People with some degree, or in the process of getting one, involving math: What were you like before, why did you choose math and was it what you expected?
(Feel free to write more)
I've been stumped on this exercise for a while now..
[IMG]http://i.imgur.com/USLB3Vx.png[/IMG]
I make all of the values substitutes:
[IMG]http://latex.codecogs.com/gif.latex?z = x+yi[/IMG]
[IMG]http://latex.codecogs.com/gif.latex?a = 1-i[/IMG]
[IMG]http://latex.codecogs.com/gif.latex?w = -1+2i[/IMG]
So that it can be rewritten as:
[IMG]http://latex.codecogs.com/gif.latex?Arg\left ( \frac{z-w}{a-w} \right )=0[/IMG]
And this is where it gets annoying, because when the stuff inside the brackets is simplified, it becomes:
[IMG]http://latex.codecogs.com/gif.latex?Arg\left ( \left ( \frac{2x-3y+8}{13}\right ) +\left (\frac{3x+2y-1}{13}\right )i\right )=0[/IMG]
And the LHS is
[IMG]http://latex.codecogs.com/gif.latex?tan(\theta)=\left ( \frac{3x+2y-1}{2x-3y+8}\right )[/IMG]
[IMG]http://latex.codecogs.com/gif.latex?\theta=arctan\left ( \frac{3x+2y-1}{2x-3y+8}\right )[/IMG]
This is where I am stuck.
I've tried doing it with one of those identities..
[IMG]http://latex.codecogs.com/gif.latex?Arg(\frac{z}{w})=Arg(z)-Arg(w)[/IMG]
Still end up just as stuck.
Any pointers?
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[editline]attempt[/editline]
This was an attempted solution.
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[IMG]http://latex.codecogs.com/gif.latex?LHS \Leftrightarrow \theta=arctan\left ( \frac{3x+2y-1}{2x-3y+8} \right )[/IMG]
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[IMG]http://latex.codecogs.com/gif.latex?RHS = 0[/IMG]
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[IMG]http://latex.codecogs.com/gif.latex?\theta = 0[/IMG]
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[IMG]http://latex.codecogs.com/gif.latex?0=arctan\left ( \frac{3x+2y-1}{2x-3y+8} \right )[/IMG]
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[IMG]http://latex.codecogs.com/gif.latex?tan(0)=\left ( \frac{3x+2y-1}{2x-3y+8} \right )[/IMG]
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[IMG]http://latex.codecogs.com/gif.latex?0=\left ( \frac{3x+2y-1}{2x-3y+8} \right )[/IMG]
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[IMG]http://latex.codecogs.com/gif.latex?0=3x+2y-1[/IMG]
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[IMG]http://latex.codecogs.com/gif.latex?y=-\frac{3}{2}x+\frac{1}{2}[/IMG]
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Does that mean all values on the line?
This topology problem has me working with a collection of collections of sets.
INSETPTION
[QUOTE=JohanGS;40137963]People with some degree, or in the process of getting one, involving math: What were you like before, why did you choose math and was it what you expected?
(Feel free to write more)[/QUOTE]
I chose maths partly just because I was/am pretty good at it, but also because I really like puzzling over a good problem (and even better, finding the solution)... University level maths, at least on the pure side, is massively more proof-based and abstract than at school, which suits me much more. In school I remember most problems being of the form 'Given this problem, calculate x', whereas most of my problems now are something like 'Given this problem, prove that x is true'. I'm certainly happy with my decision to do maths...
[QUOTE=Bradyns;40138770]I've been stumped on this exercise for a while now..
[IMG]http://i.imgur.com/USLB3Vx.png[/IMG]
I make all of the values substitutes:
[IMG]http://latex.codecogs.com/gif.latex?z = x+yi[/IMG]
[IMG]http://latex.codecogs.com/gif.latex?a = 1-i[/IMG]
[IMG]http://latex.codecogs.com/gif.latex?w = -1+2i[/IMG]
So that it can be rewritten as:
[IMG]http://latex.codecogs.com/gif.latex?Arg\left ( \frac{z-w}{a-w} \right )=0[/IMG]
[/QUOTE]
So far so good...
[QUOTE=Bradyns;40138770]
I've tried doing it with one of those identities..
