\( \newcommand{\abs}[1]{\left| \, {#1} \, \right| } \)

You CAN Ace Calculus

Topics You Need To Understand For This Page

Related Topics and Links

external links you may find helpful

youtube playlist

Wikipedia - Taylor Series

17Calculus Subjects Listed Alphabetically

Single Variable Calculus

Multi-Variable Calculus

Differential Equations

Search Practice Problems

Do you have a practice problem number but do not know on which page it is found? If so, enter the number below and click 'page' to go to the page on which it is found or click 'practice' to be taken to the practice problem.

free ideas to save on books

The main idea of Taylor Series Expansion is to replace a complicated function with a series in the form of an infinite polynomial. Polynomials are very easy to work with and have very clearly known properties. If a complicated function can be expressed as a polynomial, then there are many things that can be done like integration, which is very easy with a polynomial.

Difference Between Power Series and Taylor Series

Taylor series are a specific case of Power series where the constants ( usually functions of n ) are related to the derivative of the function. Many of the same techniques that work for one will work for the other.

A Power Series is based on the Geometric Series using the equation \(\displaystyle{ \sum_{n=0}^{\infty}{ar^n} = \frac{a}{1-r} }\) which converges for \( \abs{r} < 1 \). We may also use the ratio test and other tests to determine the radius and interval of convergence.
Power Series are discussed on a separate page.

A Taylor Series is based on the recurring formula
\(\displaystyle{ \sum_{n=0}^{\infty}{\frac{f^{(n)}(a)}{n!}(x-a)^n} = f(a) + \frac{f'(a)}{1!}(x-a) + \frac{f''(a)}{2!}(x-a)^2 + . . . }\)
We use the Ratio Test to determine the radius and interval of convergence.
Taylor Series are discussed on this page.

Taylor Series or Taylor Polynomial?

There is a subtle difference in terminology on this page that you need to notice right away.
A Taylor Series is an infinite power series and involves a radius and interval of convergence.
A Taylor Polynomial is a polynomial with a finite number of terms. It is an approximation and involves a remainder or error bounds. The polynomial is formed from the series by taking a finite number of terms.

Before we get started with the details of finding Taylor and Maclaurin Series, let's get an intuitive idea of what we are talking about from this video. This video contains some really cool graphics to give you an overview of what Taylor Series means and what it is like. This is a great video.

3Blue1Brown - Taylor series - Chapter 10, Essence of calculus [22min-19secs]

video by 3Blue1Brown

Taylor Series

Taylor Series Expansion is done around a specific point and within a specified interval. Within that interval (called the interval of convergence) the infinite series is equivalent to the function. Here is how it works.

To express a function \(f(x)\) as a polynomial about a point \( x=a\), we use the series

\(\displaystyle{ f(x) = \sum_{n=0}^{\infty}{\frac{f^{(n)}(a)}{n!}(x-a)^n} = }\) \(\displaystyle{ f(a) + \frac{f'(a)}{1!}(x-a) + }\) \(\displaystyle{ \frac{f''(a)}{2!}(x-a)^2 + . . . }\)

where we define \( 0! = 1 \) and \( (x-a)^0 = 1 \).

Maclaurin Series - A Maclaurin Series is a special case of a Taylor series where expansion is done about the point \( x=0 \), i.e. in the above equation, \(a=0\).

Interval

In order to have a complete definition, we also need to know the interval on which this equality holds.

This first video clip is a quick introduction to the formula given above. It is not as comprehensive as the second video but it will get you started or you can watch it to refresh your memory. If you are just learning about Taylor series, this video will not give you enough detail about the entire topic. So it is better to watch the second video after watching this one or instead of this one.

MIT OCW - Lec 38 | MIT 18.01 Single Variable Calculus, Fall 2007 [7min-45secs]

video by MIT OCW

Here is a great video explaining Taylor series and how to use it.

Dr Chris Tisdell - What is a Taylor series? [47min-32secs]

video by Dr Chris Tisdell

Here is a video with an informal proof of the Taylor Series Expansion.

PatrickJMT - Taylor / Maclaurin Series Expansion - Proof of the Formula [13min-44secs]

Patricks Comments:
In this video, I show how to find the Taylor series expansion for a function, assuming that one exists! It is nothing too heavy: we just take derivatives and plug in the value at which we are centering the function. I do not find any Taylor series of specific functions in this video, nor do I justify when a Taylor series expansion is valid (not all functions have power series expansion!).

video by PatrickJMT

Radius and Interval of Convergence

This section is part of the discussion found on the radius and interval of convergence page, where you will find more information, videos and practice problems.

