4

Let $f: \mathbb{N} \rightarrow \mathbb{R}$ be a function and $a \in \mathbb{R}$ such that $$f(m+n) = f(m) + f(n) + a$$ $$f(2) = 10, f(20) = 118$$ Find $a$ and $f$.

I found this exercise at the beginning of a Real Analysis textbook. I've never solved a functional equation before, but here's my solution (attempt):

i) Using induction it's easy to verify that for $m, n \in \mathbb{N}$ we have $f(m \cdot n) = m (f(n) + a)$, since $$f((m+1)n) = f(mn + n) = f(mn) + f(n) + a$$

ii) Then $118 = f(20) = f(10 \cdot 2) = 10 (f(2) + a) = 10 (10 + a) \Rightarrow a = \frac{9}{5}$

iii) Then $f(0) = f(0 + 0) = f(0) + f(0) + a \Rightarrow f(0) = -a = -\frac{9}{5}$.

We also get $10 = f(2) = f(1+1) = f(1) + f(1) + \frac{9}{5} = 2 f(1) + \frac{9}{5} \Rightarrow f(1) = \frac{41}{10}$

iv) Then finally we can define $f$ recursively by $f(0) = -\frac{9}{5}$ and $$f(n+1) = f(n) + f(1) + \frac{9}{5} = f(n) + \frac{41}{10} + \frac{9}{5} = f(n) + \frac{59}{10}$$

EDIT

Then thanks to the user lulu, the real pattern should be $f(m \cdot n) = m f(n) + (m-1) a$ instead. Using this in ii) then gives $a = 2$. Then we get in iii) that $f(0) = -2$ and $f(1) = 4$, so $f$ is defined by $f(n+1) = f(n) + 6$. And now it's pretty obivous that $f(n) = 6n - 2$ solves the equation.

  • According to your claim $f(m \cdot n) = m (f(n) + a)$, we would have $f(0)=0$. However, given the original equation, we can say $f(0)=2f(0)+a$, that is $f(0)=-a$.So unless $a=0$, your first formula is wrong. (actually, the heridity step of your induction reasoning is ok, but the initialization step is wrong !) – Suzet Aug 08 '18 at 14:10
  • 1
    Your first line is incorrect. Taking $m=2$ we have $f(2n)=f(n+n)=f(n)+f(n)+a=2f(n)+a\neq 2(f(n)+a)$ – lulu Aug 08 '18 at 14:11
  • I'm not convinced about $(i)$. Please explain elaborately. – Anik Bhowmick Aug 08 '18 at 14:12
  • Ok Wow, I was completly off even right at the beginning. –  Aug 08 '18 at 14:12
  • I think it should be $f(m.n)=m.f(n)+a$. In this way, you'd get $a=18$ !! – Anik Bhowmick Aug 08 '18 at 14:14
  • 1
    @AnikBhowmick That's wrong too. Taking $m=3$ we write $f(3n)=f(2n+n)=f(2n)+f(n)+a=2f(n)+a+f(n)+a=3f(n)+2a$. (I think that case should point to the actual pattern) – lulu Aug 08 '18 at 14:17
  • Then coefficient of $a$ can be anything $\leqslant m-1$ !! – Anik Bhowmick Aug 08 '18 at 14:20
  • @lulu Could you repost your hints as an answer so that I can accept your answer? –  Aug 08 '18 at 14:35
  • Oh, thanks for the thought but I think the posted solutions are fine. – lulu Aug 08 '18 at 14:39

3 Answers3

1

$f(m+n)+a=f(m)+f(n)+2a$, Let$g(x)=f(x)+a\rightarrow g(m+n)=g(m)+g(n)$.

So $g$ satisfies Cauchy's functional equation and its solution is $g(n)=kn\rightarrow f(n)=kn-a$

$f(2)=10,f(20)=118 \rightarrow k=6,a=2,f(n)=6n-2$.

It can be sloved as same method if $f:\mathbb R\rightarrow \mathbb R$ and $f$ is lebesgue measurable.

Details about Cauchy's functional equation could refer to https://en.wikipedia.org/wiki/Cauchy%27s_functional_equation

HQR
  • 508
  • 3
  • 13
0

HINT:

We have $$f(20)=2f(10)+a=10f(2)+9a\implies a=2$$

Now consider $$f(m-n)=f(m)+f(-n)+2\\f(m+n)=f(m)+f(n)+2$$ and adding gives $$f(m-n)+f(m+n)=2f(m)+f(-n)+f(n)+4$$ or $$f(2m)-2=2f(m)+f(0)-2+4$$ Now what do we know about $f(0)$?

ə̷̶̸͇̘̜́̍͗̂̄︣͟
  • 27,005
0

We have $$f(m+n) = f(m) + f(n) + a$$ Hence $$f(2n)=2f(n)+a$$ Thus $$f(0)=2f(0)+a$$ Then $$f(0)=-a$$

Now \begin{align} f(20) &= 2f(10) + a \\ &= 2(f(8)+f(2)+a) +a \\ &= 2(2f(4)+a) +2f(2) +3a \\ &= 4(2f(2)+a) +2f(2) +5a \\ &= 10f(2) + 9a \\ 118 &= 100 + 9a \\ a & =2 \end{align}

So $$f(0)=-2, \ f(1)=\frac{10-a}{2}=4$$ And $$f(2n)=2f(n)+2=2(f(n)+1)$$

Karn Watcharasupat
  • 4,323