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 Ask a Mathematical Physicist

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v1n0d
post Mar 4 2014, 10:00 AM

Another roof, another proof.
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QUOTE(ninty @ Mar 4 2014, 09:57 AM)
Easiest way is to simply use the projective real line/extended reals. Or consider the undefined points as removable singularities.

Nothing wrong with flipping both sides. Some might frown on it, but there are some very good justifications for it.
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The frown is due to the origin of infinity. Computer application in mathematics has blurred the lines of what is and isn't acceptable when working with infinity.
v1n0d
post Mar 4 2014, 12:13 PM

Another roof, another proof.
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QUOTE(Critical_Fallacy @ Mar 4 2014, 12:03 PM)
In physicists’ terms, we call it Singularity. laugh.gif

user posted image
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Tumbling down this rabbit hole will lead us to discussing the existence of God himself. tongue.gif
v1n0d
post Mar 5 2014, 04:42 PM

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chocobo7779 what level are you sitting for?
v1n0d
post Mar 5 2014, 04:50 PM

Another roof, another proof.
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QUOTE(chocobo7779 @ Mar 5 2014, 04:42 PM)
STPM. blush.gif
Though I have completed it and I'm looking to hone my skills. brows.gif
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If you're looking to hone your skills, I suggest the following:
1. Be a little more proactive. Pick a topic, and work on it first. The forum should be your last resort, not the first place you reach for help.
2. Don't be afraid to make mistakes - it's our best tool for learning.

Last but not least, a reminder to all who ask for help in this thread - some of the methods here aren't exactly what your teacher/lecturer prefers. If you're sitting for an internal examination (college course etc.), do check if your lecturer accepts the given solutions (some lecturers can be fussy about these kind of things, so be wary).
v1n0d
post Mar 5 2014, 06:30 PM

Another roof, another proof.
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QUOTE(chocobo7779 @ Mar 5 2014, 04:57 PM)
OK.  Then how would I solve this though?
If y=mx+c is a tangent to the circle (x-a)^2 +(y-b)^2 =r^2, show that
(1+m^2)r^2=(c-b+ma)^2.

I substituted y=mx+c, but I have to deal with expansion of trinomials.  sweat.gif

Should I expand (y-b)^2 instead?
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Expansion of trinomials was part of your STPM syllabus, so it should be manageable, no? Just use Pascal's triangle!
v1n0d
post Mar 5 2014, 06:40 PM

Another roof, another proof.
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QUOTE(chocobo7779 @ Mar 5 2014, 06:35 PM)
? But I didn't work it out before... sad.gif
That would be (mx+c-b)^2...  and should I do it this way?
(mx+c)^2 -2(mx+c)(b)+b^2? sweat.gif
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Just multiply one by one, don't group them together. You should get user posted image.
v1n0d
post Mar 5 2014, 06:42 PM

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QUOTE(chocobo7779 @ Mar 5 2014, 06:41 PM)
A normal line passing through the centre?
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Tangent = intersects at only ONE point.
Please recall your coordinate geometry.
v1n0d
post Mar 5 2014, 06:47 PM

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QUOTE(chocobo7779 @ Mar 5 2014, 06:46 PM)
Trinomial expansion looks complicated for me... and I didn't study about that before. blush.gif
I just rewrite it as (a+b)^2 and expand from there. tongue.gif
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Trinomial expansion is covered just before Binomial expansion in the STPM syllabus. sweat.gif
v1n0d
post Mar 5 2014, 06:50 PM

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QUOTE(chocobo7779 @ Mar 5 2014, 06:48 PM)
But I'm taking the new (modular) syallabus... sweat.gif
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It hasn't been removed. It follows the same basic principles as regular expansion, just more elements. You'll be fine as long as you're careful with the multiplications.
v1n0d
post Mar 5 2014, 07:01 PM

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QUOTE(chocobo7779 @ Mar 5 2014, 06:53 PM)
Care to elaborate on this? hmm.gif
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I'll give you a simple example: Addition!
1+1=2, but what is 1+2?
1+2 is in fact 1+1+1!

Mathematics operates on extensibility, i.e. the ability to extend existing results to form newer ones. Just like the addition example above, your expansion is extensible, i.e. you proceed in the same fashion as you would regular binomial expansion, just extended to more variables.

QUOTE(Critical_Fallacy @ Mar 5 2014, 06:53 PM)
Not a good reason for skipping to learn new things. Here is a free expansion for you. sweat.gif

user posted image
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Listen to Critical_Fallacy. Being proactive is very important, be it in studies, or at the workplace.

On a separate note, Wolfram shows you step-by-step computational work (you can see the button in Critical's picture). If you have problems with understanding the expansion, play around with the variables and see what the results are.

QUOTE(Critical_Fallacy @ Mar 5 2014, 06:54 PM)
v1n0d is not arguing with you. You tend to drift from the problem you face. nod.gif
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Agreed.
v1n0d
post Mar 5 2014, 07:15 PM

Another roof, another proof.
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QUOTE(chocobo7779 @ Mar 5 2014, 07:05 PM)
That feature require WolframAlpha Pro... sad.gif
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My bad. It used to be free, looks like they've slapped a subscription cost on it now.
v1n0d
post Mar 5 2014, 08:54 PM

Another roof, another proof.
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QUOTE(chocobo7779 @ Mar 5 2014, 07:19 PM)
v1n0d Critical_Fallacy
I simplified it as
(1+m^2)x^2 +(2mc-2mb-2a)x+a^2 +b^2 -2bc-c^2 -r^2=0. 

