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
$R=3.$
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.
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.
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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?