# Tag Info

## New answers tagged education

0

My opinion is that in the ultimate form of understanding, "logical reasoning by means of syntactical rules" and "contentual reasoning" are the same thing. In the beginning, there is a disconnect. You see syntax, but have not yet learned the point. You have ideas about how the subject works, but have not yet learned how to apply those ideas to answer ...

2

I’m an econ major, and I have been self-studying pure mathematics along the way for 3 years, from analysis and algebra years ago, to functional analysis, differential geometry, algebraic topology, and algebraic geometry now. I have never take any math courses beyond calculus, linear algebra and probability that are required for econ major. To assure you, ...

1

We use a brace to show cases, so you would say $$f(b)=\begin {cases} J14 & b \gt 0\\K14 & b \le 0 \end {cases}$$ where you substitute in the formulas that generate J14 and K14. Then if you have a bunch of points $x_i$ you could write $$sum=\sum_{i=1}^nf(x_i)\\avg=\frac 1n\sum_{i=1}^nf(x_i)$$

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My original answer is on Quora at https://www.quora.com/What-would-be-the-most-useful-base-for-number-system-instead-of-10/answer/Timothy-Bahry. Different bases are best for different purposes. I can think of advantages of many other bases. Base 2 is good for multiplying numbers of a given number of digits because 2 is so small. Base 6 is good because every ...

1

As a child, I learned logic with the book "The Game of Logic", by Lewis Carroll. This does not use the modern symbols, but I think that it is a great book to introduce the basic concepts.

1

Use vectors or complex numbers. In this case, they are essentially the same. If a point is at $(x, y)$, to rotate it about the origin by an angle $t$, do a multiplication (switching between points and complex numbers) $\begin{array}\\ (x+iy)e^{it} &= (x+iy)(\cos t+i \sin t)\\ &=(x, y)(\cos t, \sin t)\\ &=(x\cos t-y\sin t)+i(x\sin t+y \cos t)\\ &... 0 Coming back to this question a few years later, I've found an example of a game out there that does exactly what you intend your "wager system" to do, and it's even specifically designed for true-false questions. It's called "the credence calibration game", which is a game where you place confidences on how confident you are of a particular answer, from 50% ... 1 Claim: There is an algebraic procedure that terminates in finite steps to obtain the eigenvalues of a general$n\times n$matrix. This is true for$n \leq 4$but fails for$n \geq 5$, thanks to Abel-Ruffini. 1 Recently, I come up with the following conjecture, which is proved to be true for small matrices and false for large matrices. Conjecture. Let$E \in \mathbb{R}^{d\times d}$be the matrix of ones and$M \in \mathbb{R}^{d\times d}$be a bounded matrix representing a metric. That is,$M$satisfies the following conditions:$1 \geq M_{ij} \geq 0\ $... 6 My dear Dino, here is a story for you how arithmetic helps me. Me and my family go to the store often. Wife sees a purse. Usually, I think it is way too much money and she has enough purses anyway. Store always have discounts that go like "take an extra 25% off of the already reduced ticketed price" and then "take an extra 15% off if you pay with the store's ... 1 The best way to prove Euler's relation $$\exp(i\theta) = \cos \theta + i\sin \theta\tag{1}$$ is to use the following definition of$\exp(z)$: $$\exp(z) = \lim_{n \to \infty}\left(1 + \frac{z}{n}\right)^{n}\tag{2}$$ We will use the following simple lemma: Lemma: If$a_{n}$is a sequence of real or complex terms such that$n(a_{n} - 1) \to 0$as$n \to \...

2

You may be able to find material that "concentrate fully on geometry and trigonometry", but I don't think it's a good idea to limit your focus so absolutely. Everything in mathematics is connected ... okay, that is too strong, but many things in mathematics are connected, and if your goal is understanding, then you shouldn't spurn other areas that might give ...

