# 6.3 Taylor and maclaurin series

 Page 1 / 13
• Describe the procedure for finding a Taylor polynomial of a given order for a function.
• Explain the meaning and significance of Taylor’s theorem with remainder.
• Estimate the remainder for a Taylor series approximation of a given function.

In the previous two sections we discussed how to find power series representations for certain types of functions––specifically, functions related to geometric series. Here we discuss power series representations for other types of functions. In particular, we address the following questions: Which functions can be represented by power series and how do we find such representations? If we can find a power series representation for a particular function $f$ and the series converges on some interval, how do we prove that the series actually converges to $f?$

## Overview of taylor/maclaurin series

Consider a function $f$ that has a power series representation at $x=a.$ Then the series has the form

$\sum _{n=0}^{\infty }{c}_{n}{\left(x-a\right)}^{n}={c}_{0}+{c}_{1}\left(x-a\right)+{c}_{2}{\left(x-a\right)}^{2}+\text{⋯}.$

What should the coefficients be? For now, we ignore issues of convergence, but instead focus on what the series should be, if one exists. We return to discuss convergence later in this section. If the series [link] is a representation for $f$ at $x=a,$ we certainly want the series to equal $f\left(a\right)$ at $x=a.$ Evaluating the series at $x=a,$ we see that

$\begin{array}{cc}\hfill \sum _{n=0}^{\infty }{c}_{n}{\left(x-a\right)}^{n}& ={c}_{0}+{c}_{1}\left(a-a\right)+{c}_{2}{\left(a-a\right)}^{2}+\text{⋯}\hfill \\ & ={c}_{0}.\hfill \end{array}$

Thus, the series equals $f\left(a\right)$ if the coefficient ${c}_{0}=f\left(a\right).$ In addition, we would like the first derivative of the power series to equal ${f}^{\prime }\left(a\right)$ at $x=a.$ Differentiating [link] term-by-term, we see that

$\frac{d}{dx}\left(\sum _{n=0}^{\infty }{c}_{n}{\left(x-a\right)}^{n}\right)={c}_{1}+2{c}_{2}\left(x-a\right)+3{c}_{3}{\left(x-a\right)}^{2}+\text{⋯}.$

Therefore, at $x=a,$ the derivative is

$\begin{array}{}\\ \\ \hfill \frac{d}{dx}\left(\sum _{n=0}^{\infty }{c}_{n}{\left(x-a\right)}^{n}\right)& ={c}_{1}+2{c}_{2}\left(a-a\right)+3{c}_{3}{\left(a-a\right)}^{2}+\text{⋯}\hfill \\ & ={c}_{1}.\hfill \end{array}$

Therefore, the derivative of the series equals ${f}^{\prime }\left(a\right)$ if the coefficient ${c}_{1}={f}^{\prime }\left(a\right).$ Continuing in this way, we look for coefficients c n such that all the derivatives of the power series [link] will agree with all the corresponding derivatives of $f$ at $x=a.$ The second and third derivatives of [link] are given by

$\frac{{d}^{2}}{d{x}^{2}}\left(\sum _{n=0}^{\infty }{c}_{n}{\left(x-a\right)}^{n}\right)=2{c}_{2}+3·2{c}_{3}\left(x-a\right)+4·3{c}_{4}{\left(x-a\right)}^{2}+\text{⋯}$

and

$\frac{{d}^{3}}{d{x}^{3}}\left(\sum _{n=0}^{\infty }{c}_{n}{\left(x-a\right)}^{n}\right)=3·2{c}_{3}+4·3·2{c}_{4}\left(x-a\right)+5·4·3{c}_{5}{\left(x-a\right)}^{2}+\text{⋯}.$

Therefore, at $x=a,$ the second and third derivatives

$\begin{array}{cc}\hfill \frac{{d}^{2}}{d{x}^{2}}\left(\sum _{n=0}^{\infty }{c}_{n}{\left(x-a\right)}^{n}\right)& =2{c}_{2}+3·2{c}_{3}\left(a-a\right)+4·3{c}_{4}{\left(a-a\right)}^{2}+\text{⋯}\hfill \\ & =2{c}_{2}\hfill \end{array}$

and

$\begin{array}{cc}\hfill \frac{{d}^{3}}{d{x}^{3}}\left(\sum _{n=0}^{\infty }{c}_{n}{\left(x-a\right)}^{n}\right)& =3·2{c}_{3}+4·3·2{c}_{4}\left(a-a\right)+5·4·3{c}_{5}{\left(a-a\right)}^{2}+\text{⋯}\hfill \\ & =3·2{c}_{3}\hfill \end{array}$

equal $f\text{″}\left(a\right)$ and $f\text{‴}\left(a\right),$ respectively, if ${c}_{2}=\frac{f\text{″}\left(a\right)}{2}$ and ${c}_{3}=\frac{f\text{‴}\left(a\right)}{3}·2.$ More generally, we see that if $f$ has a power series representation at $x=a,$ then the coefficients should be given by ${c}_{n}=\frac{{f}^{\left(n\right)}\left(a\right)}{n\text{!}}.$ That is, the series should be

