6.1 Power series and functions  (Page 2/8)

 Page 2 / 8
$|{c}_{n}{x}^{n}|=|{c}_{n}{d}^{n}|{|\frac{x}{d}|}^{n},$

we conclude that, for all $n\ge N,$

$|{c}_{n}{x}^{n}|\le {|\frac{x}{d}|}^{n}.$

The series

$\sum _{n=N}^{\infty }{|\frac{x}{d}|}^{n}$

is a geometric series that converges if $|\frac{x}{d}|<1.$ Therefore, by the comparison test, we conclude that $\sum _{n=N}^{\infty }{c}_{n}{x}^{n}$ also converges for $|x|<|d|.$ Since we can add a finite number of terms to a convergent series, we conclude that $\sum _{n=0}^{\infty }{c}_{n}{x}^{n}$ converges for $|x|<|d|.$

With this result, we can now prove the theorem. Consider the series

$\sum _{n=0}^{\infty }{a}_{n}{x}^{n}$

and let S be the set of real numbers for which the series converges. Suppose that the set $S=\left\{0\right\}.$ Then the series falls under case i. Suppose that the set S is the set of all real numbers. Then the series falls under case ii. Suppose that $S\ne \left\{0\right\}$ and S is not the set of real numbers. Then there exists a real number $x*\ne 0$ such that the series does not converge. Thus, the series cannot converge for any x such that $|x|>|x*|.$ Therefore, the set S must be a bounded set, which means that it must have a smallest upper bound. (This fact follows from the Least Upper Bound Property for the real numbers, which is beyond the scope of this text and is covered in real analysis courses.) Call that smallest upper bound R . Since $S\ne \left\{0\right\},$ the number $R>0.$ Therefore, the series converges for all x such that $|x| and the series falls into case iii.

If a series $\sum _{n=0}^{\infty }{c}_{n}{\left(x-a\right)}^{n}$ falls into case iii. of [link] , then the series converges for all x such that $|x-a| for some $R>0,$ and diverges for all x such that $|x-a|>R.$ The series may converge or diverge at the values x where $|x-a|=R.$ The set of values x for which the series $\sum _{n=0}^{\infty }{c}_{n}{\left(x-a\right)}^{n}$ converges is known as the interval of convergence    . Since the series diverges for all values x where $|x-a|>R,$ the length of the interval is 2 R , and therefore, the radius of the interval is R . The value R is called the radius of convergence    . For example, since the series $\sum _{n=0}^{\infty }{x}^{n}$ converges for all values x in the interval $\left(-1,1\right)$ and diverges for all values x such that $|x|\ge 1,$ the interval of convergence of this series is $\left(-1,1\right).$ Since the length of the interval is 2, the radius of convergence is 1.

Definition

Consider the power series $\sum _{n=0}^{\infty }{c}_{n}{\left(x-a\right)}^{n}.$ The set of real numbers x where the series converges is the interval of convergence. If there exists a real number $R>0$ such that the series converges for $|x-a| and diverges for $|x-a|>R,$ then R is the radius of convergence. If the series converges only at $x=a,$ we say the radius of convergence is $R=0.$ If the series converges for all real numbers x , we say the radius of convergence is $R=\infty$ ( [link] ).

To determine the interval of convergence for a power series, we typically apply the ratio test. In [link] , we show the three different possibilities illustrated in [link] .

Finding the interval and radius of convergence

For each of the following series, find the interval and radius of convergence.

1. $\sum _{n=0}^{\infty }\frac{{x}^{n}}{n\text{!}}$
2. $\sum _{n=0}^{\infty }n\text{!}{x}^{n}$
3. $\sum _{n=0}^{\infty }\frac{{\left(x-2\right)}^{n}}{\left(n+1\right){3}^{n}}$
1. To check for convergence, apply the ratio test. We have
$\begin{array}{cc}\hfill \rho & =\underset{n\to \infty }{\text{lim}}|\frac{\frac{{x}^{n+1}}{\left(n+1\right)\text{!}}}{\frac{{x}^{n}}{n\text{!}}}|\hfill \\ & =\underset{n\to \infty }{\text{lim}}|\frac{{x}^{n+1}}{\left(n+1\right)\text{!}}·\frac{n\text{!}}{{x}^{n}}|\hfill \\ & =\underset{n\to \infty }{\text{lim}}|\frac{{x}^{n+1}}{\left(n+1\right)·n\text{!}}·\frac{n\text{!}}{{x}^{n}}|\hfill \\ & =\underset{n\to \infty }{\text{lim}}|\frac{x}{n+1}|\hfill \\ & =|x|\underset{n\to \infty }{\text{lim}}\frac{1}{n+1}\hfill \\ & =0<1\hfill \end{array}$

