Tag Info

New answers tagged

0

Not for any typical meaning of the terms in your question. But you really should be more precise with what you mean. When talking about the size of infinite sets, we'll typically be considering their cardinality, and there are infinite sets of different cardinalities.


0

To prove that $\mathbb{R}$ and $\mathbb{R^2}$ both have same size, it's sufficient to show that there is a bijection between these two. consider $f : \mathbb{R} \rightarrow \mathbb{R^2}$ which images each $x\in \mathbb{R}$ to $(x,0)$. this function is clearly one to one. Assume another function $g : \mathbb{R^2} \rightarrow \mathbb{R}$.The function formula ...


0

Let $a,b,c,d$ be positive; let also $L=\max (a,b,c,d)$. Now consider $\lim_{x\to+\infty}\frac{a^x+b^x}{c^x+d^x}=\lim_{x\to+\infty}\frac{(a/L)^x+(b/L)^x}{(c/L)^x+(d/L)^x}$ and conclude. The limit $x\to-\infty$ is done similarly.


2

For example $1/3=0.333\ldots$ is not in the image, and so the function is not onto.(This is already contained in the answer by MooS).


8

There are no integers with infinite length (in decimal system). The image of your map is contained in the rational numbers, hence can not be all of the real interval. Note that the image is not even all of the rational numbers between $0$ and $1$. For instance $\frac{1}{3}$ does not appear.


1

This can be obtained with the following theorem of Kuratowski: $\omega\leq^*\mathfrak a\iff\omega\leq2^\mathfrak a$ Namely, there is a surjection from $A$ onto $\omega$ if and only if there is an injection from $\omega$ into $\mathcal P(A)$, which in turn is to say that there is an injection from $\mathcal P(A)$ into itself which is not a surjection. ...


1

Revised, since there was an oversight in my original argument. I’ve now looked at Tarski’s paper; the very old-fashioned notation is a worse problem than the French. I’ve updated it and filled in some details. The proof in question is on pages $94$ and $95$. In essence he starts by assuming that there is an injection $\varphi:\wp(X)\to\wp(X)$ such that ...


1

We have $\sqrt[3]{n^3+an}=n\sqrt[3]{1+\frac{a}{n^2}}$. We expand the latter using Newton's Binomial Theorem to get $$\sqrt[3]{n^3+an}=n\sum_{k\ge 0} {1/3 \choose k} \left(\frac{a}{n^2}\right)^k=n(1+\frac{a}{3n^2}-\frac{a^2}{9n^4}+O(n^{-6}))$$ Repeating, we have $\sqrt{n^2+3}=n\sqrt{1+\frac{3}{n^2}}$, so $$\sqrt{n^2+3}=n\sum_{k\ge 0}{1/2 \choose ...


2

Your argument using L'Hopilat rule is correct you need just to add the condition $a>1$ For $a> 1$ it's true that $a^x$ is very larger then $x^n$ to see this you compose with a logarithm: $$\lim_{x\to \infty} \frac{a^x}{x^n}=\lim_{x\to \infty} e^{\displaystyle x\ln(a)-n\ln(x)} =e^{+\infty}=+\infty$$ because the linear functions are always larger than ...


3

To tell someone what is infinity ..take A4 paper then divided it to two half. After that , leave one half and take the other half and divided again and repeat this step more and more.Here ask question is this process will finish??


3

The problem here is: Division by $0$ or $\pm\infty$ is not generally defined! It is not entirely mathematically correct to write $\dfrac{x}{\infty}=0$ when $x$ is real. One thing you can do to mathematically justify your initial statement is to write it in the form of a limt: $$\lim_{n\to\infty}\frac{x}{n}=0~\forall~x\in(-\infty,+\infty)$$ And then, ...


0

This comes from thinking of the one-point compactification of the complex plane, called the Riemann sphere, which is basically $\mathbb{C}$ with a point called "$\infty$" added to it, and a topology so that sequences $(z_n)$ such that $|z_n|$ eventually becomes larger than any positive number you choose (i.e. the usual definition of limit) are said to "tend ...


1

To say that $1/0$ is "positive and negative infinity" is at best misleading without further context. One may say that $$ \lim_{x\downarrow 0} \frac 1 x = +\infty\quad\text{and}\quad\lim_{x\uparrow\infty} \frac 1 x = -\infty. $$ And that was probably what was meant if you read that $1/0$ is "positive and negative infinity". However, there is also an ...


