What are the functions satisfying f(x+y)=f(x)+f(y)+1 and f(1)=0?

  • Thread starter Thread starter iceblits
  • Start date Start date
  • Tags Tags
    Functions
Click For Summary

Homework Help Overview

The problem involves finding all functions f: R->R that satisfy the functional equation f(x+y)=f(x)+f(y)+1 with the condition f(1)=0. Participants are exploring the nature of the solutions and the implications of continuity and differentiability on the function's form.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss specific values of the function, such as f(1/2) and f(1/n), and consider the implications of continuity on the solutions. There is a focus on whether the function can be linear and how to prove the uniqueness of the solution.

Discussion Status

The discussion is active, with participants sharing insights and questioning assumptions about the nature of the function. Some suggest that there may be multiple solutions, while others are trying to derive a general formula for f(nx) and explore the implications of differentiability.

Contextual Notes

Participants note the challenge of proving the uniqueness of the function and the difficulty in conceptualizing the problem. There is an ongoing exploration of whether the function can take forms other than y=x-1, and the discussion includes attempts to manipulate the functional equation to derive further insights.

iceblits
Messages
111
Reaction score
0

Homework Statement



With proof, Find all functions f: R->R such that f(x+y)=f(x)+f(y)+1 and f(1)=0

Homework Equations


The Attempt at a Solution



First I cried...then...

Since f(1)=0, f(2)=1, f(3)=2...= x-1
I'm having a really hard time conceptualizing this problem and I'm not sure what the answer is suppose to look like. Is it a finite or infinite set of functions? I'm not sure. The proof part is kinda throwing me off too. Any help would be great..
 
Last edited:
Physics news on Phys.org
f(1) = 0 and f(x + y) = ?
 
You can find the values of more than just the integers. E.g. what's f(1/2)? Can you find the value of f(1/n) where n is an integer? Can you find the value of f(m/n) where m and n are integers? Are you given that f is continuous or anything like that?
 
yes f is continuous, and yes i can find f(1/2), but what i don't understand is why this problem has more than one solution (y=x-1)
 
iceblits said:
yes f is continuous, and yes i can find f(1/2), but what i don't understand is why this problem has more than one solution (y=x-1)

Why do you think it has more than one solution?
 
And if you assume it is differentiable you can think about its derivative...
 
@flyingpig: f(x+y)=f(x)+f(y)+1..it isn't given as anything else
 
iceblits said:
@flyingpig: f(x+y)=f(x)+f(y)+1..it isn't given as anything else

Looking at all the post, this question is probably out of my league. So I will just erase my answer.

Sorry about that
 
LCKurtz said:
And if you assume it is differentiable you can think about its derivative...

You don't need to assume it's differentiable. I think continuous is enough, isn't it?
 
  • #10
Dick said:
You don't need to assume it's differentiable. I think continuous is enough, isn't it?

It might be, I don't know. I haven't worked through it. But if it is differentiable that gives an easy way to find a family of functions that work, which must be included in the "all" functions.
 
  • #11
LCKurtz said:
It might be, I don't know. I haven't worked through it. But if it is differentiable that gives an easy way to find a family of functions that work, which must be included in the "all" functions.

I did work through it. To use differentiability, which I DON'T think is given, you need to find the value of something like f(1/n). Then it's straightforward. But once you've found f(1/n) why not find f(m/n) (m and n integers) and use continuity and skip the differentiability assumption? That's what I'm trying to push iceblits to do.
 
  • #12
oh I see...I'll try. I'm just having a hard time conceptualizing this because I feel like y=x-1 is the only functions that can satisfy all the points. Thank-you for all the responses though..I will keep trying this problem until I solve it
 
  • #13
iceblits said:
oh I see...I'll try. I'm just having a hard time conceptualizing this because I feel like y=x-1 is the only functions that can satisfy all the points. Thank-you for all the responses though..I will keep trying this problem until I solve it

y=x-1 IS the only function that works. You just have to prove it.
 
