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# System prepended metadata

title: "\n    Oral exam on Math Analysis\n"

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<h1>
    Oral exam on Math Analysis
</h1>

**QUESTION 1:** 
**Explain countability and dense subset of the rationals. Show an exam-
ple of irrational number and prove its irrationality.**
    1. Densely ordere:d
    - **Proposition**: For $\forall p, q \in Q, p \lt q \exists r \in Q: p \lt r \lt q$
    - **Proof**: $p = \frac{m}{n}, q = \frac{k}{l} : \frac{m}{n} \lt \frac{m+k}{n+l} \lt \frac{k}{l}$ 
    2. Proof with the $\sqrt{2}$
    - Let's say that $\sqrt{2} \in Q$, so we can assume that $\sqrt{2}$ can be represented as $\frac{p}{q}, \exists p, q, p \in Z, q \in N$
    - $p = p_1^{l_1}p_2^{l_2}p_3^{l_3}...p_n^{l_n}, p_i$ - is a prime number
    - $q = q_1^{m_1}q_2^{m_2}q_3^{m_3}...q_k^{m_k}, q_i$ - is a prime number
    - then we have 2*$q^2 = p^2$, in the left - odd number of 2, in the right - even number of 2. Then we obtain contradiction.
    
**QUESTION 2:** 
**Definitions of a bounded and convergent sequence, examples**
    1. Bounded sequence - for $\forall n \lt N, N \in N \exists$ such $M, |x_n| < M$ 
    - A sequence is bounded below if all its terms are greater than or equal to a number, K, which is called the lower bound of the sequence. The greatest lower bound is called the **infimum**.
    - A sequence is bounded above if all its terms are less than or equal to a number K', which is called the upper bound of the sequence. The smallest upper bound is called the **supremum**.
    - A sequence is bounded if it is bounded above and below, that is to say, if there is a number, k, less than or equal to all the terms of sequence and another number, K', greater than or equal to all the terms of the sequence. Therefore, all the terms in the sequence are between k and K'.
    - for example, $\frac{1}{n}$ is monotonic decreasing sequence, lower bound is 0, upper bound is 1;
    2. A sequence is said to be convergent if it approaches some limit.
    - Formally, a sequence S_n converges to the limit S
    - $lim_{n \to \infty}S_n=S$
    - if, for any $\epsilon \gt 0$, there exists an N such that $|S_n - S| \lt epsilon$. If $S_n$ does not converge, it is said to diverge. This condition can also be written as $lim_(n \to \infty)S_n=lim_(n \to \infty)S_n=S$
    - Every bounded monotonic sequence converges. Every unbounded sequence diverges.
    
**QUESTION 3:**
**Theorem about existence of only one limit of a convergent sequence.Show proof of this theorem.**
Let ${a_n}$ be a convergent sequence
Let $M = lim_{n \to \infty}{a_n}$ and $L = lim_{n \to \infty}{a_n}$
Suppose that $L$ != $M$
Let $\epsilon = \frac{|L-M|}{10}$
Then there exists such $N_1$, so that for any $n \gt N_1: |a_n - L| \lt \epsilon$
Then there exists such $N_2$, so that for any $n \gt N_2: |a_n - M| \lt \epsilon$
Let $N = max(N_1, N_2)$, if $n \gt N$, then both $|a_n - M| \lt \epsilon$ and $|a_n - L| \lt \epsilon$
**Trianlge inequality**:
$|L - M| \lt |a_n - L| + |M - a_n| \lt \frac{2}{10}|L - M|$
Then we obtain a contradiction, hence **L = M**

**QUESTION 4:**
**Theorem about limit of bounded monotonously growing sequence. Show
proof of this theorem.**
Any monotone bounded sequence ( $x_n$ ) has a finite limit equal to the exact upper bound, sup ( $x_n$  ) for non-decreasing and exact lower bound, inf ( $x_n$  ) for a non-increasing sequence.
Any monotonic unbounded sequence has an infinite limit equal to plus infinity for a non-decreasing sequence and minus infinity for a non-increasing sequence.
- if ($x_n$) is increasing and bounded above, then ($x_n$) converges.
**Proof**:
Consider $S = \{s_n | n \in N\}$
Since $s_n \lt M$ for all $M, S$ is bounded above, hence $S$ has a least upper bound, $s = sup(S)$
**Claim:** $lim_{s \to \infty} s_n = s$
Let $\epsilon \gt 0$ be given, then we need sych $N$, that $n \gt N$, then $|s_n - s| \lt \epsilon$
Consider $s - \epsilon < s$. By the defenition of $sup$, this means that there is such a $s_n \in S$ such that $S_n \gt s - \epsilon$
But then for that $N$, since $n \gt N$ as $s_N$ is increasing we have $s_n - s \gt s_N - s \gt \epsilon$
On the other hand, since $s = sup(S)$ by definition of $sup$, then we have $s_n \lt s for \forall s$ and so:
$s_n - s \lt s - s = 0 \lt \epsilon$
Therefore, we get:
$-\epsilon \lt s_n - s \lt \epsilon, |s_n - s| \lt \epsilon$
**So, $s_n$ converges to $s$**
**Corollary:** $s_n$ is decreasing and bounded below, then $s_n$ converges

**QUESTION 5:**
**The Bolzano-Weierstrass theorem about existence of the convergent
subsequence for any bounded sequence. Show proof of this theorem.**
Bolzano Weierstrass Theorem for Sequences:
- A convergent subsequence exists for every bounded sequence of real numbers.
- A convergent subsequence exists for any bounded sequence in $R^n$.
- A convergent subsequence exists for every sequence in a closed and bounded set S in $R^n$. (which eventually converges on a point in S).
**Definition** Any bounded sequence $\{a_n\}$ contains a convergent subsequence
**Proof**: Let $a = inf\{a_n\}, b = sup\{a_n\}$. Consider $[a, \frac{a+b}{2}]\ and\ (\frac{a+b}{2}, b]$, then one of those intervals contains a subsequence. Repeating the process we obtain convergent sequence of intervals which contains subsequence.These subsequences are converhent due to the convergency of the intervals.

