Johnny5

Jacques' date=' I looked at the link, and saw the diagram, all the circles. Does this bear any relationship to the Mach effect... Mach cone? Let me see if I can find something on it. Mach cone Let the Mach number be 1.22 in the Java applet, and you see the Mach cone forming. Can this effect be tied into the Doppler effect formulas? Regards

It will take me awhile to analyze this, but I will. In the meantime, what does kiesel mean? Geist means ghost? Regards

Yes, I saw that Dr. Swanson.

I think permanent magnetism has something to do with some kind of regularity in the way all the dipoles are aligned. In non-magnetic matter, the orientation must be somewhat random, and so magnetic effects just tend to cancel out, which would explain why a powerful refridgerator magnet does not attract a piece of paper. A normal wire is non-magnetic. Once you "pass current" through it, it will demonstrate very weak, mild magnetic effects. For example, Oersted noticed that if he placed a compass near a wire, and then "passed current" through the wire, the compass needle moved. So a changing electric field leads to a magnetic field. I don't know how else to say that. And there are some very simple formulas, from classical electrodynamics, which would tell you the B field created by a current carrying wire, in the region of space external to the wire. Directly from the Biot-Savart law we have: [math] \vec B \equiv \frac{\mu_0}{4 \pi} \int \vec J \times \frac{\hat r}{r^2} d \tau [/math] mu_0 is the pearmeability of free space. J is the volume current density. There is also line current density, which is represented by lambda, and surface current density, which is represented by lower case sigma. Which you use, depends upon your particular configuration. In modelling your current carrying wire, you have to decide whether or not you are going to use J or lambda. A real wire is three dimensional, so J would be more appropriate, if you are going to investigate field effects near the wire. On the other hand, if you are mainly concerned with field effects far from the wire, lamba is fine, and I think the mathematics is simpler that way. But I am not sure what you are trying to do. Now, the most elementary problem in Magnetostatics, would be an infinitely long wire carrying a constant current i. And then to a good approximation, you will use V=voltage = iR. Again I am not sure exactly what you are out to do. You can use ampere's law to figure out the B field quite easily, and you can also go through the mathematics using the Biot-savart law to get the field of the current carrying wire. Both approaches yield the same answer, so find the B field using both methods. The end result, from memory, is that the B field wraps around the wire in a circular fasion, and decreases as you move away from the wire. B is inversely proportional to the distance you are away from the wire, so the field effects fall off as 1/r. Now, in the case of a permanent magnet, the field does not wrap around in circles. The shape is famous, it's toroidal. There is a north pole, and a south pole. The magnetic field lines go from one pole to the other, and have a sort of elliptical shape to them. Now, if you have some good intuition about magnetic fields, you can "build up", so to speak. Once you understand the formula for the magnetic field of an infinitely long wire, then you can progress to the solenoid, as Dr. Swanson suggested. I recommend you do this too. Solve the following problem next, wrap a wire around a cylinder, N times, in circles, going up to the top of the cylinder, and back down, and back up. You will know the current in the wire, you will know the magnetic field contribution from tiny portions of the wire, and because the wire has been wrapped around a cylinder, there will be a NET magnetic field, all due to the wire geometry. From memory, the approximate answer is: [math] B = \mu N i [/math] Along the axis. Find a clear presentation. But my point is, your intuition can tell you what the field will be, if you already understand the answer to the infinite current carrying wire. And then, you can wrap a solenoid into a huge circle, to get a toroid. And again, your intuition and the mathematics should agree. I would suggest developing your model in stages. Each stage being more accurate than the previous one. Regards

What do you want to make a superconductor for? Regards

And to add to what Dr. Swanson said, mathematically, this happens due to usage of the Dirac delta function. See this link here: Measurement in quantum mechanics In particular, read this: The usage of the mathematical theory of probability in quantum mechanics leads to certain "conceptual problems." These problems arise for any number of reasons. Regards

Citation? I was very recently looking at a picture of an enormous solar flare, and the caption read something about an intense magnetic field. I don't remember where I read it though. I'll google it right now. Here is a link: Magnetic pole reversal of sun Here is another link: The magnetic sun And one more link: The sun is a big magnet.

