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#102 Re: Help Me ! » Poisson equation » 2006-10-06 03:58:38

I guess you are working in Cartesian coordinates?

Then

But I'm not exactly sure what you mean by 'how can I do it in a square?'.


In the meantime, until someone here with expertise on this topic posts, you could also post this on another forum where advanced things are routinely answered:

http://www.chatarea.com/Mathematics

#103 Re: Help Me ! » multiplying and dividing fractions » 2006-10-06 03:46:54

Well you're right some of it is confusing at first, but one can get used to it very quickly.. for example for fractions:

the "\frac" means 'make a fraction' and you give it two parameters, the numerator and denominator, and latex makes a fraction out of them. The notation is a little odd but logical!

#104 Re: Help Me ! » multiplying and dividing fractions » 2006-10-06 03:39:11

Well at least the basics, I know it is actually a very large language.

#105 Re: Help Me ! » multiplying and dividing fractions » 2006-10-06 03:37:26

Hi Devanté, Latex is very easy, I learned it from Dross' post in 15 minutes. big_smile

#106 Re: Help Me ! » multiplying and dividing fractions » 2006-10-06 03:29:21

(a)

The larger fraction bars act as brackets -- you must carry out the operations on the top first.

Convert to improper fractions.

Multiply the numerators and denominators.


divide 630 and 12 by 6:

#107 Re: Help Me ! » the nasty old chain rule » 2006-10-05 13:43:37

Indeed! Although, wasn't the prime notation due to Lagrange?

And I see your point about the correct interpretation.

#108 Re: Help Me ! » the nasty old chain rule » 2006-10-05 13:41:09

It's "\partial" I think .
Here is a test :

#109 Re: This is Cool » I disagree with » 2006-10-05 13:37:20

George,Y wrote:

The poor of induction
Russel once said this story
a hen saw her master giving her food one day.
she saw the same thing the 2nd day, the 3rd day...
after many many days, she concluded an ultimate truth based on induction
- he will always give me food.
the next day, she was killed by her master.

That's hilarious!
Good thing the natural numbers don't have such killer abilities!

big_smile

Seriously though, this thread is fascinating.

#110 Re: Help Me ! » the nasty old chain rule » 2006-10-05 13:20:11

Oh, I see.. so it's the derivative of v(t) with respect to the 'variable' p(t), and p(t) itself depends on t.

I guess in the prime notation this would be "v'(t) " where the prime is understood to mean "differentiate with respect to whatever v depends on", in this case p(t) ... I think the prime notation is limited in such cases, since for example say i want to write   d/ds f(t) = 0. The prime notation doesn't have a way to indicate partial derivatives.

As for the physical significance... I guess a(p(t)) means acceleration is dependent on a position function which is dependent on time?

#111 Re: Help Me ! » the nasty old chain rule » 2006-10-05 13:03:47

So the statement to prove would be:

If:

Then:

Which would be the fundamental 'from first principles' proof of the chain rule. smile

#112 Re: Help Me ! » the nasty old chain rule » 2006-10-05 12:49:24

I think a truly fundamental understanding of the chain rule would require writing it out using the limit definition, and then writing the definition of that limit using the delta-epsilon definition of limit.

Which should be equal to:


Now, how to prove that with delta-epsilon is beyond me, I always hated those proofs, but I think that would give a sufficiently deep understanding.

#113 Re: Help Me ! » convert » 2006-10-05 08:54:20

diagramzb3.jpg

Look at the diagram (yeah I know its kindof ugly, lol):

We want the angle between the x-axis and the line from the origin (0, 0) to the point (-4.51, 3.7).

We have found that

by using the inverse tangent.

BUT, this is not the angle we want !!

The inverse tangent on the calculator gives you only the *smallest angle* which has a tangent of 3.7/-4.51.

This is because *many angles* can have the same tangent -- this is because the trig functions are periodic, that means their values repeat!

So, the calculator is not giving you exactly the angle you need, just an angle with the same tangent.

But this answer is very important anyway, since it lets us figure out what the actual angle is.

Looking at the diagram, we see that the angle we want will be pi/2 + something, which I've marked as "???".

We don't know what this is yet, but we do know that the angle from the negative x-axis to the line is 0.6871, because we know the angle on the other side, from the inverse tangent.

And we know that the unknown angle ("??") and 0.6871 must add to pi/2.