0

2SAT is in P, yet whatever you said about 3SAT holds also for 2SAT.


0

Consider sorting. It runs in polynomial time, yet chooses one correct answer from $n!$ possibilities. This is greater than the potentially $2^n$ possibilities for the SAT problem over $n$ variables, for example. So if I understand the outlines of your argument, this would be a counterexample.


0

If I take a graph with vertex set $\mathbb{R}$, and connect 0 to every $x \in \mathbb{R} \setminus \{ 0 \}$, then I have uncountably many finite paths from 0. If on the other hand you require the vertex set to be countable, then the number of finite paths from any given point is countable. (As it is a countable union of countable sets). As Tryss says, if ...


0

Take $a_n=n$ and $b_n=(-1)^n$. You can choose for $b_n$ any bounded sequence that doesn't converge, and $a_n$ as a diverging sequence.


3

Assume it's true, then $$\lim_{n\to\infty}b_n=\lim_{n\to\infty}\left(a_n+b_n\right)-\lim_{n\to\infty}a_n$$ and then...


1

The extended real line is not a metric space. You are correct in noting that $d(0,+\infty)$ must be larger than any $d(0,x)$ for any $x\in\Bbb R$, and therefore cannot be a real number. However the extended real line is metrizable. This means that we can redefine the metric completely, while preserving the same topology (and therefore the same convergent ...


0

A systematic way of assigning finite values to divergent sums is a summability method. Consider the situation for convergent series. Let $V$ be the set of all convergent sequences $(z_1,z_2,,z_3,\cdots)$ of complex numbers. Then $V$ is a complex vector space, and there is a linear map $V\to\Bbb C$ given by the description $(z_1,z_2,z_3,\cdots)\mapsto ...


0

I believe that the problem with your argument is that $(-1)+(-2)+(-3)+\cdots$ should not be $1 \over 12$.


2

If you want to assign this series a sum, a more reasonable way is to consider $$ \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^s} = (1-2^{1-s}) \sum_{n=1}^{\infty} \frac{1}{n^s} = (1-2^{1-s}) \zeta(s). $$ The series on the left converges for $\Re{(s)}>0$, and as $s \downarrow 0$, it turns into Grandi's series $ \sum (-1)^{n-1}$, which we know is assigned the ...


0

All of your series are divergent, so the answer is your series has no finite sum by definition of a divergent series.


0

Your proof (after the edit, with the separate even and odd sequences of non-repeating and repeating decimals) seems correct, except for the claim (just before the edit) about one "extra" rational number in the interval $[0,1]$. (Remember how you can always book one more guest in the Hilbert Hotel.) The part that you added to the proof to handle the ...


2

Firstly, you should not say "infinite integers" if you mean "infinitely many integers". If one admits such a thing as an infinite integer, and $n$ is an infinite integer, then $n$ and $n+1$ are infinite integers, but they are not infinitely many, since there are only two of them. This is standard usage in mathematics, regardless of what usages may prevail ...


0

Integers can be broken up into 3 subsets: $\{ 0 \}$, ${+,\mathbb{N}}$ and $\{-, \mathbb{N}\}$, i.e. 0 and two instances of the set of naturals, one for positive integers and the other for negatives. The integer $0 $ corresponds with the rational $0$. If we can construct a correspondence between the positive rationals and the naturals, then we can use it ...


0

Like KittyL said in the comments, it's wrong to say that it equals $1^\infty$. It depends on what $x$ you look at first. If you take a look at the inner x first you indeed get that $\frac{1}{\infty}=0$ so that $1+\frac{1}{\infty}=1$ and $1^\infty=1$. But if you take a look at the $x$ in the exponent first, you see that $1+\frac{1}{x} > 1, \forall x>0$ ...


2

L'Hôpital works perfectly: $$\lim_{x\to\infty}x^2e^{-2x} = \lim_{x\to\infty}\frac{x^2}{e^{2x}} = \lim_{x\to\infty}\frac{2x }{2e^{2x}}=\lim_{x\to\infty}\frac{2 }{4e^{2x}}=0.$$


2

It's just one of those complications that comes up when you deal with infinity. Here's how I think about it. When you want the monkeys to type out Hamlet, you can figure that Hamlet only has a finite length and you can compute the probability of randomly typing it. Since the monkeys are typing out an infinite string of text, you can divide that up into ...


