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?

 

[iceblits]

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??

ahh yes I see that...f(m/n)=mf(1/n)+(m-1)..and then f(m/n)=m(1/n-1)+m-1...so then f(m/n)=m/n-1

 

[Dick]

ahh yes I see that...f(m/n)=mf(1/n)+(m-1)..and then f(m/n)=m(1/n-1)+m-1...so then f(m/n)=m/n-1

Ok, you are almost there. So f(m/n)=m/n-1. That means f(q)=q-1 for all rational numbers q. How do you prove it's also true for irrational numbers? It's time to use the given that f is continuous.

 

[iceblits]

Ok, you are almost there. So f(m/n)=m/n-1. That means f(q)=q-1 for all rational numbers q. How do you prove it's also true for irrational numbers? It's time to use the given that f is continuous.

Oh boy..haha...maybe...since its continuous the limit as x approaches c from the right and left equals f(c)..where c is any real number?

 

[Dick]

Oh boy..haha...maybe...since its continuous the limit as x approaches c equals f(c)..where c is any real number?

Suppose c is irrational. Then there is a sequence of rational numbers that approach c, say q_i->c (yes?), and you've already shown f(q_i)=q_i-1. What conclusion can you draw from that?

 

[iceblits]

does q_i mean "q sub i" if so then, if I have shown that f(q_i)=q_i-1.. and since the function is continuous for numbers..then a sequence of rational numbers exists to form an irrational number as the number of terms in the sequence approach infinity?

edit: and this would show that..as i>>infinity...f(q_i)=q_i-1...which means the formula is true for the irrational numbers

 

[Dick]

does q_i mean "q sub i" if so then, if I have shown that f(q_i)=q_i-1.. and since the function is continuous for numbers..then a sequence of rational numbers exists to form an irrational number as the number of terms in the sequence approach infinity?

Yeah, q_i means "q sub i", I'm often too lazy to TeX, sorry. The ideas are more important than the format in my opinion. Do you know why you can find a sequence of rationals approaching any irrational? This may mean just pointing to a page in your book. And given q_i->c can you use continuity to show f(c)=c-1?

 

[iceblits]

Yeah, q_i means "q sub i", I'm often to lazy to TeX, sorry. The ideas are more important than the format in my opinion. Do you know why you can find a sequence of rationals approaching any irrational? And given q_i->c can you use continuity to show f(c)=c-1?

Thats ok :) i don't even really know what TeX is..and...Im not sure why..I do know that certain irrational numbers can be written as a sequence of rational numbers..like (e=sum(1/n))..I didn't know that it extended to all irrational numbers (I do now that you've told me)..is there a name for such a theorem?

edit: is it because an irrational number is a sum of decimals?

so like pi=3+.1+.04+.001..

 

[Dick]

Thats ok :) i don't even really know what TeX is..and...Im not sure why..I do know that certain irrational numbers can be written as a sequence of rational numbers..like (e=sum(1/n))..I didn't know that it extended to all irrational numbers (I do now that you've told me)..is there a name for such a theorem?

The statement of the theorem usually referred to as the statement that "the rationals are DENSE in the real numbers". Meaning given any interval (however small) around an irrational number, there is a rational number in it. So you can find a sequence of rationals converging to any irrational. Makes sense, yes?

 

[Dick]

edit: is it because an irrational number is a sum of decimals?

so like pi=3+.1+.04+.001..

Yes, that's basically what it is. Your book should have a theorem that tells you something very like that. But it's more like, pi is the limit of the sequence {3,3.1,3.14,3.141,3.1415,...}.

 

[LCKurtz]

And now for extra credit

What happens if you remove the assumption that f(1) = 0?

 

[iceblits]

So now can I use induction to prove that the formula is true for all real?

 

[iceblits]

And now for extra credit

What happens if you remove the assumption that f(1) = 0?

haha.. does it become an infinite number of functions?...like since a linear function is f(x+y)=f(x)+f(y)...then would removing the assumption give me all linear functions +1?

 

[Dick]

haha.. does it become an infinite number of functions?...like since a linear function is f(x+y)=f(x)+f(y)...then would removing the assumption give me all linear functions +1?

Why don't you finish the first question first?

 

[iceblits]

Why don't you finish the first question first?

right..ok so here's what I have so far (thanks to you :) ) :
I show that f(nx)=nf(x)+(n-1), where n is an integer. Then, setting x=1 and using the initial condition that f(x)=0, I show that f(n)=n-1. Then, setting x=m/n I show that f(m)=nf(m/n)+n-1..since f(m)=m-1, this becomes m-1=nf(m/n)+n-1.. so f(m/n)=m/n-1 (m and n are integers). Since the function is continuous, it exits for all points and every irrational number has a sequence of rational numbers that approach it and so f(q_i)=q_i-1, where q_i is the sequence of rational numbers approaching an irrational number. Since this formula can be derived for all real numbers i can use induction to show that f((n+1)x)=(n+1)f(x)+n equals the original formula

 

[Dick]

right..ok so here's what I have so far (thanks to you :) ) :
I show that f(nx)=nf(x)+(n-1), where n is an integer. Then, setting x=1 and using the initial condition that f(x)=0, I show that f(n)=n-1. Then, setting x=m/n I show that f(m)=nf(m/n)+n-1..since f(m)=m-1, this becomes m-1=nf(m/n)+n-1.. so f(m/n)=m/n-1 (m and n are integers). Since the function is continuous, it exits for all points and every irrational number has a sequence of rational numbers that approach it and so f(q_i)=q_i-1, where q_i is the sequence of rational numbers approaching an irrational number. Since this formula can be derived for all real numbers i can use induction to show that f((n+1)x)=(n+1)f(x)+n equals the original formula

It gets a little garbled towards the end. Given you have f(q)=q-1 for q a rational (which I think you get), WHY can you say f(c)=c-1 if c is irrational? Say it in your own words as clearly as you can. I'm not seeing it in that explanation.