It's been a long time since I've solved a problem like this. Start off with the definition of a rectangle. Length times width. Now, the lines which are the sides of the rectangle, have to have a slope in the XY plane, which is perpendicular to the slope of the hypotenuse, which you have drawn. So the slope will be the negative reciprocal. Now the slope of the given line is -2/3. The negative reciprocal is 3/2. Now, here is the given function: [math] y = f(x) = \frac{-2x}{3} + 4[/math] That is in slope intercept form, m=slope=-2/3, and y intercept is 4. The x intercept is the point the line crosses the x axis, and the point where that occurs has a y coordinate of zero. So in order to find the x-intercept, you have to find the point (x,0), where 0=-2/3x+4. 0=-2/3x+4 2/3x=4 1/3 x =2 x=6 just as you have labeled.

d/dx eU = eU dU/dx in this case U(x)=sin x' date=' so provided you already know that d/dx(sin x) = cos x, then the answer, in your head will be: cos x e[sup']sin x[/sup] Chain rule. Regards

Well I just used L'Hopital's rule' date=' I didn't actually prove that the limit is -1, but it is plain to see. But to prove it, some would ask for an epsilon-delta proof, though I don't feel that's necessary, when you have something like L'Hopitals rule at your disposal. I have to think about this... absolute value of ratio of consecutive terms in the series, as the index tends to infinity. Let me try to interpret that formula clearly. [math] (1+x)^{1/2} = \sum_{n=0}^{n=\infty} x^n \prod_{k=1}^{k=n} \frac{3/2 -k}{k} [/math] The series is going to be composed of terms. Power series have the following form: [math] C_0+C_1x+C_2x^2+C_3X^3+...C_nx^n+... [/math] The first term of the series is C0, the second term of the series is C1x, and so on. The nth term of the series is Cn xn. Now, pick a term of the series at random. How about n=p=999493245932459934593495 Forget about all other terms in the series but this one. So ignore the decimal representation of this term, and call it p. So the pth term of the series is given by: [math] x^p \prod_{k=1}^{k=p} \frac{3/2 -k}{k} [/math] Now, I am going to think about what you said again, very carefully. The absolute value of my f(k) is the radius of convergence... being the ratio of the absolute value of consecutive terms of the series as the index k tends to infinity. hmm I don't see how you get "absolute value" from anything which I've done. Actually, let me let p go from 1 to 4, to see the pattern you say is there. Case 1: p=1 [math] x^1 \prod_{k=1}^{k=1} \frac{3/2 -k}{k} = x \frac{3/2 -1}{1} = \frac{x}{2} [/math] Case 2: p=2 [math] x^2 \prod_{k=1}^{k=2} \frac{3/2 -k}{k} = x^2 \frac{3/2 -1}{1}\frac{3/2 -2}{2} = x^2 (\frac{1}{2}) (\frac{-1}{4})[/math] Case 3: p=3 [math] x^3 \prod_{k=1}^{k=3} \frac{3/2 -k}{k} = x^3 \frac{3/2 -1}{1}\frac{3/2 -2}{2} \frac{3/2 -3}{3} = x^3 (\frac{1}{2}) (\frac{-1}{4})(-\frac{1}{2}) [/math] [math] x^4 \prod_{k=1}^{k=4} \frac{3/2 -k}{k} = x^4 (\frac{1}{2}) (\frac{-1}{4})(-\frac{1}{2})(\frac{-5}{8}) [/math] now you say... "ratio of absolute value of consecutive terms of series." Yes. I went off and thought about it Matt. My f(k) is the ratio of consecutive terms of the series. Yes. Disregard the powers of x, and focus on two consecutive coefficients of terms in the series, say the coefficient of the pth term, and the coefficient of the p+1th term. Now, take the ratio of the coefficients of the pth term, and the p+1th term, like so: Coefficient of pth term: [math] \prod_{k=1}^{k=p} \frac{3/2 -k}{k} [/math] Coefficient of p+1th term: [math] \prod_{k=1}^{k=p+1} \frac{3/2 -k}{k} [/math] [math] \text{Ratio of two consecutive coefficients} [/math] [math] \frac{\prod_{k=1}^{k=p+1} \frac{3/2 -k}{k}}{\prod_{k=1}^{k=p} \frac{3/2 -k}{k}} = \frac{\frac{3/2-(p+1)}{p+1}\prod_{k=1}^{k=p} \frac{3/2 -k}{k}}{\prod_{k=1}^{k=p} \frac{3/2 -k}{k}} =\frac{3/2-(p+1)}{p+1} [/math] And p is arbitrary. So letting k=p+1, implying k is arbitrary as well, we have, as the ratio of two consecutive terms of the series: [math] \frac{3/2-k}{k} [/math] which is my f(k), exactly as you said. SOOOOOOOOOOOOOO In taking the limit, as k tends to infinity, of f(k), we are in fact performing the ratio test, precisely because f(k) is the ratio of the nth term, and the n-1th term. Of course I still don't see where the absolute value bit comes in at, but I see exactly what you meant. Just the f(k) is the ratio of two consecutive terms of the series. Uh huh.

