Hrodelbert
• Member for 7 years, 10 months
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Although I agree with the answers already provided that in this situation (and indeed in most other ones in mathematics) there is no difference between the two notations, I would like to add the ...

Although this might be better asked at the physics SE, the simple answer is as follows: $$W=\int \vec{F}\cdot \vec{ds} = q\int \left(\vec{B}\times \vec{v} \right)\cdot \vec{ds} = q\int \left(\vec{B}\... View answer Accepted answer 3 votes Here is an idea of a proof: For any curve we can choose a reparametrisation such that its speed is constant, such that we actually have L(\gamma)^2 = 2 E(\gamma) (this follows from going through ... View answer 3 votes Yes, matrix multiplication: given two square matrices A,B with non-zero determinant (so elements of GL(n,\mathbb{R})) of any size, we know that AB does not have to be equal to BA, but A(BC) = ... View answer 3 votes Yes this is possible. Simply let a,b,c,d be your unit vectors and define the inner product of a,b,c,d to be 0 ( so (a,b) = 0 etc., but of course we still need (a,a) = 1). The inner product ... View answer 2 votes The simplest counter example for the case of positive integers is probably$$ A=\{1,3\} \qquad B=\{1,2,3\}, $$both with arithmetic mean 2. The geometric mean of A is \sqrt{3}\approx 1.73, while ... View answer 2 votes Let r be your real number and I the set of integers, i.e.$$ I = \{ n\in \mathbb{Z} | n\geq r\}. $$It is clear that I is nonempty. Suppose that I does not have a minimum: this means that ... View answer 2 votes For small a, a good approximation of the term e^{-ay} is given by$$e^{-ay} = 1 - ay + \frac{(-ay)^2}{2} + \mathcal{O}(a^3), $$which, if you plug it in to the right-hand side of your equation, ... View answer Accepted answer 2 votes Only if b is an orthonormal basis. In that case we have$$ ||a|| = ||\sum_{i=1}^d c_i b_i || = \sum_{i=1}^d |c_i| ||b_i|| = \sum_{i=1}^d |c_i | = ||c||. $$If b is not orthonormal, either ||b_i||... View answer 2 votes Since X must be invertible, you can solve the set of equations given by$$XA = BX.$$This gives 4 linear equations for the 4 unknown entries of X in terms of the coefficients of A and B. ... View answer 2 votes As you have stated yourself, the integers we are looking for must be a multiple of 30. Now it is simply a matter of counting all the multiples of 30 below 1000, which is found by dividing 1000 ... View answer Accepted answer 2 votes Concerning your second question: if A is 3x3 and satisfies A^T A = D for some diagonal matrix D, it satisfies 6 additional constraints (namely the equations from the off-diagonal terms in A^T ... View answer 1 votes Hint: consider the ratio f/(A \exp(Bz)) with A and B chosen such that A \exp(Bz) has the same quasiperiods as f, i.e. the ratio is elliptic. Remember Liouville's theorem View answer 1 votes The simplest approach in my opinion is to consider the function$$ w(d) =\frac{1}{d}, $$where d is the number of days difference. It is obvious that w(1) = 1 and that 0<w(d)<1 for all d\... View answer Accepted answer 1 votes Let me first treat part one, since this is the easiest part Part 1: the number of subsets of a given set X is 2^{|X|}-1, where |X| is the number of elements in X. I had to substract the 1 ... View answer 1 votes If a matrix is invertible, its columns seen as vectors form an independent set. Since the matrix is d\times d, where d is the dimension of the space, this means that the columns of your matrix ... View answer Accepted answer 1 votes How about$$ A(n) = n2^{n-1} $$and$$ B(n) = 2^n. $$These reproduce your sequences. To prove that they work for general n though, you would have to use mathematical induction. View answer 1 votes This sum is certainly not larger than$$ (d-1)K^{d-1}, $$since that is the value of the summation$$ (d-1) \sum_{i_1,i_2,\cdots,i_{d-1}=1}^K 1, $$which is certainly greater then sum you proposed. ... View answer Accepted answer 1 votes There is no univeral way to find conserved quantities of a given physical system. The most common way they are found is by looking at symmetries of the system. In this case, noticing that the ... View answer Accepted answer 1 votes After some extra reading, I think I can answer this now, although not from the perspective I originally intended. We use the Plemelj-Sokhotksi formula, in the following way: in the limit \epsilon \... View answer 1 votes Inspired by the discussion with Kelenner, I believe I can answer my own question: We have a function F, which is given by a convergent series for |Im(p)| <2\kappa (call this open ... View answer 1 votes I'll post this just as an answer I suppose: \frac{d}{dt}(1+y) = \frac{d}{dt} 1 + \frac{dy}{dt} =0 + (1+y)=1+y by the differential equation. Replacing x(t)=1+y(t) yields a differential equation ... View answer Accepted answer 1 votes From the two vectors, define the plane spanned by these two vectors. Choose the orientation to be such that the normal vector to this plane has positive z-coordinate. Then set polar coordinates on ... View answer Accepted answer 0 votes The easy part was already known to the OP, but since I wrote it let me keep it here: if u-v = \omega_i with \omega_i one of the periods, then it follows immediately that$$ \wp(u) = \wp(\omega_i + ...

Yes, your setup is correct. Note that the base of your fence is a circle, but the height of the fence is not periodic: at $0$, the height of the fence has a jump from 0 to $(2\pi)^2$, which does not ...

Heuristically, if the volume $V_b$ has spherical symmetry, we can proceed as follows: we are free to orient our original cartesian coordinates in any way we want, since the integral does not depend on ...

No you cannot do this. The approximation you mention is only true for very small $x$ (compared to $1$), so $x <<1$. If we take $x$ to be close to one, it is easy to see that $(1-x)^{20}$ is ...
The equation you are interested in is the following: $$\int_{-\infty}^M N e^{-(x-\mu)^2/(2\sigma^2)} = 0.85,$$ where $N$ is a normalization such that the result of the integral is $1$ when $M\... View answer 0 votes I think your own explanation is already a mathematical proof that both probabilities are indeed equal. Your products of probabilities are correct and hence the result follows. Intuitively, this is due ... View answer Accepted answer 0 votes After thinking some more, I think I have found an answer to my own question: Let us assume for simplicity that$f$has exactly one pole at the origin. The case$F_{\mathbb{R}}\$: it is clear that ...