Once the Taylor series or power series is calculated, we use the ratio test to determine the radius convergence and other tests to determine the interval of convergence. Without knowing the radius and interval of convergence, the series is not considered a complete function (This is similar to not knowing the domain of a function. As we say in the page on domain and range, a fully defined function always contains information about the domain, either implicitly or explicitly stated.) In this case, we can't leave off information about where the series converges because the series may not hold for all values of \(x\).

Using The Ratio Test

The ratio test looks like this. We have a series \( \sum{a_n} \) with non-zero terms. We calculate the limit \( \displaystyle{\lim_{n \to \infty}{\left| \frac{a_{n+1}}{a_n} \right|} = L} \).
There are three possible cases for the value \(L\).
\( L < 1 \): The series converges absolutely.
\( L = 1 \): The ratio test is inconclusive.
\( L > 1 \): The series diverges.

So we use the first case (\( L < 1 \), since we want convergence) and we set up the inequality \( \displaystyle{\lim_{n \to \infty}{\left| \frac{a_{n+1}}{a_n} \right|} < 1} ~~~~~ [ 1 ]\)

Key: Do not drop the absolute values.

Radius of Convergence

To find the radius of convergence, we need to simplify the inequality [1] to the point that we have \( \left| x-a \right| < R \). This gives the radius of convergence as \(R\). This seems very simple but you need to be careful of the notation and wording your textbooks. Some textbooks use a small \(r\). Some textbooks ask for the ratio of convergence in which case you need to give the answer as \( \rho = 1/R \). This has always seemed kind of strange to me but there must be some reason for it. However, on this site, whenever we talk about the radius of convergence in this context, we will use \(R\) as defined above.

There are three possible cases for the radius of convergence.

\(R = 0\)

series converges only at the point \(x = a\)

\( 0 < R < \infty \)

series converges within the interval

\(R = \infty \)

series converges for all \(x\)

The interesting thing is that we have a strict inequality in \( 0 < R < \infty \) and, because of the definition of the ratio test, we have no idea what happens when \( \left| x-a \right| = R \). The series could converge or diverge. The ratio test doesn't give us a clue on what happens in that case. That's where we need to find the interval of convergence, which we discuss next.

Interval of Convergence

We use the radius of convergence, \(R\), to calculate the interval of convergence as follows
\(\begin{array}{rcccl} & & \left| x-a \right| & < & R \\ -R & < & x-a & < & R \\ -R + a & < & x & < & R + a \end{array}\)
So now we have an open interval \( (-R+a, R+a) \) in which the series converges. Now we need to look at the endpoints to determine what goes on. To do that we substitute each endpoint individually for \(x\) into the series and then use the other series test to determine convergence or divergence.

Notice when we substitute \(x=-R+a\) into the \((x-a)^n\) term, we end up with \((-R)^n\) which can be simplified to \((-1)^n R^n\). Now we have an alternating series. So often, the alternating series test can be used to determine convergence or divergence. The point is that, using other tests, we need to definitively determine convergence or divergence at each endpoint. The result will be an open interval, a half-open interval or a closed interval. We call this interval, the interval of convergence.

Notice the difference between the terms radius of convergence and interval of convergence. The radius of convergence gives information about the open interval but says nothing about the endpoints. The interval of convergence includes the radius of convergence AND information about convergence or divergence of the endpoints.

See the radius and interval of convergence page page for more detail, videos and practice problems.

Taylor Polynomial

To approximate a complicated function with a polynomial, we choose a finite number of terms from the Taylor series, usually between \( 2\) and \(5\) and, as long we stay 'near' \( x=a \), then the Taylor Series approximation will work for most purposes. The further you go away from that point, the less accurate the approximation, until the polynomial veers off. However, you can compensate, up to a point, by building a higher order approximation. How close 'near' is depends on the problem as well as the number of terms required. We use the word 'order' to specify the number of terms we want. For example, if we specify a 'second order' approximation, then the highest power is \(2\).

Tangent Line

The tangent line at a point is just the first order Taylor polynomial.