5

Here is a similar example: let $x$ be a number chosen uniformly from $[0,1]$. For every $y \in [0,1]$, the probability that $x = y$ is $0$; and yet $x$ must be equal to something! What goes wrong here is that there are uncountably many numbers in $[0,1]$, while the uniform measure over $[0,1]$ is only countably additive. Measures with stronger additivity ...


3

The probability of typing out, for example, the digits of pi, is $0$. But, there are an uncountably infinite number of strings of characters the monkey could type out, so the total probability of all events is still equal to $1$. It's a bit like saying if you could throw a dart at the real number line, there is no chance you could hit one specific number. ...


1

Just use the fact that $\ln(n) < n$, so $\ln(\ln(n)) < n$ (for n big enough) Then $$\ln(\ln(n))^{\ln(n)} = e^{\ln(n) \ln( \ln (\ln(n))) } \leq e^{ \ln(n) \ln(n) } $$ And as $\sqrt{n} \geq (\ln(n))^2$ (i assume this is known), $$\ln(\ln(n))^{\ln(n)}\leq e^{\sqrt{n} \ln(7) } = 7^{\sqrt{n}}$$


2

Start with $$7=(\log(\log(n)))^{\log_{\log(\log(n))}(7)} = (\log(\log(n)))^{\log(7)/\log(\log(\log(n)))}.$$ So $$7^{\sqrt{n}} = (\log(\log(n)))^{\log(7) \sqrt{n}/\log(\log(\log(n)))}.$$ Now you just have to prove that the exponent is asymptotically larger than $\log(n)$. This is a good way to start a lot of these sorts of problems: the idea was to make ...


2

Suppose $\;a\ge b>0\;$ , then $$L:=(a^x+b^x)^{1/x}=b\left(1+\left(\frac ab\right)^x\right)^{1/x}$$ Now observe that $$\lim_{x\to\infty}\frac{\log\left(1+\left(\frac ab\right)^x\right)}x\stackrel{\text{l'H}}=\lim_{x\to\infty}\frac{\left(\frac ab\right)^x\log\frac ab}{1+\left(\frac ab\right)^x}\xrightarrow[x\to\infty]{}\log\frac ab$$ so that ...


1

We assume that $a\ge 0$ and $b\ge 0$ and write $$\begin{align} \lim_{x \to \infty} \left(\frac{a^x+b^x}{2}\right)^{1/x} &=\,a\lim_{x \to \infty} \left(\frac{1+(b/a)^x}{2}\right)^{1/x} \end{align}$$ If $a=b$, then the limit is obviously equal to $a$. Now, we assume without loss of generality that $a>b$. Then, it is easy to see that ...


1

$$\lim_{x\rightarrow \infty }\frac{(a^x+b^x)^{1/x}}{2^{1/x}}$$ now we have two cases as follows if $a\geq b$ $$\lim_{x\rightarrow \infty }\frac{(a^x(1+(b/a)^x)^{1/x}}{2^{1/x}})=\lim_{x\rightarrow \infty }\frac{a(1+(b/a)^{x})^{1/x}}{2^{1/x}}=a$$ and when b>a the limit becomes $b$


5

HINT: We have $$\dfrac{\max(a^x,b^x)}2 \leq \dfrac{a^x+b^x}2 \leq \max(a^x,b^x)$$ Hence, $$\dfrac{\max(a,b)}{2^{1/x}} \leq \left(\dfrac{a^x+b^x}2\right)^{1/x} \leq \max(a,b)$$


6

There is no largest "number" (i.e., cardinality) possible. Cantor showed that for any cardinal $\kappa$, $2^\kappa > \kappa$, where $2^\kappa$ is the power set of $\kappa$ (the set of all subsets of $\kappa$). The value of $2^{\aleph_0}$ is not fixed by the usual axioms of set theory. That is, in standard set theory (Zermelo–Fraenkel with choice, ZFC for ...


0

Note that since $x\to\infty$, it is meaningless to think about the left and right limits because the neighborhood of $\infty$ cannot be approached. We can however use the definition of this limit to show that it is positive, unbounded and ultimately divergent. First note that $$ \lim\limits_{x\to\infty} f(x) =\infty $$ Is defined by the following statement ...


