Samadhi's school help thread

[Samadhi]

I'll ask questions about math every now and then, and I'll post my java assignments for review.

No math questions yet, but I do have a java assignment. I'd just like someone who is familiar with how anal professors can be to critique it given the posted requirements. To wit:
[instructions]
Write your 2nd Java program that reads in three whole numbers and outputs the average of the three numbers.

• You will use comments on top of your program. The comments will cover the following information:

1. Programmer’s name (your name in this case)
   
2. Program name.
   
3. Programming date.
   
4. Modification date if any.
   
5. Programming language used (Java).
   
6. Computer used to compile the program (PC? Mac? Or?)
   
7. The purpose of this program.
   
8. Any special requirements for the input?
   
9. What will be the output?
   
10. Variable names and their usages

[/instructions]
I think it's a little silly to put the comments for the variables at the top of the program but ~shrug~

[Talzor]

For all normal purposes your program looks fine, but heres a few suggestion for those really anal professors:

You could specify the range of the input (-2,147,483,648 to 2,147,483,647). In comparison the range of a long is -9,223,372,036,854,775,808 to 9,223,372,036,854,775,807.

nextInt() throws 3 exceptions:
InputMismatchException - if the next token does not match the Integer regular expression, or is out of range.
NoSuchElementException - if input is exhausted
IllegalStateException - if this scanner is closed.
You can setup a try - catch statement to handle these (read: printout a telling error message).

[Samadhi]

Error catching has been specifically avoided as of yet. I'd love to throw it in, but I'd probably get marked off for it.
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[Samadhi]

This is a beginner class BTW. I've never messed with java, but I know programming and this should be easy. But I've never messed with a programming professor.....
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[Chuck]

Java is supposed to be used to program video games. The program is incorrect unless you have to kill something to see the answer.

[Samadhi]

Give me your address there
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[Samadhi]

Math question 1:
This is from the chapter titled Matrices, Determinants and the Cross Product.

Find the intersection of the planes x + (y - 1) + z = 0 and -x + (y + 1) - z = 0.

Since this is in the above titled chapter I figured it should have something to do with that. Since the line needs to be perpindicular to the normals for both planes, I figured I'd take the cross product of the normals and use that to get the line, but it doesn't work.

Anyone know this one? (it should be in the form of x = at, y = bt, z = ct)
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[mith]

That should work (the cross product of the normals is parallel to the line, then find a point on the line). If you want to post your work here, I can look through it.

[Samadhi]

Ok, here's what I had...

[Samadhi]

I guess I should trust my instincts more.
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[The Ragin' South Asian]

unless your instinct is to not trust your instincts

[Samadhi]

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[mith]

That's not quite correct; t=0 would give (0, 0, 0), which is clearly not on either plane, for example.

[ogm*]

Just by inspection you can add the equations and get y=0. Subsitute that back in and you get x+z=1. So the answer would be x=t, y=0, z=1-t, I think.

[Samadhi]

[CzarJ]

You need two things to describe a line in Ranything: a position vector and a direction vector. Call the position vector r0, the direction vector a, and the line r(t), and you have r(t) = r0 + ta. This begs the question: What does that mean? Well, a direction vector is any vector parallel to the line. A position vector is any point on the line (technically any vector with its initial point at the origin and its terminal point on the line, of course, but why bother saying all that?). The former you have; the latter you're missing.

[Samadhi]

I see it now. And since a point on that line is (0, 0, 1) I would add the zero and the one to the tx and tz in the formula, but since it's a negative slope, then ogm is exactly correct.
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[CzarJ]

Yessir. I'm glad there's this way, though; I wouldn't want to ogm's way every time =P.

[Samadhi]

Here's one that's a symptom of rustiness

If x ₁ + kx ₂ = c and x ₁ + lx ₂ = d have the same solution set, show that the equations are identical.

It's rather obvious, but it doesn't discuss it in the book and I'm don't remember the methodology for this kind of thing at all.

I mean I can subtract the second from the first and get kx ₂ - lx ₂ = c - d......but I don't know what to do with it after that.
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[mith]

What happens when x ₂ is 0?

[Samadhi]

Ah c = d. thanks
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[Samadhi]

I'm missing it on this one. "Prove the identity without evaluating the determinants"

With this text, the vertical lines on the outside mean determinant of the system.

I'm guessing (1 + t) ( 1 - t) has something to do with it but I'm drawing a blank.
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[Antrax]

To get from left to right, you do three things:
R1 <- R1 - t*R2 (doesn't change determinant)
R1 <- R1 / (1-t^2) (multiplies determinant by 1-t^2
R2 <- R2 - t*R1 (doesn't change determinant)
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[Samadhi]

Cool, thanks. I figured there had to be row adds but just couldn't see it for whatever reason.
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[Samadhi]

So here's what I thought of mentioned in this thread.