However, since we are talking about approximating a function, we often want to know how close of an approximation we have. It is nearly impossible to know exactly how close we are. So, we use the idea of error bounds to get a known upper limit on the error, i.e. we can get a value that we know the error is not above. Of course, this is approximate too but we do have a theorem that tells us we can find an upper bound and how to get it. This is discussed in the following video.

Here is a great video explaining Taylor polynomials and how to use them. Although the second part contains information about the remainder and error bounds, which are discussed on a separate page, it will help you to watch all of this video to get a good introduction to this topic.

Dr Chris Tisdell - What is a Taylor Polynomial? [41min-25secs]

video by Dr Chris Tisdell

Remainder and Error Bounds

Once you have found the infinite Taylor Series, you will often be asked to approximate the original function with a finite number of terms from the Taylor Series. Doing so introduces error since the finite Taylor Series does not exactly represent the original function. To handle this error we write the function like this.

\(\displaystyle{ f(x) = f(a) + \frac{f'(a)}{1!}(x-a) + \frac{f''(a)}{2!}(x-a)^2 + . . . + \frac{f^{(n)}(a)}{n!}(x-a)^n + R_n(x) }\)

where \(R_n(x)\) is the remainder. Notice we are cutting off the series after the n-th derivative and \(R_n(x)\) represents the rest of the series.

Lagrange's formula for this remainder term is

\(\displaystyle{ R_n(x) = \frac{f^{(n+1)}(z)(x-a)^{n+1}}{(n+1)!} }\)

This looks very similar to the equation for the Taylor series terms . . . and it is, except for one important item. Notice that in the numerator, we evaluate the \(n+1\) derivative at \(z\) instead of \(a\). So, what is the value of \(z\)? \(z\) takes on a value between \(a\) and \(x\), but, and here's the key, we don't know exactly what that value is. So this remainder can never be calculated exactly. However, since we know that \(z\) is between \(a\) and \(x\), we can determine an upper bound on the remainder and be confident that the remainder will never exceed this upper bound. So how do we do that?

Upper Bound on the Remainder (Error)

We usually consider the absolute value of the remainder term \(R_n\) and call it the upper bound on the error, also called Taylor's Inequality.

\(\displaystyle{ \abs{R_n(x)} \leq \frac{\abs{f^{(n+1)}(z)(x-a)^{n+1}} }{(n+1)!} }\)

How To Choose z

To find an upper bound on this error, we choose the value of \(z\) using these rules.
1. If the \(n+1\) derivative contains a sine or cosine term, we replace the sine or cosine term with one, since the maximum value of sine or cosine is one. This seems somewhat arbitrary but most calculus books do this even though this could give a much larger upper bound than could be calculated using the next rule. [ As usual, check with your instructor to see what they expect. ]
2. If we do not have a sine or cosine term, we calculate \(\abs{f^{(n+1)}(z)}\) and then choose the value of \(z\) between \(a\) and the \(x\)-value that we are estimating that makes this term a maximum. Sometimes, we need to find the critical points and find the one that is a maximum. Since we have a closed interval, either \([a,x]\) or \([x,a]\), we also have to consider the end points. Many times, the maximum will occur at one of the end points, but not always.

Okay, so what is the point of calculating the error bound? The point is that once we have calculated an upper bound on the error, we know that at all points in the interval of convergence, the truncated Taylor series will always be within \(\abs{R_n(x)}\) of the original function \(f(x)\). This \(\abs{R_n(x)}\) is a mathematical 'nearness' number that we can use to determine the number of terms we need to have for a Taylor series.

This section is part of the discussion found on the remainder and error bounds page, where you will find more information, videos and practice problems.

List of Common Power Series

This panel contains a list of common power and Taylor series. These are often used to build other series. Videos and practice problems related to determining Taylor series using these series can be found on the power series page.

\(\displaystyle{ \frac{1}{1-x} = \sum_ {n=0}^{\infty}{x^n} }\)

converges \( -1 < x < 1 \)

power series

\(\displaystyle{ e^x = \sum_{n=0}^{x^n}{\frac{x^n}{n!}} }\)

converges \(-\infty < x < \infty\)

Maclaurin series

\(\displaystyle{\ln(x) = \sum_{n=0}^{\infty}{ \frac{(-1)^{n+1} (x-1)^n}{n} } }\)

Taylor series about \( x = 1 \)

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

converges \(-\infty < x < \infty\)

Maclaurin series

\(\displaystyle{ \cos(x) = \sum_{n=0}^{\infty}{ \frac{(-1)^n x^{2n}}{(2n)!} } }\)

converges \(-\infty < x < \infty\)

Maclaurin series

Practice

Conversion Between A-B-C Level (or 1-2-3) and New Numbered Practice Problems

Please note that with this new version of 17calculus, the practice problems have been relabeled but they are MOSTLY in the same order. So, Practice A01 (1) is probably the first basic practice problem, A02 (2) is probably the second basic practice problem, etc. Practice B01 is probably the first intermediate practice problem and so on.

