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#51 Re: Help Me ! » derivatives in multiple variables » 2008-09-26 01:21:23
Perhaps I'm using the wrong name, but I seem to remember, as one of the limit laws:
I'm just not sure if we can apply this as a general rule, particularly, in the context of vectors.
That's what you want to show, so plug it in then prove that the linear mapping you're given is the derivative by having the limit go to 0.
you mean plug in the linear mapping they presented me with in the first place and show it works, without deriving it?
well that would work, certainly. However the solution in the answer key of my textbook, obtains it from scratch by the same process I did. But it does what made me uncomfortable, taking the limit in the denominator, and not the numerator. Thats why I wondered if they were applying some sort of quotient rule for limits.
#52 Re: Help Me ! » derivatives in multiple variables » 2008-09-25 23:53:22
I'm not sure I follow you. As far as I can tell, i don't have a mapping L to put in the limit until I bring the h's in the denominator to 0.
Oh, and i was kind of wondering, can we use the quotient rule for limits with vectors?
#53 Re: Help Me ! » derivatives in multiple variables » 2008-09-25 09:28:25
I think I may have a weak understanding of limits thats causing me to struggle with derivative mappings as a whole.
Let f : R[sup]n[/sup] -> R, n ≥ 1, f(x) = ||x|| be the usual norm in R[sup]n[/sup]
prove that:
using simple algebra (particularly be multiplying above and below by the conjugate of the numerator), I can show:
now I can see the idea is that the denominator approaches 2||x|| but in order to complete the proof, we need to kill the h's in the denominator (by setting the equal to zero) and leave h in place in the numerator, giving us the form:
which is of the form:
L(h) + o(||h||), with L being a linear mapping. However, we took h to zero in the denominator, and left it in place in the numerator. This makes me uncomfortable and I'm not sure if this is even legal. I seem to remember a quotient rule for limits in real numbers, but i don't know if this is legal for dot products of vectors in R[sup]n[/sup]
#54 Re: Help Me ! » derivatives in multiple variables » 2008-09-24 10:33:13
That was awesome! And easy to understand!
My only problem is where you used the triangle inequality. Shouldn't there be a plus in between the norms on the right side, and not a minus?
#55 Re: Help Me ! » derivatives in multiple variables » 2008-09-24 02:06:17
I think i follow you...
You're providing various reasons why the proof doesn't make sense, if i'm not mistaken. (though I don't think I fully understand them all)
If you could show me, or link me too a correct version of this proof, that would be nice! 
#56 Help Me ! » derivatives in multiple variables » 2008-09-23 18:25:50
- mikau
- Replies: 12
I'm having trouble understanding this proof in my book. The book is written by my teacher and free to copy and distribute, so i just scanned the portion I'm stuck on.
The proof is to show that the linear map which we call the derivative is unique.
Here are the relevant portions from the book (sorry for the low quality):
if those pics don't look right, try these URLS:
http://i21.photobucket.com/albums/b299/mikau16/derivative0.jpg
http://i21.photobucket.com/albums/b299/mikau16/derivative2.jpg
What throws me is the part I put question marks around, when they suddenly plug v into the linear map, and claim that the difference between D[sub]a[/sub](f)(v) - L(v) = etc...
I have no idea where that observation came from! No idea whatsoever.
#57 Re: Maths Is Fun - Suggestions and Comments » Whats with LQ? » 2008-09-23 12:19:16
He's sad? 
I'm sorry LQ if I hurt your feelings, I just never understand what you're saying anymore.
#58 Re: Help Me ! » ants » 2008-09-23 10:07:38
Ooh, I guess 'how long' means 'how far'. I thought it meant 'for how much time'.
#59 Re: Help Me ! » ants » 2008-09-23 08:50:15
aren't you given the ants speed?
#60 Re: Maths Is Fun - Suggestions and Comments » Whats with LQ? » 2008-09-23 04:07:13
"Just ask and I will remove the whole topic."
I didn't start this topic to make fun of LQ. He's acting weird, and its just a little annoying! When's the last time he made a post related to math?
#61 Help Me ! » Limits problem in two variables » 2008-09-22 14:21:54
- mikau
- Replies: 2
let
if x ≠ 0 and y ≠ 0, and otherwise
prove that:
exists, but that the iterated limits:
and
do not exist.
