# Tag Info

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Isomorphisms define the mathematical concept of sameness. If I say G and H are isomorphic I mean they are the same up to renaming. In a sense, all groups that are isomorphic to G are really G in disguise. A homomorphism from G to H that is injective will define an isomorphism from G to a subgroup of H which means that H has a subgroup that, from a group ...

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The reason we care at all is that these structures we're talking about have an operation on them. An arbitrary function between two groups is rarely interesting. We want functions that have some relation to the operations. Homomorphisms are the most useful such functions because they literally preserve the operation. For another example an antihomomorphism ...

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Kaczor and Nowak Problems in Real Analysis series is pretty good. It lists all the problems and has the solutions in a separate area. Also, if you search around online you can find people who have written their own solutions to various texts. I've seen partial solution sets to Spivak and Rudin floating around online, for example. They can be hit or miss ...

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To be honest, I think it depends what you mean by Algebra. In my experiences with some of my classes, the professors stop caring about complicated arithmetic. If there comes some point of a proof where complicated arithmetic, or the solution of a complicated system of equations is needed, we are allowed/encouraged to use computer algebra software. In ...

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Another thought. If $0<a<b$, $n\ne -1$, then $$\int_a^b x^n \;dx = \frac{b^{n+1}-a^{n+1}}{n+1}$$ and the limit of that as $n \to -1$ is $\ln b - \ln a$. Not infinity. So, in a certain sense, your confusion comes from ignoring the constant of integration.

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My answer is a very "soft" answer, but it's a soft question, and the point is to explain to laymen, so I'll go ahead. As you mentioned, you could of course show them the proofs, which might convince them that it is true but wouldn't explain why. I would simply skip the calculus involved and point out that $e$ itself can be arrived at by the formula: ...

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Personally, I think back to times I previously thought this, and subsequently learnt the material anyway. It's especially noticeable in my current project of working through Awodey's Category Theory. Every so often I flick through the future chapters of the book and ask myself, "Can I understand the words on this page yet?" The answer is always no. Yet after ...

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I guess we all encounter some "heavy" material from time to time and feel pressure for not being able to understand everything perfectly. This is perfectly normal, or to rephrase it in the immortal words of von Neumann: "Young man, in mathematics you don't understand things. You just get used to them." Personally, as long as I can remember, I've ...

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I am experiencing precisely what you are describing - I am approaching a realm of mathematics new to me, in fact, I did not know it existed until 3 weeks ago. It will probably sound a bit cliched, but I have found some good effective strategies for developing the positive focused mindset that you describe involve: Starting from an application based ...

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If it's a book on mathematics, a person wrote it. If a person wrote it, another person (you) can understand it. At least, you can understand it at least as well as the person who wrote it. In some cases that might end up being "not very well" ;) , but that's when it's time to dive for a more original, more basic textbook...if possible, one from the ...

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As $∝$ looks like c and × together, A similar symbol could look like c and + together:$\Subset\hspace{-0.78em}\color{white}{\subset}\hspace{-0.4em}\large\shortmid$

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If $X$ is a connected metric space such that every real valued continuous function on $X$ is uniformly continuous, then $X$ is compact. In particular, it is bounded. Proof: If $X$ is not compact, there is a sequence $S = \{x_n\}_{n=1}^\infty$ with no convergent subsequence. Then for each $n$ there is $\epsilon_n > 0$ such that the ball ...

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Chrystal's Algebra Volume Two. See this note for a link to an online electronic copy: http://recursed.blogspot.com/2009/12/chrystals-algebra-is-available.html For something more recent, try Doug Hensley's Continued Fractions.

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The conjecture of Birch and Swinnerton-Dyer? Ok, this is very advanced, but it can (almost) be presented in a very elementary way, and being a millennium prize problem, it can be appealing for them. You can define elliptic curves in an intuitive way and/or giving their equations, both ways are simple (you are in characteristic 0). You can justify the ...

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One of logarithmic inequalities says (http://www.jstor.org/stable/3615890?seq=1#page_scan_tab_contents): $$\frac{x}{1+x} < \ln (1 + x) < x, \forall x > -1$$ Because $\lim_{x\rightarrow a} f(x)=0$, then $f(x)>-1$, for all $x$ in some vicinity of $a$. So $$\frac{f(x)}{1+f(x)} < \ln (1 + f(x)) < f(x)$$ from all $x$ in some vicinity of $a$. As ...

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I'm suprised that the following two haven't shown up: What is the smallest Riesel number? What is the smallest Sierpiński number? In both cases we know they exist because they are smaller than or equal to 509,203 and 78,557 respectively.

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You will likely get as many different opinions on books and US programs as you get Comments and Answers. And you will likely have no way to assess the experience or qualifications of those who give their opinions. So I will give you some information that you can verify for yourself and discuss possible strategies. First, you need to consider your grades in ...

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I like Lorenz Halbeisen's "Combinatorial Set Theory" book. It also gives some basic introduction to logic, and how it is used in set theory. The book itself is very thorough and the parts I have read were mostly well-written. Let me also add, that if you felt a bit shaky on the way logic was used in the set theory proofs, then perhaps it's best not to skip ...