  • #14
oh! i see!
 
  • #15
ok now i have some direction lol...I was thinking there was an infinite set of functions or something (like some weird sine wave going diagonally up)..or like the line y=x-1 except if you imagine it like a string, push the string together to get various functions..or something like that
 
  • #16
Dick said:
y=x-1 IS the only function that works. You just have to prove it.

No, it isn't the only one.

[Edit]Woops, ignore that. I agree. I forgot f(1) = 0. :frown:
 
  • #17
LCKurtz said:
No, it isn't the only one.

[Edit]Woops, ignore that. I agree. I forgot f(1) = 0. :frown:

That makes me feel better.
 
  • #18
gahhh..ok so if I let x and y both equal n, then, f(2n)=2f(n)+1. if I let x and y equal 1/n I get f(2/n)=2f(1/n)+1 and if I let x and y equal m/n, I get f(2m/n)=2f(m/n)+1...I don't think that shows anything (sorry, I haven't taken a proof class or anything yet :( ). Should I be trying to manipulate f(x+y)=f(x)+f(y)+1 to achieve y=x-1 or should I assume y=x-1 is the only solution and then show that any if any other function exists to satisfy the condition, it would have to be y=x-1? I guess I'm still lost haha..I thought I would be able to prove that y=x-1 is the only solution really easily but i guess i cant...
 
  • #19
iceblits said:
gahhh..ok so if I let x and y both equal n, then, f(2n)=2f(n)+1. if I let x and y equal 1/n I get f(2/n)=2f(1/n)+1 and if I let x and y equal m/n, I get f(2m/n)=2f(m/n)+1...I don't think that shows anything (sorry, I haven't taken a proof class or anything yet :( ). Should I be trying to manipulate f(x+y)=f(x)+f(y)+1 to achieve y=x-1 or should I assume y=x-1 is the only solution and then show that any if any other function exists to satisfy the condition, it would have to be y=x-1? I guess I'm still lost haha..I thought I would be able to prove that y=x-1 is the only solution really easily but i guess i cant...

You found f(1/2), right? How would you find f(1/3)? Does that help you to find a general way to find f(1/n)? Here's a hint. Can you figure out a formula for f(nx) where n is an integer and x is any real? Formally, it's an induction proof, but if you haven't done induction you should still be able to figure out the formula just by thinking about it.
 
  • #20
ye f(1/2+1/2)=f(1/2)+f(1/2)+1...f(1)=2f(1/2)+1, f(1/2)=-1/2..wouldn't the formula just be y=x-1?
 
  • #21
iceblits said:
ye f(1/2+1/2)=f(1/2)+f(1/2)+1...f(1)=2f(1/2)+1, f(1/2)=-1/2..wouldn't the formula just be y=x-1?

Sure, y=x-1 works for x=1/2. But we are trying show it works at any point WITHOUT assuming it's 1/n. Have you figured out f(nx) (see previous post)?
 
  • #22
Would it be f(2nx)=2f(nx)+1?
 
  • #23
Ohh wait let me try getting 1/3..
 
  • #24
Ok I'm still working on finding f(1/3), but I noticed that the function f(x+y)=f(x)+f(y)+1 is really close to the definition of a linear function (f(x+y)=f(x)+f(y)). Can I use this to prove that the function is of the form y=ax+1?
 
Last edited:
  • #25
iceblits said:
Ok I'm still working on finding f(1/3), but I noticed that the function f(x+y)=f(x)+f(y)+1 is really close to the definition of a linear function (f(x+y)=f(x)+f(y)). Can I use this to prove that the function is of the form y=ax+1?

I don't see how you could do it that way. f(2x)=f(x+x)=f(x)+f(x)+1=2f(x)+1. Next f(3x)=f(x+2x)=f(x)+f(2x)+1. You know f(2x) from the first equation, put it in. Now do f(4x). By now you should be able to guess the form for f(nx).
 
