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0

The phrase "is reduced by e.g. 10% in infinity" is perhaps ambiguous, but let me try to make it more precise: If you reduce a quantity $A$ by $10\%$ (i.e., a proportion of $\frac{1}{10}$, then the remaining value is $$\left(1 - \frac{1}{10}\right) A = \frac{9}{10} A,$$ and, inductively, if you do so $n$ times, then the remaining values is ...


0

In a sense, it's equal to zero in the same way your example is equal to 1. Both converge on to their target, but only after an infinite number of digits/reductions.


4

The aleph numbers can certainly be multiplies, but not to give the results you mention. In fact, $2\aleph_0=\aleph_0$ and there is no such thing as fractional alephs so $0.5|A|$ is meaningless. For the sets $A$ and $B$ you mention it is very easy to establish a bijection and thus $|A|=|B|$.


0

No. The symbol $\infty$ as used in analysis usually means either positive infinite element of affinely extended real line (in this case you also would have $-\infty$ as another element) or (more rarely) the infinite element of the projectively extended real line. In the later case it is the multiplicative inverse of zero ($\frac 10=\infty$). So in fact you ...


0

The mathematical side of the question has been dealt with in detail. But since you were applying for a job as a C# programmer, maybe they were (also) looking for a programming solution? Floating point numbers have limited precision, i.e. they can only represent a finite number of different values. Therefore, the open interval $(0,1)$ does have a maximum: ...


3

The set has a supremum (equal to 1), but not a maximum.


-1

This is an interesting question. In terms of the mathematics we are usually taught, there is no maximum. In particular, for any putative maximum, you can find a bigger element in the set. If you are asked this question, that would be the answer to give. But, there are other kinds of mathematics. In these different kinds of mathematics, we have different ...


3

Mostly likely, the response they were looking for was a computer science answer, not a mathematician answer. As usual in interviews, they were probably more interested in the way you handled the question than your actual answer. I would have handled it something like this: "Let's suppose we're storing 1.000... in a floating point variable. Now, we want to ...


0

It matters if you compactify the real line with $2$ infinities, as you implicity did. If you use the real line Alexandroff Compactification (which seems like what you were naively assuming when you said "Aren't both numbers equal by all measurable accounts?"), you can actually define $1/0=\infty$ in order to make this function continuous in the whole line ...


1

On the surface, it would appear that there is a maximum. However an open set is just that - open. No matter how close you get to the upper limit (which is 1) you can never touch it - but you can get a teeny bit closer each time. Now, in (your) practice, there is a maximum. The maximum would be whatever data type the decimal was cast as. It would be ...


-1

For every open interval of real numbers, of course including the real number set itself, there exist both maximal and minimal element which lie outside that interval. In other words, for every open interval the question of asking "what is the biggest number that you can imagine" is equivalent to asking the same question restricted to that interval. Note that ...


13

That's a very creative answer, but infinity is tricky. If $$ 1 - \frac{1}{10^\infty} < 1 $$ would you not have to admit that $$ 1 - \frac{1}{10^\infty} < 1 - \frac{1}{2} \cdot \frac{1}{10^\infty} < 1 $$ as well? This would contradict your assertion that $1 - \frac{1}{10^\infty}$ was the maximum value of $(0,1)$. You are right that this seems ...


4

For every $x\in(0,1)$ we can find an $y\in(0,1)$ (e.g. $y=\frac12+\frac12 x$) such that $x<y$. So no $x\in(0,1)$ is maximal in $(0,1)$.


28

Your answer doesn't make much sense. It's equal to $1$ if anything (see Is $.999999999... = 1$?). You have to look at the definition: $m$ is the maximum if $m\in(0,1)$ and $\forall x\in(0,1),\ x\le m$. Clearly, this fails for any element $m\in(0,1)$ because $m<(m+1)/2<1$.


