- #1
- 4,796
- 32
I'm puzzled by this question: Show that for all function f:R-->R. there exists an even function p and an odd function i such that f(x) = p(x) + i(x) forall x in R.
I got nothing.
I got nothing.
Proving the Existence of Even and Odd Functions in f:R-->R
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Homework Help Overview
The discussion revolves around proving the existence of even and odd functions for any function f: R → R, specifically expressing f as the sum of an even function p and an odd function i.
Discussion Character
- Exploratory, Assumption checking
Approaches and Questions Raised
- Participants explore the relationship between f(-x) and the proposed functions p(x) and i(x), questioning how to derive insights from this relationship.
Discussion Status
Some participants have offered guidance on examining f(-x) in relation to p and i, while others express uncertainty about the implications of assuming the theorem is true. The discussion reflects a mix of insights and confusion, with no clear consensus reached.
Contextual Notes
Participants note the challenge of proving the theorem without additional information or assumptions, highlighting the complexity of the problem.
- #2
Physics Monkey
Homework Helper
- 1,363
- 34
Try looking at f(-x) and relating it to p(x) and i(x). Do you notice anything?
- #3
- 4,796
- 32
But there is nothing to look at. f(-x) = ......?
The only thing would be SUPPOSING the result of the thorem is true, then it would implies that there exist p and i such that f(x) = p+i and hence f(-x) = p(x)-i(x), but that's as far as that goes.
The only thing would be SUPPOSING the result of the thorem is true, then it would implies that there exist p and i such that f(x) = p+i and hence f(-x) = p(x)-i(x), but that's as far as that goes.
- #4
Hurkyl
Staff Emeritus
Science Advisor
Gold Member
- 14,922
- 28
No it's not.
- #5
- 4,796
- 32
Oh I see. That was very insightful Hurkyl. 