$\sum _{n=0}^{\infty }\frac{{f}^{\left(n\right)}\left(a\right)}{n\text{!}}{\left(x-a\right)}^{n}=f\left(a\right)+{f}^{\prime }\left(a\right)\left(x-a\right)+\frac{f\text{″}\left(a\right)}{2\text{!}}{\left(x-a\right)}^{2}+\frac{f\text{‴}\left(a\right)}{3\text{!}}{\left(x-a\right)}^{3}+\text{⋯}.$

This power series for $f$ is known as the Taylor series for $f$ at $a.$ If $x=0,$ then this series is known as the Maclaurin series for $f.$

## Definition

If $f$ has derivatives of all orders at $x=a,$ then the Taylor series    for the function $f$ at $a$ is

$\sum _{n=0}^{\infty }\frac{{f}^{\left(n\right)}\left(a\right)}{n\text{!}}{\left(x-a\right)}^{n}=f\left(a\right)+{f}^{\prime }\left(a\right)\left(x-a\right)+\frac{f\text{″}\left(a\right)}{2\text{!}}{\left(x-a\right)}^{2}+\text{⋯}+\frac{{f}^{\left(n\right)}\left(a\right)}{n\text{!}}{\left(x-a\right)}^{n}+\text{⋯}.$

The Taylor series for $f$ at 0 is known as the Maclaurin series    for $f.$

Later in this section, we will show examples of finding Taylor series and discuss conditions under which the Taylor series for a function will converge to that function. Here, we state an important result. Recall from [link] that power series representations are unique. Therefore, if a function $f$ has a power series at $a,$ then it must be the Taylor series for $f$ at $a.$

how do they get the third part x = (32)5/4
can someone help me with some logarithmic and exponential equations.
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
I don't understand what the A with approx sign and the boxed x mean
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
oops. ignore that.
so you not have an equal sign anywhere in the original equation?
Commplementary angles
hello
Sherica
im all ears I need to learn
Sherica
right! what he said ⤴⤴⤴
Tamia
hii
Uday
what is a good calculator for all algebra; would a Casio fx 260 work with all algebra equations? please name the cheapest, thanks.
a perfect square v²+2v+_
kkk nice
algebra 2 Inequalities:If equation 2 = 0 it is an open set?
or infinite solutions?
Kim
The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
Al
y=10×
if |A| not equal to 0 and order of A is n prove that adj (adj A = |A|
rolling four fair dice and getting an even number an all four dice
Kristine 2*2*2=8
Differences Between Laspeyres and Paasche Indices
No. 7x -4y is simplified from 4x + (3y + 3x) -7y
is it 3×y ?
J, combine like terms 7x-4y
how do you translate this in Algebraic Expressions
Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
what's the easiest and fastest way to the synthesize AgNP?
China
Cied
types of nano material
I start with an easy one. carbon nanotubes woven into a long filament like a string
Porter
many many of nanotubes
Porter
what is the k.e before it land
Yasmin
what is the function of carbon nanotubes?
Cesar
I'm interested in nanotube
Uday
what is nanomaterials​ and their applications of sensors.
what is nano technology
what is system testing?
preparation of nanomaterial
Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
what is system testing
what is the application of nanotechnology?
Stotaw
In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
Azam
anybody can imagine what will be happen after 100 years from now in nano tech world
Prasenjit
after 100 year this will be not nanotechnology maybe this technology name will be change . maybe aftet 100 year . we work on electron lable practically about its properties and behaviour by the different instruments
Azam
name doesn't matter , whatever it will be change... I'm taking about effect on circumstances of the microscopic world
Prasenjit
how hard could it be to apply nanotechnology against viral infections such HIV or Ebola?
Damian
silver nanoparticles could handle the job?
Damian
not now but maybe in future only AgNP maybe any other nanomaterials
Azam
Hello
Uday
I'm interested in Nanotube
Uday
this technology will not going on for the long time , so I'm thinking about femtotechnology 10^-15
Prasenjit
can nanotechnology change the direction of the face of the world
At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
the Beer law works very well for dilute solutions but fails for very high concentrations. why?
how did you get the value of 2000N.What calculations are needed to arrive at it
Privacy Information Security Software Version 1.1a
Good
Got questions? Join the online conversation and get instant answers!