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title: '[Notes](#fn1): Problem Class, 9 Oct 2020'

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# [Notes](#fn1): Problem Class, 9 Oct 2020
> Group 11 -- Problem Class, MATH40001/40009 IUM
> Stergios M. Antonakoudis (stergios@imperial.ac.uk)

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Welcome to IUM Problem Class Group 11. \
Today's class: <u>**Problem Sheet 1 of Part III.**</u>

[^1]: <small> *These notes[^1] are typed up and created during and for the purposes of the discussion in the online session of Problem Class Group 11 for IUM on MS Teams, similarly to the use of a 'whiteboard' in an actual class.* </small>

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## A brief overview

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## Problems to discuss 

### 1. The parallelogram law states geometrically that: the sum of the squares of the lengths of the sides of a parallelogram equals the sum of the squares of the length of its diagonals. 
(a) Using vectors, state the parallelogram law.(b) Prove the parallelogram law.

### Solution to 1. 
(a)
Let $\vec{a}$ and $\vec{b}$ be two non-zero vectors in the plane. There are points $A$ and $B$ in the place, so that $\vec{a}= \vec{OA}$ and $\vec{b}=\vec{OB}$, where $O$ denotes the origin in the plane. We could describe any parallelogram as having four vertices $O, A, B$ and $C$ -- we'd like to describe these points in terms of $\vec{a}, \vec{b}$.

We know that the edge $OA$ is parallel to the edge $BC$. Therefore the vector represented by $\vec{BC}$ is equal to the vector $\vec{OA}$. Hence, $\vec{OC} = \vec{OB} + \vec{BC} = \vec{OB} + \vec{OA} = \vec{a} + \vec{b}$, which is one of the diagonals. Likewise, the vector $\vec{AB} = \vec{b} - \vec{a}$ is the vector corresponding to the other diagonal.

==From now one, I will denote $\vec{a}$ simply by $a$ etc (i.e. no arrow notation)==

Hence, **the parallelogram law** states that the following equation is true for all vectors $a,b$:$$| a + b |^2 + | b-a|^2 = 2(|a|^2 + |b|^2)$$.

(b) 
In order to prove this equations, we **need** to use the fact $$|u|^2 = u \cdot u$$ for all vectors $u$. This is useful because the **dot** product is a bilinear operation on vectors!

$|a+b|^2 = (a+b) \cdot (a+b) = a \cdot a + a \cdot b + b \cdot a + b \cdot b$
$|b-a|^2 = (b-a) \cdot (b-a) = b \cdot b - b \cdot a - a \cdot b + a \cdot a$
If we add up the above two equations, the terms $a  \cdot b$ and $b \cdot a$ cancel out and  we are left with $| a + b |^2 + | b-a|^2 = 2(|a|^2 + |b|^2)$\
QED

### Problem 9 (The Cauchy-Schwarz inequality)
Consider $u$ and $v$ two vectors of $\mathbb{R}^n$, then $|u\cdot v|\leq |u||v|$, with equality if and only if $u$ and $v$ are linearly dependent. 

### Solution 9 
If $u=0$, then both the inequality and the statement in the case of equality are *obviously* true. Likewise, if $v=0$ then both statements are also true. So we may assume that both $u$ and $v$ are non-zero vectors!

($\star$) *A vector $u$ is non-zero if and only if its length $|u| \neq 0$.* 

We can consider the (**unit**) vectors $\frac{u}{|u|}$ and $\frac{v}{|v|}$; we know that the following is always true for all vectors $u,v$.

$$|\frac{u}{|u|} - \frac{v}{|v|}|^2 \geq 0$$

But $|\frac{u}{|u|} - \frac{v}{|v|}|^2 = (\frac{u}{|u|} - \frac{v}{|v|}) \cdot  (\frac{u}{|u|} - \frac{v}{|v|}) = \ldots = 1 - \frac{(u \cdot v)}{|u||v|} - \frac{(v \cdot u)}{|u||v|} + 1$ which is equal to $2(1- \frac{(u \cdot v)}{|u||v|})$, because $u\cdot v = v \cdot u$. However, LHS was always non-negative, so the RHS is also non-negative. I.e. 
$$ 1 - \frac{u \cdot v}{|u||v|} \geq 0 $$
We conclude that $u \cdot v \leq |u| |v|$ is true for all vectors $u$ and $v$. Replacing the vector $v$ with the vector $-v$, we conclude that $- u \cdot v \leq |u| |v|$ is also true, and therefore  $|u \cdot v| \leq |u||v|$.

QED (for the proof of the inequality)

Let me skip the "equality" part of the question (HINT: go through the steps in the proof/argument given above when the equality occurs, and use basic observation $\star$ about the length of a non-zero vector).

## Solution to problem 10

Let $R>0$. We consider the set $\{(x,y) \in \mathbb{Z}^2: x^2 + y^2 < R^2 \}$ and denote by $f(R)$ the size, i.e. number of elements, of the set.

==You should think of areas!==

Points with integer coordinates are **_self-isolating_** in the plane! The distance between any two of the points is at least 1.
In particular, for every such point in our set we can consider and associate to it a square that is centered at that point and unit length edges parallel to the $x-y$ axes in the plane. 
Since the vertices of all such squares have coordinates in the set $(\frac{1}{2},\frac{1}{2})+ \mathbb{Z}^2$, their distance from any  point in our set is at least $\frac{\sqrt{2}}{2}$; hence, none of their vertices can be in the interior of any one of the squares. We conclude that there is no overlap between any two distinct squares, and therefore the union of all these squares covers a region in the plane with total area $f(R)$.
It follows from elementary geometry and the observations about the distance between the different sets of points given in the paragraph above, that the region covered up but all the unit squares centered at the points with integer coordinates inside the circle of radius $R$ contains the all the (interior) points of the circle with the same centre and radius $R - \sqrt{2}$ and it is entirely contained in the interior of the circle with the same centre and radious $R + \sqrt{2}$. In particular, we deduce that the area enclosed by the smaller of the two circles is at most $f(R)$ and the area enclosed by the bigger of the two circles is at least $f(R)$; we conclude $\pi (R - \sqrt{2})^2 \leq f(R) \leq \pi (R + \sqrt{2})^2$.
QED