GOT IT. THANKS!

Instructions - - Unless otherwise instructed, find the Taylor series of these functions about the given point (if no point is given, find the Maclaurin series). Use the Taylor Series Expansion to work these problems unless it is explicitly stated to use a known series. Give your answers in exact terms and completely factored.

Basic Problems

Find the Maclaurin polynomials of degrees 1, 2, 3 and 5 for \(f(x)=e^x\). Plot \(f(x)\) and all 4 polynomials on the same set of axes, clearly labeling each function.

Problem Statement

Find the Maclaurin polynomials of degrees 1, 2, 3 and 5 for \(f(x)=e^x\). Plot \(f(x)\) and all 4 polynomials on the same set of axes, clearly labeling each function.

Solution

The Taylor polynomial we want is of the form

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

where n is the order of the polynomial.

The table below shows the information we need to build the Taylor polynomial.

\(k\)

\(f^{(k)}(x)\)

\(f^{(k)}(0)\)

0

\(f(x)=e^x\)

\(f(0)=1\)

1

\(f'(x)=e^x\)

\(f'(0)=1\)

2

\(f''(x)=e^x\)

\(f''(0)=1\)

3

\(f^{(3)}(x)=e^x\)

\(f^{(3)}(0)=1\)

4

\(f^{(4)}(x)=e^x\)

\(f^{(4)}(0)=1\)

5

\(f^{(5)}(x)=e^x\)

\(f^{(5)}(0)=1\)

First Order Polynomial
\(\displaystyle{p_1(x) = f(0) + \frac{f'(0)x}{1} = 1+x}\)

Second Order Polynomial
\(\displaystyle{p_2(x) = f(0) + \frac{f'(0)x}{1} + \frac{f''(0)x^2}{2} = 1+x+\frac{x^2}{2}}\)

Third Order Polynomial

\(\displaystyle{ p_3(x) = f(0) + \frac{f'(0)x}{1} + \frac{f''(0)x^2}{2} + \frac{f^{3}(0)x^3}{3!} = 1+x+\frac{x^2}{2} +\frac{x^3}{6} }\)

Fifth Order Polynomial
\(\displaystyle{ p_5(x) = f(0) + \frac{f'(0)x}{1} + \frac{f''(0)x^2}{2} + \frac{f^{(3)}(0)x^3}{3!} + \frac{f^{(4)}(0)x^4}{4!} + \frac{f^{(5)}(0)x^5}{5!} = 1+x+\frac{x^2}{2} +\frac{x^3}{6} + \frac{x^4}{24} + \frac{x^5}{120} }\)

close solution

Write the Maclaurin series for \( f(x) = 3x^3+4x^2-2x+1 \)

Problem Statement

Write the Maclaurin series for \( f(x) = 3x^3+4x^2-2x+1 \)

Solution

353 solution video

video by MIT OCW

close solution

Use a known series to build the Maclaurin series for \(e^{-3x}\).

Problem Statement

Use a known series to build the Maclaurin series for \(e^{-3x}\).

Final Answer

\(\displaystyle{ \sum_{n=0}^{\infty}{ \frac{(-3x)^n}{n!} } }\)

Problem Statement

Use a known series to build the Maclaurin series for \(e^{-3x}\).

