How do bessel functions relate to drums?
490529
For circular membranes, the radial part of the vibrations is a bessell function while the angular part is an ordinary trigonometric function. This comes from writing the wave equation in polar coordinates and doing a separation of variables. The radial function turns out to satisfy Bessell's equation. The frequencies then turn out to correspond to the zeros of the bessell functions.
763829
I'm not familiar with that particular book, but the differential equations of simple harmonic motion are interesting.  You can make the system as complex as you like.  I've studied SHM in both maths and physics.  I simulated SH oscillators with operational amplifiers; in other words an analogue computer.  First you simulate a frictionless pendulum, then you introduce friction so the oscillations die away, then you introduce forcing terms of various frequencies to see what happens; like pushing a child on a swing.  If you're really clever, you can introduce hysteresis (delay), and extra degrees of freedom (flexible joints in the pendulum), and vary the Q of the system (how sharply you can tune it to the resonant frequency).  It's about the most open-ended project I ever saw.  I did it in second-year physics.  At its simplest, it's just sixth form maths, but you can turn it into an impossibly complex exercise in chaotic systems if you're clever.
525579
A parabola has a focal point. At this focal point "on" a Parabolic Reflector Microphone is, well, a microphone. The "dish" gathers sound waves (lots and lots) and directs them toward the focal point where the microphone collects/converts the sounds waves into electrical signals. It's similar to cupping your hand behind your ear to hear someone. Your hand reflects more sound into your ear than just your outer ear alone.\n\nFor the mathematics behind how it all works, see any analytic geometry text and search for focus, parabola. It's actually quite simple yet very effective.
71115
You need some theory background first. Integration is most rigorously defined as a sum of an infinite number of rectangles that start on the x-axis and stretch up or down to the function (called a Riemann sum). The rectangles are infinitely thin, but each one still has area length * width, and integration is merely the sum of the areas of all those rectangles.\n\nImagine a rectangle is attached to the x-axis, and the x-axis acts as an axle, so you can spin the rectangle around that axle. The spinning rectangle will look like a cylinder whose radius is the height of the rectangle and whose width is the width of the rectangle (the amount touching the axis). \n\nThe area of a cylinder is pi * r^2 * w. As I stated, the radius is the height of the rectangle, which stretches from the x-axis (y=0) to the function (y=f(x)). Subtract those y-values to get the radius: f(x)-0=f(x). So, r=f(x).\n\npi * r^2 * w = pi * [f(x)]^2 * w\n\nThe width of the cylinder is how much is touching the x-axis and is usually represented by delta-x.  However, when an infinite number of rectangles are used, you use dx instead.\n\npi * [f(x)]^2 * dx\n\nSo, just like the definite integral from a to b of f(x)dx gives you the sum of the areas of the infinite rectangles beneath f(x), the definite integral from a to b of pi * [f(x)]^2 * dx is equal to the volume of the cylinders created when those rectangles are rotated.
165255
Take a look at the error function, and modify it so that you are dealing with distributions in 2 dimensions, using polar coordinates.\n\nFor example, the counterpart of the error function in 2 dimensions, in polar coordinates can actually be integrated, to produce the comulative distribution function\n\nF(R,s)=1-e^[-(R^2)/(2*s^2)]\n\nwhere s is the standard deviation, and R is the radius of a circle centered on the origin. \n\nAs R--> infinitity, then F(R,s)----> 1\nF(0,s)=0\n\nSuppose that the target has an edge R0 such that\n\nF(R0,s)=.5 for a given value of s. \n\nThen, you have to find the value of S for which F(R0,S)=0.9.
913325
If we assume SHM, then\n\n(a) for a given amplitude A, the max velocity and max acceleration is greater for higher frequency sounds, because SHM is given by the following equations:\n\ny(t) = Asin(2πft + γ) where γ is a phase factor\nv(t) = dy/dt = 2πfAcos(2πft + γ)\na(t) = dv/dt = -(2πf)²Asin(2πft + γ)\n\n(b) For A = 1.0 x 10^-8 m and f = 20.0 Hz\n\nv(max) = 2πfA = 1.26 x 10^-6 m/s\na(max) = -(2πf)²A = - 1.58 x 10^-4 m/s^2
819996
T1=m1*a, no matter are there m2 & m3 or not.\nT2=(m1+m2)*a, no matter whether m1 & m2 are connected with a string or are in one lump, the presence of m3 being unimportant too.