When my teacher mentioned the cross product of vectors he said that it can only be used in R3. Well, that bugged me. I know that the cross happens to be the area of the parallelagram, but the way I really thought of it was a vector that is orthogonal to both inputs. And since in R3 any plane needs two vectors that are non scalar to describe it, the cross product is therefore the normal for the plane. In the books (so far) and by my teacher, using a matrix to calculate the cross has been described only as a pneumonic.

Anyway, I thought, "Ok, getting the normal in R3 uses a 3x3 matrix, would a 4x4 work for R4?" So, I spent a couple of minutes plugging in non parallel vectors and seeing that it worked and then, *ding*. Duh. Of course it works!

What's being done on the typical cross product (R3) is setting up a 3x3 matrix with ones for the top row, the scalar factors for the other two vectors in the rows below that, and then finding the dets for the top row. In any square matrix multiplying the determinants of one row by any other row of the matrix will always equal zero. And that's the definition of a dot product (vector multiply) and when the dot equals zero, the vectors are orthogonal. So, the cross is obviously the normal to the plane because it's vector multiple of any other row equals zero. Which means the dot product is zero and therefore the vectors are orthogonal. This translates to hyperplanes as well.

It wasn't easy explaining it to my prof. I started with the premise that n points with no 3 points colinear are necessary to describe a plane/hyperplane in Rn (n>1 obviously). This equates to n-1 vectors that can not be found by addition/subtraction of any of the other vectors in the set (my description of this is a bit weak). But you have to have the normal to describe any plane and this method desribes it. I'm thinking this has to be important since the differential in Rn where n >2 is a plane tangent to the point.

Anyway, it was fun getting that.
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[Samadhi]

Holy crap. By the same formulas that make U*(VxW) work for the volume of the parallelpiped, I think that works for the "volume" of a n-parrallelpiped too (using v1Xv2X....vn as the row matrix I said in the last post). I'll have to think about it in the morning (when sober).
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[MatthewV]

My calc 3 teacher briefly mentioned something of this sort in class...

[Bicho the Inhaler]

You may know this already (maybe that's what you meant in your post), but:

[Samadhi]

Well, yeah I did know that about parrallelpipeds, in fact when shown I thought it was the most beautiful thing I've seen in years. But I didn't know that that works for R4+. That's what I was wondering.

I hadn't taken the time to check the outcome of the formulas, but it holds. I love it.
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[Samadhi]

Given that you're in n space, and you have n vectors (non scalar multiples), I know how to calculate the angle between any two, but what about the angle between those two and another vector?
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[CzarJ]

I'm not sure I understand the question. To be fair, I spent about 45 minutes today proving the most mundane facts about vectors using the definition (for example, a*0=0 for any scalar a, and -u=(-1)*u), so I'm in the mode where I need everything painstakingly spelled out for me.

edit: my parentheses were in goofy places.

[Samadhi]

Seems like we're about the same level actually. I learned that about 2 weeks ago.

I'm having difficulty describing this because there really isn't any well defined terminology for anything above R3. People throw on "hyper" and think they're cool, but that doesn't do it justice.

Basically, I'm trying to say that just like in R3, if you have 3 non-planar vectors, (u, v, & w) if you take u & v or u & w or v & w you can find the area defined by that pair of vectors by multiplying one vector times the other vector times the sine of the angle between them. And you can find the angle by the inverse cosine of (the inner product of the vectors divided by the product of their lengths).

Moving onto the third vector, you can't use the angle between it and either vector, you need to use the angle between it and both to find the appropriate height of the described volume/hyperarea/whatever.

In R3, by a very sweet proof, you can use cross product to find that height, and by further observance the determinant of the 3x3 matrix to find the volume.

That doesn't translate well to R4 or higher.
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[Samadhi]

Well...it appears to work and all. I just can't prove it.
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[CzarJ]

Are you looking for the angle between the third vector and the plane defined by the first two?

[Samadhi]

Call it a surface since I'm going for R4+. But yes.
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[CzarJ]

You lost me. =P

In R3, two non-colinear vectors define a plane. To find the angle between the plane and that vector, take the cross product of the two vectors (or any two vectors) in the plane to get the normal, find that angle, and subract from pi/2.

To find the angle between a vector and the subspace defined by three other non-coplanar vectors in R4, there is an analogue of the cross-product which maps uses three vectors instead of two and produces a vector perpendicular to all three. As you might expect, it's:

[Samadhi]

See....I'm trying to prove exactly that. So, thanks for nothing.
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[Samadhi]

(seriously, read 25 and 26)
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[kevinatilusa]

As a random side note, there's a slick way of showing that "4 dimensional cross product" really is perpendicular to the vector in question:

If you think about expansion by minors along the first row you can get that

[CzarJ]

Ooh, that is slick. Does that transfer to other dimensions? I'd figure it myself, but I know next to nothing about determinants.