## Wednesday, 4 April 2018

### Two little math questions

I get math questions day in and out and some of them are quite interesting that I considered to put it here.

1) Consider the following nutrition label:

per 200g: 17.1g of carbohydrates with 17.1g sugar
per 100g: 8.6g of carbohydrates with 8.5g sugar

Can you approximate the sugar content in the carbohydrates?

If it's an olympiad question (junior level of course), we will probably end up with the sugar content being a fraction p/q for integers p and q, then you are asked to minimize q...

2) Consider the space of continuous functions $C(\mathbb{R}^n \times \mathbb{R}^n, \mathbb{R})$, and the subset:

$K = \left\{ f(x,y) = \varphi (x) \varphi (y) \mid \varphi \in C(\mathbb{R}^n , \mathbb{R}) \right\}$

where the kernel can be separated. We want to see if it is possible to relate the space $C(\mathbb{R}^n , \mathbb{R})$ with $K$ so that we can have some linear structure on it.

Define a mapping $r: C(\mathbb{R}^n , \mathbb{R}) \to C(\mathbb{R}^{2n} , \mathbb{R})$ by $\varphi \mapsto f(x,y) = \varphi (x) \varphi (y)$. Let $A \subseteq C(\mathbb{R}^n , \mathbb{R})$ be an independent set. Clearly $r(A)$ is an independent set in $C(\mathbb{R}^{2n} , \mathbb{R})$. To see this consider the equation

$\sum _{i=1}^n a_i \varphi _i (x) \varphi _i (y) = 0$

for all $x,y\in \mathbb{R}^n$. Now note that by definition $\varphi _i \neq 0$ so they do have non-empty support. If the supports are all disjoint then clearly we conclude $a_i = 0$. Otherwise pick a point $y_0$ in the intersection of at least the supports of two $\varphi _i$'s, then we have

$\sum _{i=1}^n (a_i \varphi _i (y_0)) \varphi _i (x) = 0$

for all $x$, and from the independence of $\varphi _i$'s we conclude that $a_i = 0$.

On the other hand $K$ is not a subspace of $C(\mathbb{R}^{2n} , \mathbb{R})$. To see this lets assume we can find a function $\varphi$ so that $r(\varphi ) = r(\varphi _1 ) + r(\varphi _2)$ for $\varphi _1, \varphi _2 \in C(\mathbb{R}^n , \mathbb{R})$.

By setting $x =y$ we know that $\varphi (x) = sgn (\varphi ) \sqrt{\varphi _ 1^2 + \varphi _2^2}$. Substituting back to the original equation we have

$\sqrt{(\varphi _1 ^2(x)+\varphi _2^2(x))(\varphi _1^2(y) +\varphi _2^2(y))} = \varphi _1(x)\varphi _1(y)+\varphi _2(x)\varphi _2(y)$

$\varphi _1^2(x)\varphi _2^2(y) + \varphi _2^2(x)\varphi _1^2(y) = 2 \varphi _1(x)\varphi _2(x)\varphi _1(y)\varphi _2(y)$

which is the equality version of the AM-GM equatity! We conclude that $\varphi _1^2(x)\varphi _2^2(y) = \varphi _2^2(x)\varphi _1^2(y)$ and so

$(\varphi _1(x)/\varphi _1(y))^2 = (\varphi _2 (x)/\varphi _2(y))^2 = c$

being constants for all $x,y$ in the support. Hence they must be constant functions. [I have missed the details handling the sign of $\varphi$ but it can be verified that it does not matter due to the square].

so we can actually assign a rank to $span(K)$ to the minimum dimension of $\varphi$ required so that the kernel separation trick works.