14

A maximum must be in the set. In the case of $(0,1)$ if $a \in (0,1)$ then the point $(1+a)/2$ is larger than $a$ and still in the set $(0,1).$ So there can be no max. There is a sup, namely $1,$ of the set $(0,1).$


2

I suspect there is no natural sigmoid function that will "preserve" in some sense your original function $f$. In all generality, you want to "squeeze" the real line ($\mathbb{R}$) into an (open) interval of the form $(x_1,x_2)$. Let's try to squeeze both the $x$-axis and the $y$-axis so that we may fit one graph into a rectangle. Let this rectangle be, for ...


0

Definition of infinite sums is in terms of limit of finite sums: namely the partial sums. Now, adding finitely many $0's$ only gives you $0$. So, $S_n=0$ for all $n$. The limit of a sequence whose terms are all $0$ is again $0$. So, the series is convergent, and the sum is $0$.


1

Use the definition of an infinite series, the limit of the sequence of partial sums: $$\displaystyle \sum_{k=0}^\infty 0 = \lim_{N \to +\infty} \sum_{k = 0}^N 0$$ By the definition of limits at infinity, for every $\epsilon > 0$, choose any $M$ you want, it doesn't matter because the sequence is constant: $$\displaystyle N > M \implies -\epsilon ...


4

Analytic Regularization This answer tries to explain why it is said that $$ 1+2+3+4+5+\dots=-\frac1{12}\tag{1} $$ because $$ \zeta(-1)=-\frac1{12}\tag{2} $$ and $$ \zeta(s)=\frac1{1^s}+\frac1{2^s}+\frac1{3^s}+\frac1{4^s}+\frac1{5^s}+\dots\tag{3} $$ since plugging $s=-1$ into $(3)$ and formally applying $(2)$ gives $(1)$. However, $(3)$ does not converge for ...


4

This may be to your surprise. In certain settings, the statement $$S = \sum_{k=0}^\infty 2^k = 1 + 2 + 2^2 + 2^3 + \cdots = -1$$ is true and the manipulation $$S = 1 + 2S \quad\implies\quad S = -1$$ is actually legal! The catch is the convergence of the series is not the ordinary one over real numbers! To understand this, we need to step back and ask what ...


2

For $N \in \mathbb N$, the number of powers that are $\le N$ is at most $$\lfloor N^{1/2}\rfloor + \lfloor N^{1/3}-1\rfloor + \lfloor N^{1/4}-1\rfloor + \lfloor N^{1/5}-1\rfloor + \cdots$$ (where the $-1$'s are to avoid counting $1$ itself more than once). There are at most $\log_2N$ non-zero terms here, and all of them are $ \le N^{1/2}$. So the number of ...


-2

The problem is that infinity is not a number. Infinity is a concept. It has a lot to do with numbers, which is why we write stuff like $\infty + 2 = \infty$ or $\infty \cdot 3 = \infty$, but you can't say that $\infty$ is a number like $2$ or $3.14159$ or $\pi$. Yes, idiomatically we say things like $\infty + 1 = \infty$, but as @Asaf Karagila pointed out in ...


6

The problem is the reasoning $S=1+2S\implies S=-1$. This holds only for finite $S$ (cancelling infinities doesn't preserve the limiting behavior).


3

For $x \to -\infty$, we have that $\sqrt{x^2} = |x| = -x$.


2

$\sqrt{x^2}=|x|$ for real $x$ For $x\le0, |x|=-x$


1

Hint: Break up the sum into even and odd parts. $$\sum_{i=1}^\infty\frac{1}{n^4} = \sum_{i=1}^\infty\frac{1}{(2n-1)^4}+\sum_{i=1}^\infty\frac{1}{(2n)^4}$$ You know the value of the quantity on the LHS, and can easily find the value of $\sum_{i=1}^\infty\frac{1}{(2n)^4}$. Solve for the remaining quantity and you have your answer.


2

In general we have for all $n\ge 1$ $$ \zeta(n)=\frac{2^n}{2^n-1}\sum_{k=1}^{\infty}\frac{1}{(2k-1)^n} $$ This can be proved as follows \begin{align*} \zeta(n) & = \left(\frac{1}{1^n}+\frac{1}{3^n}+\cdots\right)+\left(\frac{1}{2^n}+\frac{1}{4^n}+\cdots\right) \cr & =\left(\frac{1}{1^n}+\frac{1}{3^n}+\cdots\right)+\frac{1}{2^n}\zeta(n) \cr ...


