Power Series Expanded, Arfken 5.7.16

In summary, a power series expansion is a mathematical tool used to represent a function as an infinite sum of terms. The general form of a power series expansion is given by <em>f(x) = c<sub>0</sub> + c<sub>1</sub>(x-a) + c<sub>2</sub>(x-a)^2 + c<sub>3</sub>(x-a)^3 + ...</em>, where <em>c<sub>0</sub>, c<sub>1</sub>, c<sub>2</sub>, ...</em> are constant coefficients and <em>a</em> is the center of expansion. It is often used in mathematical analysis to approximate difficult functions
  • #1
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Homework Statement


The behavior of a neutron losing energy by colliding elastically with nuclei of mass A is described by a parameter ξ1,
ξ1 = 1 + [itex]\frac{(A-1)^2}{2A}[/itex]*ln[itex]\frac{A-1}{A+1}[/itex]

An approximation, good for large A, is
ξ2= [itex]\frac{2}{A+2/3}[/itex]

Expand ξ1 and ξ2 in powers of A−1. Show that ξ2 agrees with ξ1 through (A−1)2. Find the difference in the coefficients of the (A−1)3 term.

Homework Equations



Taylor Expansion, see wiki page

ln(1-x)= x+ [itex]\frac{x^2}{2}[/itex]+ [itex]\frac{x^3}{3}[/itex]+...+[itex]\frac{x^n}{n}[/itex]

ln(1+x)= x- [itex]\frac{x^2}{2}[/itex]+ [itex]\frac{x^3}{3}[/itex]+...+(-1)^(n+1)[itex]\frac{(x^n)}{n}[/itex]

The Attempt at a Solution



Alright so since it is in powers of A−1 I decided to substitute x=A−1. I get:

ξ1= 1+ [itex]\frac{x((1/x)-1)^2}{2}[/itex]* ln(1-x)-ln(1+x)

I use the expansions from above and substitute.
Now I do not know how to expand the entire thing from here.
 
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  • #2
citra said:

Homework Statement


The behavior of a neutron losing energy by colliding elastically with nuclei of mass A is described by a parameter ξ1,
ξ1 = 1 + [itex]\frac{(A-1)^2}{2A}[/itex]*ln[itex]\frac{A-1}{A+1}[/itex]

An approximation, good for large A, is
ξ2= [itex]\frac{2}{A+2/3}[/itex]

Expand ξ1 and ξ2 in powers of A−1. Show that ξ2 agrees with ξ1 through (A−1)2. Find the difference in the coefficients of the (A−1)3 term.

Homework Equations



Taylor Expansion, see wiki page

ln(1-x)= x+ [itex]\frac{x^2}{2}[/itex]+ [itex]\frac{x^3}{3}[/itex]+...+[itex]\frac{x^n}{n}[/itex]

ln(1+x)= x- [itex]\frac{x^2}{2}[/itex]+ [itex]\frac{x^3}{3}[/itex]+...+(-1)^(n+1)[itex]\frac{(x^n)}{n}[/itex]
Suggestion: Use LaTeX for the entire expression instead of for just bits and pieces.

You should clean up your math a bit. Those expressions above aren't true because the series on the righthand sides have an infinite number of terms. They don't terminate with the nth term.

The Attempt at a Solution



Alright so since it is in powers of A−1 I decided to substitute x=A−1. I get:

ξ1= 1+ [itex]\frac{x((1/x)-1)^2}{2}[/itex]* ln(1-x)-ln(1+x)

I use the expansions from above and substitute.
Now I do not know how to expand the entire thing from here.
Again, what you wrote is incorrect. Use parentheses (or square brackets, etc.)!
[tex]\xi_1 = 1+\frac{x}{2}\left(\frac{1}{x}-1\right)^2[\log(1-x)-\log(1+x)][/tex]You just need to substitute the series in, combine terms, and then multiply everything out and simplify. In other words, do a bunch of algebra. Note that since you're only looking for the first few terms, those are the only ones you really need to worry about. Anything that results in higher powers of x you can ignore.
 

What is a power series expansion?

A power series expansion is a mathematical tool used to represent a function as an infinite sum of terms. Each term in the series is a polynomial multiplied by a constant factor, and as the number of terms increases, the approximation of the original function becomes more accurate.

What is the equation for a power series expansion?

The general form of a power series expansion is given by f(x) = c0 + c1(x-a) + c2(x-a)^2 + c3(x-a)^3 + ..., where c0, c1, c2, ... are constant coefficients and a is the center of expansion.

What is the purpose of a power series expansion?

A power series expansion is often used in mathematical analysis to approximate a function that is difficult or impossible to evaluate directly. It can also be used to find the derivatives and integrals of a function, as well as to solve differential equations.

What is the interval of convergence for a power series expansion?

The interval of convergence is the range of values for which the power series expansion accurately represents the original function. It is determined by the values of x that make the infinite series converge, and may include the center of expansion as well as some or all of the points on the boundary.

How do you determine the coefficients in a power series expansion?

The coefficients in a power series expansion can be found by using the Taylor series formula, which involves taking the derivatives of the function at the center of expansion and evaluating them at the center. Alternatively, the coefficients can be determined using the Cauchy integral formula, which involves integrating the function along a contour surrounding the center of expansion.

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