Solving for Nondecreasing Continuous Functions: g(x+g(y))=g(g(x))+g(y)

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Homework Help Overview

The discussion revolves around finding all nondecreasing continuous functions g: R → R that satisfy the equation g(x+g(y))=g(g(x))+g(y) for all x, y ∈ R. This problem is situated within the context of functional equations and properties of continuous functions.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the nature of the function and its monotonicity. Some suggest differentiating the equation with respect to x and y as a potential approach. Others express uncertainty about the hint provided and its implications.

Discussion Status

The discussion is ongoing, with participants exploring different lines of reasoning. Some have offered hints and suggestions, while others are questioning the necessity of solving the problem outright. There is a mix of attempts to engage with the problem and requests for direct solutions.

Contextual Notes

Some participants mention time constraints due to upcoming exams, which influences their urgency in seeking solutions. There is also a reference to a friendly competition regarding who can solve the problem first, although its importance is questioned.

vip89
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Very important...

Let R denote the set of real numbers. Find all nondecreasing continuous
functions g : R → R such that

g(x+g(y))=g(g(x))+g(y) for all x,y∈R
 
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And why is that "very important"?
 
Looks monotonous to me … :rolleyes:
 
tiny-tim said:
Looks monotonous to me … :rolleyes:

I think u see it easy...

I read this problem in abook,and I liked it
so can you solve it?
 
tiny-tim said:
Looks monotonous to me … :rolleyes:

HallsofIvy said:
And why is that "very important"?


I read this problem in abook,and I liked it
so can you solve it?
 
Hi vip89! :smile:

Hint: try differentiating with respect to x, and y, separately. :smile:
 
Oh, I see. That hint didn't seem to make much sense until I did!

(Funny how often that happens.)
 
HallsofIvy said:
Oh, I see. That hint didn't seem to make much sense until I did!

(Funny how often that happens.)

Oh my goodness! Does it work? :smile:

I didn't actually try it! :rolleyes:
 
...

oh...I can't solve it.!
can u send me the answer??I need it quickly and there is no time to solve it because I have final exams


Thankx very much
 
  • #10
vip89 said:
… I have final exams

Hi vip89! :smile:

It's because you have final exams that you need to practise the technique …in case it comes up!

(and if it doesn't come up … why are you bothered? :rolleyes:)

So … differentiating with respect to x … what do you get? :smile:
 
  • #11
I have acomption with my friend ,who find the answer first
 
  • #12
I am going to stay with the topic as given in the subject line here.

Why is the competition with your friend important?
 
  • #13
ok , it is not important
can u solve it ,pls??
 
  • #14
Wouldn't that be cheating?
 
  • #15
vip89 said:
ok , it is not important
can u solve it ,pls??

Actually no, I can't. But I do have some advice for you.

In this forum, you'll pretty much never get other people to solve entire problems for you. People like to see some attempt that you tried to solve it first. So far you haven't done that.

Did you try the suggestion from tiny-tim in message #6 ? If not, then try it. If you did try it then explain how far you got with it, or why you were unable to get very far with it, or just what you don't understand.

And lastly, welcome to physicsforums.com :smile:
 
  • #16
ok
I will try
and I want to say that is not cheating because It is not homework
 
  • #17
That is wt I did:
 

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