is a geometric series that converges if
$\left|\frac{x}{d}\right|<1.$ Therefore, by the comparison test, we conclude that
$\sum _{n=N}^{\infty}{c}_{n}{x}^{n}$ also converges for
$\left|x\right|<\left|d\right|.$ 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
$\left|x\right|<\left|d\right|.$
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
$\left|x\right|>\left|x*\right|.$ 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
$\left|x\right|<R,$ 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
$\left|x-a\right|<R$ 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
$\left|x-a\right|=R.$ 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
$\left|x-a\right|>R,$ 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
$\left(\mathrm{-1},1\right)$ and diverges for all values
x such that
$\left|x\right|\ge 1,$ the interval of convergence of this series is
$\left(\mathrm{-1},1\right).$ 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
$\left|x-a\right|<R$ and diverges for
$\left|x-a\right|>R,$ 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.
for all values of
x . Therefore, the series converges for all real numbers
x . The interval of convergence is
$\left(\text{\u2212}\infty ,\infty \right)$ and the radius of convergence is
$R=\infty .$
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
$R=0.$
The ratio
$\rho <1$ if
$\left|x-2\right|<3.$ Since
$\left|x-2\right|<3$ implies that
$\mathrm{-3}<x-2<3,$ the series converges absolutely if
$\mathrm{-1}<x<5.$ The ratio
$\rho >1$ if
$\left|x-2\right|>3.$ Therefore, the series diverges if
$x<\mathrm{-1}$ or
$x>5.$ The ratio test is inconclusive if
$\rho =1.$ The ratio
$\rho =1$ if and only if
$x=\mathrm{-1}$ or
$x=5.$ We need to test these values of
x separately. For
$x=\mathrm{-1},$ the series is given by
This is the harmonic series, which is divergent. Therefore, the power series diverges at
$x=5.$ We conclude that the interval of convergence is
$\left[\mathrm{-1},5\right)$ and the radius of convergence is
$R=3.$
fullerene is a bucky ball aka Carbon 60 molecule. It was name by the architect Fuller. He design the geodesic dome. it resembles a soccer ball.
Tarell
what is the actual application of fullerenes nowadays?
Damian
That is a great question Damian. best way to answer that question is to Google it. there are hundreds of applications for buck minister fullerenes, from medical to aerospace. you can also find plenty of research papers that will give you great detail on the potential applications of fullerenes.
Tarell
what is the Synthesis, properties,and applications of carbon nano chemistry
Yeah, it is a pain to say the least. You basically have to heat the substarte up to around 1000 degrees celcius then pass phosphene gas over top of it, which is explosive and toxic by the way, under very low pressure.
Harper
Do you know which machine is used to that process?
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?