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[Ahmad]

Notice that or .





We can transfer the into since it's an odd function.





So they're really the same thing after all. 

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Thanks ahmad, though I'm still not entirely convinced why we're allowed to transfer the sign(a) into the tan-1.

[Ahmad]

is an odd function, this means that . By definition is either 1 or -1, depending on the sign of a. If it's 1 then we really aren't doing anything at all, we're just multiplying something by 1, which does nothing. If it's -1 then we're using the property that is odd.

[/0]

Ah, ok thanks heaps

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In stewart calculus they define an open region D as one "for which for every point P in D there is a disk with center P that lies entirely in D."

The definition is used in the following proof

"Suppose is a vector field that is continuous on an open connected region D. If is independent of path on D, then is a conservative vector field on D; that is, there exists a function such that "

I don't really understand the significance of the definition of an 'open' region and how it relates to vector fields though... can someone plz explain?

[zzdfa]

http://en.wikipedia.org/wiki/Open_set

"a set U is open if any point x in U can be moved by a small amount in any direction and still be in the set U." so obvious examples of open sets are R^n, the empty set, the open interval on R^1... etc.

basically by requiring that D is open we can be sure that there is room to maneuver, no matter how close x is to the boundary.

to see why this is an important property, go through the proof given in the textbook. Find out where the argument breaks down if we omit the hypothesis that D is open.

[/0]

Thank zzdfa, I was thinking it might be like an open interval (a,b) but in 2D, but wasn't completely sure. The proof doesn't make much use of the fact, but I guess there could be an extreme case that invalidates it.

[zzdfa]

i just checked, (i think) it's a pretty important part of the proof. here we have a set D, thats not open, because the point c2 is on the boundary.


 

from c2, no matter how small a step you take, to the left or right, you will end up outside D. (ok i didnt draw it very well, imagine c2 is at the tip of a upside down parabola)


thus, following along with the proof, when you get to the part where you need to choose the destination for the first part of the journey, you won't be able to do so.

[/0]

Ah, I see, thanks again. I guess it's pretty important to the proof.

But somehow the proof seems kind of contrived, you know, taking a step before getting to C2. What if there is a better proof that doesn't require this step? Then it wouldn't require an open region.

And, come to think of it, why would it? While having an open region is essential to the given proof, would it really be essential in practice? In practice what you're doing is simply moving from C1 to C2 - you're not doing any sidesteps.

It all seems very meta-mathematical if ya know what i mean :/

[zzdfa]

i agree, i didnt find the proof very natural when i first saw it. but i think the reason why openness is essential is because

we are trying to find a function f such that grad f = F. lets pretend we have the situation we have before in my above post. that is, f is to be defined on the set D that contains a boundary point c2.

suppose we have such an f such that grad f = F. specifically, df/dx at c2 equals the  x component of F at c2.

but if you look at the picture, and imagine a function defined on D, you should see that df/dx is actually undefined. why? because df/dx means

'the change in f if we keep y constant and move along the x direction by a small amount' .


but there is no way to move along the x direction by a small amount because you'd step out of D. hence the derivative is undefined and if its undefined its obviously not equal to F.

it's like trying to find the derivative of a single isolated point. it doesn't make sense.

tldr, you are right, in practice, you would just restrict the domain to not include those boundary points, or to expand the domain to an open set (not always possible).

[/0]

When a question asks you to calculate the flux through a surface



For an oriented surface, how do you decide which order to do the cross product to get the right direction?

thx

[Ahmad]

The question should specify something like "outward flux" in which case you'll want the resulting cross product to point outwards. One way to do this is just to compute the cross product arbitrarily, then sub in a particular value of u and v (doesn't matter which, as long as it's on the surface) to obtain a normal vector. If the normal vector points outwards then you chose the correct order, otherwise just switch the order you did the cross product by slapping a negative sign in front of the result you obtained.

Of course, if you're able to visualise the directions of r_u and r_v then you can set it up beforehand so that the cross product points in the right direction (use the right hand rule).

[/0]

thx ahmad

To prove:



Just wondering, is it valid to imagine the vertical strips of as components of a vector, so that it becomes ? I'm not sure how solid that line of reasoning is...

[Ahmad]

It really feels like that doesn't it! Not a bad idea at all. To make things rigorous you'd probably want to write down your idea for Riemann sums then pass to the limit. But this isn't the best way to go about it, since this is a specific form of Cauchy Schwarz inequality applied to an appropriately chosen inner product space. Do you know the cute proof for general inner product spaces? If so can you think of an appropriate inner product space?

[/0]

It really feels like that doesn't it! Not a bad idea at all. To make things rigorous you'd probably want to write down your idea for Riemann sums then pass to the limit. But this isn't the best way to go about it, since this is a specific form of Cauchy Schwarz inequality applied to an appropriately chosen inner product space. Do you know the cute proof for general inner product spaces? If so can you think of an appropriate inner product space?

Oh cool, I didn't realise you could do that...
Good thing I did some inner product stuff before xD

Let

Testing the axioms:

Conjugacy:



Linearity:



Positive Definiteness: and if





is an inner product space!!

So then by cauchy-schwarz!


(Btw I'm using "Calculus on Manifolds" by Spivak now (done with Stewart), because from what I've heard online it's a good rigorous treatment of calculus. Have you seen the book before, do you think it's too advanced?)