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If I remember correctly; this result is a corollary to result with regards to perdendicularity. Insofar as "The perpendicularity relation on a Bilinear space $(V, B)$ is symmetric if and only if $B$ is either symmetric or alternating".

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This is used to prove Prop. 2.7 A bilinear form $B$ on $V$ is reflexive if and only if it is either symmetric or alternate. Basically a bilinear form is reflexive if it satisfies: $B(u,v) = 0$ if and only if $B(v,u) = 0$. Requiring $B(u,v) = B(v,u)$ for all $u$, $v$ is too strong, but the condition given in Prop. 2.6 turns out to be necessary ...

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Edit. When the characteristic of the field is not $2$, the intersection cannot be a singleton. Since $A$ is not a scalar matrix, its size is at least $2\times2$. Now suppose the field's characteristic is not $2$. There are two possibilities: If $A=aI+bE$, where $a,b$ are scalars, $E$ is the all-one matrix and $b\ne0$ (by assumption, $A$ is not a scalar ...

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You could have a go with H-D Ebbinghaus, J Flum: Finite Model Theory and T Jech: Set Theory.

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Logic textbook "less verbose" and quite general : Dirk van Dalen, Logic and Structure (5th ed - 2013). As an alternative (with less topics covered) : Derek Goldrei, Propositional and Predicate Calculus : A Model of Argument (2005) Set theory : Derek Goldrei, Classic set theory (1996).

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I think the most direct way to answer your question is provided by infinitary logic. Given cardinals $\kappa, \lambda$ - with $\kappa\ge\lambda$ (you'll see why) - we let $\mathcal{L}_{\kappa\lambda}$ be the logic which you get by taking first-order logic and closing under $(<\kappa)$-ary conjunctions, and allowing quantification over ...

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In general, new structures are studied as the result of 1) realizing that multiple interesting objects possesses this structure, and 2) showing that this structure admits some nice results, and ideally helps tackle existing questions. So if a certain structure isn't studied, either no one has thought to study it, or it doesn't have a nice set of examples or ...

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The constant in the Berry-Esseen theorem: If we have a bunch of i.i.d. random variables $(X_j)_{j\geq 1}$ with a finite third moment, that is $E[|X_j|^3]<\infty$ (and thus they also have some mean $\mu$ and variance $\sigma^2$), then we can prove without too much trouble that their scaled average, $A_n := \frac{(\sum_{j=1}^n X_j)-n\mu}{\sigma \sqrt{n}}$ ...

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For your given $A$ we have $x=\sqrt{1-y^2}$ and $y$ ranges from $[-1,1]$. So $\phi (A) =\{\phi (\sqrt{1-y^2}, y) | y \in [-1,1]\}$ = $\{(1-y^2,0)|y \in [-1,1]\}$ However, in general it is not something that one can simply answer other than saying $\phi (A) = \{\phi (a) | a \in A\}$

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Your question is very very very very very broad. In general, given a set $A\subseteq X$ and a set $B\subseteq X$, as long as $A$ has at least as many elements as $B$, you can find some function $\phi: X\to X$ such that $\phi(A)=B$. The function may not be pretty (for example, it may not be continuous), but it does exist.

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Trying to develop an intuitive idea using images and, doing so, cutting a lot of corners. Suppose Bill Gates gets a raise of 1 dollar an hour. And suppose he decides to work forever. He will make 1$more every hour. So an infinite amount of money more than without the raise. But do you think this raise is significant? No. The amount of money he makes with ... 1 Yes. In finite dimensions , if$||z_n||=||y_n||=1$and$(z_n,y_n)\ge 1-2^{-n}$for$ n \in N$then for some strictly increasing$f:N\to N$, the sequences$(z)_{f(n)}$and$(y)_{f(n)}$will converge in norm to the same unit vector. This is because a finite-dimensional Banach space is locally compact.But in infinite dimensions, the sequences$(x_n)$and ... 3$\lim_{x\to\infty} \frac{x^{2}}{x^{2}+x} = 1$As stated, this is because$x$grows much more slowly than$x^2$. So the ratio goes to 1.$1^2\over 1^2+1$$={1\over 1+1}$$={1\over 2}$$=0.5 2^2\over 2^2+2$$={2\over 4+2}$$={4\over 6}$$=0.663^2\over 3^2+3$$={9\over 9+3}$$={9\over 12}$$=0.75 10^2\over 10^2+10$$={100\over 100+10}$$={100\over 110}$$=0.91$... 2 How would a basis for$\mathbb R$as a field over$\mathbb Q$look like? By the axiom of choice we know one must exist, but I have no idea how it would look. It would be quite odd, as it must be large enough so that every real number can be expressed as a finite sum of its elements, but it must be small enough to be linearly independent. For a simpler ... 1 $$\Gamma (1+ x) \, \left( \gamma + \psi (1+x) \right) = \gamma \, \Gamma (1+ x) + \Gamma' (1+ x)$$ https://en.wikipedia.org/wiki/Digamma_function#Relation_to_harmonic_numbers https://en.wikipedia.org/wiki/Gamma_function 3 There are many examples of objects whose existence can be proven using Probability Theory. One example is the existence of matrices with the restricted isometry property: https://en.wikipedia.org/wiki/Restricted_isometry_property. If one wants matrices of certain dimensions, the only known way to construct them is to create matrices at random using a ... 3 The change in a quantity is proportional to the amount you have. This is how interest in a bank works. The amount given to you as interest is proportional to how much money you have already$\frac{{\rm d}y}{{\rm d}t} = \alpha y$. The change in money every years equals $$\Delta y =\left( \frac{{\rm d}y}{{\rm d}t}\right) \Delta t = (\alpha y) \Delta t$$ We ... 0 Following a logarithmic curve is the same as saying$y=\frac{dy}{dt}$, since that case implies$y=e^t$, where$e$is the natural logarithm base. 4 For the limit, $$\lim_{x \to \infty} \frac{x^2}{x^2+x}$$ The$x$term becomes insignificant and$x^2$is the dominant term as$x$gets bigger. 100 Vadim's answer handles the math (and I've upvoted it), so I will try to provide intuition. The idea is the word "closer" is relative. That is, in some sense,$100{,}000$is closer to$100{,}010$than$1$is to$0$. Of course, in an absolute sense,$1$is$1$away from$0$while$100{,}000$is$10$away from$100{,}010$. But, say you had$\$1$ and I took ...

