When can we take that $f(1)=1$?

Question

I have been doing some functional equations and in some of them they just say " WLOG let $f(1)=1$ ", but I don't get why they can do that... Can someone please help me?

I can't find the example of $f(1)=1$ but here is one where they take $f(0)=0$: $f(x^3)-f(y^3)=(x^2+xy+y^2)(f(x)-f(y))$.

At the beginning they just say WLOG $f(0)=0$.

It depends on the functional equation there are many functions who doesn't satisfy $f(1)=1$ — kingW3, Jan 06 '15 at 20:26
It baffles me why someone would immediately vote to close this question that was asked 2 minutes ago without giving the OP direction with regard to how to improve the question. — layman, Jan 06 '15 at 20:28
Can you please add more detail to the question, such as explaining in which context you've seen "WLOG let $f(1) = 1$"? That way, your question will be more clear. — layman, Jan 06 '15 at 20:30
@SoulEater Great, thanks for making the question more clear! — layman, Jan 06 '15 at 20:36
I am sorry i didn't make my question clear, I am still getting used to this community. — CryoDrakon, Jan 06 '15 at 20:37

Henry · Accepted Answer · 2015-01-06T21:01:15.253

3

If $f(x^3)-f(y^3)=(x^2+xy+y^2)(f(x)-f(y))$ and $g(x)=k\; f(x)$

then $g(x^3)-g(y^3)=(x^2+xy+y^2)(g(x)-g(y))$ because $kf(x^3)-kf(y^3)=(x^2+xy+y^2)(kf(x)-kf(y))$ by multiplying both sides of the original expression by $k$.

so you lose very little by assuming that $f(1)=1$ and remembering that you can then multiply any result by a constant.

Added: If you let $h(x)=f(x)-f(0)$ then $h(x^3)-h(y^3)=(x^2+xy+y^2)(h(x)-h(y))$ because $f(x^3)-f(0)-f(y^3)+f(0)=(x^2+xy+y^2)(f(x)-f(0)-f(y)+f(0))$. You know $h(0)=f(0)-f(0)=0$ so again you lose very little by assuming that $f(0)=0$ and remembering that you can then add a constant to any result.

edited Jan 06 '15 at 21:01

answered Jan 06 '15 at 20:45

Henry

157,058

I am sorry for the missunderstanding, but in that example they took WLOG f(0)=0 not f(1)=1. I can't find the example I had for f(1)=1. – CryoDrakon Jan 06 '15 at 20:48
They tell you $f(0)=0$ at the start of your question. You can use it to find $f(x^3)=x^2f(x)$ – Henry Jan 06 '15 at 20:50
1

@Henry: If that was the case the WLOG should not be there ... – String Jan 06 '15 at 20:52
But what if $f(0)$ is not equal to $0$, what then... with that you can't get $f(x^3)=x^2f(x)$... – CryoDrakon Jan 06 '15 at 20:53
See my addition. An obvious solution is $f(x)=x$ and this implies $f(x)=ax+b$ is also a solution to $f(x^3)-f(y^3)=(x^2+xy+y^2)(f(x)-f(y))$ – Henry Jan 06 '15 at 21:01
@Henry: Great observation! – String Jan 06 '15 at 21:03
Thanks a lot. Now I know on what I have to take a look to make such decisions, thank you. – CryoDrakon Jan 06 '15 at 21:42

Mark Bennet · Answer 2 · 2015-01-06T23:42:11.547

2

In the example, because the functional equation depends only on the difference between two values of $f$, if the function $g$ satisfies the equation, so will $f=g+c$ where $c$ is an arbitrary constant. Choose $c=-g(0)$ and you will find that $f(0)=0$.

Since you can always choose such a value, you can't lose any generality if you do.

edited Jan 06 '15 at 23:42

answered Jan 06 '15 at 21:40

Mark Bennet

100,194

When can we take that $f(1)=1$?

2 Answers2