Therefore, this unknown angle is pi/2 - 0.6871

So we almost have the angle we want.. now we just have to see that the total angle from the x-axis all the way to the line to our point (-4.51, 3.7) is:

pi/2 + the unknown angle = pi/2 + (pi/2 - 0.6871) = pi - 0.6871 = 2.4545 (approximately)

All the measurements are in Radians of course.

#114 Re: Help Me ! » Helpppppppppppppppp with decimals » 2006-10-05 08:12:58

@ cambridge:

as espeon said, just remember to *align the decimal points* for addition and subtraction, make sure that you have them in the same spot in the top and bottom number.
Also remember that you can write as many 0s as you want after the final digit behind the decimal point: for example 1.2 = 1.20 = 1.200 = 1.2000 ...

so say you need to add 1.23 and 2.23432

write it like this for the addition:

2.23432
1.23000
---------

so that the decimals align!

#115 Re: Help Me ! » convert » 2006-10-05 07:05:02

Oops the angle should be (pi - 0.6871) = 2.45454

That's what I get for not drawing a diagram!

#116 Re: Help Me ! » lim » 2006-10-05 07:02:57

sin and lim , not lin big_smile

lim means limit.. for example a number that some sequence is approaching.

What does this approach?
0.9, 0.99, 0.9999, 0.9999999, 0.99999999999, 0.999999999999999999, 0.9999999999999999999 ...

#117 Re: Help Me ! » convert » 2006-10-05 06:43:35

Yes that's right.

To see what the polar angle is, it's best to draw a diagram with a line from (0, 0) to the original point (-4.51, 3.7), and you can see that you need to add pi/2 to 0.6871 to reach the point. smile

#118 Re: Help Me ! » convert » 2006-10-05 06:21:55

The order of operations of something like this is *innermost operation first*, inside brackets.

So, we first divide, and then take the inverse tangent of the result:

(But remember in which quadrant the point is!)

#119 Re: Help Me ! » convert » 2006-10-05 05:19:33

You have the correct magnitude (r), but for the angle, use:

But keep in mind in which quadrant the point lies!

Also, are you supposed to express the result using the polar coordinate unit vectors?

#120 Re: Help Me ! » lim » 2006-10-05 01:58:47

Is n a real constant? Or a natural number?
Also, how did you get 6x George?

#121 Re: Help Me ! » i need help with this proof » 2006-10-04 11:50:34

The theorem that we need here is this:

If W is a set of one or more vectors from a vector space V, then W is a subspace of V iff:

Here we need to show that:

and:


For the first one:

Using the definitions of V and W:

(where s, t, x, y, are real numbers)

This addition simplifies to merely:

Now to show:

We let

(By the properties of real numbers)

Therefore:

So the first part is complete.

Now must show that:


which is pretty much the same process, except the truth of that depends on the fact that k(x, y) = (kx, ky) and by the properties of real numbers, (kx, ky) is an element of (V + W).

I hope I'm going about this the right way, this is sort of simple yet confusing.

#122 Re: Help Me ! » Past exam paper!!!! » 2006-10-04 03:45:44

Well I'm not sure, it seems to me that the constants c are arbitrary, meaning that whatever we set them to, we will get an identical curve, but shifted up or down by the amount c.

For example the equation  f = cr - b  defines a "family of lines", and unless you are given some specific points that a specific member of this family must cross, it is not possible to solve for a specific c.

If we knew for example that f must pass through (1, 2) and (3, 4),  then
c - b = 2
4 = 3c - b

and we could solve these to get theconstants. But we don't know any specific points at all here, so I don't think we can solve for specifc constants.
Unless I'm not understanding the question correctly, in which case hopefully someone else can help. smile

#123 Re: Help Me ! » equidistant point » 2006-10-04 03:38:07

Haha sorry it's already done ! oops smile

#124 Re: Help Me ! » equidistant point » 2006-10-04 03:37:05

We need a point, call it P, which is on the x-axis, and therefore of the form P = (x, 0)

The distance formula between two points is:

We need the distance between (10, 4) and P to be the same as the distance between (6, 2) and P.

Ie, we need:

to be equal to:

Setting these to be equal:

Square both sides:

Expanding:

Re-arranging:

So the point we need is:

#125 Re: Help Me ! » triangle » 2006-10-04 03:14:09

Alright, so in that case I think the solution should be

and not

smile

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