What is the relationship between power series and exponential functions?

In summary, this equation is wrong because it doesn't take into account the exponential nature of the function.
  • #1
tigigi
38
0
1. I this from my homework solution.

(1-t/s)^n = exp(-t/s)
as n goes to infinity

I don't understand. I checked the exponential power series. It should be :
exp(x) = summation (x^n / n!)
n=0 to infinity

How come it could be a exponential function ?

2. another is that why

<t> = integral from 0 to infinity (t*P(t) dt) ?
average t

P(t)dt = probability that an electron has no colission till time t *
probability that it has a collision between time t

probablitiy has no collision is exp(-t/s)

t+dt = exp(-t/s) *dt/s

Thanks a lot !
 
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  • #2
1. This is obviously wrong:

[tex]\sum_{n=0}^{\infty}(1-t/s}^n \ne \exp(-t/s)[/tex]

You don't need to expand the series. Use t/s=0.5. The series evaluates to 2, which is obviously not [itex]1/\sqrt e[/itex].

I suspect you are missing something here.

2. In general, the expected value of some random variable [itex]X[/itex] with respect to some probability density function [itex]p(t)[/itex] is defined as [itex]\int X p(t)dt[/itex], where the integration is performed over the domain of the PDF. Here, you want the expected value of the time until a collision for an exponentially distributed collision time. BTW, the PDF is [itex]p(t)=\exp(-t/s)/s[/tex], not [itex]\exp(-t/s)[/itex].
 
  • #3
oops, I should make the notation more clear.


the ans says that

( 1- dt/s )^n = exp ( -t/s ) as n goes to infinity

there's no summation here.

I'm wondering where this comes from ?
 
  • #4
This is still not right. This is:

[tex]\exp\left(-\;\frac t s\right) \equiv \lim_{n\to\infty} \left(1-\frac 1 n\;\frac t s\right)^n[/tex]
 
  • #5
It's not a matter of "more clear"- you didn't say anything about a limit before!

Oh, and now you have "dt/s" where before you had "t/s". Was that a typo?

Of course, for fixed t and s, 1- t/s is just a number. If that number is larger than 1, the limit is [itex]\infty[/itex], if that number is less than 1, the limit is 0, if that number is equal to 1, the limit is 1. It certainly is not "e-t/s".

It is true that [itex]\lim_{n\rightarrow \infty} (1+ \frac{x}{n})^n= e^{x}[/itex]
Replacing x by -t/s, we get
[tex]\lim_{n\rightarrow \infty} (1- \frac{x}{ns})^n= e^{-t/s}[/tex]

Is it possible that you are missing an "n" in the denominator?
 
Last edited by a moderator:
  • #6
I got it. Thanks a lot ! :)
 
  • #7
yes, I think so. There's an n missing in the denominator.
 
  • #8
hye D H...i need ur help...do check your private inbox...plssss...
 

What is a power series?

A power series is an infinite series of the form ∑n=0∞ an(x-c)n, where an are coefficients, x is the variable, and c is a constant. It is used to represent a function as a sum of infinitely many terms, each involving powers of the variable x.

What is the difference between a power series and an exponential function?

While both a power series and an exponential function involve the variable x raised to a power, the difference lies in the fact that a power series can include terms with non-integer powers, while an exponential function is defined as f(x) = ax where a is a constant. Additionally, a power series is an infinite series while an exponential function is a single term.

How do you determine the radius of convergence for a power series?

The radius of convergence for a power series is determined by finding the values of x for which the series converges. This can be done using the ratio test, which compares the absolute value of the ratio of consecutive terms to a limit. If the limit is less than 1, the series will converge, and the radius of convergence is the value of x for which the limit is equal to 1.

What is the relationship between a power series and its derivatives?

A power series can be differentiated term by term, meaning that the derivative of a power series is equal to the sum of the derivatives of each term. This is useful for finding the derivatives of functions represented by power series, as it allows us to easily find higher order derivatives.

How are power series used in calculus and physics?

Power series are used in calculus and physics to represent and approximate functions that cannot be expressed in closed form. They are also used in differential equations to find solutions, and in physics to model natural phenomena such as oscillations and wave functions. Additionally, power series are used in numerical analysis to approximate solutions to equations and functions.

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