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

[iceblits]

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..

 

[flyingpig]

f(1) = 0 and f(x + y) = ?

 

[Dick]

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?

 

[iceblits]

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)

 

[Dick]

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?

 

[LCKurtz]

And if you assume it is differentiable you can think about its derivative...

 

[iceblits]

@flyingpig: f(x+y)=f(x)+f(y)+1..it isn't given as anything else

 

[flyingpig]

@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

 

[Dick]

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?

 

[LCKurtz]

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.

 

[Dick]

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.

 

[iceblits]

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

 

[Dick]

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.

 

[iceblits]

oh! i see!

 

[iceblits]

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

 

[LCKurtz]

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.

 

[Dick]

No, it isn't the only one.

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

That makes me feel better.

 

[iceblits]

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...

 

[Dick]

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.

 

[iceblits]

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?

 

[Dick]

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)?

 

[iceblits]

Would it be f(2nx)=2f(nx)+1?

 

[iceblits]

Ohh wait let me try getting 1/3..

 

[iceblits]

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?

 

[Dick]

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).

 

[iceblits]

ok so...f(nx)=nf(x)+(n-1) right?

and: f(1)=0 so the above is...f(n)=n-1

 

[iceblits]

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?

 

[Dick]

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?

 

[iceblits]

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

 

[Dick]

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).

 

[iceblits]

setting x to 1/n i could get...f(1)=nf(1/n)+n-1...(1-n)/n=f(1/n) ? ..which is f(1/n)=1/n-1

 

[Dick]

So can i use the formula f(nx)=nf(x)+n-1 and let n=m/n..and setting x=1 to obtain f(m/n)=m/n-1?..probably not because all I am doing there is replacing n with another variable (m/n) which would no doubt result in the same equation

So far you only have an equation you can prove holds where n is a positive integer. So you can't substitute m/n for n. That's not an integer. What you can do is put x=1/n, since x is any real number. What does that tell you about f(1/n)?

 

[iceblits]

yep I just saw that :) f(1/n)=1/n-1

 

[Dick]

setting x to 1/n i could get...f(1)=nf(1/n)+n-1...(1-n)/n=f(1/n) ? ..which is f(1/n)=1/n-1

Oh, you're getting ahead of me. So using f(nx)=nf(x)+(n-1). And putting n to m and x to 1/n??

 

[iceblits]

and then...setting x to m/n I get..f(m)=nf(m/n)+n-1..and f(m)=m-1..so then
m-1=nf(m/n)+n-1..so then f(m/n)=m/n-1?