Orthogonal Functions | Homework Statement

Click For Summary

Homework Help Overview

The discussion revolves around the concept of orthogonal functions, specifically focusing on the conditions under which functions are orthogonal to the constant function 1. The participants are exploring integral properties and continuity requirements related to these functions.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants are examining the integral condition for orthogonality and questioning the necessity of certain boundary conditions, such as f(a) = f(b) = 0. There is also a discussion about the relevance of derivatives and the properties of antiderivatives in this context.

Discussion Status

The conversation is ongoing, with participants sharing their thoughts on integral representations and properties of functions. Some guidance has been offered regarding the evaluation of integrals, but there is no explicit consensus on the implications of the conditions being discussed.

Contextual Notes

There is an emphasis on the continuity of functions and the implications of boundary conditions, which are currently under examination. The discussion reflects a mix of foundational calculus concepts and their application to the problem at hand.

dirk_mec1
Messages
755
Reaction score
13

Homework Statement




http://img55.imageshack.us/img55/8494/67023925dy7.png



The Attempt at a Solution



All functions orthogonal to 1 result in the fact that: [tex]\int_a^b f(t)\ \mbox{d}t =0[/tex]

Now the extra condition is that f must be continous. (because of the intersection).

But where does the fact that f(a)=f(b)=0 comes from? And why look at the deratives?
 
Last edited by a moderator:
Physics news on Phys.org
Remember way back in first year calc, when you learned that to do that integral you find an antiderivative F(x) and evaluate F(b)-F(a). This is that same problem in disguise.
 
Well I thought of this: [tex]\int_a^b \int_a^t f(s)\ \mbox{d}s \mbox{d}t =0[/tex]
 
dirk_mec1 said:
Well I thought of this: [tex]\int_a^b \int_a^t f(s)\ \mbox{d}s \mbox{d}t =0[/tex]

Fine. What are you going to do with it? Why don't you just define [tex] F(x)=\int_a^x f(s)\ \mbox{d}s[/tex]
What are some of the properties of F(x)?
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 5 ·
Replies
5
Views
5K
  • · Replies 22 ·
Replies
22
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 15 ·
Replies
15
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 19 ·
Replies
19
Views
2K
  • · Replies 5 ·
Replies
5
Views
5K
  • · Replies 11 ·
Replies
11
Views
3K