[IMG]http://latex.codecogs.com/gif.latex?Arg(\frac{z}{w})=Arg(z)-Arg(w)[/IMG]
Still end up just as stuck.[/QUOTE]
So when you use that, you end up with
[IMG]http://latex.codecogs.com/gif.latex?Arg\left ({z-w} \right )=Arg\left ({a-w}\right )[/IMG]
When you subtract w from everything like that, you are in some sense shifting the origin to be at w rather than 0. Then just use the geometric interpretation of Argument - namely the angle around the origin. So what you deduce is
'The angle of z around w' = 'The angle of a around w'
In other words, you get all the lines on a [i]ray[/i] coming out from w through a. (Things get a little tricky for z=w since arg(0) isn't really defined)
That feel when you spend a bunch of time on a really involved proof and you're almost done and then you spot a completely different way to prove the statement that only takes a few lines.
[editline]3rd April 2013[/editline]
Tried to prove that a compact subset of a metric space contains a pair of points such that the distance between them is greater than or equal to the distance between any other pair of points in the space and I started doing it with open covers and showing that the subspace is bounded and then trying to figure out how to prove it from there using the sequential compactness of compact metric spaces when I noticed that the result follows almost immediately from the extreme value theorem. brb killing self
[QUOTE=Joey90;40142723]
So far so good...
So when you use that, you end up with
[IMG]http://latex.codecogs.com/gif.latex?Arg\left ({z-w} \right )=Arg\left ({a-w}\right )[/IMG]
When you subtract w from everything like that, you are in some sense shifting the origin to be at w rather than 0. Then just use the geometric interpretation of Argument - namely the angle around the origin. So what you deduce is
'The angle of z around w' = 'The angle of a around w'
In other words, you get all the lines on a [I]ray[/I] coming out from w through a. (Things get a little tricky for z=w since arg(0) isn't really defined)[/QUOTE]
I worked out the general form:
It's a straight line passing through the points w and a.
[IMG]http://latex.codecogs.com/gif.latex?y = mx+b[/IMG]
[IMG]http://latex.codecogs.com/gif.latex?w=(-1, 2)[/IMG]
[IMG]http://latex.codecogs.com/gif.latex?a(1, -1)[/IMG]
[IMG]http://latex.codecogs.com/gif.latex?m=\left ( \frac{2 - (-1)}{-1 - 1} \right )= -\frac{3}{2}[/IMG]
Subbing (1, -1) in.
[IMG]http://latex.codecogs.com/gif.latex?(-1) = \left ( -\frac{3}{2} \right )(1)+b[/IMG]
[IMG]http://latex.codecogs.com/gif.latex?b = \frac{1}{2}[/IMG]
[IMG]http://latex.codecogs.com/gif.latex?y = -\frac{3}{2}x +\frac{1}{2}[/IMG]
Cheers for the help ^_^
[QUOTE=Bradyns;40151691]I worked out the general form:
It's a straight line passing through the points w and a.
[IMG]http://latex.codecogs.com/gif.latex?y = -\frac{3}{2}x +\frac{1}{2}[/IMG]
Cheers for the help ^_^[/QUOTE]
That's only half right... If you try subsituting that in, you get
[IMG]http://latex.codecogs.com/gif.latex?Arg\left ( \frac{\frac{1}{2} (2-3i)(x+1)}{(2-3i)} \right )[/IMG]
which leaves you with
[IMG]http://latex.codecogs.com/gif.latex?Arg\left ( \frac{1}{2} (x+1)}) \right )[/IMG]
The thing inside the argument is then a real number (a good start) but the argument of a real number is only zero if it's [i]positive[/i]... If x=-2 for example, you get Arg(-1/2) = pi which is no good!
The problem is, when you're taking complex numbers z which make the same angle as some other number a (around w), z has to be in the same direction from w as a.
[img]https://dl.dropbox.com/u/4081470/ray.png[/img]
The solution is only the darker line to the bottom right, whereas you've included the pale line at the top left as well. (The equation for the line is right, but you need the extra condition that x>-1)
[QUOTE=Joey90;40137237]I don't know many ways to analyse things to do with composing a function with itself (pretty much only the contraction mapping theorem)... I'd usually interpret periodic to mean something like f(x+c) = f(x).
I had a look at the problem and couldn't really make much progress. It might be worth trying it for dyadic rationals first (ones of the form n/2[sup]k[/sup]) since it at least simplifies the problem of fractions in lowest form etc.
Certainly if there is such a function, it can't be differentiable (look at f(1/n!)) so I don't know if you can really talk about hyperbolic fixed points and stuff...[/QUOTE]
What are some necessary conditions for invertibility of a real valued function?
[QUOTE=Krinkels;40160518]What are some necessary conditions for invertibility of a real valued function?[/QUOTE]
The function being both injective and surjective is necessary and sufficient.