Solution

357 solution video

video by Krista King Math

Final Answer

\(\displaystyle{ \sum_{n=0}^{\infty}{ \frac{(-3x)^n}{n!} } }\)

close solution

Find the 3rd degree Maclaurin polynomial for \( f(x)=e^{4x} \)

Problem Statement

Find the 3rd degree Maclaurin polynomial for \( f(x)=e^{4x} \)

Final Answer

\(\displaystyle{ 1+4x+8x^2+(32/3)x^3} \)

Problem Statement

Find the 3rd degree Maclaurin polynomial for \( f(x)=e^{4x} \)

Solution

360 solution video

video by PatrickJMT

Final Answer

\(\displaystyle{ 1+4x+8x^2+(32/3)x^3} \)

close solution

Find the Maclaurin series of \(\displaystyle{ f(x) = e^{5x} }\)

Problem Statement

Find the Maclaurin series of \(\displaystyle{ f(x) = e^{5x} }\)

Final Answer

\(\displaystyle{ \sum_{n=0}^{\infty}{ 5^n\frac{x^n}{n!} } }\)

Problem Statement

Find the Maclaurin series of \(\displaystyle{ f(x) = e^{5x} }\)

Solution

362 solution video

video by PatrickJMT

Final Answer

\(\displaystyle{ \sum_{n=0}^{\infty}{ 5^n\frac{x^n}{n!} } }\)

close solution

Find the Maclaurin series of \(\displaystyle{ f(x) = \sin(x) }\)

Problem Statement

Find the Maclaurin series of \(\displaystyle{ f(x) = \sin(x) }\)

Final Answer

\(\displaystyle{ \sum_{n=0}^{\infty}{ \frac{(-1)^nx^{2n+1}}{(2n+1)!} } }\)

Problem Statement

Find the Maclaurin series of \(\displaystyle{ f(x) = \sin(x) }\)

Solution

364 solution video

video by PatrickJMT

Final Answer

\(\displaystyle{ \sum_{n=0}^{\infty}{ \frac{(-1)^nx^{2n+1}}{(2n+1)!} } }\)

close solution

Compute the first four non-zero terms of the Maclaurin polynomial for \(f(x)=\sqrt{1-x}, x \leq 1\).

Problem Statement

Compute the first four non-zero terms of the Maclaurin polynomial for \(f(x)=\sqrt{1-x}, x \leq 1\).

Solution

1867 solution video

video by Dr Chris Tisdell

close solution

Intermediate Problems

Calculate the second order Maclaurin series polynomial for \(\displaystyle{ f(t) = \frac{1}{\mu} \ln\abs{\cosh(\omega t)} }\).

Problem Statement

Calculate the second order Maclaurin series polynomial for \(\displaystyle{ f(t) = \frac{1}{\mu} \ln\abs{\cosh(\omega t)} }\).

Final Answer

\(\displaystyle{ f(t) = \frac{1}{\mu} \ln\abs{\cosh(\omega t)} \approx \frac{\omega^2 t^2}{2\mu} }\)

Problem Statement

Calculate the second order Maclaurin series polynomial for \(\displaystyle{ f(t) = \frac{1}{\mu} \ln\abs{\cosh(\omega t)} }\).

Solution

The Taylor polynomial we want is of the form
\(\displaystyle{ \sum_{k=0}^{2}{\frac{f^{(k)}(a)}{k!}(t-a)^k} = f(0) + f'(0)t + \frac{f''(0)t^2}{2} }\)

\(\displaystyle{ f(t)=\frac{1}{\mu}\ln\abs{\cosh(\omega t)} }\)

\(\displaystyle{ f'(t)=\frac{1}{\mu}\frac{\omega\sinh(\omega t)}{\cosh(\omega t)} }\)

\(\displaystyle{ f''(t)=\frac{\omega}{\mu\cosh^2(\omega t)} }\) \(\displaystyle{ \left[ \omega\cosh^2(\omega t)-\omega\sinh^2(\omega t) \right]}\)

When we plug in \(t=0\) in each of the above equations, we get these values.

\(f(0)=0\)

\(f'(0)=0\)

\(\displaystyle{f''(0)=\frac{\omega^2}{\mu}}\)

Final Answer

\(\displaystyle{ f(t) = \frac{1}{\mu} \ln\abs{\cosh(\omega t)} \approx \frac{\omega^2 t^2}{2\mu} }\)

close solution

Compute the first few terms of the Maclaurin series for \( f(x)=\sec(x) \)

Problem Statement

Compute the first few terms of the Maclaurin series for \( f(x)=\sec(x) \)

Final Answer

\(\displaystyle{ f(x) = \sec(x) = 1 + \frac{x^2}{x} + \frac{5x^4}{24} + \frac{61x^6}{720} + \cdots }\)

Problem Statement

Compute the first few terms of the Maclaurin series for \( f(x)=\sec(x) \)

