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  1. B

    Is this considered as a function?

    Oops! You win. "Relation." Apparently I forgot "relation."
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    Linear momentum conservation vs angular momentum conservation

    Somewhat unclear, but that sounds good. Maybe? Are you trying to say that the conservation of linear momentum implies the conservation of angular momentum? In that case, I could buy into that. It might be preferable to show this more rigorously by making arguments that rigid rotators are...
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    A challenging sum series

    Ah. Seems like you've got this one. I'm curious, could you post the solution when you get it? I'd be much obliged.
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    1/f(x) graph

    The easiest way to get this is to flip the x and y axes around. So, you could draw an imaginary line (like a dashed line) along the x-y line (45 degrees from the x-axis in quadrant 1 going through the origin). Then, take the lines under the dashed line, draw them on top like they'd be seen in...
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    Is this considered as a function?

    It's the "general math" forum! The answer should be general! ;-)
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    Is this considered as a function?

    I mean function like "mapping elements from a set A to a set B." (Rudin, 3rd ed.) You know? No bijectivity required.
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    Is this considered as a function?

    Right, true. Yes, ok. Fine, but functions don't by definition need to be one-to-one. That's just an injective ("one-to-one") function. Restricting the domain to pass the vertical line test doesn't need to happen to define it as a function...just a one-to-one function. I thought if...
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    Particle picture is misleading. True?

    Things also go faster than the speed of light. You know: from a contradiction, you can get anything. Best of luck.
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    Particle picture is misleading. True?

    "Hilbert space" if you please.
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    Underdamped response

    Right. The zero values don't help much. Pick a pair of peaked values.
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    A challenging sum series

    That's neat! The square root of the imaginary term can be rewritten as i=e^{i\tfrac{\pi}{2}} \sqrt{i}=e^{i\tfrac{\pi}{4}} \sqrt{i}=\cos\tfrac{\pi}{4}+i\sin\tfrac{\pi}{4} \sqrt{i}=\tfrac{\sqrt{2}}{2}\left(1+i\right)
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    De Broglie wavelength neutrons

    Time to celebrate.
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    Polar equation problems

    Wait. Hang on. No. dv=-(1/2)\cos 2\theta d\theta
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    Polar equation problems

    Yes, you're correct. But it is of no consequence because we don't need the differential of v.
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    Check my working for finding if vectors are parallel ?

    Ok, we can use the same idea then. See if you can find a number that gets from one line to the other. I wouldn't use a cross product, but if I did, then I'd need to keep the constants. I'll also hazard that posting sort of opens you up to questions about process and understanding. The...
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    Polar equation problems

    Try u=2\theta \Rightarrow du=2d\theta v=\sin 2\theta \Rightarrow dv=-2\cos 2\theta Then plug into your expression of uv-\int v du Don't forget to evaluate that first part with limits.
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    De Broglie wavelength neutrons

    No troubles. You've done quite well. Great picture. I believe you're a good student for being so patient yourself. Answer: We need to halve it. Then do what you were saying.
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    Particle picture is misleading. True?

    What? 1. Mathematics that solves the double slit was done quite a bit ago. 2. The wave/particle duality refers to particles having wave like properties and particle like properties (like momentum). 3. Who's "Neumaier"? 4. Why does a search for this guy return religious poetry? 5. What were...
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    Check my working for finding if vectors are parallel ?

    Let's simplify because this is straight forward without the parametrization. Define a vector between A and B (taking the components and subtracting them) and the same for C and D. Then take the coefficients of one vector and see if we can multiply all of them by one number to get the other...
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    De Broglie wavelength neutrons

    Excellent! That's the second step. We're missing the first one. Nearly there. Think about where the "angle between incoming neutrons and reflected beam" is.
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    Underdamped response

    Interesting problem. Nothing will ever change the frequency, right? Right. Take the value after going 2\pi and plug in the time that took. Since the \sin(\omega-\phi will be the same, we can write \rm{Value}\, \rm{beginning}\,=\rm{Value}\, \rm{end}\, 2\pi \exp[-\sigma t] Finding \kappa...
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    De Broglie wavelength neutrons

    Well done. Now think on that for a bit.
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    Is this considered as a function?

    For clarification, we're talking about "one to one" functions.
  24. B

    De Broglie wavelength neutrons

    Perform an internet search for "Bragg Diffraction."
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    Quazar beam

    Well, this is a bit here and there. If you can give a cleaner account, then maybe we can go further, but I'll leave it with a simple explanation: Yes, you're right in that nothing gets out of a black hole because 1) Nothing goes faster than the speed of light (it would violate causality) 2)...
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    De Broglie wavelength neutrons

    Right, real simple. Try drawing a picture and then revisit what they mean by "angle between incoming neutrons and reflected beam" and make sure you drew it correctly. You're almost there.
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    A challenging sum series

    I think I sort of see how to do this...Let's give it a go. I'll post more later if something else comes to me: First, there's that pesky arctangent in there. So, let's use the relation you have to get rid of it. That is, we need to propose two variables a and b that give us...
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    Linear attenuation coefficient.

    That's true, something like "exp[x]" is not linear, but think of the logarithmic plot. When we plot this function by taking a logarithmic plot, we see that we get a linear trend. That is, the more particles we send at our attenuator, the more are absorbed. This is characterized by a...
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    Trig question

    Miike012, don't think that your classes are holding you back. If you'd like to poke around in a library, then check out a calculus book, muddle through that for a bit, and then get a book called: Basic Complex Analysis by Marsden This book shows a lot of cool things about trig identities in...
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    B and H field for a long rod and a disk

    Usually we speak of magnetization as being inside of some material. That directly relates to the "H field" (or magnetic field inside a medium) by a factor \chi. The "B field" then follows directly from that using basic relations; however, can you provide more context? Why do you ask?
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