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0

I see Ramanujans formula here, but the one by Chudnovsky Brother's is still missing. Lets change that: $${\pi =1/12\, \left( \sum _{k=0}^{\infty }{\frac { \left( -1 \right) ^{k} \left( 6\,k \right) !\, \left( 545140134\,k+13591409 \right) \\ \mbox{}}{ \left( 3\,k \right) !\, \left( k! \right) ^{3}{640320}^{3\,k+3/2}}} \right) ^{-1}}$$ If you do it for the ...


0

The fastest known formula for calculating the digits of pi is Chudnovsky formula: $$\frac{1}{\pi}=12 \sum_{k=0}^\infty \frac{(-1)^k (6k)! (163 \cdot 3344418k + 13591409}{(3k)! (k!)^3 640320^{3k+1.5}}$$ This formula is used to create world record for the most digits of pi. This formula rapidly converges and it needs 3-4 terms to yield good approximation of pi ...


0

Here i get another example to 100: (1x2)+(3x4)+5+6+(7x8)+9+10


1

For the $100$ part, $$ 1-2+(3\times 4\times5)-6+(7\times 8)-9=100. $$


1

Here's an informal definition: Suppose $X$ is a metric space and $A$ is a subspace of $X$. Let $A_t$ be the $t$-neighborhood of $A$. Then, $$\partial A = \lim_{\hspace{.6em}t\rightarrow 0^+}\frac{1}{t}(A_t\setminus A).$$ For full disclosure, I thought I had first seen this on Stack Exchange, but a quick search didn't turn up anything. Regardless, this ...


2

$$x^2\gt y^2\implies 1\gt\left({y\over x}\right)^2\implies1\gt{y\over x}$$ with no condition on $x$. But if $x\gt0$, then $$1\gt{y\over x}\implies x\gt y$$


2

$x^2>y^2\Rightarrow x^2-y^2>0\Rightarrow (x+y)(x-y)>0$, now if $y\geq x$ both factors has to be negative and therefor $x<0$. Hence $x>0\Rightarrow y<x$.


5

Noting that $|x|=x$ for $x\gt 0$, we have$$x^2\gt y^2\Rightarrow |x|\gt|y|\Rightarrow x\gt |y|\ge y\Rightarrow x\gt y.$$


1

Well, if $y \ge 0$, then since $x \mapsto x^2$ is an increasing function on the nonnegative real numbers, $x^2 > y^2 \implies x > y$. On the other hand, if $y < 0$, then information about $x^2$ and $y^2$ is unneded, since we have immediately that $x > 0 > y$.


2

Due to my nearly unreadable handwriting, I prefer Im and Re instead of the calligraphic symbols, as noone would be able to read them. I've never seen anyone use the calligraphic symbols in real life either.


2

I personally write Re and Im on my preprints, too. It seems to me that Re $z$ is much more elegant than $\Re z$.


1

Here, $$f(t), \quad t \in [0, 1]$$ is a parameterization of a curve, that is, it's an instruction for how to draw a curve on the space $X$: At each time $t$, the pencil is at the point $f(t)$ in $X$. Continuous means that you don't have to lift your pencil up when following the instructions, that is, that the pencil doesn't suddenly "jump" from one point to ...


1

The important word was continuous. Roughly speaking, a function $f$ from $[0,1]$ to $\mathbb{R}$ is continuous if and only if you can draw its values with pencil, without lifting it from $X = 0$ to $X = 1$. When you do this, the figure you create looks like a path. Here you can see an example of a continuous function (left), and a discontinuous function ...


1

You can always think of a path as a journey. It is common in texts to define a path as you give, which you can think of as of "time" $1$, but some texts, for example this one, define a path as a map $f:[0,r] \to X$ which can be thought of as a journey of time $r \geqslant 0$ from $f(0)$ to $f(r)$. If also $g:[0,s] \to X$ is a path and $f(r)=g(0)$ then you ...


1

Imagine Bob the bug walking from $0$ to $1$ on the real number line. When Bob is at $x\in [0,1]$, Bob's sister bug, Brenda is at $f(x)$ in $X$. Because $f$ is continuous, Brenda traces out a path in $X$ from $f(0)$ to $f(1)$ while Bob goes from $0$ to $1$ in $[0,1]$. So $f(0)$ and $f(1)$ are connected by a path.


-1

I've found this to be rather surprising: $\displaystyle\sum\limits_{n=1}^{\infty}\frac{1}{2^n}=1$ $\displaystyle\sum\limits_{n=1}^{\infty}\frac{1}{2^n\ln(2^n)}=1$ As it essentially yields the identity: $$\sum\limits_{n=1}^{\infty}\frac{1}{2^n}=\sum\limits_{n=1}^{\infty}\frac{1}{2^n\ln(2^n)}$$ It is surprising because obviously: ...


0

Very basic method based on A = πr2, A = π when r = 1. Get some really big piece of graph paper with a grid of fine lines. Draw either a circle centered on the intersection of two lines, or at least a quadrant of a circle. Then count squares at least 50% inside the circle. Normalize to the number of grid squares in a quadrant square. The finer the grid with a ...