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The reason is that $$\frac{x^2}{x^2+x}=\frac{x^2}{x^2+x}\frac{1/x^2}{1/x^2}=\frac{1}{1+1/x}$$ The fraction is always less than $1$ (for positive $x$), as the denominator is bigger than $1$. However, as $x\to\infty$, we have $\frac{1}{x}\to 0$, so the fraction gets closer and closer to $1$.

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As lhf explained, every rational function grows like some integer power of $x$ as $x\to\infty$. Since $H_n\sim\log n$ as $n\to\infty$, there can be no rational function $f$ with $f(n)=H_n$. This can be generalized to algebraic functions, which are functions satisfying an equation $$\tag{\star} \sum_{i=0}^n a_i(x) f(x)^i =0$$ for some polynomials ...

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I have been quite surprising when I discovered that the elliptic curves of the shape $X^3+Y^3=AZ^3$ , A cube-free integer, does not have nonzero rational point if and only if all its quadratic points (there are always an infinity) are equivalent to (I called it) a conjugate point which have the form $(X,Y,Z)= (a+b\sqrt m, a-b\sqrt m, c)$ where $a,b,c,m$ ...

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It can be proved, for instance,that $x^2+x+1$ is irreducible in $\mathbb F_p[x]$ for $p=29$ which gives two "irrational" elements (the two roots of the polynomial) in a quadratic extension of $\mathbb F_{29}$. What kind of object is each of these two roots? Absolutely non idea. And for all finite field there are in general infinitely many of these ...

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If we raise $\sqrt{2}$ to the $\sqrt{2}$ power, and raise the result to the $\sqrt{2}$ power, then we have raised an irrational number to an irrational power and gotten a rational, but we don't know in which step we did it...

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This is a partial answer. Since $H_n \sim \log n$, there is no formula for $H_n$ using a rational function of $n$ because $p(x)/q(x) \sim x^k$, for some integer $k$, and $\log x$ is never asymptotic to $x^k$. Here, $f(x) \sim g(x)$ when $\displaystyle \lim_{x\to\infty} \dfrac{f(x)}{g(x)}=1$.

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$$H_n=\frac1{n!}\left(\sum_{k=1}^n\prod_{1\le j\le n, j\neq k}j\right)$$ This is a closed form: there are no trascendental operations involved, and it can be computed in finte time. Ugly, but closed.

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Let $p=2n+1$ be an odd prime, and consider the $2^n$ expressions you get by all possible choices of signs in $$\pm1\pm2\pm3\pm\cdots\pm n$$ They can't be perfectly uniformly distributed among the $p$ residue classes modulo $p$, since $p$ doesn't divide $2^n$, but the are distributed as uniformly as they can be, in that each nonzero residue class comes up the ...

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I was about 10 when I called my dad to the blackboard to see my theorem that if the square root of an integer was not also an integer then the decimal expansion did not terminate. It was a proof by cases based on the least significant digit of the square root, so didn't prove the square root was irrational.

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I haven't looked at all the other answers to be sure I'm not duplicating their suggestions, but I'd definitely try to get into software development. There are some awesome applications for mathematics within the computer science domain. I personally working on one of the hottest applications (to my knowledge): mapping knowledge into a $n$-dimensional ...

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This is really obvious but I realized something I had not been tought, from this : Let $P(x)$ be a polynomial of degree $n$, than: $P(x+1)-P(x)$ is degree $n-1$. This was not surprising, in fact this just came from the binomial theorem. What was surprising to me is the fact that we can model quantitative data that is taken in constant intervals just ...

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I was teaching myself about tensor algebra and playing around with Einstein summation notation, and I surprised myself by deriving a closed formula for the principal invariants of a second-order tensor, using the generalized Kroenecker deltas. \det (A-\lambda ...

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