**QUESTION 6:**    
**Definition of the superior limit for the sequence. Examples.**
**Definition:** A limit superior is a maximum of subsequential limits:
$a_n = (1 - \frac{1}{n})cos(n)$
$\limsup_{n \to \infty}a_n = 1$

**QUESTION 7:**
**The Bernuolly theorem about the number e. Proof.**
$e = \lim_{x \to \infty}(1 + \frac{1}{x})^x = \sum_{k = 0}^{\infty}\frac{1}{k!}$
**Proof**
From the binomial theorem we have:
$\lim_{x \to \infty}(1 + \frac{1}{x})^x = \sum_{k = 0}^{\infty}C_n^k\frac{1}{n^k} = \sum_{k = 0}^{n}\frac{n}{n}\frac{n-1}{n}\frac{n-2}{n}...\frac{n-k+1}{n}\frac{1}{k!}$
As $n \to \infty$, each term in the sum increases towards a limit of $\frac{1}{k!}$ and the number of terms to be summed increases, so 
$(1 + \frac{1}{x})^x \to \sum_{k = 0}^{\infty}\frac{1}{k!}$

**QUESTION 8:**
**Different forms of functions definitions: explicit, implicit and paramet-
ric.**
Explicitly defined functions: ordered pair of $(x, y)$ can be represented as $(x, f(x))$
Implicit function: An implicit function is a function, written in terms of both dependent and independent variables: $x + y = 1$
Parametric functions: The parametric function takes the form: p(t) = (f(t), g(t)) for a < t < b.

Implicit form of a circle: $x^2 + y^2 = r^2$
Paramteric form of a circle:
$x = rcost$
$y = rsint,\ 0 \lt t \lt 2\pi$

**QUESTION 9:**
**Two different definitions for a limit of a function. Clarify its equiva-
lence. Examples.**
1st definition of a limit of a function:
Function $f(x)$ has a limit A when x approaches a, if for $\forall\ \epsilon \gt 0\ \exists\ \delta \gt 0$, such that:
    $|f(x) - A| \lt \epsilon$ only if $|x - a| \lt \delta$

2nd definition:
There's also the Heine definition of the limit of a function, which states that a function $f(x)$ has a limit $L$ at $x = a$, if for every sequence $\{x_n\}$, which has a limit at $a$ the sequence $f(x)$ has a limit L. The Heine and Cauchy definitions of limit of a function are equivalent.

**QUESTION 10:**
**A rule of changing of variable into the limit: theorem, proof, example.
counterexample.**
If $\lim_{x \to a}g(x)=b, then\ \lim_{x \to a}f(g(x))=\lim_{y \to b}f(y)$
Let $a=b=0, g(x)=x^2$ and let:
$f(x) = \Biggl\{ 1, x \gt 0\ 0, x \lt 0$
Then $\lim_{y \to 0}f(y)$ does not exist, but the $\lim_{x \to 0}f(g(x)) = 1$
If we suppose that $\lim_{y \to b}f(y)$ exists, then the result holds.
First suppose that the limit exists and equal to $L$. Fix $\epsilon \gt 0$, and let $\delta$ be such that:
$|y - b| \lt \delta \to |f(y) - L| \lt \epsilon$
Now choose $\delta'$ such that:
$|x - a| \lt \delta' \to |g(x) - b| \lt \delta$
Now we have:
$|x - a| \lt \delta' \to |f(g(x)) - L| \lt \epsilon$
this shows that $\lim_{x \to a}f(g(x)) = L$

**QUESTION 11:**
**A definition of continuous function. Examples of continuous and dis-
continuous ones.**
A function f(x) is said to be a continuous function in calculus at a point x = a if the curve of the function does NOT break at the point x = a. The mathematical definition of the continuity of a function is as follows. A function f(x) is continuous at a point x = a if:
- f(a) exists
- $\lim_{x \to a}f(x)$ exists
- Both of the above values are equal: $\lim_{x \to a}f(x) = f(a)$
- if the graph doesn't have any holes or asymptotes at a point, it is always continuous at that point
Examples: $y = x^2$ (continuous function), $y = \frac{1}{x}$ (infinite discontinuity)

**QUESTION 12:**
**Asymptotic properties, definitions and examples.**
Asymptote: a line that draws increasingly nearer to a curve without ever meeting it.

There are basically three types of asymptotes: horizontal, vertical and oblique.
Assymptotic properties:
if there $\exists c \in R:\ c|f(x)| \gt |g(x)|\ \forall x \to x_0$, then we have:
$g(x) = O(f(x))$ - g is of order not exceed f
in other words:
$\forall \epsilon \gt 0\ \exists\ c \in R: c|f(x)| \gt g(x), |x - x_0| \lt \epsilon$
1) sinx = O(x), $x \to 0$

If $\frac{f(x)}{g(x)} \to 0, x \to x_0$, then we will write $f(x) = o(g(x))$ - f is of order less than g
sinx = o(1), $x \to 0$

If $\frac{f(x)}{g(x)} \to 1, x \to x_0$, then f is asymptotic to g
$x^2 + x ~ x, x \to 0$