Tom Mattson

As others have remarked: No, it isn't. Sure, you can do that. Just define the function in a piecewise manner, as follows. [math]f(x) = \left\{\begin{array}{cc}sin(x) & 0 \leq x \leq \pi \\e^x & \pi < x \leq 2\pi \\ 2 & x>2 \pi,x\neq 3\pi\end{array}\right\}[/math] The above function is a sine wave from 0 to pi, an exponential function from pi to 2pi, and a constant function from 2p onwards, with a hole at 3pi. edited to add: OK, can one of you computer geeks tell me how to write a piecewise-defined function in LaTeX without putting the right curly brace there? If I leave it out then the rendering software says that I have a LaTeX error.

Well, if that's true then that's pretty bad. It should be common knowledge that a theory has to start with someone making a conjecture that has empirical content. I did make that a separate point. It's #1 on the list.

Wow, this is a blast from the past. It doesn't violate rule #2. The lone premise (call it P) is also a prediction, so the trivial derivation "P, therefore P", while valid, is also trivial. Certainly, the premises of a theory need not be derived. It's the subsequent predictions that I was talking about. They can be known to be logically valid, which is the type of validity I was talking about.

Not really. There are patents to be had in applied physics, engineering, and even applied mathematics too. When I was in grad school (for physics) I knew of a grad student in applied math who had a patent for an algorithm she developed. While the ideas that get tossed around in pure mathematics and the pure sciences don't directly translate into monetary gain, they do so indirectly. When someone comes up with a good idea, they certainly want to be credited with the discovery. At the research level people typically form collaborations that result in joint authorship of a publication in a refereed journal. I doubt that there are too many people out there who would be willing to invest a lot of time and energy in a collaboration with someone whose only intention is to take the other's input, and publish it without crediting him. So "Sometimes" is my vote.

Yes, I supposed they do. But I immediately have the converse from the definitions of addition and of equality. If [math]a=b[/math], then I have: [math]a+c=a+c[/math] (tautology) [math]a+c=b+c[/math] (substitution) So there we have the converse, and everything is now logically closed.

Here's another proof, not by contradiction. Let [math]a,b,c\in\mathbb{R}[/math]. Proposition 1 [math]a+c=b+c[/math] implies [math]a=b[/math]. (Actually this could be an "iff" statement, but I don't need the converse to get to what I want.) Proof [math]a+c=b+c[/math] (by hypothesis) [math](a+c)+(-c)=(b+c)+(-c)[/math] (add [math]-c[/math] to both sides) [math]a+(c+(-c))=b+(c+(-c))[/math] (by associativity under addition) [math]a+0=b+0[/math] (by the additive inverse property) [math]a=b[/math] (by the additive identity property) Proposition 2 [math]0a=a0=0[/math] Proof [math]a0+0=a(0+0)[/math] (by the additive identity property) [math]a0+0=a0+a0[/math] (by the left distributive property) [math]0=a0[/math] (by proposition 1) [math]0=0a[/math] (by commutativity under multiplication) Proposition 3 [math](-a)b=-(ab)[/math] Proof [math]a+(-a)=0[/math] (by the additive inverse property) [math](a+(-a))b=0b[/math] (multiply both sides by [math]b[/math]) [math]ab+(-a)b=0b[/math] (by the right distributive property) [math]ab+(-a)b=0[/math] (by proposition 2) [math](-a)b=(-ab)[/math] (by the additive inverse property) Proposition 4 [math](-a)(-b)=ab[/math] Proof [math]a+(-a)=0[/math] (by the additive inverse property) [math](a+(-a))(-b)=0(-b)[/math] (multiply both sides by [math](-b)[/math]) [math]a(-b)+(-a)(-b)=0(-b)[/math] (by the right distributive property) [math]a(-b)+(-a)(-b)=0[/math] (by proposition 2) [math]-ab+(-a)(-b)=0[/math] (by proposition 3) [math](-a)(-b)=ab[/math] (by the additive inverse property) Note: At the end of the proofs for Propositions 3 and 4, I relied on the fact that the additive structure of a field is a group, and that every element in a group has a unique inverse element.

I don't know about anyone else, but speaking for myself I cannot figure out what you're trying to say. Let's say that [math]F(s)[/math] is the transfer function. Are you trying to say that [math]F(s)[/math] has poles at [math]f_{p1}[/math] and [math]f_{p2}[/math]? If so then why not just call the poles [math]p_1[/math] and [math]p_2[/math]? And if not then what do you mean? And where do the numbers "3dB and "6%" come from?

Here's another puzzle for you, along the same lines. Say you have a pure electric field in one inertial frame. Can you perform a Lorentz transformation to some other inertial frame in which the field is purely magnetic? Hint: You can get the answer fairly quickly by considering the field strength tensor in Post #2.

Just scream the whole thing at the top of your lungs, like so: THERE ONCE WAS A MAN FROM NANTUCKET...

Off the top of my head the only one I can think of are tangent lines to a parabola. Specifically, the tangent line to a parabola at point P makes equal angles with: 1. the line passing through P and the focus, and 2. the axis of the parabola. This is a theorem of course, so it requires a proof. But if you're learning this stuff in the sequence algebra/geometry-->trigonometry-->calculus, I suppose you could just define the tangent line to a parabola at P to be the line that satisfies those properties, and later show that it is identical to the tangent line from calculus. You mean perpendicluar lines, right? The reason I say to prove it is because anyone who has studied mathematics through trigonometry can prove it. So why not do it?