I don't get it. I get 2 for all of them! 
Also, I notice this function defines f(x,y) = 0 if x or y = 0, but since we're dealing with limits approaching zero, don't we not really need to know what happens at exactly 0?
#62 Re: Help Me ! » Is this rubbish? » 2008-09-20 03:49:38
The text explicitly states in that solution, that sign(t) := 0 where t = 0, 1 where t > 1 and -1 where t < 0
I left out that remark because I figured we all knew that anyway.
Here is how i think the problem needs to be solved, replace y with sqrt(x^2 + y^2) in the original equation:
observe now that if
then
and so
or just
0 = 1*0
therefore, in the equation,
it is safe to assume sign(...) is not zero, for if it is, we could have replaced sign(...) with 1 where it originally appeared.
for brevity, let
So now, since we can assume sing(t) != 0,
so
his solution technically did the same thing, but only after he masked out the sign(y - a) term by squaring. Before that, the equation remains incorrect.
#63 Re: Help Me ! » Is this rubbish? » 2008-09-20 03:48:05
The text explicitly states in that solution, that sign(t) := 0 where t = 0, 1 where t > 1 and -1 where t < 0
I left out that remark because I figured we all knew that anyway.
Here is how i think the problem needs to be solved, replace y with sqrt(x^2 + y^2) in the original equation:
observe now that if
then
and so
or just
0 = 1*0
therefore, in the equation,
it is safe to assume sign(t) is not zero, for if it is, we could have replaced sign(t) with 1 in its original location.
for brevity, let
So now, since we can assume sing(t) != 0,
so
his solution technically did the same thing, but only after he masked out the sign(y - a) term by squaring. Before that, the equation remains incorrect.
#64 Re: Maths Is Fun - Suggestions and Comments » Whats with LQ? » 2008-09-19 11:32:12
lol...
no seriously, what is with the guy? He's freaking me out!
#65 Maths Is Fun - Suggestions and Comments » Whats with LQ? » 2008-09-19 06:49:45
- mikau
- Replies: 29
Can we get him to stop talking like a fortune cookie?
#66 Re: Help Me ! » Is this rubbish? » 2008-09-19 05:43:16
Sorry Mikau, what I did above was wrong. The z-coordinates of P and P′ are the same, but their y-coordinates are not. I was posting in a hurry this morning and didnt check my equations.
Its okay, Jane! 
Thanks for taking the time to help! It appears my teacher made the same mistake you did.
When both you and my teacher said the same thing, I figured it had to be me, but now that you agree, I feel more confident that I'm right.
However, I did run into my teacher at school today, and tried to explain my gripe. I had little success communicating my reasoning, and he still insists he's correct.
I'm pretty sure he's wrong now... but what am i going to do... this might show up on an exam. 
#67 Re: Help Me ! » Is this rubbish? » 2008-09-19 01:47:21
what am I doing with the solution? Well the point of the problem is just to find a Cartesian equation of the torus, thats all. Nothing that follows is contingent on this, its just an example problem.
#68 Re: Coder's Corner » those who speak binary.. » 2008-09-19 01:28:22
but ONLY useful if we spread the word so everyone understands it!
Everybody, TELL YOUR FRIENDS!
#69 Re: Help Me ! » Is this rubbish? » 2008-09-18 23:51:16
If P lies on the torus, let P′ be the corresponding point on the torus in the yz-plane. That is, P′ is the point with coordinates (0,y,z). Then P and P′ have the same distance from the z-axis. Does this also make sense?
no it doesn't. I might be too tired right now, so forgive me if I'm way off, but that seems to be stating that
for any point on the torus, which clearly implies x = 0, and thus the entire torus is on the yz plane.
Lets see... let a = 3 and r = 1, you would agree that the point (0,a+r,0) = (0,4,0) is a point on the torus that lies on the yz axis, right? It even satisfies that equation.
But this is a torus, revolving around the z axis, therefore (4,0,0) is a point on the torus, no?
Plug x = 4, y = 0, z = 0, r = 1, a = 3 into the equation, and you get:
4^2 + 0^2 = (3 + sign(0-3)sqrt(1^2-0^2))^2
~ 16 = 4
(4,0,0) is a point on the torus, thus it should satisfy the equation, no?
What am I missing here?