725972
You mean the graph of the function?  The second term is a sine wave with frequency 2x, meaning it goes through a full cycle in pi radians instead of 2*pi radians.  The max amplitude of the second term is +/- 1/2, since the max of the sine is +/- 1.  That's the second term.  Now, offset that by 4 units down at every point.  Got it?  Okay, your turn.
1139180
Yes.\nThe closure relation on a function space produces a dirac delta function and the dirac delta function can be expanded in any basis set, in particular in a complete orthonormal family. This can be used to solve PDEs by using eigenfunction expansions.\n\nFor example, the dirac delta function has a representation in the sine basis functions, a representation in the cosine basis functions and also a representation in the fourier series basis functions.
272748
Okay I will prove it.\n   there are 2 extreme cases of kite:\n  (i)when both sides become equal:\n      then it becomes a rhombus and area is .5*d1*d2\n  (ii) when one side is zero :\n      then it becomes a triangle and its area 1/2*b*h\nsatisfies the formula for kite (try drawing it)\n   Now the variation of area with the diff of sides is a smooth function must have a formula that satisfies both extremes. 1/2d1d2 satisfies the condition.\n   Assume that there is one more function satisfying the given criteria.It must have same value at all pts since area  \nis a unique entity.2 functions have the same graph and hence they must be identical.This proves that no other answer exists.
219785
f(x) = 2 sin[x - pi/3]\n\nI can see that x is not multiplied by another coefficient, so the wave graph is not altered by shrinking or expanding, and thus the period will not change, and the period is still 2pi.(it may be translated to the right by pi/3, but the size will not change)\n\n^_^
881328
this requires way more than anyone can do here, but, i know of a cool java program that will do it for you: \n\nhttp://www.math.rutgers.edu/courses/ODE/sherod/phase-local.html\n\nyou can even set your initial conditions and print it out!
343400
Visit the Pearl Drums website and they got free instructions. And try learnin by drumming along to a song or playing it slow motion to get the beat. Good luck
744417
My answer assume that you have a good knowledge of integral calculus. If it's not the case, then use a soft like Matlab...\n\nNotations :  I will use SIGMA for the 1 to infinite sum and INTEGRAL for the integral between -T/2 and T/2.\n\nIf you consider a periodic signal x(t) with a T = 1/f0 period, its Fourier development is :\n\nx(t) = a0/2 + SYGMA[k=1 -> infinite]  (ak*cos(2*PI*f0*t)) +  SYGMA[k=1 -> infinite] (bk*sin*(2*PI*f0*t))\n\nwhere f0 = 1/T is the fundamental frequency  of the signal, a0/2 the mean value and ak, bk are the Fourier coefficients in cosinus and sinus.\n\nThe Fourier coefficients are calculated like this :\n\nak = 2/T * INTEGRAL[-T/2 -> T/2] (x(t)*cos(2*PI*k*f0*t) dt)    for k >= 0\n\nbk = 2/T * INTEGRAL[-T/2 -> T/2] (x(t)*sin(2*PI*k*f0*t) dt)     for k >= 1      \n\nAfter developpement, you will find the values  of all a and b coefficients in function of k.\n\nI hope this will help :)
654455
Two tutorials:\n\nhttp://users.rowan.edu/~polikar/WAVELETS/WTtutorial.html\nhttp://perso.wanadoo.fr/polyvalens/clemens/wavelets/wavelets.html\n\nAnd a list of  Internet Sources:\nhttp://www.cosy.sbg.ac.at/~uhl/wav.html
873172
u can get them with the help of second and higher derivatives.\ninflection point is there where the second derivative is zero.
908707
a.\nThere are few observation in X(t)\n-constant w set\n-cos and sin are periodic functions, this means cos(t) and cos(t+T) would behave identical over the time period T, 2T,3T... [consider sameargument for sin too]\n\nFor your refference let us consider the indices as n=1,2,..\n\nSo, for each n, X(t) will have a joint distribution with independent unit normal random variables Y1,Y2. And, this would be nothing but linear combination of two unit notmal distirbutions.\n\nSince, the behaviour if sine and cosine remains same over time intervals (nT), this joint distribution will remain same for all n=1,2,...\n\nThis indicated that the process {X(t)} is stationary.\n\nSince, the mean & variance of the each distribution occuring at eact time interval may not remain constant or same, This is the reason why {X(t)} is a wekly stationary process.\n\nI hope now you understand.\n\nAll the best.