Nice quotes there by Dr Einstein. ... interaction of matter with the cosmic microwave background radiation... the electromagnetic fields of the quantum vacuum. i.e. the photons right? Quantum electrodynamics by Feynman? Sort of not disagreeing with me actually. The sun has an enormous magnetic field associated with it. Toroidal in shape. The "magnetic field lines" run through the ecliptic plane. So we already know the shape of the B-field. So, in your own words, what is the B-field composed of? There aren't real magical lines running through space. You have E-fields, and B-fields in your model, but otherwise I don't see much difference.

Agreed.

Well, as you seem to know exactly what's going on, I will let you answer any further questions he has. But to understand the answer, he needs to be shown Coulomb's formula. But you know that. Kind regards 5614

I was wondering what that means. What's GCSE stand for? General science something or other. Also 5614, IMO, you should use some mathematics in your explanation. Just an opinion this time though, I see clearly he wants a clear verbal explanation.

Can someone show me how to compute the radius of convergence of the following power series? [math] \sum_{n=0}^{n=\infty} \frac{nx^n}{2^{n+1}} [/math] There are different ways to do it. Regards The first term is zero, so [math] \sum_{n=1}^{n=\infty} \frac{nx^n}{2^{n+1}} = \sum_{n=1}^{n=\infty} \frac{n(n-1)!x^n}{(n-1)!22^{n}} = \frac{1}{2}\sum_{n=1}^{n=\infty} x^n \prod_{k=1}^{k=n} \frac{k}{2(k-1)} [/math] So [math] f(k) = \frac{k}{2(k-1)} [/math] In the sum, n goes to infinity, hence k goes to infinity. The product from k=1 to n, of f(K), gives the nth coefficient of the power series. So in the limit we have: [math] \lim_{k \to \infty} f(k) = \lim_{k \to \infty} \frac{k}{2(k-1)} [/math] And we can use L'Hopital's rule here, so: [math] \lim_{k \to \infty} \frac{k}{2(k-1)} = \frac{1}{2} [/math] Actually, let me put in x, so that we are looking at f(x,k), instead of just f(k). [math]\sum_{n=1}^{n=\infty} \frac{nx^n}{2^{n+1}} = \sum_{n=1}^{n=\infty} \frac{n(n-1)!x^n}{(n-1)!22^{n}} = \frac{1}{2}\sum_{n=1}^{n=\infty} \prod_{k=1}^{k=n} \frac{xk}{2(k-1)} [/math] Hence: [math] f(x,k) = \frac{xk}{2(k-1)} [/math] Actually let's look at the product: [math] f(n,x,k) = \prod_{k=1}^{k=n} \frac{xk}{2(k-1)} [/math] The formula above, is the n-th term of the series. Now, consider the limit as n approaches infinity: [math] f(n,x,k) = \prod_{k=1}^{k=\infty} \frac{xk}{2(k-1)} [/math] This is now the limit as k approaches infinity of f(x,k). [math] \lim_{k \to \infty} f(x,k) = \lim_{k \to \infty}\frac{xk}{2(k-1)} =\frac {x}{2} [/math] for a fixed x, which does not depend upon k. So as long as x<2, the nth term of the series is less than 1, as n approaches infinity. I am considering series whose terms are positive right now. The limit as k approaches infinity of f(k), is one over the radius of convergence of the series. Right now, I am looking for a proof of this. Consider the harmonic series: [math] H_n = \sum_{n=1}^{n=\infty} \frac{1}{n} = 1+1/2+1/3+1/4+...+1/n+... [/math] Clearly, the nth term goes to zero, as n approaches infinity, but that does not mean that the series converges. What has to be done, is to consider the sequence of partial sums, and ask whether or not that goes to zero, as n tends to infinity. When n=2, we have: 1+1/2=2/2+1/2=3/2 when n=3, we have 3/2+1/3=9/6+2/6=11/6 when n=4 we have 11/6+1/4 = 44/24+6/24=50/24=25/12 when n=5 we have: 25/12+1/5=125/60+12/60=137/60 Thus, the first few terms of the sequence of partial sums is given by; (1,3/2,11/6,25/12,137/60,...) Let us list the first few terms, with each term having a common denominator: (60/60,90/60,110/60,125/60,137/60,...) Here is some stuff on limits of sequences: limit of a sequence First find f(k). [math] H_n = \sum_{n=1}^{n=\infty} \frac{1}{n} = 1+ \sum_{n=2}^{n=\infty} \frac{(n-1)!}{n(n-1)!} = 1+ \sum_{n=2}^{n=\infty} \frac{(n-1)!}{n!}[/math] [math] 1+ \sum_{n=2}^{n=\infty} \frac{(n-1)!}{n!} = 1+ \sum_{n=2}^{n=\infty} \prod_{k=2}^{k=n} \frac{k-1}{k} [/math] Hence: [math] f(k) = \frac{k-1}{k} [/math] In the limit as k goes to infinity, f(k) approaches 1. Here is Wolfram on the concept of limit. Wolfram on limit Now, the series Hn converges if and only if the sequence of partial sums converges. Let me branch again in this discussion, to consider whether or not the following series converges (because I know something about the partial sums of this series, is particularly simple): [math] \sum_{n=1}^{n=\infty} \frac{1}{n(n+1)} = 1/(1*2) + 1/(2*3) +1/(3*4) +... [/math] Now, we can use partial fractions to rewrite the summand. Now, I know of two ways to do this, the coefficient method, and the Heaviside cover up method. I will use the coefficient method. [math] \frac{1}{n(n+1)} = \frac{A}{n}+\frac{B}{n+1} [/math] the goal is to now solve for A,B. [math] \frac{1}{n(n+1)} = \frac{A}{n}+\frac{B}{n+1} = \frac{A(n+1) + nB}{n(n+1)} [/math] From which it follows that: [math] A(n+1) + nB = 1 [/math] Hence: [math] An+A + nB = 1 [/math] Hence: [math] (A+B)n+A = 1 [/math] Hence A=1, and A+B=0, from which it follows that B=-1. Therefore: [math] \frac{1}{n(n+1)} = \frac{1}{n}-\frac{1}{n+1} [/math] So we can now rewrite the orginal series, with an equivalent summand, in a different form: [math] \sum_{n=1}^{n=\infty} \frac{1}{n(n+1)} = \sum_{n=1}^{n=\infty}\frac{1}{n}-\frac{1}{n+1} [/math] Now, here is the formula for the NTH partial sum: [math] \sum_{n=1}^{n=N}\frac{1}{n}-\frac{1}{n+1}= (1-1/2)+(1/2-1/3)+...+(\frac{1}{N}-\frac{1}{N+1}) [/math] The series above is an example of a telescoping series. The intermediate terms cancel each other out. So, we have a simple formula for the NTH partial sum, which is: [math] 1-\frac{1}{N+1} [/math] So this is the formula for the sequence of partial sums: [math] S(N) = 1-\frac{1}{N+1} [/math] And in the limit as N approaches infinity, S(N) approaches 1. Thus, the sequence of partial sums converges, thus the series converges. So back to the harmonic series Hn. The harmonic series converges if and only if its sequence of partial sums converges, so if the sequence of partial sums diverges then the harmonic series diverges. [math]H_n = \sum_{n=1}^{n=\infty} \frac{1}{n} = 1+ \sum_{n=2}^{n=\infty} \frac{(n-1)!}{n!} = 1+ \sum_{n=2}^{n=\infty} \prod_{k=2}^{k=n} \frac{k-1}{k} [/math] The next step is to focus on the sequence of partial sums, as in the previous example.

