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Notes to a video lecture on http://www.unizor.com Double Limits Sometimes we are interested in sequences that depend on two natural numbers like {am,n}. For example, am,n=arctan(m)+2−n. Now, if m→∞ and n→∞, how to calculate the limit of am,n? It might be limm→∞ [limn→∞ (am,n)] or limn→∞ [limm→∞ (am,n)] depending on which index, m or n, we will use to go to a limit first, and there is no guarantee that the answers will be the same. Let's check both methods. limn→∞ (am,n) = = limn→∞ (arctan(m) + 2−n) = = arctan(m) limm→∞ (arctan(m)) = π/2 If we change the order of limits, limm→∞ (am,n) = = limm→∞ (arctan(m) + 2−n) = = π/2 + 2−n limn→∞ (π/2 + 2−n) = π/2 As you see, the results are the same. In some way, the equality of these two limits might be considered similar to a standard accounting procedure of checking the calculations. Imagine an M⨯N matrix with numbers. If you summarize the numbers within each of M rows with row #i having a sum Ri and summarize all these row totals by i from 1 to M, you will get a total of all numbers in an original table. If you summarize the numbers within each of N columns with column #j having a sum Sj and summarize all these column totals by j from 1 to N, you should get exactly the same total of all numbers. If these two calculated totals are not equal, that is if Σi∈[1,M] Ri ≠ Σj∈[1,N] Sj then your calculations are wrong. The equality of two limits in the example above is not coincidental. We shall prove the following theorem. Theorem IF limn→∞ (am,n) = bm limm→∞ (bm) = B limm→∞ (am,n) = cn limn→∞ (cn) = C THEN B = C Proof Assume, B≠C and |B−C|=d. Since limm→∞(bm)=B, there exists some natural number M1 such that |B−bm| < d/4 for all m>M1 Since limn→∞(am,n)=bm, there exists some natural number N1 such that |am,n−bm| < d/4 for all n>N1 Then, for all pairs m,n such that m>M1 and n>N1 both following inequalities are true: |B − bm| < d/4 and |bm − am,n| < d/4 Therefore, |B−am,n| < d/2 That is, our double sequence am,n with both indices m and n sufficiently large (m>M1 and n>N1) is closer to number B than d/2. Absolutely similarly we can prove that this same sequence am,n with both indices sufficiently large (m>M2 and n>N2) is closer to number C than d/2. Here is a proof in mathematical symbolics. limn→∞ (cn) = C ∴ ∴ ∃M2: |C−cn| < d/4 ∀m>M2 limm→∞ (am,n) = cn ∴ ∴ ∃N2: |cn−am,n| < d/4 ∀n>N2 m>M2 ∧ n>N2 ∴ ∴ |C−cn| < d/4 ∧ |cn−am,n| < d/4 ∴ ∴ |C−am,n| < d/2 Choosing M=max(M1,M2) and N=max(N1,N2) we see that all am,n with m>M and n>N should be closer to B than d/2 and at the same time should be closer to C than d/2. With the distance between B and C equal to d it's impossible. We came to contradiction that signifies that our initial assumption that B≠C is wrong. Therefore, B=C. End of proof.

 We are going to prove that Lagrangian Mechanics of a closed (no external forces) mechanical system with its components acted among themselves with some conservative forces produces differential equations of motion that are equivalent to Newtonian equations, but easier to deal with.

However, using the Langrangian mechanics, this problem can be analyzed with much less efforts and the corresponding differential equation can be constructed relatively easy.

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The usefulness of these definitions is in our ability to differentiate by parameter t, assuming that derivative should be zero at points of local minimum or maximum.But that is a subject of the next lecture.

SQRT{N + N/[(N-1)(N+1)]}=N + SQRT{N/[(N-1)(N+1)]}

#VectorField #ScalarField #Gradient #Divergence #Curl #Circulation #Conservative

#Maxwell #magnetic #electric #nabla #Faraday

A new lecture on UNIZOR.COM - “Physics 4 Teens” - “Waves” - “Field Waves” - “E-M Equations 2″ has been released. It explains in a relatively simple format the Maxwell’s equation that describes the Gauss Law for magnetic fields, which in differential form states that there is no magnetic monopole. The site is totally free, no ads, no strings attached.

Every single wavelength of an oscillating rope carries a constant potential and equal to it kinetic energy. Watch a lecture on UNIZOR.COM - Physics 4 Teens - Waves - Light Energy - Rope Energy. Why? Because sin^2 + cos^2 = 1. #physics #waves #energy #oscillations #amplitude #frequency #angular  #elasticity  #kinetic

New lecture on UNIZOR.COM - "Physics 4 Teens" - "Waves" - "Light Energy" - "Rope and Springs" is about modeling transverse waves on a rope with longitudinal motion of infinite number of infinitesimal springs attached to unlinked tiny pieces of a rope.It allows simpler calculation of energy of transverse waves. #physics #waves #energy #oscillations #amplitude #frequency #angular  #elasticity  #kinetic

The oscillations of a rope is a simple example of transversal waves. In the lecture on UNIZOR.COM - "Physics 4 Teens" - "Waves" - "Light Energy" - "Rope Waves" we derive the wave equations for rope oscillations and the expression for speed of these waves. #physics #waves #energy #oscillations #amplitude #frequency #angular  #elasticity  #kinetic

Check your understanding of phenomena of light by taking exam on UNIZOR.COM - "Physics 4 Teens" - "Waves" - "Phenomena of Light" - "Exam". It's totally free, no ads, take it as many times as you want, but DO GET A PERFECT SCORE.

There are some substances, called polaroids, naturally occurring crystals or artificially created, that, when placed on the trajectory of the light ray, selectively allow to pass through only oscillations in one particular plane. This process of selecting only one particular plane of oscillations of electromagnetic waves to go through, while suppressing all others is polarization of light.