Solution

354 solution video

video by MIT OCW

Final Answer

\(\displaystyle{ f(x) = \sec(x) = 1 + \frac{x^2}{x} + \frac{5x^4}{24} + \frac{61x^6}{720} + \cdots }\)

close solution

Find the third order Taylor polynomial at \(x=1\) for \( f(x)=x^3-3x^2+2x+1 \)

Problem Statement

Find the third order Taylor polynomial at \(x=1\) for \( f(x)=x^3-3x^2+2x+1 \)

Final Answer

\( f(x) = x^3-3x^2+2x+1 = 1-(x-1)+(x-1)^3 \)

Problem Statement

Find the third order Taylor polynomial at \(x=1\) for \( f(x)=x^3-3x^2+2x+1 \)

Solution

This problem is solved in two videos. Most of the work is done in the first video and the answer is finally determined in the first few minutes of the second video. The rest of the second video shows her checking her work. This is good to do when you can since you can often find mistakes and learn a lot when you check your answers.

358 solution video

video by Krista King Math

358 solution video

video by Krista King Math

Final Answer

\( f(x) = x^3-3x^2+2x+1 = 1-(x-1)+(x-1)^3 \)

close solution

Find the Maclaurin series of \(\displaystyle{ \frac{1}{1+x} }\).

Problem Statement

Find the Maclaurin series of \(\displaystyle{ \frac{1}{1+x} }\).

Final Answer

\(\displaystyle{ \frac{1}{1+x} = \sum_{n=0}^{\infty}{ (-1)^n x^n } }\)

Problem Statement

Find the Maclaurin series of \(\displaystyle{ \frac{1}{1+x} }\).

Solution

359 solution video

video by PatrickJMT

Final Answer

\(\displaystyle{ \frac{1}{1+x} = \sum_{n=0}^{\infty}{ (-1)^n x^n } }\)

close solution

Find the third order Maclaurin polynomial of \(\displaystyle{ \frac{1}{5x+1} }\).

Problem Statement

Find the third order Maclaurin polynomial of \(\displaystyle{ \frac{1}{5x+1} }\).

Final Answer

\(\displaystyle{ 1-5x+25x^2-125x^3 }\)

Problem Statement

Find the third order Maclaurin polynomial of \(\displaystyle{ \frac{1}{5x+1} }\).

Solution

361 solution video

video by PatrickJMT

Final Answer

\(\displaystyle{ 1-5x+25x^2-125x^3 }\)

close solution

Find the Taylor series for \( f(x)=\ln(x) \) at \(x=2\)

Problem Statement

Find the Taylor series for \( f(x)=\ln(x) \) at \(x=2\)

Final Answer

\(\displaystyle{ \ln(2)+\sum_{n=1}^{\infty}{ (-1)^{n+1}\frac{(x-2)^n}{n\cdot 2^n} } }\)

Problem Statement

Find the Taylor series for \( f(x)=\ln(x) \) at \(x=2\)

Solution

363 solution video

video by PatrickJMT

Final Answer

\(\displaystyle{ \ln(2)+\sum_{n=1}^{\infty}{ (-1)^{n+1}\frac{(x-2)^n}{n\cdot 2^n} } }\)

close solution

Find the cubic approximation for \( f(x) = \sin(x) \) at \( x = 3\pi/2 \)

Problem Statement

Find the cubic approximation for \( f(x) = \sin(x) \) at \( x = 3\pi/2 \)

Final Answer

\(\displaystyle{ \sin(x) \approx -1 + \frac{1}{2}(x- 3\pi/2)^2 }\)

Problem Statement

Find the cubic approximation for \( f(x) = \sin(x) \) at \( x = 3\pi/2 \)

Solution

365 solution video

video by PatrickJMT

Final Answer

\(\displaystyle{ \sin(x) \approx -1 + \frac{1}{2}(x- 3\pi/2)^2 }\)

close solution

Find the quadratic approximation for \( f(x) = \sqrt{4+x} \) at \(x=0\)

Problem Statement

Find the quadratic approximation for \( f(x) = \sqrt{4+x} \) at \(x=0\)

Final Answer

\(\displaystyle{ \sqrt{4+x} \approx 2 + x/4 - x^2/64 }\)

Problem Statement

Find the quadratic approximation for \( f(x) = \sqrt{4+x} \) at \(x=0\)

