Representing Airy's function as a power series

Click For Summary
SUMMARY

The discussion focuses on finding the first five non-zero terms of the power series solution to the differential equation d²y/dx² - xy = 0 about the point x = -2, with initial conditions y(-2) = 1 and y'(-2) = 1/2. The Taylor expansion is utilized, leading to the transformation x' = x + 2. A recurrence relation a_(n+3) = [a_n - 2a_(n+1) - 1]/(n + 3)(n + 2) is derived to find additional terms. The need for convergence analysis is also highlighted as a critical aspect of the solution process.

PREREQUISITES
  • Understanding of second-order ordinary differential equations (ODEs)
  • Familiarity with Taylor series and power series expansions
  • Knowledge of recurrence relations in mathematical analysis
  • Basic calculus concepts, including derivatives and initial value problems
NEXT STEPS
  • Study the derivation and application of Taylor series for solving ODEs
  • Learn about convergence criteria for power series solutions
  • Explore methods for solving recurrence relations in mathematical contexts
  • Investigate the properties of Airy functions and their applications in differential equations
USEFUL FOR

Students and researchers in mathematics, particularly those studying differential equations, power series solutions, and mathematical analysis. This discussion is also beneficial for anyone seeking to deepen their understanding of recurrence relations and convergence in series.

Ratpigeon
Messages
52
Reaction score
0

Homework Statement


Find the first five non-zero terms of the power series solution to
d2y/dx2-xy=0 about x=-2; y(-2)=1;y'(-2)=1/2

Homework Equations



... calculus in general?
and the taylor expansion of y(x) is - assuming remainder term is zero:
[itex]\sum[/itex]y(n)(-2)/n! *(x+2)n (n from 0 to infinity)
and y''(x) is:
[itex]\sum[/itex]y(n)(-2)/(n-2)! *(x+2)n-2 (n from 2 to infinity

The Attempt at a Solution


I let x'=x+2, giving:
[itex]\sum[/itex]y(n)(-2)/(n-2)! *(x')n-2-x'[itex]\sum[/itex]y(n)(-2)/n! *(x+2)n +2[itex]\sum[/itex]y(n)(-2)/n! *(x+2)n
(parameters of first sum are from 2 to infinity, of second two sums from 0 to infinity)
But I'm not sure what to do from here - do I need to find y?
 
Physics news on Phys.org
You know already two terms, namely: y(-2) , y'(-2)(x+2).

Now in order to find the other non-zero terms, just use your ODE, equate terms with the same powers of x^n to zero, and from there you can find a recurrence relation that will help you find the other terms.

Post back if your'e still stuck.
 
I get a recurrence relation:
a_(n+3)=[a_n-2a_(n+1)-1]/(n+3)(n+2)
but I don't think that can be right, because I need to comment on the convergence...
 

Similar threads

Replies
8
Views
2K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 9 ·
Replies
9
Views
3K
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
8
Views
3K
  • · Replies 22 ·
Replies
22
Views
4K