What should I do next?  Using sum and product of roots?  But the constant part looks scary... sweat.gif
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No. Use the discriminant and the fact that y=mx+c is the tangent. This information was given in the question for a reason.

QUOTE(ninty @ Mar 5 2014, 07:53 PM)
I know for a fact that Cramer's Rule is not covered in the SPM syllabus. It was in the STPM syllabus, but I don't know the current situation.

I do see SPM candidates using it though, probably from their tutors. In fact I've never used it, always used Gaussian Elimination.
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You're right, it's not covered in the SPM syllabus. As I've mentioned earlier, using methods which aren't specifically mentioned in the syllabus isn't recommended for exams (it's fair game elsewhere). I know for a fact that many of the examiners these days are new, and tend to be very strict when it comes to following the answer scheme.

QUOTE(chocobo7779 @ Mar 5 2014, 08:33 PM)
Critical_Fallacy v1nod  Is it possible for me to make the solution in part (b) more elegant?  tongue.gif

https://www.dropbox.com/s/ms1vne8uuvahf3a/If%20secA.pdf
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I've seen too many students pressing on shortcuts and pretty answers that they end up making mistakes and/or not understanding what they're actually doing. When you become comfortable with your workings, that elegance you seek will come naturally. smile.gif
v1n0d
post Mar 5 2014, 09:15 PM

Another roof, another proof.
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QUOTE(chocobo7779 @ Mar 5 2014, 09:10 PM)
Wasn't using the discriminant would make it even more complicated? sweat.gif
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I've not tried computing it, but some of the terms should cancel out. wink.gif
v1n0d
post Mar 5 2014, 11:18 PM

Another roof, another proof.
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QUOTE(chocobo7779 @ Mar 5 2014, 09:47 PM)
The expression looks very complicated... sweat.gif  Especially this part:
(2mc-2mb-2a)^2 -4(1+m^2)(a^2 +b^2 -2bc+c^2 -r^2)=0
*
Looks like you're right.

We'll go back to using the sum and product of roots then.

Based on the quadratic equation you gave in Post #1074,
Sum of roots = 2alpha = -(2mc-2mb-2a)
Product of roots = alpha^2 = a^2 +b^2 -2bc-c^2 -r^2

Thus alpha = a+mb-mc and (a+mb+mc)^2 = a^2 +b^2 -2bc-c^2 -r^2. Does this reduce to the answer you seek?
v1n0d
post Mar 5 2014, 11:32 PM

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QUOTE(chocobo7779 @ Mar 5 2014, 11:28 PM)
Now, how would I factor it into
(c-b+ma)^2? hmm.gif

Also, wasn't the sum of roots is -b/a?  I think you left out 1+m^2 here... tongue.gif
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+1 for noticing that. I almost always reply to this thread in between games of DotA, so you'll have to excuse my tardiness. sweat.gif

If you can't get the final answer, look back to the quadratic equation to make sure there are no errors there that may have been carried on.
v1n0d
post Mar 6 2014, 02:05 PM

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Here's a good one, courtest of /k/.
user posted image

maximR you may find this one interesting.
v1n0d
post Mar 8 2014, 07:34 PM

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QUOTE(chocobo7779 @ Mar 8 2014, 06:10 PM)
Critical_Fallacy v1n0d Can anyone explain why:
user posted image

I know it has something to do with reference angles but I always get confused with that. rclxub.gif
Notes: angles are in degrees... tongue.gif
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It is because of the periodic nature of trigonometric functions, as kingkingyyk has demonstrated. Do you remember your acute/obtuse angles?
v1n0d
post Mar 9 2014, 04:46 PM

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QUOTE(ystiang @ Mar 9 2014, 02:38 PM)
Let me try this also:

» Click to show Spoiler - click again to hide... «

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Well done! rclxms.gif

QUOTE(Critical_Fallacy @ Mar 9 2014, 04:11 PM)
Have you ever wondered about a rip in the space-time continuum that create a Dimensional Vortex? sweat.gif

user posted image
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Never really gave much thought to it. I do however have something to ask you:

Suppose we have a manifold M immersed in space-time. What would the physical interpretation of the manifold's tangent space be?
v1n0d
post Mar 10 2014, 12:54 AM

Another roof, another proof.
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QUOTE(Critical_Fallacy @ Mar 9 2014, 11:04 PM)
There are few ways to define a tangent space. The cleanest and most intrinsic definition of a tangent vector at point p in a manifold M is as a point-derivation. If you ask me, the intuitive physical interpretation of the manifold's tangent space would be the velocity vector of a smooth curve at p in manifold M. Is Algebraic Topology readily accessible to the average math students in your university? happy.gif
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I think sometimes you forget that I'm doing my PhD in maths. sweat.gif

We teach general topology as an undergraduate course, and point set topology in the postgraduate classes.
v1n0d
post Mar 10 2014, 04:07 PM

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QUOTE(Critical_Fallacy @ Mar 10 2014, 03:38 PM)
Good! By the way, are you familiar with Pseudo-Spectral Method? It is a a very powerful joint theoretical-computational method for solving optimal control problems that involves partial differential equations. Even for one-dimensional problems, spectral methods will amaze you with their power and efficiency. icon_rolleyes.gif
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Not really. I rarely deal with numerical methods as I'm an information geometrist. I generally work with the differential-geometric properties of statistical manifolds.

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