Solution

366 solution video

video by PatrickJMT

Final Answer

\(\displaystyle{ \sqrt{4+x} \approx 2 + x/4 - x^2/64 }\)

close solution

Find the third degree Taylor polynomial for \(\displaystyle{\frac{1}{2+x}}\) at \(x=1\)

Problem Statement

Find the third degree Taylor polynomial for \(\displaystyle{\frac{1}{2+x}}\) at \(x=1\)

Final Answer

\(\displaystyle{ \frac{1}{2+x} \approx \frac{1}{3} + \frac{x-1}{9} + \frac{(x-1)^2)}{27} + \frac{(x-1)^3)}{81} }\)

Problem Statement

Find the third degree Taylor polynomial for \(\displaystyle{\frac{1}{2+x}}\) at \(x=1\)

Solution

367 solution video

video by PatrickJMT

Final Answer

\(\displaystyle{ \frac{1}{2+x} \approx \frac{1}{3} + \frac{x-1}{9} + \frac{(x-1)^2)}{27} + \frac{(x-1)^3)}{81} }\)

close solution

Find the second degree Taylor polynomial for \(\displaystyle{ f(x) = \cot(x) }\) at \(x=\pi/4\)

Problem Statement

Find the second degree Taylor polynomial for \(\displaystyle{ f(x) = \cot(x) }\) at \(x=\pi/4\)

Final Answer

\(\displaystyle{ \cot(x) \approx 1 - 2(x-\pi/4) + 2(x-\pi/4)^2 }\)

Problem Statement

Find the second degree Taylor polynomial for \(\displaystyle{ f(x) = \cot(x) }\) at \(x=\pi/4\)

Solution

368 solution video

video by PatrickJMT

Final Answer

\(\displaystyle{ \cot(x) \approx 1 - 2(x-\pi/4) + 2(x-\pi/4)^2 }\)

close solution

Use a known series to find the sum \(\displaystyle{ \sum_{n=0}^{\infty}{ (-1)^n \frac{x^{4n}}{n!} } }\)

Problem Statement

Use a known series to find the sum \(\displaystyle{ \sum_{n=0}^{\infty}{ (-1)^n \frac{x^{4n}}{n!} } }\)

Solution

1331 solution video

video by Krista King Math

close solution

Use a known series to find the sum \(\displaystyle{ \sum_{n=0}^{\infty}{ \frac{(-1)^n\pi^{2n+1}}{4^{2n+1}(2n+1)!} } }\)

Problem Statement

Use a known series to find the sum \(\displaystyle{ \sum_{n=0}^{\infty}{ \frac{(-1)^n\pi^{2n+1}}{4^{2n+1}(2n+1)!} } }\)

Solution

1332 solution video

video by Krista King Math

close solution

Advanced Problems

Approximate \( \int_{0}^{1}{ x~\cos(x^3) ~dx } \) to within 3 decimal places.

Problem Statement

Approximate \( \int_{0}^{1}{ x~\cos(x^3) ~dx } \) to within 3 decimal places.

Final Answer

\(\displaystyle{ \int_{0}^{1}{ x~\cos(x^3) ~dx }\approx 0.440 }\)

Problem Statement

Approximate \( \int_{0}^{1}{ x~\cos(x^3) ~dx } \) to within 3 decimal places.

Solution

369 solution video

video by PatrickJMT

Final Answer

\(\displaystyle{ \int_{0}^{1}{ x~\cos(x^3) ~dx }\approx 0.440 }\)

close solution

Use the Maclaurin series to evaluate \(\displaystyle{ \lim_{x\to 0}{ \frac{x-\arctan(x)}{x^3} } }\)

Problem Statement

Use the Maclaurin series to evaluate \(\displaystyle{ \lim_{x\to 0}{ \frac{x-\arctan(x)}{x^3} } }\)

Final Answer

\(\displaystyle{ \lim_{x\to 0}{ \frac{x-\arctan(x)}{x^3} } = 1/3 }\)

Problem Statement

Use the Maclaurin series to evaluate \(\displaystyle{ \lim_{x\to 0}{ \frac{x-\arctan(x)}{x^3} } }\)

Solution

370 solution video

video by PatrickJMT

Final Answer

\(\displaystyle{ \lim_{x\to 0}{ \frac{x-\arctan(x)}{x^3} } = 1/3 }\)

close solution
Real Time Web Analytics