I've got a few suggestions for that document. You've got a typo with the y-coordinates there. You shouldn't use the symbol [math]n[/math] in both the index and the upper limit of the index. Let the index be something else, such as [math]j[/math]. Then we have the following. [math]\sum_{j=1}^nar^{j-1}=\frac{a(r^n-1)}{r-1}[/math] I think you meant 359 degrees. But even so, why stop at 359 degrees? 359.5 degrees is also in Quad. 4, no? I think it would be better to express the angular range as either an inequality or an interval. 1st Quad: (0,90) 2nd Quad: (90,180) 3rd Quad: (180,270) 4th Quad: (270,360) That one's wrong. You got it right two lines down from there though. Sure, but then what? I think it would be useful here to say: Second derivative test fails. Resort to first derivative test. Only if [math]f(x) \geq 0[/math] for all x in the interval of integration. Otherwise the integral doesn't give the area between f and the x-axis. Continuing... Never seen that notation before. Did you perhaps mean this? [math]Area=[ F(x) ]_{a}^{b}[/math], where [math]F^{\prime}(x)=f(x)[/math]. Continuing... You're subtracting something from itself. That would be zero, no matter what f is.

"Nice shoes, wanna f**k?"

Since you like those two subjects you may enjoy studying the synthesis of the two: analytic geometry. Here's an outline of what I think would be a great problem for you to work on. Start with the equation of a circle in general form: [math]Ax^2+Bx+Cy^2+Dy+E=0[/math] Problem: Find the line tangent to the circle at some arbitrary point [math](x_1,y_1)[/math] on the circle. Proceed as follows. 1.) Put the equation into standard form: [math](x-h)^2+(y-k)^2=r^2[/math] (Research "completing the square" for this). 2.) Find the equation of the line joining the center (which is the point [math](h,k)[/math]) and the point [math](x_1,y_1)[/math]. 3.) Show that perpendicular lines have negative reciprocal slopes (requires trig). 4.) Find the equation of the line that is perpendicular to the line found in step 2 and that passes through [math](x_1,y_1)[/math]. You will at this point have found the line that is tangent to the circle at [math](x_1,y_1)[/math]. This problem is interesting because it is one of the few tangent line problems that can be solved without calculus.

Hey, I don't know who YT was talking about stuffing but it had better NOT be me! :-O

What?? You mean that wasn't a joke? Try to see it from my point of view. You're asking this question about formaldehyde--you know, that stuff they use to preserve dead people. Then you announce that your mother in law is coming. Come on, you don't see the obvious joke there?

I am totally laughing my ass off.

Did you mean for the reputation points to be private? If so then I should probably tell you that the reputation points of any member are visible in their profile.

In that case, no wavefunctions do not ever satisfy the equations of classical physics. However the expectation values of operators can satisfy them (more on this later). Here's what does happen. You've got a wavefunction that is a normalized superposition of the eigenstates of the observable you're measuring. Let's say we're measuring some generic observable with operator [math]\hat{A}[/math] and let's say that the eigenstates of [math]\hat{A}[/math] are denoted by [math]\phi_n(\vec{x},t)[/math] with eigenvalue [math]a_n[/math]. So we have the following eigenvalue equation. [math]\hat{A}\phi_n(\vec{x},t)=a_n\phi_n(\vec{x},t)[/math] And we can write down the wavefunction [math]\psi(\vec{x},t)[/math] of our system as a linear combination of the basis eigenstates as follows. [math]\psi(\vec{x},t)=\sum_{n=1}^{\infty}c_n\phi_n(\vec{x},t)[/math], where [math]\sum_{n=0}^{\infty}|c_n|^2=1[/math]. Now it's important to remember that not only is [math]\psi(\vec{x},t)[/math] a quantum state (that is, it satsifies the Schrodinger equation), but so are all of the [math]\phi_n(\vec{x},t)[/math]. When you make a measurement of observable [math]\hat{A}[/math], all you do is force the system into one of the eigenstates. Let's say the system jumps into the [math]k^{th}[/math] one. So instead of finding our system in the above linear combination, we find that our system jumps: [math]\psi(\vec{x},t)\longrightarrow\phi_k(\vec{x},t)[/math], and the probability for doing so is [math]|c_k|^2[/math]. So the system is in a pure eigenstate of [math]\hat{A}[/math]. But it's still a quantum state.

A blonde is told by her doctor that she's pregnant. She promptly asks, "Is it mine?" ba-dum-bump.

I'm a moderator at Physics Forums. I also post at sciforums. Only I don't visit there in that "ooh let's see what new and interesting things those chaps are discussing" sort of way, but rather in that "let's satisfy my morbid curiosity" sort of way.

Oh and by the way: I like your "Speculations" section. Kudos on improving the way you handle cranks. It's nice to not have to look at that stuff in the regular physics section.

Wow, I must have been away too long because I had no idea that this went down. I just tracked down Matt's final posts to see what the hubbub was about. Heh. I know everyone has a breaking point, but I thought for sure that Matt's was at least another couple dozen "0.999...=1" threads in the future.

I don't know what you mean by "classical state". To me that suggests that the motion of a particle satisfies Newtonian/Maxwellian equations. If that's what you mean than no, wavefunctions cannot collapse to such states, even in principle. What they collapse to is an eigenstate of the observable being measured. Pure eigenstates of QM operators are still quantum states.

You can do it like this. [math]a^mb^m=(a \cdot a \cdot\cdot\cdot a)(b \cdot b \cdot\cdot\cdot b)[/math], where each of the parentheses contains m factors. You can then use the commutative and associative properties of multiplication to rearrange the various factors of a and b as follows. [math](ab)\cdot(ab)\cdot\cdot\cdot(ab)[/math] Now you have m factors of (ab). But this is precisely (ab)m. So we arrive at the rule: [math]a^mb^m=(ab)^m[/math] So back to your example... [math]9^5 \cdot 8^5=(9 \cdot 8)^5=72^5[/math]