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1

Well, the first problem is the circle is an uncountable set, so the usual way of defining infinite sums with sequences and partial sums doesn't really work. There's a way to extend sums to uncountable sets, but the sum will diverge unless all but a countable subset of the terms are zero. There are also alternative summation methods which are interesting ...

1

Often, when we use the $\infty$ sign, it is convenient shorthand for something else. In calculus, we might write $\lim_{x\rightarrow a} f(x) = \infty$, as shorthand for saying that the limit fails to exist, and in particular that it fails to exist in such a way that as $x\rightarrow a$, the values of $f$ become larger than any finite amount. The ancient ...

2

However it is written at the outset we are given the sequence $$x_n:={a\,n\over b}\qquad(n\geq1)\ ,$$ with the tacit assumption that $b\ne0$. If $a=0$ then $x_n=0$ for all $n$, hence $\lim_{n\to\infty}x_n=0$. If $a\ne0$ then we all know that the $x_n$ converge to $\infty$ if $ab>0$, and to $-\infty$, if $ab<0$. This means that the sequence is divergent ...

0

Well, $$\lim_{n \to \infty} f(n)g(n) = \left( \lim_{n \to \infty} f(n) \right) \left(\lim_{n \to \infty} g(n) \right)$$ only holds if both $\lim_{n \to \infty} f(n)$ and $\lim_{n \to \infty} g(n)$ exist. In this case, $f(n)=a$ and $g(n) = \frac{n}{b}$. It is clear then that $\lim_{n \to \infty} g(n)$ does not exist.

3

Let me modify the parametrization in William's answer and give another sense in which the answer is yes. In his answer, he parametrized the arc for a given $r$ by the function $$f_r(s)=\left(r \cos\left(\frac{s}{r}\right), r \sin\left(\frac{s}{r}\right)\right)$$ where $s\in[0,a]$. These functions do not converge pointwise as $r\to\infty$, essentially ...

0

Alright, make $r$ as big as you want. Let's say, for example, $r$ = $10^{10}a$. It turns out that the arc length $a$ actually remains an arc if you didn't change your viewport's size (i.e. the distance from an hypothetical orthographic camera to the paper in which you are drawing). So to supply you with an answer, let's say your (orthographic) camera is $d=... 2 The arc will flatten out as$r\to \infty$. Specifically, it flattens out closer and closer into a line because its curvature approaches zero. By this we mean that we parametrize the arc in$\mathbb{R}^2$by $$f(s)=\left(r \cos\left(\frac{s}{r}\right), r \sin\left(\frac{s}{r}\right)\right)$$ for$s\in [0,a]$for some$a>0. Notice then that this is an ... 1 In order to evaluate the limit, we must retain terms of second order in the expansion of the complex exponential terms. Note that we can write $$e^{\pm it/\sqrt{m+n}}-1= \pm it\,\frac{1}{\sqrt{m+n}}-\frac12 \frac{t^2}{m+n}+O\left(\frac{1}{(m+n)^{3/2}}\right)$$ Then, we have \begin{align} n\left(e^{ it/\sqrt{m+n}}-1\right)+m\left(e^{- it/\sqrt{m+n}}-1\... 6 I think that you could do it faster considering the implicit functionF=\log(x)-\frac 1y \log(y)=0\implies F'_x=\frac{1}{x}\qquad F'_y=\frac{\log (y)}{y^2}-\frac{1}{y^2}$$Now, from the implicit function theorem$$\frac{dy}{dx}=-\frac{F'_x}{F'_y}=\frac{y^2}{x-x \log (y)}$$If now you replace x by y^y, you effectively get$$\frac{dy}{dx}=\frac{y^{2-y}}{... 0 Since you mean to ask why\int_{-\infty}^{0}\frac{1}{x} + \int_{0}^{\infty}\frac{1}{x}$doesn't equal zero; it simply follows since these integrals do not converge. Consider$\int_{0}^{\infty}\sin(x)$; as we approach infinity the areas always cancel out after the fixed period and the value of the integral at any point larger than zero never exceeds$\int_{...

0

The limiting argument for the improper integral is clearly the way to go. Note that $x=0$ is the only point of infinite discontinuity of the integrand $\frac{1}{x}$ in $(-\infty,\infty)$. Thus, $\displaystyle\int_{-\infty}^{\infty}\frac{1}{x}\,dx$ $=\displaystyle\lim_{\substack{\epsilon \to 0+ \\A \to -\infty }}\displaystyle\int_{A}^{0-\epsilon}\frac{1}{x}... 0 This would be an interesting experiment in psychology or human behavior. What is the distribution of natural numbers "randomly" picked by volunteers? I suspect you'll see a bell-like distribution with a long tail where the 1st standard deviation is < 1000. Once you have that curve defined with experimental data, real probabilities can be calculated. ... 0 The intuition is a common one. It may be intuitive to say $$\int_{-\infty}^{\infty} \frac{1}{x} dx=\int_{0}^{\infty} \frac{1}{x} dx+\int_{-\infty}^{0} \frac{1}{x} dx=$$ $$\lim_{a \rightarrow 0^+} \lim_{b \rightarrow \infty} \left(\int_{a}^b \frac{1}{x} dx +\int_{-b}^{-a} \frac{1}{x} dx\right)=\lim_{a \rightarrow 0^+} \lim_{b \rightarrow \infty} \left(\int_{... 0 I think what you said make sense. It is a time to get more understanding of improper integration. By definition improper integral is the limit of proper integral. In this case, you have four integral.$$\int_{1}^{\infty} \frac{1}{x} \ \mathrm{ dx}\int_{-\infty}^{-1} \frac{1}{x} \ \mathrm{ dx}\int_{0}^{1} \frac{1}{x} \ \mathrm{ dx}\int_{-1}^{0}... 6 What is not possible is to pick a random natural number in a way such that every natural number is equally likely. If you don't require every number to have the same probability, then one can certainly come up with a well-defined probabilistic experiment where every natural is a possible outcome -- for example, keep flipping a fair coin until you get tails, ... 0 As already mentioned that's just shorthand for limits. Anyways suppose we have$1$pizza and we want to divide this pizza equally into$n$friends. Then each person gets a proportion:$\frac{1}{n}$of the pizza. Obviously the more friends you have the less each person is going to eat. And as your number of friends approaches$\infty\$ then you get closer ...

4

So first of all this is really a type error. Cardinals don't let you divide. Perhaps instead you mean "an infinite natural number" but there is no such thing. In particular, there is no uniform distribution on an infinite set (in the sense that a given positive probability is assigned to every point). Those last two sentences assume that we are working in ...

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