0

You need a method to calculate exact $\pi$ by hand by an iteration? Or just some good numerical approximations, which easy to evaluate? In the first case you can have this. $$ \pi = \left( \frac{2\sqrt{2}}{9801} \sum^\infty_{k=0} \frac{(4k)!(1103+26390k)}{(k!)^4 396^{4k}} \right)^{-1}. $$ This is a formula by Ramanujan and this gives 8 correct decimal ...


2

Jean-Claude Arbaut has reminded us of the identity $$ \frac\pi4=4\arctan\frac15-\arctan\frac1{239}. $$ Let us examine that. You learned in high school that $\tan\dfrac\pi4=1$, and that \begin{align} \tan(\alpha+\beta) & = \dfrac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta} \tag 1 \\[10pt] & =\frac{c+d}{1-cd} \end{align} Thus $$ \arctan c+\arctan ...


0

One easy-to-understand improvement to your method, which I I don't see used much, is: $$\pi/6 = \arctan \left ( \frac{\sqrt{3}}{3} \right ) \\ = \int_0^{\frac{\sqrt{3}}{3}} \frac{1}{1+x^2} dx \\ = \sum_{n=0}^\infty \frac{(-1)^{n} \left ( \frac{\sqrt{3}}{3} \right )^{2n+1}}{2n+1} \\ = 3^{-1/2} \sum_{n=0}^\infty \frac{(-1)^{n} 3^{-n}}{2n+1}.$$ Consequently ...


6

By hand, it's relatively easy to use the development of the arctangent, and a Machin-like formula: Machin: $$\frac\pi4=4\arctan\frac15-\arctan\frac1{239}$$ Gauss: $$\frac\pi4=12\arctan\frac1{18}+8\arctan\frac1{57}-5\arctan\frac1{239}$$ I have done it once with Machin's formula and 24 decimals, in a few hours. It's recommended to do it by two methods, to ...


0

For some fast-converging ideas, I recommend looking at this part of the wikipedia page. For example, the first option they present is $$ \frac{\pi}{2} = \sum_{k=0}^\infty\frac{k!}{(2k+1)!!}=\\ \frac{1}{1} + \frac{1}{3 \cdot 1} + \frac{2\cdot1}{5\cdot3\cdot1} + \frac{3 \cdot2 \cdot 1}{7\cdot 5 \cdot 3 \cdot 1} + \cdots $$ Taking this out to the eighth step ...


1

Lang's Algebra is very nice, and van der Waerden (vol1, vol2) is the best I know of, together with Mac Lane & Birkhoff. I find the latter significantly harder to follow. There is also Dummit & Foot, at a level equivalent to Lang. I don't own Artin's Algebra, but I'd say it's also at the level of Lang, and it has a very good reputation. Edit: the ...


0

there is only one way to study maths...LEARN THEM! this is the bit that is always missing in maths. having acquired a certain amount of knowledge and mental dexterity is is necessary to 'learn' how to do a new problem. this ,learning is done by get an example go through it and through it ( with pencil in hand) until you understand it and you have ...


0

I agree with RR and would just add that students in Europe who have been identified early-on for a math/science education might also take 2 or even 3 math subjects per year in high school and subsequently would not be bothered with requirements covering much history, social studies, health, etc. So they may learn more math in one year than the average ...


0

I'm in a same situation as you, in the sense that I'm teaching myself math. When I was in school I never paid much attention to math and I simply lost track, didn't build my foundation that subsequent math study depended on and hence really struggled. I made it through but just barely. Stuff such as calculus, integral and differential equations didn't mean ...


0

In abstract algebra, we don't refer to it as Polish notation, but "prefix notation" and "postfix notation" do seem to be apt names for talking about these two things. The two notations $(x)f$ and $f(x)$ have both been used historically; however, I think the latter is far more prevalent today. There may be some concrete anthropological reason we prefer the ...


0

Consider it (and call it, if you need a name) the composition of functions (after all, permutations are functions). Remember: $$(\phi\circ \psi)(x) = \phi(\psi(x))$$ We work from inside out to determine $(\phi\circ \psi)(x) = \phi(\psi(x))$: First determine $\psi(x)$, then you can determine $\phi(\psi(x))$. This is no different than how we approach the ...


0

I'm not sure I understand the question "does the word 'ten' have a base?" but what your friend says is correct. You can think of 'ten' mapping to $||||||||||$, and then that maps to various representations like $10$, $1010$, and $\text{A}$ depending on the base. And if the base is known the mapping can be reversed from the representation to the word. If it ...


2

Your friend is right. "Ten" is indeed a baseless number, which exists whether you write it down in digits or not. It's worth mentioning, however, that the names we give to numbers are based on the significance of this specific one. I guess the reason for that is the number of fingers each one of us has on both their hands. Having said that, still, your ...


3

I would recommend Charles W. Trigg, Mathematical Quickies: $270$ Stimulating Problems with Solutions.


