A calculating ruler is a ruler for doing algebric calculations. This ruler has three arms, two of them are sationary and one can move freely right and left. Each of arms is gradient. Gradation of each arm depends on the algebric operation ruler does. For eaxample the ruler below is designed for multiplying two numbers. Gradations are logarithmic. Invalid image file For working with ruler, (e.g for calculating $x.y$) we must move the middle arm that the arrow at the beginning of its gradation locate above the $x$ in the lower arm. We find $y$ in the middle arm, and we will read the number on the upper arm. The number written on the ruler is the answer. 1) Design a ruler for calculating $x^{y}$. Grade first arm ($x$) and ($y$) from 1 to 10. 2) Find all rulers that do the multiplication in the interval $[1,10]$. 3) Prove that there is not a ruler for calculating $x^{2}+xy+y^{2}$, that its first and second arm are grade from 0 to 10.
Problem
Source: Iranian National Math Olympiad (Final exam) 2006
Tags: function, vector, algebra, functional equation, algebra proposed
21.09.2006 11:49
If we want to calculate the value of the monotonically increasing with respect to each variable function $t(x,y)$ when $x,y\in [a,b]$ with such a rule we need to find three monotonic increasing funtions $f,g,h$, so that $f(x)+g(y)=h(t(x,y))$, where the value each one returns is the distance in the corresponding sub-rule between the origin and the tick signalling the If we want to calculate the value of the monotonically increasing with respect to each variable function $t(x,y)$ when $x,y\in [a,b]$ with such a rule we need to find three monotonic increasing funtions $f,g,h$, so that $f(x)+g(y)=h(t(x,y))$, where the value each one returns is the distance in the corresponding sub-rule between the origin and the tick signalling the number we introduced as argument. Without loss of generality we can suppose $f(a)=g(a)=h(t(a,a))=0$. 1) We can take $f(x)=h(c)=ln(lnx)$, $g(y)=lny$, so that $f(x)+g(y)=ln(ln(x))+lny=ln(yln(x))=ln(ln(x^{y}))=h(x^{y})$. I don't know if there is something more to be said. 2) We have $f(x)+g(y)=h(xy)$, $x,y\in [1,10]$. By the assumption made at the beginnign we can let $f(1)=g(1)=h(1)=0$. This leads by alternatively letting $x=1$ and $y=1$, to $f=g=h$, and hence Cauchy's functional equation $f(x)+f(y)=f(xy)$ which in this case, being $f(x)$ monotonic, has only solution $f(x)=c (lnx)$ for arbitrary constant $c$; here $c>0$. [ To see this result is just necessary to make the change $g(x)=f(e^{x})$ in orther to get the first Cauchy f.e.: $g(x)+g(y)=g(x+y)$ ]. 3) By alternatively letting $x=0$, $y=0$ we get that $f(x)=g(x)=h(x^{2})$, so we should have $h(x^{2})+h(y^{2})=h(x^{2}+xy+y^{2})$. On one hand making the change $x=y=\sqrt{z}$, leads to $2h(z)=h(3z)$. On the other $x=\frac{y}{\sqrt{3}}=\sqrt z$ leads to $h(z)+h(3z)=h((4+\sqrt{3})z) \Rightarrow 3h(z)=h((4+\sqrt{3})z)$. Calling $q(x)=ln (h(e^{x}))$ (also monotonically increasing), and letting $S$ the set of vectors such that $(z,w)\in S\leftrightarrow q(z)=w$ we get from the previous two equations: $(v\in S \leftrightarrow v+a\in S) \wedge (v\in S \leftrightarrow v+b\in S)$ where $a=(ln3,ln2)\;\;b=(ln(4+\sqrt{3}),ln3)$. But $a$ and $b$ are linearly indipendent, so $\forall N\in \mathbb{R}^{+},\;\exists (a,b) \in (S\cap [0,ln 10]\times \mathbb{R})/b>N$ which is clearly impossible. (I know this last part should be written more in detail; sorry.) Being so long, there must be many mistakes.