0

When I'd read it I found the book Infinity and mind of Rudy Rucker much instructive. It is pretty broad and, if I recall correctly, discusses also the problem of mixing the property of being infinite and the extrapolation from natural numbers to an infinite number , which in my view is a frequent source of "paradoxa" and "perplexing ideas" for the beginner ...


2

One thing you could do is to consider this as a limit. Note first that: $$33333=\frac {10^5-1}3$$ Then $$333338=10\frac {10^5-1}3+8=\frac{10^6-10+24}3=\frac {10^6+14}3$$ Then $$0.333338=10^{-6}\cdot\frac {10^6+14}3=\frac {1+14\times 10^{-6}}3$$ Now if we have $r$ threes followed by an eight, the equivalent formula is $$\frac {1+14\times ...


1

can't you use $$ x^2 - 3 \le x^2 + 3\sin x \le x^2 + 3$$ and that $\lim_{x \to \infty} x^2 = \infty$ to argue that $$\lim_{x \to \infty } x^2 + 3 \sin x = \infty$$


2

It’s a standard result that the set of infinite sequence of zeroes and ones is uncountable: there’s a bijection between it and $\wp(\Bbb N)$. Each equivalence class of $R$, however, is countable: given a sequence $\sigma=\langle x_n:n\in\Bbb N\rangle$ of zeroes and ones, there are only countably many finite $F\subseteq\Bbb N$ and hence only countably many ...


0

I should clarify a few things you got wrong in the original post. Particularly the edited part. First, you appear to only be using only finite decimal expansions on all your real numbers. But this will not even get you out of the rational numbers as the finite decimal $a_n...a_1a_0.d_1d_2...d_m$ can be written as $a_n...a_1a_0d_1d_2...d_m/10^m$. This ...


2

Your argument seems to revolve around the length of the members being finite. Note that in a set of finite strings, each string must have a finite length, but this length ($N$) need not be the same for all members. This means: Given a number $N$, the set $\Sigma^{\le N}$ of strings with length at most $N$ is finite (you argue about such sets) The set ...


3

Belief has nothing to do with it. Either you accept the "classical" (and somewhat implicit) rules of mathematics, which permit an infinite set of finite strings. Or you reject them, in which case you have to tell us what you think is true or false. Your argument completely fails, because there is no "maximal length finite string" of digits, since in that ...


5

Even simpler,$$\{A, \quad AA,\quad AAA,\quad AAAA,\quad\ldots\}$$ Compare this to the observation that $\mathbb{N}=\{1,2,3,4,\ldots\}$ is an infinite set of numbers, each of which is nevertheless finite. There is no reason that a set of numbers must have a maximum number, just like there is no reason that a set whose elements are finite length strings must ...


2

This is a duplicate. I copy below my answer Using Taylor expansion we have $$\sin t=t (1+o(1))\text{ when }t\to 0.$$ At this point if $x$ goes to infinity, then $$\left(1+sin \frac{1}{x}\right)^x \sim \left(1+\frac{1}{x}\right)^x\to e.$$


0

$$\lim_{n\to\infty}\left(1+\sin\frac1n\right)^n$$ $$=\left[\lim_{n\to\infty}\left(1+\sin\frac1n\right)^\frac1{\sin\frac1n}\right]^{\lim_{n\to\infty}n\cdot\sin\frac1n}$$ Set $\sin\frac1n=1/m$ to get $\lim_{n\to\infty}\left(1+\sin\dfrac1n\right)^\frac1{\sin\frac1n}=\lim_{m\to\infty}\left(1+\dfrac1m\right)^m=e$ Finally for the exponent, $\lim_{n\to\infty} ...


-2

Please not though that $x = 0.\bar 9$ is not an admissible way of writing a real number. The decimal representation of a real number may not end with repeating $9$'s, and the reason is that if we allow that, then we do not have uniqueness of representation, as the real number $1$ would have two different representation, $1$ and $0.\bar 9$ (I point this ...