Last edited:
  • #26
ok so...f(nx)=nf(x)+(n-1) right?

and: f(1)=0 so the above is...f(n)=n-1
 
Last edited:
  • #27
and I just searched up proof by induction.. so i would just show that this holds true for some particular n..like n=1..then assume that its true for all n=k..then show that its true for all n=k+1?..but how do i go about proving that its of the form y=x-1?..like by proving the above formula do i automatically prove that y=x-1 is the only function?
 
Last edited:
  • #28
iceblits said:
and I just searched up proof by induction.. so i would just show that this holds true for some particular n..like n=1..then assume that its true for all n=k..then show that its true for all n=k+1?..but how do i go about proving that its of the form y=x-1?..like by proving the above formula do i automatically prove that y=x-1 is the only function?

No, you don't prove y=x-1 by induction. Where you would use induction is to find a formula for f(nx) in terms of f(x) for all integers n. You already know f(2x)=2f(x)+1. What's f(3x) in terms of f(x)? If you can find this formula, it will be really useful. I promise. Your next goal is to prove f(m/n)=m/n-1 for all integers m and n. That would mean you've shown f(q)=q-1 for all rational numbers, right?
 
  • #29
isnt f(3x)=3f(x)+2?...ok let me work on f(m/n)

edit:
so far I have: f(nx)=nf(x)+(n-1)..and if f(1)=0, this simplifies to f(n)=n-1...
using induction i can say that f((n+1)x)=(n+1)f(x)+n...the left side equals: f(nx+x)=f(nx)+f(x)+1 and setting this equal to the right side I get: f(nx)+f(x)+1=nf(x)+f(x)+n..which simplifies to f(nx)=nf(x)+n-1...which was the original formula
 
Last edited:
  • #30
iceblits said:
isnt f(3x)=3f(x)+2?...ok let me work on f(m/n)

edit:
so far I have: f(nx)=nf(x)+(n-1)..and if f(1)=0, this simplifies to f(n)=n-1...
using induction i can say that f((n+1)x)=(n+1)f(x)+n...the left side equals: f(nx+x)=f(nx)+f(x)+1 and setting this equal to the right side I get: f(nx)+f(x)+1=nf(x)+f(x)+n..which simplifies to f(nx)=nf(x)+n-1...which was the original formula

Let's talk about induction later. But yes, f(nx)=nf(x)+(n-1). You would use induction to prove that. Can you use the formula to say what f(1/n) is? Then, by all means, work on f(m/n).
 

Similar threads

Find f s. t. ||f||=1 and f(x) < 1 with ||x||=1
  • · Replies 20 ·
Replies
20
Views
3K
Proof given ##x < y < z## and a twice differentiable function
  • · Replies 8 ·
Replies
8
Views
2K
Find g(x)/h(y) for a given F(x,y)
  • · Replies 3 ·
Replies
3
Views
1K
Is f(z) =(1+z)/(1-z) a real function?
  • · Replies 17 ·
Replies
17
Views
4K
Find a function satisfying these conditions: f(x)f(f(x)) = 1 and f(2020) = 2019
  • · Replies 14 ·
Replies
14
Views
3K
Differentiable function proof given ##f''(c) = 1##
  • · Replies 1 ·
Replies
1
Views
1K
Find f(x,y) given partial derivative and initial condition
  • · Replies 6 ·
Replies
6
Views
2K
Problem 1-23 and 1-24 from Spivak's Calculus on Manifolds
  • · Replies 6 ·
Replies
6
Views
2K
Prove that ##\lim\limits_{x \to \infty} f(x) = 0##
  • · Replies 3 ·
Replies
3
Views
1K
Prove that f is a homeomorphism iff g is continuous, fg=1 and gf=1
  • · Replies 5 ·
Replies
5
Views
2K