Need quick help with Series, it will only take a few seconds

  • Context: Graduate 
  • Thread starter Thread starter twisted079
  • Start date Start date
  • Tags Tags
    Seconds Series
Click For Summary
SUMMARY

The discussion focuses on solving the differential equation y'' - xy = 0 using power series, specifically at the point x0 = -1. The user initially struggles with distributing the term x across the series but successfully applies a mathematical trick to simplify the expression. The key transformation involves rewriting xƩan(x+1)n as (x+1-1)Ʃan(x+1)n, allowing for the distribution of x+1 to yield -Ʃan(x+1)n+1. This method is confirmed as valid by other participants in the discussion.

PREREQUISITES
  • Understanding of differential equations
  • Familiarity with power series expansions
  • Knowledge of series manipulation techniques
  • Basic calculus concepts
NEXT STEPS
  • Study the method of power series solutions for differential equations
  • Learn about series manipulation techniques in calculus
  • Explore the application of generating functions in solving differential equations
  • Review the theory behind convergence of power series
USEFUL FOR

Mathematics students, educators, and anyone involved in solving differential equations using power series methods.

twisted079
Messages
25
Reaction score
1
So the differential equation I have to solve using power series is
y''-xy=0 when x0 = -1

So i set it up
Ʃ(n+2)(n+1)an+2(x+1)n - x Ʃ an(x+1)n

I know how to generally solve equations like this, but I never solved one like this, where I have to distribute the x ... x(x+1)n ... I just need to figure this part out (I know I left out the n=0 to ∞)
 
Physics news on Phys.org
Ok I figured it out using a cheap (but valid) math trick... in case anyone is wondering...

xƩan(x+1)n = (x+1-1)Ʃan(x+1)n

Now the x+1 can be distributed to give -Ʃan(x+1)n+1

...anyone care to double check me on this?
 
twisted079 said:
Ok I figured it out using a cheap (but valid) math trick... in case anyone is wondering...

xƩan(x+1)n = (x+1-1)Ʃan(x+1)n

Now the x+1 can be distributed to give -Ʃan(x+1)n+1

...anyone care to double check me on this?

Yep, that's the trick you want to use.
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
561
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 2 ·
Replies
2
Views
4K
  • · Replies 65 ·
3
Replies
65
Views
8K
  • · Replies 2 ·
Replies
2
Views
4K