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سيمينار المسبر - أمس الثلاثاء 22 أبريل 2014 - جامعة القاهرة - الساعة 10 صباحاً = = = = = = https://www.facebook.com/photo.php?fbid=836472786367041&set=gm.708787645810694&type=1&theater = = = = = = ...

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There are also two books by the St.-Peterburg school which could be worth looking at: L.A. Takhtajan, Quantum Mechanics for Mathematicians and an older one L.D. Faddeev, O.A. Yakubovskii, Lectures on Quantum Mechanics for Mathematics Students Takhtajan's book is more advanced and modern: he covers inter alia supersymmetry and Feynman path integrals in ...

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1.Since $[,]$ is non-degenerate, there should be such an eigenvector $\eta$ corresponding to the eigenvalue $\lambda'$ for the eigenvector $\xi$ corresponding to the eigenvalue $\lambda$ that $[\xi, \eta] \neq 0$, which is possible only if $\lambda' = \bar{\lambda}$. Thus, the two-dimensional invariant plane $\pi_\lambda$ is nonnull. 2.Since $\xi$ is a ...

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Because cycles and oscillations and things with periodicity, are all intimately related to the circle. And $sin$ and $cos$ are defined based on the circle.

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Another reason is that we perceive time as always advancing, and we see many examples of rotation with consistent revs, so it is natural for us to express oscillations in terms of time to angle plus radius and from that x,y position. Sin and Cosine by definition give us the x,y coordinates given an angle and radius of 1, so it is convenient to use them.

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One of the big reasons not discussed above is Fourier theory -- any function $f(x)$ can be expressed in the form $f(x) = \int dk\, A(k)e^{ikx}$, which basically means that any function can be decomposed into an infinite sum of sines and cosines. Since this is the case and dealing with sine and cosine is mathematically simpler than the general case of ...

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Yes, there are alternatives. But a big part of the reliance on sines and cosines is historical. The analysis of oscillating mechanical systems naturally focused on sines, since that's the way things vibrate. With that framework in place, it turned out the electrical and magnetic systems also respond. Plus (or perhaps you might say, alternatively) circular ...

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Part of it is that since Newtonian mechanics is described in terms of calculus. When we consider vibrational motions, we're talking about some particle that tends to not be displaced from some equilibrium position. That is, the force on the particle, at displacement $x$, $F(x)$, is equal to some function of displacement $x$, $g(x)$. There are two ways ...

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A literal response to your title question would simply be "because in the physical world, oscillations behave in ways consistent with $\sin$ and $\cos$." Of course, one then wonders why these functions are so ubiquitous. Depending on your level of physics background, you may be familiar with the harmonic oscillator - that is, a system for which there exists ...

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For linear algebra regarding matrices (gaussian elimination, eigenvectors, laplace transforms etc.) try the MIT opencourseware playlist on youtube. http://www.youtube.com/playlist?list=PLE7DDD91010BC51F8 In case of link rot in the future : Course Title : MIT 18.06 Linear Algebra, Spring 2005 Instructor: Prof. Gilbert Strang Abstract : This is a basic ...

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Higher maths for beginners is ana amzing little book on all the subjects you mentioned, written by one of the fathers of Soviet nuclear bomb, and theoretical phsyicists. On math physics, the best introductory test is Elements of applied math physics, it has dufferential equations and complex analysis and other cool topics. Unfortunately, it may not have ...

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Tthis is again probably a duplicate. See here for a full list of links to book questions. By 'alrebra book (college level)' what do you mean exactly? Algebra in maths is a HUGE area. If you mean Linear Algebra, then maybe I.M. Gelfand is a great book, but probably not the best for a first read. Anton & Roerrs would be better for a first book, ...

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We can consider area as for a small surface and the direction of the area vector is in the direction normal of to that surface. But for a large surface area we cannot consider it as a vector e.g. Earth.

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I) First some terminology. Consider a symplectic manifold $(M;\omega)$. In a local chart $U\subseteq \mathbb{R}^{2n}$, the symplectic two-form reads $$\tag{1} \omega~=~\frac{1}{2} \omega_{ij}~\mathrm{d}x^i \wedge \mathrm{d}x^j ,$$ and the corresponding Poisson bi-vector $$\tag{2} \pi~=~\frac{1}{2} \pi^{ij}~\partial_i \wedge \partial_i,$$ where $$... 4 If you are interested in physical applications you could also include: Bratteli-Robinson: Operator algebras and quantum statistical mechanics It is a two-volume quite complete book, mathematically minded, discussing lots of applications of operator algebras theory to several physical systems, especially arising from statistical mechanics. Haag: Local ... 0 Hints: Prove that a one-parameter group (\Phi_t)_{t\in I} of diffeomorphisms \Phi_t: M \to M is generated by a vector field X\in\Gamma(M). Prove that if the one-parameter group (\Phi_t)_{t\in I} preserves the a form \omega then {\cal L}_{X}\omega =0. Prove that {\cal L}_{X}\omega =0 together with the fact that \omega is a symplectic ... 1 There are many physically fundamental properties that would not hold if the completeness requirement were dropped. First of all the spectral theorem for self-adjoint operators would not hold. So, an observable (let us assume to deal with observables with pure point spectrum) would not have a complete set of eigenstates. There is a fundamental idea in quantum ... 1 Zeta function regularization is used in other fields, and even in pure mathematics to obtain finite answers from otherwise divergent integrals. In bosonic string theory, the mass of states in lightcone gauge is,$$M^2 = \frac{4}{\alpha'} \left[ \sum_{n>0} \alpha^{i}_{-n}\alpha^{i}_n + \frac{D-2}{2}\left( \sum_{n>0} n\right) \right] where $\alpha'$ ...

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" i.e how does having applied a unit current and removed the same current between the points on which one wants to determine the effective resistance between, imply that the currents at the other nodes are zero." It doesn't. It implies that the net current at each of the other nodes is zero. In your example, node D, for instance, will have considerable ...

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