[QUOTE=Krinkels;40160518]What are some necessary conditions for invertibility of a real valued function?[/QUOTE]
Assuming you mean a [i]continuous[/i] real valued function... If it's defined on the whole of R I [i]think[/i] that it is both necessary and sufficient that the function is continuous (of course), unbounded above and below, and either strictly increasing or strictly decreasing.
Inverting a real valued function y = f(x) constitutes flipping the graph in the line x=y, so you need something that will still be the graph of a continuous function when you do that.
In other news, I am unreasonably happy with a diagram I made for my current essay/paper/thing, just makes me happy for some reason :v:
[img]https://dl.dropbox.com/u/4081470/3D%21.png[/img]
What's the paper on?
[QUOTE=JohnnyMo1;40162627]What's the paper on?[/QUOTE]
The title is 'Factorisation Systems'
One definition involves taking morphisms f in a category C and factorising each as f = m[sub]f[/sub] e[sub]f[/sub] in a functorial way (In other words a functor from C[sup]2[/sup] to C[sup]3[/sup]). I'm then looking at what nice properties you get given certain additional conditions.
I imagine I'll post a link when I'm done, deadline is the 3rd May.
[QUOTE=JohanGS;40137963]People with some degree, or in the process of getting one, involving math: What were you like before, why did you choose math and was it what you expected?
(Feel free to write more)[/QUOTE]
Nearly at the end of my first year of a straight maths degree.
Before: Found A-level maths (dunno where you're from guessing not England since you said math not maths, A-level = age 16-18 level), really easy, didn't like how whenever you asked "why is this true" the teachers just said "you'll find out if you do a maths degree", was expecting that at University there'd be less of that and more derivation.
Was originally going to do Physics, but then I realized 2 weeks before I submitted my application that I wasn't actually that interested in it, and preferred Maths. Re-did the application and personal statement I'd spent the last couple of months on and sent it off.
Now: Incredibly glad I didn't go for Physics. I've found that at University everything's much more formal and rigorous. Everything's clearly defined in strict definitions, there's no ambiguity (like there kind of was at A-level), and they do show you the derivation/proof of what they're teaching you, even if it's beyond the level you're at, in which case they say "we don't expect you to reproduce this, it's just for your own interest". There's a few things which they still do say "proof of this too long and complicated to spend time on, but you'll see it later", but a) it doesn't happen that often and b) they do often tell you roughly how the proof works, so if you really want you can easily look it up yourself.
Personally I'm still finding it really easy. You also find that with time stuff you find hard now will become really easy. For example I remember being taught proof by induction 2 years ago and it was the hardest thing on the exam, no-one in my class really understood it. After revisiting it at University it's now so easy I have no idea why I ever thought it was hard.
[QUOTE=Drakehawke;40163638]Personally I'm still finding it really easy. You also find that with time stuff you find hard now will become really easy. For example I remember being taught proof by induction 2 years ago and it was the hardest thing on the exam, no-one in my class really understood it. After revisiting it at University it's now so easy I have no idea why I ever thought it was hard.[/QUOTE]
I feel like math is kind of easy, at least comparative to physics. Probably because nature is ugly as hell.
Do you have to study much to get stuff, Johnny? How much?
[QUOTE=JohanGS;40165618]Do you have to study much to get stuff, Johnny? How much?[/QUOTE]
I do, but most of my "studying" either comes from doing homeworks, or just when I have a question so I read up on the math related to it, so I couldn't really tell you how much I study. I don't exactly have a set study time. :P
That's how I was, until this semester when I took nothing but straight math courses. Now, even doing the homework isn't enough; and I barely have time to finish all of my homework for each class... :(
Calc II, Analytic Geo, Linear Algebra, Math. Structures & Discrete Math., and Physics II all at once.... I'm an idiot...
[QUOTE=Drakehawke;40163638]Nearly at the end of my first year of a straight maths degree.
Before: Found A-level maths (dunno where you're from guessing not England since you said math not maths, A-level = age 16-18 level), really easy, didn't like how whenever you asked "why is this true" the teachers just said "you'll find out if you do a maths degree", was expecting that at University there'd be less of that and more derivation.
Was originally going to do Physics, but then I realized 2 weeks before I submitted my application that I wasn't actually that interested in it, and preferred Maths. Re-did the application and personal statement I'd spent the last couple of months on and sent it off.
Now: Incredibly glad I didn't go for Physics. I've found that at University everything's much more formal and rigorous. Everything's clearly defined in strict definitions, there's no ambiguity (like there kind of was at A-level), and they do show you the derivation/proof of what they're teaching you, even if it's beyond the level you're at, in which case they say "we don't expect you to reproduce this, it's just for your own interest". There's a few things which they still do say "proof of this too long and complicated to spend time on, but you'll see it later", but a) it doesn't happen that often and b) they do often tell you roughly how the proof works, so if you really want you can easily look it up yourself.