#70 Help Me ! » Is this rubbish? » 2008-09-18 17:38:54
- mikau
- Replies: 11
here is something taken directly from my teachers lecture notes in my Vector Calc class. It didn't make any sense to me, and i won't get a chance to ask him about it until next week, so I was wondering if someone can verify that this is wrong:
The circle
on the yz plane (a,r are positive real numbers) is revolved around the z-axis, forming a torus T. Find the equation of this torus.
Solution: Let (x,y,z) be a point on T. If this point were on the yz plane, it would be on the circle, and the of the distance (notice the odd typo, i think it was supposed to read 'and the distance') to the axis of rotation would be
where sign(t) is the sign function. Anywhere else, the distance from x,y,z to the z axis is the distance of this point to the point
(that sentence doesn't seem syntactically correct)We must have:
STOP! the proof continues by applying more algebra, but that statement seems utterly senseless. How the devil can we assume that? if the point is no longer on the yz plane, how can we equate the distance from the z axis (the axis of rotation) to the point (x,y,z) with the distance from the z axis of the point (0,y,z), using an equation which we derived by assuming (0,y,z) was on that circle, (and thus a-r <= y <= a + r) ???
further, if I'm not mistaken, since the torus cuts through the points (0, a+r, 0) and (0,a-r,0) on the yz plane, and its being revolved around the z axis, then it should also cut through the points (a+r, 0,0) and (a-r,0,0) on the xz plane.
So now let a = 5, and r = 1, consider the point (a+r,0,0) = (6,0,0)
inserting these values into the equation above:
which reduces to
36 = 16.
Furthermore, the proof continues with the expression
by multiplying out on the right side, solving for sign(y-a) on side, and squaring again to obtain sign(y-a)^2 on one side. The book then claims:
"sign(y-a)^2 = 1, it could not be 0 (why?)"
...why indeed? is not the point (0, a,+-r) a part of this torus? At those points, sign(y - a) = 0, certainly.
Either this is absolute rubbish, or I'm missing something crucial here!
#71 Re: Coder's Corner » those who speak binary.. » 2008-09-18 16:44:55
Why not just keep it the same?
10010 is ten thousand and ten.
exactly as luca said! The whole point of this discussion is to escape the habitual base 10 mindset. If we start using base 10 names to describe them, that just defeats the point of this whole discussion. (in my oppinion? )
I agree with you, mathsisfun, I noticed that. I was just doing that so it sounded exactly like how we verbalize numbers in base 10. But yeah, thats easier and actually less confusing. So, better in every way.
LOL! I love your 'ary' theme idea, chewy! awesome! However, you need to be thinking of only names for the values 2^(3m) with m an integer greater or equal to 1.
another thing worth noting, is that, like i said, in base 10 we always use something between 1 and 999 before one of the big names, and here we're using something between 1 and 2^3-1 before each big name, (1 to 7)
groups of 7 or less are much easier to visualize in your head than groups of say, 788. I mean i can easily picture 7 cookies in my head 
but 788 is just hard.
#72 Re: Help Me ! » "Learn by Doing" - what's it mean and is it the only way? » 2008-09-16 12:41:38
well i took the time to read it, for your sake!
I feel I should point out that math was invented entirely through the ingenuity of people, which means every topic was originally conceived in someone's mind through logic/reasoning/intuition, not by having someone show it to them. Therefore it is possible to understand it without having someone who can 'interface with you', you can come to understand it just through reason alone, thats where math comes from.
As for this statement "So, from what I understand, a lot of lower level Math, in classes like CALC1,2,3, Linear Algebra, and Differential Equations, is all about algorithms."
if by 'algorithm', you mean like a monotonous routine, a step by step process to solve a problem, that claim is absolutely absurd. Many problems in all of those subjects can only be solved through ingenuity. You are given problems you have never seen before, and you have to use not only your knowledge, but also your wits. Let me tell you, many of the problems you will face in those courses are TOUGH and they give you no direct algorithm to solve them. This doesn't mean mean you can't solve it yourself, it means you actually have to think about it for a while.
The goal of these courses is to teach you not only how to solve common, and ordinary problems that have straightforward solutions, but also how to combine knowledge and experience, with your own ingenuity/creativity to find your way through problems you've never seen before.
Any mathematician that cannot solve a problem he's never seen before without help, would be absolutely worthless.
