Overlord_Prime

Hey, how exactly did you reach |z+z0| < (2+δ)|z0| for |z-z0| < δ. I can't seem to understand..

I have this sum: [LATEX] lim_{(x,y)->(0,0)} tan^{-1} (\frac{|x|+|y|}{x^2+y^2}) [/LATEX] and here's my attempt at a solution using the limit definition: I'll work with whats inside the arctan first, then dump the limit in there..I guess my limit as 0 (for fun--how to make intuitive guesses for proposed limits, I do not know). So [LATEX] lim_{(x,y)->(0,0)}\frac{|x|+|y|}{x^2+y^2} [/LATEX] [LATEX]|\frac{|x|+|y|}{x^2+y^2}|<\epsilon\ \ whenever\ \ \sqrt{x^2+y^2}<\delta [/LATEX] [LATEX]\frac{|x|+|y|}{x^2+y^2}< \frac{\sqrt{x^2}+\sqrt{y^2}}{x^2+y^2}[/LATEX] [LATEX]<\frac{\sqrt{x^2+y^2}+\sqrt{y^2+x^2}}{x^2+y^2}[/LATEX] [LATEX]=\frac{2 \sqrt{x^2+y^2}}{x^2+y^2} = \frac{2}{\sqrt{x^2+y^2}}=\frac{2}{\delta}[/LATEX] So, taking [LATEX] \delta=\frac{2}{\epsilon}[/LATEX] The limit should be 0. so my initial functions limit would be arctan(0)=0. Now wolfram alpha tells me that my limit does not exist, and my textbook tells me that my limit exists and is equal to [LATEX]\frac{\pi}{2}[/LATEX] What's going on ??!

I think I got this figured out. At the end of the day, you need as many unknowns as there are equations, so for the two body case, we start out with 8 unknowns as follows: [math]x_a=Acos(\omega_{-}t+\phi_1)+Bcos(\omega_{+}t+\phi_2)[/math] [math]x_b=Ccos(\omega_{-}t+\phi_3)+Dcos(\omega_{+}t+\phi_4)[/math] namely, A, B, C, D, ϕ1, ϕ2, ϕ3,and ϕ4. These being normal modes, dictates that ϕ1= ϕ3,and ϕ2= ϕ4. It also just so happens that in all our equation grinding, we found the ratio of A:C and B:D as 1 and -1 respectively. and so, the equations boil down to: [math]x_a=Acos(\omega_{-}t+\phi_1)+Bcos(\omega_{+}t+\phi_2)[/math] [math]x_b=Acos(\omega_{-}t+\phi_1)-Bcos(\omega_{+}t+\phi_2)[/math] 4 unknowns, 4 initial conditions, and we're done. If there were 3 masses, then there would be 3 normal modes, each with a frequency and phase associated with it. What I previously mentioned as (q1+q2+q3) actually represent the superposition of normal modes IF the ratio of the amplitudes were 1. But it may be the case that A:B:C is 4:2:1 in which case it would change the general equation. For example: for normal mode 1, we might get: [math]x_a=Acos(\omega_{-}t+\phi_1)[/math] [math]x_b=0.71Acos(\omega_{-}t+\phi_1)[/math] [math]x_c=-Acos(\omega_{-}t+\phi_1)[/math] i.e: the ratio of A:B:C (in normal mode 1) is 1:0.71:-1 and in normal mode 2, A:B:C=-2:1:5, and in normal mode 3, A:B:C=4:1:-3 so the general equation for the body is: [math]x_a=Acos(\omega_{-}t+\phi_1)-2Bcos(\omega_{+}t+\phi_2)+4Ccos(\omega_{++}t+\phi_3)[/math] etc. 6 unknowns, 3 initial positions, 3 initial velocities, and we're set for life. Please correct me if I am wrong.

So I have my differential equations for 2 bodies coupled with 3 springs. [LATEX]\ddot{x_{a}}+2\omega_{o}^2x_{a}-\omega_{o}^2x_{b}=0[/LATEX] and [LATEX]\ddot{x_{b}}+2\omega_{o}^2x_{b}-\omega_{o}^2x_{a}=0[/LATEX] I use 2 trial functions [LATEX]q_{1}=Acos(\omega t)[/LATEX]and [LATEX] q_{2}=Bcos(\omega t)[/LATEX] and get two solutions for the normal mode frequencies: [LATEX]\omega_{-}=\omega_{o}[/LATEX]and [LATEX] \omega_{+}=\omega_{o}\sqrt{3}[/LATEX] Now to get the motion for [LATEX]x_{a},x_{b}[/LATEX], I am supposed to take: [LATEX]\frac{q_{1}+q_{2}}{2}[/LATEX]and [LATEX] \frac{q_{1}-q_{2}}{2}[/LATEX] My questions: -Why do I have to divide by 2 ? Since the solution is a superposition of normal modes, why not just add q1 and q2. Shouldn't the constants just take care of any initial conditions I impose? -If there were 3 bodies, and therefore, 3 trial solutions [LATEX]q_{1}=Acos(\omega t), q_{2}=Bcos(\omega t), q_{3}=Ccos(\omega t)[/LATEX] what would be the solutions to the equations of motion of the three particles ? Would it be (q1+q2+q3)/3, (q1-q2+q3)/3 and (q1-q2-q3)/3 ? What about (q1+q2-q3)/3 ? Don't we have to take all linear combinations, or is it based on the ratios of amplitudes ? (for example, in the 2 bodied one, we got A/B=1 for one normal mode and then A/B=-1 for another normal mode, therefore q1+q2 and q1-q2)