1

There's something called the "extended reals," which is $\mathbb R\cup\{-\infty,+\infty\}$ (that is, the real numbers with the addition of infinity and negative infinity). ($\infty-\infty$ is left undefined, as are others.) Limits are usually thought of, at least informally, as being in the extended real number system. In any case, to be formal, the ...


8

If you read the lines in your textbook clearly, then you will find there must be somewhere explaining the notation "$= \infty$", which merely stands for, say "gets arbitrarily large", and the symbol $\infty$ does not denote any number. Usually in calculus we work with the real numbers, and there is no real number called infinity. To avoid unnecessary ...


0

There are many excellent proofs given above. Another approach would be to note that the summation (LHS) is actually a binomial expansion of the result (RHS). (NB - This would only be applicable if you have to prove that the result is true, instead of having to find the result.) Hence $$\begin{align} \color{blue}{\frac 1{1-x}}&=(1-x)^{-1}\\ ...


0

Let $S=1+x+x^2+\dotsb$. Now, take a look at this: $\phantom xS=1+x+x^2+x^3+\dotsb$ $xS=\phantom{1+{}}x+x^2+x^3+\dotsb$ Thus, we have: \begin{align} S&=1+xS\\ S-Sx&=1\\ S(1-x)&=1\\ S&=\frac1{1-x} \end{align} Moving on to the second one: This one is a bit trickier, but we'll use the identity $(n+2)-2(n+1)+n=0$ in a clever way. Let ...


1

Others have answered on how to justify that $S=1+x+x^2+\cdots+x^n+\cdots$ equals $\frac{1}{1-x}$ for $|x|<1$. To get the second identity without differentiation, note that \begin{align*} 1+2x+3x^2+4x^3+\cdots&=1+(x+x)+(x^2+x^2+x^2)+(x^3+x^3+x^3+x^3)+\cdots\\ &=(1+x+x^2+x^3+\cdots)+(x+x^2+x^3+\cdots)+(x^2+x^3+\cdots)+\cdots\\ ...


0

we have $\sum_{i=0}^nx^i=\frac{x^{n+1}-1}{x-1}$ and the limit for $n$ tends to infinity exists if $|x|<1$


3

$\frac{1-x^{n+1}}{1-x}=1+x+x^2+\dots+x^n$, now if $n\to\infty$ and $|x|<1$ we get the first one. Differentiate the first to get the second.


0

Hints: $$(x^0+x^1+x^2+x^3+...x^n)(1-x)=x^0-x^1+x^1-x^2+x^2-x^3+x^3\cdots+x^{n}-x^{n+1}\\ =1-x^{n+1}$$ and $$(x^0+2x^1+3x^2+4x^3+...(n+1)x^n)(1-2x+x^2)\\ =x^0-2x^1+x^2+2x^1-4x^2+2x^3+3x^2-6x^3+3x^4\cdots\\+(n+1)x^n-2(n+1)x^{n+1}+(n+1)x^{n+2}\\ =1-(n+2)x^{n+1}+(n+1)x^{n+2}.$$


2

Here's a nifty way to do this problem. Recall that, by the infinite geometric series formula, $$f(x) = \frac{1}{1-x} = 1 + x + x^2 + x^3 + ...$$ From here, you can figure out all of your derivatives. $$f'(x) = 1 + 2x + 3x^2 + 4x^3 + ...$$ $$f''(x) = 2 + 6x + 12x^2 + ...$$ $$f'''(x) = 6 + 24x + ...$$ Now, just remember the formula for a Maclaurin ...


2

. The third derivative is $-2(1-x)^{-3}$ nd the 4th is hence $6(1-x)^{-4}$


1

If you have covered the Limit Comparison Test, you could use the facts that $\hspace{.3 in}\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^2}$ converges and $\hspace{.27 in}\displaystyle\lim_{n\to\infty}\frac{1-\cos\frac{1}{n}}{\frac{1}{n^2}}=\lim_{t\to 0^{+}}\frac{1-\cos t}{t^2}=\lim_{t\to 0^{+}}\frac{\sin t}{2t}=\frac{1}{2}\lim_{t\to 0^{+}}\frac{\sin ...



Top 50 recent answers are included