2

this also proves that two different numbers can be equal. The point is that they are NOT different numbers, just two different representations. You've already said that $0{,}\overline{3}$ (The overline is common notation for a repeating set of digits) is $\frac{1}{3}$. To find out what the other repeating representations are equal to, use the same ...


5

The idea is the same, just solve for $x=0{.}\overline{8}$: $$10x=8{.}\overline{8}=8+x\implies 9x=8\implies x=\frac 89$$ Surely you can do the other fractions on your own, or just note that $0{.}\overline{1}=\frac19$ by the same argument, and therefore $0{.}\overline{2}=2\cdot\frac19=\frac29$ and so on. Note that the "simpler" proof relies on the fact ...


0

It's just that $1$ has $2$ different radix representations in base $10$: $\{1, 0.\bar9\}$. In fact, all integers have $2$ different radix representations in base $10$. The integer itself, and the previous integer $+ \space0.\bar 9$. $13$ can be written as either $13$ or $12.\bar 9$. These two representations point to the same value. $0.\bar8$ is a radix ...


0

Your espression $0.3333...x$ is meaningless. You can imitate the proof you gave like this: $$ x = 0.333..., \\ 10x = 3.333... \\ 10x-x = 3.000... = 3 \\ 9x = 3 \\ x = 3/9 = 1/3 $$ Your other questions can be answered similarly. $0.7777... = 7/9$ and so on.


1

The question of exact iterations has been settled. Let us look at it from a numerical standpoint. Assuming that the current iterate is $\epsilon$ away from Dottie's number, $$D+\epsilon'=cos(D+\epsilon)=\cos D\cos\epsilon-\sin D\sin\epsilon\approx\cos D-\frac{\epsilon^2}2\cos D-\epsilon\sin D\approx D-\epsilon\sin D.$$ The $\epsilon^2$ term quickly ...


0

I think if you proceed by contradiction, you can figure it out: let's say $n$ is finite: so we can write $\cos^{(n)}(A) = d$ (taking the cosine $n$ times) where $d$ is Dottie number, i.e. $d = cos(d)$. So, what we're saying is $\cos^{(n)}(A) = cos(d)$. One solution of that is $\cos^{(n-1)}(A) = d$. So, if you can reach $d$ is $n$ steps, then you can reach it ...


3

It will either require $0$, $1$, $2$, or infinitely many steps. Here's why. Denote the Dottie number by $x$, and the initial point of our iteration by $y$. If $y=x$ then we are already at the Dottie number after $0$ iterations. Define $$A=\{ y : \cos(y) = x \text{ and } y \neq x \} = \{ x + 2 \pi k : k \in \mathbb{Z} \setminus \{ 0 \} \} \cup \{ -x + 2 ...


0

In exact arithmetic it would never get there, but because there is roundoff error in your calculator you end up after finitely many steps with some approximation of the correct number.


0

If you mean the real number corresponding to $0.00\dots$ the name is $0$, but if you mean the representation of the real number as a sequence of digits the "name" could be $(0\cdot10^{-i})_{i\in\mathbb N}=(0\cdot10^{-0},0\cdot10^{-1},0\cdot10^{-2},\dots)$. If you represent the decimal numbers with functions $\mathbb N\to D$ (which is equivalent with ...


1

In some contexts one adjoins two objects to the real line called $\infty$ and $-\infty$ so that one is working in the set $\mathbb R\cup\{\infty\}\cup\{-\infty\}$, and the less-than relation among the members of this set is just what you would expect. In that case the only things less than $\infty$ are real (in particular, finite) numbers and $-\infty$. ...


5

When dealing with a sum of nonnegative terms $\sum_{n=1}^\infty a_n$, there are only two things such a sum can do: it converges to a finite value $L$, in which case we say $\sum_{n=1}^\infty a_n = L$, or it diverges to $+\infty$, in which case we say $\sum_{n=1}^\infty a_n = \infty$. In case (1.), if we don't care to mention (or don't know) the value ...


1

Each guest was hungry and wanted two portions, so after the ice cream was served, guest $n$ moved to chair $2 \cdot n$ and ate both his own ice cream and the ice cream at the place to his right.



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