As for studying, if you want to take these courses in school, that works fine! You have a book, you have a teacher, you will have homework problems that are challenging and not just boring algorithms to follow. And yes, you will have them in Calc 1,2,3, Linear, and Differential Equations.
If you want to study these on your own, then buy a textbook, read the chapters, try to understand, and do the problems! The problems are designed to build up your skill, and let me tell you, they work! If you get stuck, checkout alternate explanations of the topic online/other textbooks, or ask questions here.
I actually studied math up to and including calculus, entirely on my own before college, just by reading books, and doing the problems. I believe a really advanced member here, Jane Fairfax, learns only through self study as well. Let me tell you from experience, it works, in fact I personally think it works better than formal schooling, but its important to have official degrees if you're hoping to find a job!
#73 Re: Help Me ! » try it....! » 2008-09-16 05:53:11
i THINK what you're doing is saying, let y = x^2 - 2^x and trying to minimize y so that they become equal. but.. minimizing doesn't find the pair closest to zero. For that, you'd need to minimize the absolute value of y, and test to see if its zero.
also, looks like you forgot that the derivative of log2(x) is not 1/x, its 1/(ln(2)x ).
#74 Help Me ! » Karnaugh Maps » 2008-09-15 15:32:21
- mikau
- Replies: 1
Karnaugh maps are not covered in my discrete math text, but my teacher is required to teach them in accordance with an agreement with the computer science department.
Further, my teacher stated that he hates Karnaugh maps, and only went over them briefly because he has to. And I only grasped so much of his lecture.
I can sort of see the logic behind it, but i don't understand the exact rules for how to connect the 1's in the map. Wikipedia says: "the groups may be 4 boxes in a line, 2 boxes high by 4 boxes long, 2 boxes by 2 boxes"
they don't seem to include single 1's in this description. What about if you have something like this:
1 0 1 0
0 1 0 1
is it illegal to connect (1, 0, 1) in the top row, or do you just need the number of 1's to be a power of 2? do the groupings have to consist entirely of 1's?
another question, the map that looks like this:
1 1 1 1
0 1 1 0
can be grouped in many ways. How do you know when you've got the one that most simplifies the resulting boolean expression?
Note thats three question.
#75 Coder's Corner » those who speak binary.. » 2008-09-13 10:15:25
- mikau
- Replies: 12
So the other night I was thinking, why is it so much easier for me to think in base 10 than in base 2? Is it just because I'm always translating base 10 so I feel more comfortable? If we rewrote history such that we always used based 2, all of us should be perfectly comfortable with seeing a number like 10010100 and instantly get a sense of the size of the number. So why don't I?
Is it just unfamiliarity? Or is there some other difference that makes it seem awkward?
The only answer I came up with is I know of no way to actually verbalize a number in binary.
In base 10 we have special names for the first 3 powers of 10 (one, ten, hundred) and then a new name for every 3rd power of 10 from then on (thousands, millions, billions, etc) which we combine with the first three terms to express every power of 10.
So if we'd like to verbalize a binary number, shouldn't we have special names for say, 1, 2, 4, and then special names for 2^3, 2^6, 2^9, ... etc? rather than calling them the '128's' column, which is describing the number in base 10?
So let me try giving intuitive names to these values:
1: one
2: pair
4: quad
and some arbitrarily selected names to the following:
8: spider
64: chess
512: hawk
I chose Hawk for Tony Hawk, world renowned skateboarder born on 5/12. (I'm open for better suggestions! )
Anyway, using the terms defined thus far, we can take any number between 0 and 2^10 - 1, (any 9 digit binary number) and read it! like we do with base 10.
examples:
1001
add commas after every third digit like we do in base 10:
1,001
we now have 'one spider, one.
or say,
111,001
can be read:
one quad pair one spider, one.
lastly, take the biggest number available:
111,111,111
this is:
one quad pair one hawk, one quad pair one spider, one quad pair one.
The only trouble I see, is that you need a lot more names just to cover every number between 1 and a million, than you do in base 10. In base ten we only need 2, in base two, we need 7, for the following values:
8
64
512
4096
32768
262144
2097152
Still, it can be fun to make up your own names for these numbers. I tried to use names that were in some way related, so they were easier to remember, but that pretty much stopped working after 64. But you could use some other sort of naming convention. You can also rename 1, 2, and 4.
What names would you give these numbers?