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If $f:R\to R$ satisfies $f(x+2xy)=f(x)+2f(xy)\forall x,y\in R$ and $f(10)=11$ then show that

  • A) $f$ is odd
  • B) $f(x)=2f(\frac x2)$
  • C) $f(x+y)=f(x)+f(y)$
  • D) $f(11)=12.1$

Putting $y=-1$, I get $f(-x)=-f(x)$

Putting $y=\frac12$, I get $f(2x)=f(x)+2f(\frac x2)$

Putting $y=\frac1x$, I get $f(x+2)=f(x)+2f(1)$

I have tried other substitutions too but not reaching anywhere.

A similar question exists here. But the answers posted there are way too complicated for me. I am looking for something simpler.

aarbee
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  • It's complicated in the sense that there are many steps. But, so long as you are comfortable with induction, each step is straightforward and elementary. There's no mathematical heavy machinery being used. I would advise looking back over the answer, slowly and carefully, trying to replicate every step. I think you'll find that you'll get the answers to your questions. And if not, then you can edit this question and tell us which step trips you up. Good luck! – Theo Bendit Jul 05 '21 at 19:54
  • @TheoBendit Thanks, I'll do that. – aarbee Jul 05 '21 at 19:55
  • Just to point out: You've only shown $$f(x + 2) = f(x) + 2f(1)$$ for $x \neq 0$. – Aryaman Maithani Jul 05 '21 at 20:00
  • @AryamanMaithani oh yes, thanks. – aarbee Jul 05 '21 at 20:01
  • Also, note that instead of taking $y = 1/x$, you could've taken $y = \frac{1}{2x}$. This is a somewhat more natural thing to do because it then associates $f(x + 1)$ directly with $f(x)$. And we like this because it then lets us successively calculate $f(2), f(3), \ldots$ in terms of $f(1)$ and $f(1/2)$. This is what the answer in the link has done. Then that answer also shows how to get $f(1/2)$ in terms of $f(1)$. Thus, $f$ is determined at each positive integer in terms of $f(1)$. Finally, using $f(10)$, you can calculate $f(1)$ and you are done. – Aryaman Maithani Jul 05 '21 at 20:25
  • Substitute $y$ by $1+\frac yx$ in the given equation and then put $x=0$ to obtain condition (B). – RiverX15 Jul 05 '21 at 20:26
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    @TheoBendit You were right. The steps there were straightforward and elementary. I needed to have been more patient. I have understood it now, thanks. – aarbee Jul 05 '21 at 20:39
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    @AryamanMaithani Your explanation helped a lot. In fact, you summarized it completely. Thanks. – aarbee Jul 05 '21 at 20:40
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    @RiverX15 I could have never thought of that substitution but it's working very nicely, thanks. – aarbee Jul 05 '21 at 20:41
  • @aarbee Good on you for giving it another try. Not everyone would. Have a +1 from me. – Theo Bendit Jul 06 '21 at 01:33

1 Answers1

3

$A)$ You got it

$B)$ Start from $x=0$ to get

$$f(0)=f(0)+2f(0)\Rightarrow f(0)=0$$

Then at $y=-\frac{1}{2}$ we have

$$0=f(0)=f(x)+2f\left(-\frac{x}{2}\right)$$

Combined with $A)$ this gives us

$$f(x)=2f\left(\frac{x}{2}\right)$$

$C)$ Set $y=\frac{z}{2x}$ (for all $x\neq 0$). Then

$$f(x+z)=f(x)+2f\left(\frac{z}{2}\right)$$

Combined with $B)$ this gives us

$$f(x+z)=f(x)+f(z)$$

For $x=0$, the equation is trivially true since

$$f(0+y)=f(y)=0+f(y)=f(0)+f(y)$$

$D)$ From $C)$ we have

$$11=f(10)=f(1)+f(1)+...+f(1)=10f(1)\Rightarrow f(1)=\frac{11}{10}$$

Thus

$$f(11)=f(10)+f(1)=11+\frac{11}{10}=\frac{121}{10}=12.1$$

QC_QAOA
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