0

A lot of us share the same feeling towards math. Maybe our high school teachers were terrible. Or maybe we also wasn't that interested in math at that point in time. Time passed by and now you found yourself in a position that you really need to master the fundamentals of math in order to advance in your academic/professional life. I'll share a secret ...


2

I'm no topologist, I just like to sketch things. We know that the $ca$ loop can be pinched apart like that because the endpoints of the other $a$ and $c$ segments are identified already. Now it just remains to attach two handles to connect $a$ to $a$ and $c$ to $c$.


2

Generally, people tend to think of distances between normed spaces in multiplicative terms, because it fits the way how composition of operators works. That is, the smallest value of the distance is $1$ and the triangle inequality has multiplication instead of addition. If you don't like this, take the logarithm. The Banach-Mazur distance does not directly ...


6

I had the same question as you when I was studying topological spaces: in particular, it annoyed me that the definition didn't look like "preserves some structure" in the sense that I'd become familiar with in abstract algebra, e.g. preserving a group operation in the case of morphisms of groups. Here were my thoughts at the time. Here are two proposals I ...


0

Here is one way to answer the question. The data of a continuous map between sober topological spaces is equivalent to the data of a geometric morphism of the associated toposes of sheaves. Though the restriction to sober spaces may seem unsatisfying, keep in mind that the full category of all topological spaces is full of pathologies.


1

We usually define a circle of radius $r$ about a point $(x_0,y_0)$ in an $x,y$ coordinate plane by using the Euclidean metric $\|x,y\|_2 = \sqrt{x^2 + y^2}:$ we set $\|x-x_0, y-y_0\|_2 = r.$ If instead we use the taxicab metric, $\|x,y\|_1 = |x| + |y|,$ and define a circle of radius $r$ about $(x_0,y_0)$ as the set of points $(x,y)$ satisfying $\|x-x_0, ...


1

First, in the theorem, you have to assume that manifolds are connected. Next, it is a theorem (Rado and Caratheodory, I think), that every topological surface has a smooth structure and that two surfaces are homeomorphic if and only if they are diffeomorphic. See here for more references than you probably wanted. As for Riemann surfaces: if you want to ...


1

Although this is hardly the reason that cause confusion because people have common sense, "..." is not well defined if you want to be picky. For example $ \{1,2,3,4,...,100\} $ could mean set of natural number less or equal to 100 or the image of that set under function $f(n)=(n-1)(n-2)(n-3)(n-4)(n-100)+n$ Define things using recursive remove the ...


1

Prestige matters very little. Going to a smaller university where you can get more attention from your professors can be especially beneficial if they get to know you very well and write you good recommendation letters. Sometimes in better schools the professors are too busy doing their own research, and although the department is quite renowned in its ...


3

Yes. On the ordinary sphere of radius $1,$ there are countably many pairs of "squares" and circles with equal areas, where both the (geodesic) radius of the circle and the length of the four sides of the square can be constructed. The limit of this is when both are a hemisphere, regarded as a square with four angles all equal to $\pi.$ Similar in the ...


3

If $a,b,c,d$ are in arithmetical progression, then $$\frac{d^2-a^2}{c^2-b^2}=3.$$


0

There is probably more than one way to answer this question, so I make no claim to mine being the definite one. A morphism of topological spaces should be one that is somehow compatible with the topological structure, and indeed as you say there are two natural notions of "compatibility" to consider. To decide which one to use, you should ask yourself what ...


1

Of course heart curves are really nice, or roses or cycloids. But if you are looking for some really cool stuff, then what about Albert Einstein curve? This parametric equation really gives 2Pac. Gauss is also an interesting one. WolframAlpha can plot other person curves. My favorite one is Nicolas Cage.


0

I had run into the exact same situation when I took a course in Abstract Algebra 1: called Group Theory. I also had a "Real" math course, it was a very rigorous proof oriented linear algebra class. But I ended up dropping the group theory course. The problem was the prof taught graduate level courses for 10 years, then after 10 years was asked to teach an ...


4

Because only multiplication of two numbers is defined at that point. Also, if it's not yet defined to be associative, that expression may be ambiguous.


6

$n^k = \underbrace{n\cdot n \cdots n}_{k\ \textrm{times}}$ only works when $k$ is a positive integer. In order to define exponentiation more generally, we must refine the definition of exponentiation several times: First, we must address a suitable definition for $k = 0$ and $k \in \{-1, -2, \ldots \}$, including operations of the form $n^{k_1}n^{k_2}$ ...


2

Why do we write $ab$ instead of $$\underbrace{b+b+\cdots+b}_{a \textrm{ times}}~~~?$$


2

The AMS, american mathematical society has publications that are useful, eg "Notices of the American Mathematical Society"; there are others, link here, there are a slew of journals at AMS, and could help. Also there is another publication "Girls' Angle", link here, this is a girls math education magazine and could serve as a good knowledge base for ...


2

I'm a fan of Math Horizons, although it might be a bit breezy if you're looking for more "serious" math that's still accessible to undergraduates.



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