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77

I think that I'd say that one of the underlying themes of analysis is, really, the limit. In pretty much every subfield of analysis, we spend a lot of time trying to control the size of certain quantities, with taking limits in mind. This is especially true in PDEs, when we consistently desire norm estimates on various quantities. Let's just discuss the "...

31

As an example: $f(x)=6|x|$ does not have a derivative at $x=0$ $g(x)=3x|x|$ does have a first derivative everywhere of $f(x)$ but not a second derivative at $x=0$ $h(x)=x^2|x|= |x^3|$ has a first derivative everywhere of $g(x)$ and a second derivative of $f(x)$ but not a third derivative at $x=0$

25

Traditionally, the sum of a sequence is defined as the limit of the partial sums; that is, for a sequence $\{a_n\}$, $\sum{a_n}$ is that number $S$ so that for every $\epsilon > 0$, there is an $N$ such that whenever $m > N$, $|S - \sum_{n = 0}^ma_n| < \epsilon$. There's no reason we can't define it like that for uncountable sequences as well: let $\... 23 It's a metric that is bounded above by$1$, while maintaining the same topology. This means that bounded metrics are just as powerful as general metrics (which is arguably interesting in itself). More concretely, there's a commonly used construction for turning a countable product of metric spaces into a metric space itself. Specifically, if we have spaces$...

20

Why solving a differentiated integral equation might eventually lead to erroneous solutions of the original problem? The reason is that taking a derivative is not an invertible operation. So the new equation you obtain is true, but not equivalent to the original one -- the set of solutions has increased. The simplest example is trying to solve an ordinary ...

19

You can just squeeze it: $$1\leq(n!)^{1/n^2}\leq (n^n)^{1/n^2}=n^{1/n}\to 1.$$ So $\lim_{n\to\infty} (n!)^{1/n^2}=1$.

18

From my viewpoint, Real Analysis is a study of functions of (one or several) real variable. Everything else (limits, derivatives, integrals, infinite series, etc.) is a tool serving this purpose. [There is a mild exception one has to make here for sequences and series of real numbers/vectors; these are functions defined on the set of natural numbers and ...

17

A concrete example Consider the ring $\mathbb{Q}[x]$ of polynomials over $\mathbb{Q}$. These can be linearly ordered as in this answer, in a way that preserves the ordering of the rationals. Now we have the "regular" naturals in this field: 1, 2,3, and so on. But the polynomial $x$ is larger than all of these, and $x^2$ is larger than $x$, and so on. So in ...

17

Suppose for some $x_0$ you have $f(x_0)\neq0$. Then you can solve the differential equation $$\frac{f'(x)}{f^2(x)}=1$$ with the initial condition $f(x_0)$, which gives $$\frac{-1}{f(x)}+\frac{1}{f(x_0)}=x-x_0\iff f(x)=\frac{1}{c-x}$$ where $c$ is a constant, and this holds for every $x$ that is in the same component of $\mathbb R\backslash\{c\}$ with $x_0$, ...

15

The bisection method only cares if the function changes sign, so it goes straight past the 'fake' root without noticing. If the coefficients have a slight error in them, then perhaps the 'fake' root should have been a root.

15

If $\{z_i:i\in I\}$ is any indexed set of complex numbers, then the series $\sum_{i\in I}z_i$ is said to converge to the complex number $z$ if for every $\epsilon>0$ there is a finite subset $J_\epsilon$ of $I$ such that, for every finite subset $J$ of $I$ with $J_\epsilon\subseteq J$, $\vert \sum_{i\in J}z_i-z\vert<\epsilon$. In other words, to say ...

15

Long ago someone told me this. I still remember it ... Sometimes I find myself just pushing symbols around, and I wonder, "Am I really doing analysis?" But when an argument begins, "Let $\varepsilon > 0$," then I know it really is analysis.

14

No, $f(x)=1/x$ is the only solution. The proof consists of many simple steps. part 1: $f$ is strictly decreasing: First, we show that $f$ is surjective. Let $x>0$. Then $$f(f(f(1/x)))(1/x)=1 \quad\Rightarrow\quad f(f(f(1/x)))=x.$$ Thus $f$ is surjective. Next, we show that $f$ is injective. Let $x,y>0$ with $f(x)=f(y)$. Then $$x f(f(f(y))) = ... 14 Such a function exists. To construct it, we need an auxiliary function w: \mathbb R \to [0,1] that is Continuous. Rational almost everywhere. Irrational at 0. Zero outside the interval (-1,1). This can be obtained from the Cantor function c: [0,1] \to [0,1] by taking some u \in (0,1) for which c(u) is irrational and by assigning$$w(x) = \...

13

Consider the functions, $f:[0,1]\rightarrow [-1,1]$ and $g:[0,1]\rightarrow[-1,1]$ where $$f = \begin{cases}\sin\frac{1}{x},& x>0 \\ 0, & x = 0\end{cases}$$ and $$g = \begin{cases}-\sin\frac{1}{x},& x>0 \\ 1, & x = 0\end{cases}$$

13

I realised this question has been asked before as you can see here. Anyway I will write down my solution here again. First of all consider Ramanuajan's Master Theorem. Ramanujan's Master Theorem Let $f(x)$ be an analytic function with a MacLaurin Expansion of the form $$f(x)=\sum_{k=0}^{\infty}\frac{\phi(k)}{k!}(-x)^k$$then the Mellin Transform of ...

