Can a Neumann Series Be Proposed for the Resolvent Kernel in Iterative Methods?

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Discussion Overview

The discussion revolves around the proposal of a Neumann series for the resolvent kernel in the context of solving an integral equation using iterative methods. Participants explore the validity of this approach and its implications for the solution of the equation.

Discussion Character

  • Exploratory, Technical explanation, Debate/contested

Main Points Raised

  • One participant presents an integral equation involving a kernel and a small eigenvalue, proposing an iterative scheme to solve it.
  • The same participant questions whether a Neumann series can be proposed for the resolvent kernel in the form of a series involving iterated kernels.
  • Other participants engage in light-hearted commentary, focusing on a misunderstanding regarding the term "lambda," which is clarified as an eigenvalue rather than a reference to a dance.

Areas of Agreement / Disagreement

The main technical question regarding the Neumann series remains unresolved, as the discussion has not reached a consensus on its applicability. The humorous exchanges indicate a lack of focus on the original mathematical inquiry.

Contextual Notes

The discussion includes a mix of technical inquiry and off-topic humor, which may distract from the mathematical considerations. The validity of the proposed Neumann series and the conditions under which it may apply are not fully explored.

Who May Find This Useful

Readers interested in integral equations, iterative methods, and the application of Neumann series in mathematical analysis may find this discussion relevant.

eljose
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let be the integral equation:

[tex]f(x)=\lambda \int_{0}^{1}dyK(x,y)f(y)[/tex]

where the Kernel is known and "lambada" is an small eigenvalue... the question is if i want to solve i propose the iterative scheme:

[tex]f_{n+1} (x)=\lambda \int_{0}^{1}dyK(x,y)f_n(y)[/tex]

My question is if in this case i can propose a "Neumann series" for the resolvent Kernel R(x,y) in the form...

[tex]R= \sum_{n=0}^{\infty} \lambda ^ n K^{n}[/tex]

for K^n the n-th iterated kernel since the equation is HOmogeneus (all the functions involved except the Kernel are unknown) or if we have enough with the iteration procedure to solve the equation..thanks.
 
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"lambda" is an eigenvalue. "Lambada" is a dance!
 
HallsofIvy said:
"lambda" is an eigenvalue. "Lambada" is a dance!
:smile: :smile: :smile:
 
- Ooooh...¡what a wonderful and marvelous joke¡¡..I ask myself daily why you are here wasting your time when you could be a "millionaire" Hollywood comediant like Eddie Murphy or appear on "Saturday Night LIfe"...
 
I keep asking myself that!
 

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