for all values of x . Therefore, the series converges for all real numbers x . The interval of convergence is $\left(\text{−}\infty ,\infty \right)$ and the radius of convergence is $R=\infty .$
2. Apply the ratio test. For $x\ne 0,$ we see that
$\begin{array}{cc}\hfill \rho & \hfill =\underset{n\to \infty }{\text{lim}}|\frac{\left(n+1\right)\text{!}{x}^{n+1}}{n\text{!}{x}^{n}}|\\ & =\underset{n\to \infty }{\text{lim}}|\left(n+1\right)x|\hfill \\ & =|x|\underset{n\to \infty }{\text{lim}}\left(n+1\right)\hfill \\ & =\infty .\hfill \end{array}$

Therefore, the series diverges for all $x\ne 0.$ Since the series is centered at $x=0,$ it must converge there, so the series converges only for $x\ne 0.$ The interval of convergence is the single value $x=0$ and the radius of convergence is $R=0.$
3. In order to apply the ratio test, consider
$\begin{array}{cc}\hfill \rho & =\underset{n\to \infty }{\text{lim}}|\frac{\frac{{\left(x-2\right)}^{n+1}}{\left(n+2\right){3}^{n+1}}}{\frac{{\left(x-2\right)}^{n}}{\left(n+1\right){3}^{n}}}|\hfill \\ & =\underset{n\to \infty }{\text{lim}}|\frac{{\left(x-2\right)}^{n+1}}{\left(n+2\right){3}^{n+1}}·\frac{\left(n+1\right){3}^{n}}{{\left(x-2\right)}^{n}}|\hfill \\ & =\underset{n\to \infty }{\text{lim}}|\frac{\left(x-2\right)\left(n+1\right)}{3\left(n+2\right)}|\hfill \\ & =\frac{|x-2|}{3}.\hfill \end{array}$

The ratio $\rho <1$ if $|x-2|<3.$ Since $|x-2|<3$ implies that $-3 the series converges absolutely if $-1 The ratio $\rho >1$ if $|x-2|>3.$ Therefore, the series diverges if $x<-1$ or $x>5.$ The ratio test is inconclusive if $\rho =1.$ The ratio $\rho =1$ if and only if $x=-1$ or $x=5.$ We need to test these values of x separately. For $x=-1,$ the series is given by
$\sum _{n=0}^{\infty }\frac{{\left(-1\right)}^{n}}{n+1}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\text{⋯}.$

Since this is the alternating harmonic series, it converges. Thus, the series converges at $x=-1.$ For $x=5,$ the series is given by
$\sum _{n=0}^{\infty }\frac{1}{n+1}=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\text{⋯}.$

This is the harmonic series, which is divergent. Therefore, the power series diverges at $x=5.$ We conclude that the interval of convergence is $\left[-1,5\right)$ and the radius of convergence is $R=3.$

find the 15th term of the geometric sequince whose first is 18 and last term of 387
The given of f(x=x-2. then what is the value of this f(3) 5f(x+1)
hmm well what is the answer
Abhi
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?
hmm
Abhi
is it a question of log
Abhi
🤔.
Abhi
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
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
Leaves accumulate on the forest floor at a rate of 2 g/cm2/yr and also decompose at a rate of 90% per year. Write a differential equation governing the number of grams of leaf litter per square centimeter of forest floor, assuming at time 0 there is no leaf litter on the ground. Does this amount approach a steady value? What is that value?
You have a cup of coffee at temperature 70°C, which you let cool 10 minutes before you pour in the same amount of milk at 1°C as in the preceding problem. How does the temperature compare to the previous cup after 10 minutes?
Abdul