Ok, here is the power series; [math] (1+x)^r = \sum_{n=0}^{n=\infty} x^n \prod_{k=1}^{k=n} \frac{r+1-k}{k} [/math] Now things are coming back to me, about radius of convergence. I recall that all I have to do is analyze f(k), where: [math] f(k) = \frac{r+1-k}{k} [/math] Let me see what happens as k approaches infinity. Let's fix r, at 1/2, for this, hence: [math] (1+x)^{1/2} = \sum_{n=0}^{n=\infty} x^n \prod_{k=1}^{k=n} \frac{3/2 -k}{k} [/math] So if r=1/2, then [math] f(k) = \frac{3/2 - k}{k} [/math] Now, we have to evaluate the limit as k approaches infinity of f(k). [math] \lim_{k \to \infty} f(k) = \lim_{k \to \infty} \frac{3/2 - k}{k} [/math] The limit can be determined by using L'Hopital's rule. Let me try and formulate L'Hopital's rule correctly. Suppose that in taking the limit of g(k)/h(k) we have either the case 0/0 or infinity/infinity. Then the limit of g(k)/h(k) will equal the limit of g`(k)/h`(k). So in our problem here, we wish to compute: [math] \lim_{k \to \infty} f(k) = \lim_{k \to \infty} \frac{3/2 - k}{k} [/math] The derivative of the numerator with respect to k is equal to -1, and the derivative of the denominator with respect to k is equal to one, and the ratio of these is -1. Hence the limit is defined. Now, I just have to remember how this ties into the radius of convergence.