Personally I'm still finding it really easy. You also find that with time stuff you find hard now will become really easy. For example I remember being taught proof by induction 2 years ago and it was the hardest thing on the exam, no-one in my class really understood it. After revisiting it at University it's now so easy I have no idea why I ever thought it was hard.[/QUOTE]
I like this rigorous aspect of math, too. In prepa, there are small oral exams every week, in which 3 pupils do exercises on a blackboard in front of a professor. As long you read your lesson beforehand, I find it to be a very efficient way of learning new techniques and eventually mastering a lesson. It also teaches rigor in mathematics, as the professor can directly detect and correct logical fallacies or laxism. This can be a bit intimidating at first, but after a while I find it to be quite enjoyable.
I also study Physics, and yeah, it isn't really as "strict" as math, as you often have to do approximatioins in order to solve a problem. Personally I find it a bit easier than math (many math exercizes require a form of "trick" to solve them, whereas a given method often works with several types of physics problems, which makes it easier to study). Of course, it isn't as satisfying to solve as math problems, but it has its share of benefits, too.
[editline]5th April 2013[/editline]
Here's the exercice I got in the oral part today, for example:
Let f be a function continuous on [a,b]. Then may for all k integer from 0 to n, [integral from a to b]f(t)*t^k dt=0
Prove that f cancels itself n+1 times on [a,b].
just a quick question; what's the fastest and shortest way to invert a 5 by 5 matrix by hand?
Laplace expansion is nice and all but not really practical to do for anything above 3 x 3 in an exam without a computer or a calculator that does matrices in any sort of reasonable time
[editline]6th April 2013[/editline]
i mean trying to find the determinant is bad enough, but the cofactor matrix is just too far man
[QUOTE=Uber|nooB;40179319]just a quick question; what's the fastest and shortest way to invert a 5 by 5 matrix by hand?
Laplace expansion is nice and all but not really practical to do for anything above 3 x 3 in an exam without a computer or a calculator that does matrices in any sort of reasonable time
[editline]6th April 2013[/editline]
i mean trying to find the determinant is bad enough, but the cofactor matrix is just too far man[/QUOTE]
Short answer is that you pretty much never want to (and really shouldn't ever have to...)
The best method is probably [url=http://en.wikipedia.org/wiki/Gaussian_elimination#Finding_the_inverse_of_a_matrix]Gauss-Jordan Elimination[/url] - Basically, you take your matrix, stick an identity matrix on the side, then use row operations to reduce the original matrix to the identity.
If the matrix has special properties (if it is symmetric, orthogonal, hermitian, unitary etc.) there are usually other tricks that might help too.
Pretty much what Joey said. We never went over anything other than Gauss-Jordan by-hand inversion, since our proff thought that we would never really need to do it any other way. Well, I take that back, we briefly went over using Cramer's Rule, but we were never tested on that specific method on our exam.
[QUOTE=Joey90;40181528]Short answer is that you pretty much never want to (and really shouldn't ever have to...)[/QUOTE]
welp, shit, our aerodynamics lecturer disagrees
Gauss-Jordan elimination seems to be alright as long as you have somewhat round numbers
just found an example in the notes (not even 5x5 but 4x4)
[img]http://i.imgur.com/eHu3ImG.png[/img]
ffffffffffffffffffff
You have to do those by hand, without any type of calculating device? Not even an abacus? Who's your teacher, Hades himself?
we get to experience the magic of a calculator, but graphing calculators (which are also the ones capable of matrix inversion) are banned from all our exams
still don't get why we have to do it like that, since this is an exam on aerodynamics, not numerical methods or linear algebra; if we weren't doing it in an exam we'd just do it in MATLAB
I guess the mentality is like our calc classes for Maths majors. You're expected to understand the principle behind everything; not rely on a calculator to do everything. Just be glad you're capable of a 4-ftn calculator, it sounds like. We're not capable of them in any of my classes except Physics, either.
Stuck on this..
[QUOTE]
[IMG]http://i.imgur.com/mK4GTft.png[/IMG][/QUOTE]
Attempt at a graphical solution.
