tag:blogger.com,1999:blog-8231784566931768362.post669308164725591843..comments2023-09-09T08:21:55.454-04:00Comments on MathNotations: The Matrix Reloaded: Deriving sin(A+B), cos(A+B) by Rotation MatricesDave Marainhttp://www.blogger.com/profile/13321770881353644307noreply@blogger.comBlogger8125tag:blogger.com,1999:blog-8231784566931768362.post-25104645342104399982010-02-02T08:22:05.426-05:002010-02-02T08:22:05.426-05:00Genial brief and this mail helped me alot in my co...Genial brief and this mail helped me alot in my college assignement. Thank you for your information.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-8231784566931768362.post-5681408997786703422007-11-13T20:45:00.000-05:002007-11-13T20:45:00.000-05:00My condolences, Eric.My condolences, Eric.Dave Marainhttps://www.blogger.com/profile/13321770881353644307noreply@blogger.comtag:blogger.com,1999:blog-8231784566931768362.post-3700130924690183662007-11-13T08:37:00.000-05:002007-11-13T08:37:00.000-05:00There had been a death in my family last month, Da...There had been a death in my family last month, Dave.<BR/><BR/>TC, I see I had exaggerated the case. Now, should I bring up the old Polish joke about stability theory?Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-8231784566931768362.post-38811024593950192552007-11-13T08:22:00.000-05:002007-11-13T08:22:00.000-05:00"In fact, positive feedback is always disastrous...."In fact, positive feedback is always disastrous."<BR/><BR/>Not entirely true. Electrical engineers have made effective use of positive feedback to make cool things like square wave generators and oscillators. When you want instability, this is one way to go. <BR/><BR/>TC.Unknownhttps://www.blogger.com/profile/06449079338919787252noreply@blogger.comtag:blogger.com,1999:blog-8231784566931768362.post-91660075752131577492007-11-13T06:05:00.000-05:002007-11-13T06:05:00.000-05:00Eric--Welcome back (of course, I haven't been supp...Eric--<BR/>Welcome back (of course, I haven't been supplying you with much fuel!) <BR/><BR/>I had to make some difficult choices in what to include or exclude here. Since my primary goal was to reveal some of the beauty of linear algebra to high schoolers, I chose to make matrices the focus. <BR/><BR/>By the way, tc appropriately pointed out that I omitted the important idea that the composite of 2 rotation transformations is equivalent to the product of their matrices which in turn is equivalent to the rotation determined by the SUM of the angles. I deleted that step right before publishing it as I felt it was too much, but I probably should have left it in. Oh well...<BR/><BR/>Eric, I really like the complex 'argument' however and I did allude to it in the last paragraph. The idea that multiplying complex numbers in trig form or polar form involves adding their 'arguments' (i.e., angles) is powerful stuff. <BR/><BR/>Consider also the form of multiplying complex numbers in rectangular form:<BR/>(a+bi)(c+di) = (ac-bd) + (ad+bc)i<BR/>The real and complex parts should remind students of that same pattern one sees in the cos(t+u) and sin(t+u) formulas. <BR/><BR/>The differential equation argument is my favorite however. Using the idea that a solution can be be expressed in multiple forms and then equated is one that students rarely appreciate. Thanks for sharing that! <BR/><BR/>Related to that, using differentiation rather than integration, is one of my favorite calculus derivations:<BR/><BR/>Assume x and a are positive.<BR/>d/dx(ln(ax)) = 1/x<BR/>d/dx(ln(x)) = 1/x<BR/>Therefore, ln(ax) = ln(x) + C<BR/>Replacing x by 1 leads to C = ln(a) and, one has the product rule for logarithms!Dave Marainhttps://www.blogger.com/profile/13321770881353644307noreply@blogger.comtag:blogger.com,1999:blog-8231784566931768362.post-70187606623300170402007-11-12T20:48:00.000-05:002007-11-12T20:48:00.000-05:00You can then relate this to Euler's formula e^{i&t...You can then relate this to Euler's formula e^{iθ} = cos θ + i sin θ. But that's too easy.<BR/><BR/>Advanced students who know some differential calculus could understand this backwards method too: consider the differential equation y″ + y = 0. Students who remember their differentiation tables will realize that y = cos x and y = sin x are both solutions, and so y = a cos x + b sin x is a solution for any real a and b.<BR/><BR/>Now, a second-order linear ODE has a 2-dimensional space of solutions; in other words, any solution will be of the form a cos x + b sin x. So, sin (x + θ) is of this form!<BR/><BR/>(1) sin (x + θ) = a cos x + b sin x.<BR/><BR/>Letting x = 0,<BR/><BR/>sin θ = a.<BR/><BR/>Differentiate equation (1) to get<BR/><BR/>(2) cos (x + θ) = -a sin x + b cos x.<BR/><BR/>Let x = 0:<BR/><BR/>cos θ = b.<BR/><BR/>Of course we usually derive the derivatives from the addition formulas, so the utility of this method isn't quite clear.<BR/><BR/>The equation y′ = y leads to e^{x + a) = e^x e^a too.<BR/><BR/>There's one final social point here. The equation y″ + y = 0 appears in science far more that the equation y″ - y = 0. Trigonometric functions are ubiquitous. Hyperbolic functions appear only rarely: the formula for the catenary. They appear as positions, not as motions. This is because equations of motion that involve hyperbolic functions blow up! These can't be physical.<BR/><BR/>If you think of the y″ as the acceleration of a particle, then the trigonometric case has the acceleration directed oppositely from the position. This is the real meaning of the term <I>negative</I> feedback. y″ = y gives <I>positive feedback</I>. Current slang defines positive feedback as 'always positive', and that is a false reading of the phrase. In fact, positive feedback is always <A HREF="http://upload.wikimedia.org/wikipedia/commons/4/46/Tacoma_Narrows_Bridge_Falling.png" REL="nofollow"> disastrous</A>.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-8231784566931768362.post-14099518837590370882007-11-12T15:18:00.000-05:002007-11-12T15:18:00.000-05:00Absolutely! I was about to do this but I felt comp...Absolutely! I was about to do this but I felt compelled to stop - I was getting overTexified! The product of matrices for the rotations is a critical piece of all this. I may decide to add more detail.Dave Marainhttps://www.blogger.com/profile/13321770881353644307noreply@blogger.comtag:blogger.com,1999:blog-8231784566931768362.post-28495606710059171782007-11-12T15:13:00.000-05:002007-11-12T15:13:00.000-05:00Looks good!For the last step, it is also worthwhil...Looks good!<BR/><BR/>For the last step, it is also worthwhile multiplying the rotation matrices for theta and alpha, and see that the resulting rotation matrix corresponds to the sum of the angles. <BR/><BR/>Similarly for the difference, have theta and -alpha.<BR/><BR/>TCUnknownhttps://www.blogger.com/profile/06449079338919787252noreply@blogger.com