13

$\log(e^x)=\int_1^{e^x}\frac{1}{t}dt=\int_0^x\frac{1}{e^u}e^u du=x$ and since it's easy to prove that $e^x$ is bijective then $\log$ is its inverse.

12

Wolfy says it is 4 times the Catalan's constant. One (not optimal) way to derive this is \def\sech{\operatorname{sech}} \begin{align*} \int_{-\infty}^\infty\arcsin\sech x\,\mathrm{d}x&=2\int_0^\infty\arcsin\sech x\,\mathrm{d}x\\ &=2\int_0^1\frac{\arcsin u\,\mathrm{d}u}{u\sqrt{1-u^2}}\quad(u=\sech x)\\ &=2\int_0^{\pi/2}\frac{\theta\,\mathrm{d}\... 12 We may adopt the technique for Ramanujan's infinite radical. Let p(x) = x^2 + 3x + 1 and define F : [0, \infty) \to [0, \infty) by F(x) = \sqrt{p(x) + \sqrt{p(x+1) + \sqrt{p(x+2) + \cdots }}} $$Then F solves$$ F(x)^2 = p(x) + F(x+1). $$Now we make an ansatz that F(x) takes the form F(x) = ax + b. Plugging this and comparing coefficients ... 12 What is the connection between the following three properties of a subset A of \mathbb{R}? (a) A is open. (b), (c) are pure maths. If you say one property implies another, you have to prove that implication. If you say that a property does not imply another, you have to give an explicit example to show that. Consider the set \mathbb{I} of ... 12 Hint: AM-GM implies$$ a_1+a_2+\cdots +a_n\ge n\sqrt[n]{a_1a_2\cdots a_n} $$and$$ \frac1{a_1}+\frac1{a_2}+\cdots +\frac1{a_n}\ge \frac{n}{\sqrt[n]{a_1a_2\cdots a_n}}. $$12 The main ingredient in my answer is the following fact: |\mathbb{R}^{\mathbb N}| = |\mathbb{R}|. Its proof isn't directly related to this question so I won't put it right here. But you can find it in the answer to this question. Due to this fact we can talk about the \mathbb R as about the disjoint union \coprod\limits_{i=1}^\infty R_i where all of ... 11 Even easier: two points in the plane. 11 and then divide through by n to give$$= \frac{\frac{1}{n}-(\frac{3}{8})^n}{\frac{n^{1728}}{2^{3n}}+\frac{1}{n}}$$My answer is zero using the following limits: \lim_{n\to\infty}\frac{1}{n} = 0 and since \frac{3}{8} <1 \implies\lim_{n\to\infty}\frac{3}{8}^n = 0 But not only the numerator tends to zero, the denominator does too.... 11 No, it is not possible. Suppose otherwise. Then, there is some N\in\mathbb N such that$$n\geqslant N\implies\lvert a_n-\varepsilon\rvert<\varepsilon\iff0<a_n<2\varepsilon.$$But this is impossible, since you are assuming that you always have a_n\leqslant0. 11 And you won't find them, since (\cdot,\cdot) is an inner product. The problem lies elsewhere: your space is not a complete metric space. 10 Let$$ F(x) = \det \begin{bmatrix} f(a) & g(a) & h(a) \\ f(b)& g(b) & h(b) \\ f(x) & g(x) & h(x) \end{bmatrix}, $$then F(a) = F(b )=0. Also note that$$ F'(x) = \det \begin{bmatrix} f(a) & g(a) & h(a) \\ f(b)& g(b) & h(b) \\ f'(x) & g'(x) & h'(x) \end{bmatrix},  which could be directly showed by ...

10

Hint: If you want to show that $\mathbb{R}^{n}$ is not compact, you just need to provide a specific example of an open cover that has no finite subcover. Try with the open cover $\{\Omega_{k}\}_{k=1}^{\infty}$, where $\Omega_{k}$ is the open ball of radius $k$, centred at $\mathbf{0}$. (By the way, this is an example of the general principle that to show ...

10

Let $L=({n!})^{\frac{1}{n^2}}$ $L=e^\frac{\log_{e}{n!}}{n^2}=e^\frac{\sum_{i=1}^{n}{\log(i)}}{n^2}$ Now $0\leq\frac{\sum_{i=1}^{n}{\log(i)}}{n^2}\leq \frac{n\log(n)}{n^2}=\frac{\log(n)}{n}$ which tends to $0$ as $n$ tends to $\infty$. Hence$\lim\limits_{n \to \infty}{\frac{\sum_{i=1}^{n}{\log(i)}}{n^2}} = 0$. $\lim\limits_{n \to \infty}{L} = e^{\lim\... 10 I'll address your points one by one. If you have further questions, please ask away: Aren't all functions$C^\infty$then? No. Examples of functions whose$k$-th order derivative is not continuous abound for every$k$. It is true that$C^1$functions are very similar to$C^\infty\$ functions, but for the time being it is better to focus on the ones we will ...

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