Find the value of $f\left(\frac{22}{7}\right)$

Question

The function $f(x)$ satisfying two properties as follows:

$$f(x+y)+f(x-y)=2(f(x)+f(y))$$ $$f(1)=1$$

Then find the value of $f\left(\frac{22}{7}\right)$.

Answer 1

Let's see.

$f(x+0)+f(x-0)=2f(x)+2f(0)$, hence $f(0)=0$.

$f(0+y)+f(0-y)=2f(0)+2f(y)=2f(y)$, hence the function is even. (Not that we'll need it much.)

$f(x+x)+f(x-x)=2f(x)+2f(x)$, hence $f(2x)=4f(x)$.

At this point we might start to suspect what function is it, but let's not run ahead.

$f(3x)=f(2x+x)+f(2x-x)-f(x)=2f(2x)+f(x)=9f(x)$.

By extending this pattern and proving it by induction, we'll express $f(nx)$ via $f(x)$ for all integer $n$.

By reversing that expression, we'll find $f(x)$ at all rational $x$.

Answer 2

Letting $x=y=0$ proves $f(0)=0$, and then letting $x=0$ proves $f(-y)=f(y)$.

Fix a $y\ne0$ and put $$x_k:=ky\>, \qquad a_k:=f(x_k)\qquad(k\in{\mathbb Z})\ .$$ We then have $$a_{k+1}-2a_k+a_{k-1}=2f(y)\qquad(k\in{\mathbb Z})\ .\tag{1}$$ This shows that the second difference of the $a_k$ is constant. The general solution of $(1)$ therefore is $$a_k= k^2 f(y)+\beta k+\gamma$$ with arbitrary $\beta$ and $\gamma$. From $a_0=0$ and $a_{-k}=a_k$ it follows that necessarily $\beta=\gamma=0$, so that $$f(ky)=a_k=k^2 f(y)\qquad(k\in{\mathbb Z})\ .$$ Since this is true for arbitrary $y\ne0$ we can proceed as follows: For arbitrary $r={k\over j}\in{\mathbb Q}$ we have $$f(r)=f\left(k\cdot{1\over j}\right)=k^2f\left({1\over j}\right)=k^2{f(1)\over j^2}=r^2 f(1)=r^2\qquad(r\in{\mathbb Q})\ .$$ In particular $f\left({22\over7}\right)={484\over 49}$.

If we want $f$ continuous then $f(x)=x^2$ is the only solution to the problem. But there are many more solutions: Let $(e_\iota)_{\iota\in I}$ be a Hamel basis of ${\mathbb R}$, the latter considered as a vector space over ${\mathbb Q}$. We may assume that $e_0=1\in{\mathbb R}$ is one of the basis vectors. Let $f(e_0)=p_0:=1$, and choose $f(e_\iota):=p_\iota\in{\mathbb R}$ arbitrarily for all $\iota\ne0$. Then extend $f$ to all of ${\mathbb R}$ as follows: For arbitrary rational $r_k$ put $$f\left(\sum_{k=1}^N r_k e_{\iota_k}\right)=\sum_{k=1}^N r_k^2 \>p_{\iota_k}\ .$$

Answer 3

Note that

$$(x+y)^2+(x-y)^2=2(x^2+y^2)$$

This means that $f(x)=x^2$ satisfies the two properties, which means that $(22/7)^2$ is a possible value for $f(22/7)$. It remains to rule out any other possibility.

The easiest way to do this is to show inductively that $f(nx)=n^2f(x)$ for all $n\in\mathbb{N}$, because we would then have

$$22^2f\left({7\over22}\right)=f\left(22\cdot{7\over22}\right)=f(7)=7^2f(1)=7^2$$

The identity $f(nx)=n^2f(x)$ is trivially true for $n=1$. It's also true for $n=0$, since setting $x=y=0$ in the main property implies $2f(0)=f(0+0)+f(0-0)=2(f(0)+f(0))=4f(0)$, which implies $f(0)=0$. If we now have $f(nx)=n^2f(x)$ and $f((n-1)x)=(n-1)^2f(x)$, then by induction, we have

$$\begin{align} f((n+1)x) &=f(nx+x)+f(nx-x)-f(nx-x)\\ &=2(f(nx)+f(x))-f((n-1)x)\\ &=2(n^2f(x)+f(x))-(n-1)^2f(x)\\ &=(2n^2+2-(n^2-2n+1))f(x)\\ &=(n^2+2n+1)f(x)\\ &=(n+1)^2f(x) \end{align}$$

Remarks: Note that the proof by induction requires two preceding cases, not just one. Also, we've only shown, in effect, that $f(r)=r^2$ when $r$ is rational. I won't swear to it, but I don't think you can say anything about $f(x)$ if $x$ is irrational, without additional hypotheses on $f$. (Continuity would certainly be enough.)