2

There are many solutions. Let $\pi$ be any permutation of $\{1,\ldots,n\}$ and let $$k_i=10^{i-1}\cdot2^{\min(\pi(i),n-1)}.$$ The $k_i$ are distinct because they are divisible by distinct powers of $5$. We have $$\begin{eqnarray*} \sum_{i=1}^n\frac{10^{i-1}}{k_i} &=&\sum_{i=1}^n\frac1{2^{\min(\pi(i),n-1)}}\\ &=&\sum_{i=1}^n\... 0 Pi is the ratio between the circumference and the diameter. If you multiply a length (diameter) by pi you get the circle that has that length has a diameter. If you multiply by 3, you don't get the whole circumference, because you will be missing a piece. In order to get a circle again, you need to "curve" the space. Depending on whether you curve the space ... 0 Without loss of generality, we can set: A=(0,0), B=(1,0) and C=(a,b), with b>0. By the tangent bisection formula one gets then the slopes m and m' of the angle bisectors through A and B:$$ m=\sqrt{{1-a/\sqrt{a^2+b^2}\over1+a/\sqrt{a^2+b^2}}}, \quad m'=-\sqrt{{1-(1-a)/\sqrt{(1-a)^2+b^2}\over1+(1-a)/\sqrt{(1-a)^2+b^2}}}. $$You can now find ... 0 Let's tackle the 3D case. Let \theta be the angle formed by vector \vec s with the z-axis, and \phi be the angle formed by the projection of \vec s onto the xy-plane with the x-axis. Then geometry tells us that:$$ \vec s=(s\sin\theta\cos\phi, s\sin\theta\sin\phi, s\cos\theta), $$where, as usual, we set s=|\vec s|. Let's (nearly) follow ... 0 As already mentioned that's just shorthand for limits. Anyways suppose we have 1 pizza and we want to divide this pizza equally into n friends. Then each person gets a proportion: \frac{1}{n} of the pizza. Obviously the more friends you have the less each person is going to eat. And as your number of friends approaches \infty then you get closer ... 0 I don't think you need to wait for inspiration, start trying things out and see if you can approach the problem rationally. You can solve this but substituting each equation into another to collapse it into one variable. I often take this approach to reduce the dimensionality or non-linearity of a problem rather than waiting for inspiration. You can then ... 7 Well it's not the answer you'd want to hear for the middle of a competition, but my solution is usually to sleep on it. It's amazing the things your brain is capable of figuring out on its own (i.e. subconsciously). John Cleese has an excellent talk about how his subconscious helps him with his creativity. I specifically like the part where he talks about ... 17 The Aha moment:$$a^2+b^2+c^2+d^2+e^2+55=2a+4b+6c+8d+10e\implies(a-1)^2+(b-2)^2+(c-3)^2+(d-4)^2+(e-5)^2=0a=1;b=2;c=3;d=4;e=5$$7 Well that's one neat problem. Adding the 5 equations you get:$$a^2 -2a + b^2 -4b + c^2 -6c + d^2 - 8d + e^2 -10 e + 55 =0$$Now completing each square and observing that 1+4 +9 +16+ 25 =55 yields:$$(a-1)^2+ (b-2)^2+ (c-3)^2+ (d-4)^2 + (e-5)^2 = 0$$So each square must be zero. I think the way I found the solution is that "I believed there's a nice ... 1 Assume we can indeed find such a that makes f continuous, we will have$$ a = f(0,0) = \lim_{(x,y) \to (0,0)} f(x,y) = \lim_{x \to 0} f(x,0) = 0 $$so the only possible value for a is zero. However, this doesn't prove that if we set a = 0, we get a continuous function f (since we checked the existence of the limit and the continuity only along ... 2 My favorite method by far for proving e^{ix}=\cos x+i\sin x, is to use Taylor- MacLaurin series for the natural exponential, sine and cosine, expand both sides and prove equality.All these series are known to converge for all real x in basic calculus, which I hope wouldn't be too difficult for your students to understand. Judging from some of the other ... 0 Five x Ten to the power of Negative Two, + Four x Ten to the power of Negative Three ... is a lot easier if you know the notation:$$5 \times 10^{-2} + 4\times 10^{-3}$$[And you should realize that 10^{-3} is less than 10^{-2}. In fact 10^{-3} = 0.001 and 10^{-2} = 0.01.] Then you can adjust the power-of-ten multipliers to get them the same, ... 3 Let n be the number of elementary particles in the universe. Then$$ \begin{aligned} e^{i\theta}& = \left(1+\frac{i\theta}{n}\right)^n \\&= 1+ n \frac{i\theta}{n}+ {n \choose 2} \left(\frac{i\theta}{n}\right)^2 + {n \choose 3} \left(\frac{i\theta}{n}\right)^3 +{n \choose 4} \left(\frac{i\theta}{n}\right)^4 +\ldots \\&= \left(1 - {n \choose 2} \...

2

I think 'probability theory' is an extremely vague topic to list as your field of interest on a PhD application, but maybe it is just right in your case. It seems clear from what you say that you have not really settled on any specific part of probability as a possible area for your PhD thesis. So you may not be ready to be more specific, and may look ...

-1

For a range $(n, m)$ both inclusive,number of multiples of a number $x$ can be calculated as $x = (m-n+1)/(given \ number)$

3

All of these topics can be understood intuitively, they just don't tend to be taught intuitively. I remember when starting linear algebra, I already knew what a vector is so it made sense to me, but I pity someone who was trying to understand what a vector is from the axioms of a vector space, which were what was being written onto the blackboard in front of ...

12

My experience is that if you spend enough time both actively and passively (subconsciously) thinking about the material, these mathematical concepts will become as natural as breathing. It is a slow process at first (it took me until my junior year of undergrad to really start to understand what math was really about) but once you have a sufficient ...

5

I'm a high school student, so I have no idea what a Jacobian or a manifold is, but as someone who's self-studied linear algebra and abstract algebra, I think it's pretty complex and takes a rather smart/dedicated person to pass these classes, so you're definitely smart enough to understand these concepts. In my opinion, these fields are kind of intuitive in ...

4

The answer, as you might guess, is b). But still, work hard. It is worth it! I have only very basic intuition and only in the area I am beginning to specialize in, but every time I gain a little insight and really feel it, the fun and the rush are worth it! Do lots of problems. Ask each and every question you have, multiple times, to multiple people. Compare ...

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