[QUOTE][IMG]http://i.imgur.com/0vjhXlT.jpg[/IMG][/QUOTE]
Algebraically I managed to whittle it down to:
[IMG]http://latex.codecogs.com/gif.latex?d = \left | Im\left ( e^{-i\phi}z \right ) \right |[/IMG]
[IMG]http://latex.codecogs.com/gif.latex?d = \left | sin(\theta)cos(\phi)-sin(\phi)cos(\theta) |[/IMG]
[IMG]http://latex.codecogs.com/gif.latex?d = \left | sin(\theta - \phi) |[/IMG]
[IMG]http://latex.codecogs.com/gif.latex?d = \sqrt{\left ( sin(\theta - \phi) \right )^2}[/IMG]
[IMG]http://latex.codecogs.com/gif.latex?d^2 = sin^2(\theta - \phi)[/IMG]
This is where I am stuck.. :/
It appears that in the picture I have drawn..
If L passes through (-1, 0), (0, i)... Where z = -1+0i, Arg(z) = 0 and |z| = 1
Then, phi is equal to (pi/4), thus,
[IMG]http://latex.codecogs.com/gif.latex?d^2 = sin^2(0 - \frac{\pi}{4})[/IMG]
[IMG]http://latex.codecogs.com/gif.latex?d = \frac{1}{\sqrt{2}}[/IMG]
I am just absolutely confused, because it holds when z = 0+i
[QUOTE=Bradyns;40192963]
[IMG]http://latex.codecogs.com/gif.latex?d = \left | Im\left ( e^{-i\phi}z \right ) \right |[/IMG]
[IMG]http://latex.codecogs.com/gif.latex?d = \left | sin(\theta)cos(\phi)-sin(\phi)cos(\theta) |[/IMG]
[/QUOTE]
I don't understand how you've got from the first line to the second... You want to write z = r e[sup]i theta[/sup], but you've lost the 'r', which means it only works for r=1 (hence why you still get the same answer for 1+0i)
However as the question says, it's much easier to do graphically. Translate the two operations into what they do geometrically:
'multiply by e[sup]-i phi[/sup]' becomes 'rotate by -phi around the origin'
'modulus of imaginary part' becomes 'distance from imaginary axis'
then think what happens if you apply these operations to (a point on) the line.
Who wants to do this?
"The length of a rectangle is 6m longer than the width. If the area of the rectangle is 91 cm squared, find the dimensions of the rectangle."
[QUOTE=UntouchedShadow;40202957]Who wants to do this?
"The length of a rectangle is 6m longer than the width. If the area of the rectangle is 91 cm squared, find the dimensions of the rectangle."[/QUOTE]
Drawing pictures can be helpful.
[QUOTE][IMG]http://i.imgur.com/e68vRvC.png[/IMG][/QUOTE]
Finding what we know, and what we need to know is also pretty helpful.
[IMG]http://latex.codecogs.com/gif.latex?Width = x[/IMG]
[IMG]http://latex.codecogs.com/gif.latex?Length = x+600cm[/IMG]
[IMG]http://latex.codecogs.com/gif.latex?Length \times Width = 91cm^2[/IMG]
[IMG]http://latex.codecogs.com/gif.latex?(x+600cm) \times (x) = 91cm^2[/IMG]
[IMG]http://latex.codecogs.com/gif.latex?x(x+600) = 91[/IMG]
[IMG]http://latex.codecogs.com/gif.latex?x^2 +600x = 91[/IMG]
[IMG]http://latex.codecogs.com/gif.latex?x^2 +600x - 91 = 0[/IMG]
Now we use the quadratic formula.
[IMG]http://latex.codecogs.com/gif.latex?x=\frac{-600\pm \sqrt{600^2 -4(1)(-91)}}{2(1)}[/IMG]
[IMG]http://latex.codecogs.com/gif.latex?x=\frac{-600\pm \sqrt{360000+364}}{2}[/IMG]
[IMG]http://latex.codecogs.com/gif.latex?x=\frac{-600\pm \sqrt{360364}}{2}[/IMG]
When you evaluate this, you get two solutions:
[IMG]http://latex.codecogs.com/gif.latex?x \approx 0.15163[/IMG]
[IMG]http://latex.codecogs.com/gif.latex?x \approx -600.15163[/IMG]
Given that we are working with length here, you would need to pick the positive value.
In exact form the solution would be:
[IMG]http://latex.codecogs.com/gif.latex?x=-300+\sqrt{90091}[/IMG]
Therefore the dimensions would be:
[IMG]http://latex.codecogs.com/gif.latex?Length =300+\sqrt{90091}[/IMG]
[IMG]http://latex.codecogs.com/gif.latex?Width =-300+\sqrt{90091}[/IMG]
You're a godsend, thank you so much.
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