) = x - 1, p_{2} (x) = (x - 1) - \frac{1}{2} {(x - 1)}^{2}, p_{3} (x) = (x - 1) - \frac{1}{2} {(x - 1)}^{2} + \frac{1}{3} {(x - 1)}^{3}$

$p_{0} (x) = 0, p_{1} (x) = 1, p_{2} (x) = (x - 1), p_{3} (x) = {(x - 1)}^{2} + \frac{1}{3} {(x - 1)}^{3}$

Problem

Approximate $l n 1.5$ to four decimal places using the third-degree Taylor polynomial for $f (x) = \ln x$ .

$0.0417$

$0.4167$

$0.125$

$0.6667$

Problem

Find the Maclaurin polynomials of order $0 comma 1 comma 2$ , and $3$ for $f (x) = \sin x$

$p_{o} (x) = 0, p_{1} (x) = x, p_{2} (x) = x, p_{3} (x) = x - \frac{1}{6} x^{3}$

$p_{o} (x) = 0, p_{1} (x) = 0, p_{2} (x) = x, p_{3} (x) = x - \frac{1}{6} x^{3}$

$p_{o} (x) = 0, p_{1} (x) = x, p_{2} (x) = x, p_{3} (x) = x - \frac{1}{6} x^{2}$

$p_{o} (x) = 0, p_{1} (x) = x, p_{2} (x) = x, p_{3} (x) = x - \frac{1}{3} x^{2}$

Problem

Approximate $sine 0.3$ to four decimal places using the third-degree Maclaurin polynomial for $f (x) = \sin x$ .

$0.0045$

$0.3000$

$0.2955$

$0.3045$

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We have more practice problems on Taylor Series & Taylor Polynomials

Here’s what students ask on this topic:

The Taylor series is a representation of a function as an infinite sum of terms, calculated from the values of its derivatives at a single point. It is derived using the formula:

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

Here, $f (n) (a)$ is the nth derivative of the function evaluated at $a$ , and $n!$ is the factorial of $n$ . The series approximates the function near $a$ .

The Taylor series is a general representation of a function as an infinite sum of terms based on its derivatives at a specific point $a$ . The formula is:

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

The Maclaurin series is a special case of the Taylor series where $a = 0$ . Its formula is:

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

Thus, the Maclaurin series is centered at $x = 0$ , while the Taylor series can be centered at any point $a$ .

The nth-degree Taylor polynomial for a function $f (x)$ centered at $a$ is a finite approximation of the Taylor series. The formula is:

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

To find it:

Compute the derivatives of $f$ up to the nth order.
Evaluate each derivative at $a$ .
Substitute these values into the formula.

This polynomial approximates $f$ near $a$ .

The remainder term in the Taylor series, also known as the Lagrange remainder, quantifies the error between the actual function and its Taylor polynomial approximation. It is given by:

$R n (x) = \frac{f^{(n + 1)} (c)}{(n + 1)!} (x - a) n^{+ 1}$

Here, $c$ is some value between $a$ and $x$ . The remainder term is important because it helps determine how well the Taylor polynomial approximates the function within a specific interval. By analyzing the remainder, we can assess the accuracy of the approximation.

The Taylor series is widely used to approximate functions in real-world applications, especially when exact computation is difficult. For example:

Physics: Approximating complex functions like sine, cosine, or exponential functions in mechanics and wave analysis.
Engineering: Simplifying control systems and signal processing equations.
Computer Science: Calculating values of transcendental functions in numerical algorithms.
Economics: Modeling nonlinear relationships in financial data.

By using a finite number of terms, the Taylor series provides a manageable approximation that balances accuracy and computational efficiency. The accuracy depends on the number of terms used and the proximity to the expansion point.

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Taylor Series & Taylor Polynomials: Videos & Practice Problems

Taylor Series

Taylor Series Video Summary

Find the Taylor Series of f(x)=cosxf\left(x\right)=\cos x centered x=πx=\pi. Then, write the power series using summation notation.

Taylor Series Example 1

Taylor Series Example 1 Video Summary

Convergence of Taylor & Maclaurin Series

Convergence of Taylor & Maclaurin Series Video Summary

Find the interval of convergence for the Taylor series for f(x)=sinxf\left(x\right)=\sin x centered at x=π2x=\frac{\pi}{2}.

Find the interval of convergence for the Maclaurin series for f(x)=tan−1xf\left(x\right)=\tan^{-1}x

Taylor Polynomials

Taylor Polynomials Video Summary

Find the Taylor polynomials of order 0,1,20,1,2, and 33 for f(x)=ln(x)f\left(x\right)=\ln\left(x\right) centered at x=1x=1.

Approximate ln1.5\ln1.5 to four decimal places using the third-degree Taylor polynomial for f(x)=lnxf\left(x\right)=\ln x.

Find the Maclaurin polynomials of order 0,1,20,1,2, and 33 for f(x)=sinxf\left(x\right)=\sin x

Approximate sin0.3\sin0.3 to four decimal places using the third-degree Maclaurin polynomial for f(x)=sinxf\left(x\right)=\sin x.

Do you want more practice?

Here’s what students ask on this topic:

What is the Taylor series, and how is it derived?

What is the Taylor series, and how is it derived?

What is the difference between a Taylor series and a Maclaurin series?

What is the difference between a Taylor series and a Maclaurin series?

How do you find the nth-degree Taylor polynomial for a function?

How do you find the nth-degree Taylor polynomial for a function?

What is the remainder term in the Taylor series, and why is it important?

What is the remainder term in the Taylor series, and why is it important?

How is the Taylor series used to approximate functions in real-world applications?

How is the Taylor series used to approximate functions in real-world applications?

Your Calculus tutors

Find the Taylor Series of $f (x) = \cos x$ centered $x = π$ . Then, write the power series using summation notation.

Find the interval of convergence for the Taylor series for $f (x) = \sin x$ centered at $x = \frac{π}{2}$ .

Find the interval of convergence for the Maclaurin series for $f (x) = \tan^{- 1} x$

Find the Taylor polynomials of order $0 comma 1 comma 2$ , and $3$ for $f (x) = \ln (x)$ centered at $x equals 1$ .

Approximate $l n 1.5$ to four decimal places using the third-degree Taylor polynomial for $f (x) = \ln x$ .

Find the Maclaurin polynomials of order $0 comma 1 comma 2$ , and $3$ for $f (x) = \sin x$

Approximate $sine 0.3$ to four decimal places using the third-degree Maclaurin polynomial for $f (x) = \sin x$ .