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 Samadhi's school help thread Goto page 12345678910111213 Previous  1, 2, 3, ... 11, 12, 13  Next
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Samadhi
+1

 Posted: Sat Mar 11, 2006 4:38 am    Post subject: 41 kevin: the top row isn't really | i j k l|, that's just a pneumonic. It actually contains the determinants of the minors. As I said in post 25, a determinant row times any other row always equals zero, which means the represented vectors are orthogonal. This is valid for any N space._________________And he lived happily ever after. Except for the dieing at the end and the heartbreak in between.
Bicho the Inhaler
Daedalian Member

Posted: Sat Mar 11, 2006 4:56 am    Post subject: 42

Here's a slight (?) generalization of the above discussion that I came to recently:

In n-dimensional space, any m points (m <= n) p 1 , ..., p m determine an m-dimensional "hyper-parallelopiped" (whose 2^m vertices are all the possible a 1 p 1 + ... + a m p m where each a k is taken from {0,1}). We can ask ourselves: what is the m-dimensional volume V of this hyper-parallelopiped? I claim:

V 2 = [Sum over m-element subsets {k_1, ..., k_m} of {1, ..., n} of] det(P {k_1, ..., k_m} ) 2 ,

where P is the m x n matrix
 Code: [   p_1   ] [    .    ] [    .    ] [    .    ] [   p_m   ]
(here, we wrote each point p k as 1 x n row vector), and P {k_1, ..., k_m} is the m x m matrix obtained from P by selecting only columns k_1, ..., k_m.

Well-known special cases:
1. m = 1. The hyper-parallelopiped is in fact just the line segment from 0 to the point p 1 . The 1-element subsets of {1, ..., n} are just {1}, ..., {n}, and the "columns" of the matrix P are just the individual components of p 1 . The formula simply states that the square of the length of the segment is the sum of the squares of the components, which we know is correct by the Pythagorean theorem.

2. m = n. We have an n-dimensional hyper-parallelopiped in n-dimensional space. The only n-element subset of {1, ..., n} is {1, ..., n}, so the formula simply states that V 2 = det(P) 2 , which is just the formula in post #28.

3. m = n - 1. We have an (n-1)-dimensional hyper-parallelopiped in n-dimensional space. Each det(P S ) is one component of the "generalized cross-product" entertained in posts #36 and #39. The formula thus says simply that the volume of this shape is the magnitude of the vector resulting from the cross-product.

I can probably prove this, but only inelegantly. Anyone who can prove this well: please do so. [edit: Or disprove it, if that's your thing.]
Samadhi
+1

 Posted: Sat Mar 11, 2006 6:40 am    Post subject: 43 Just a nitpick on your description. I think you need to tighten up your definition a bit. It's not *any* m points. It's any m points where no more than two are colinear. Thats one of the reasons I prefer to use vectors (or line segments if you wish), it's easier to say any m vectors where none are scalar multiples of each other._________________And he lived happily ever after. Except for the dieing at the end and the heartbreak in between.
Samadhi
+1

 Posted: Sat Mar 11, 2006 6:43 am    Post subject: 44 BTW I re-explained my notion to my prof the other day and he understood what I was saying. He said that someone named Harrison (IIRC) played with this idea before but was never able to prove it rigorously._________________And he lived happily ever after. Except for the dieing at the end and the heartbreak in between.
Samadhi
+1

 Posted: Sat Mar 11, 2006 7:03 am    Post subject: 45 I just figured out how to prove mine. Not sure it's quite rigorous but, we'll see._________________And he lived happily ever after. Except for the dieing at the end and the heartbreak in between.
Bicho the Inhaler
Daedalian Member

Posted: Sat Mar 11, 2006 7:10 am    Post subject: 46

 Samadhi wrote: Just a nitpick on your description. I think you need to tighten up your definition a bit. It's not *any* m points. It's any m points where no more than two are colinear. Thats one of the reasons I prefer to use vectors (or line segments if you wish), it's easier to say any m vectors where none are scalar multiples of each other.
Actually, the points can be arranged any way you want. If vectors are scalar multiples of one another, though, then the volume will be zero, and the formula will reflect it. It's not hard to show that if the list of m vectors is linearly dependent*, then all of the determinants in the formula will be zero. This meshes with intuition: if the vectors are linearly dependent, then they are all contained in some (m-1)-dimensional hyperplane, which will make the parallelopiped "flat" (volume 0).

*Vectors being scalar multiples of one another is a special case of linear dependence.
Samadhi
+1

 Posted: Sat Mar 11, 2006 7:13 am    Post subject: 47 Bah, true enough._________________And he lived happily ever after. Except for the dieing at the end and the heartbreak in between.
Samadhi
+1

 Posted: Sat Mar 11, 2006 7:36 am    Post subject: 48 Here's the basics of my proof. Take any n vectors (even parallel ones ) and assemble them in an nxn matrix A. Take any two of the vectors and do one of the following: 1. Swap their order in determining the area. This clearly does not affect the area. 2. Multiply one of the vectors by any non-zero number. This clearly increases the area by a factor equal to that non-zero number. 3. Add any scalar multiple of one vector to the other vector. Using geomotry it can be show that the area remains the same. Expanding upon this to the hyper-area (or volume or whatever) or those two vectors and another third vector, it can easily be shown that the (I'll just call it volume) volume is only affected by the change in the area from the first two, since the height of the parrallelpiped is determined by the third vector times the sine of the angle made between that vector and the surface described by the other two. And the orientation of the "plane" described by the two vectors is not affected by any sort of multiplication or addtion between them, so the angle remains the same. This directly translates to applying elementary matrices (because the volume is an absolute value so applying -1 is inconsequential) to the nxn matrix until you achieve I n . Doing this provides you with the determinant of the original matrix times I. Clearly, any I matrix representation has a volume of 1, therefore the original volume is the determinant of the original._________________And he lived happily ever after. Except for the dieing at the end and the heartbreak in between.
Bicho the Inhaler
Daedalian Member

 Posted: Sat Mar 11, 2006 7:46 pm    Post subject: 49 That proof is fine. You can argue along similar lines for mine, but it's not so simple at the end, because if m < n, there's no identity matrix to row-reduce to, so you need some other basis case. Actually, ::slaps head:: I just realized a huge simplification of my formula: Let P be the mxn matrix whose rows are p 1 , ..., p m-1 . Then we can find the volume by the following: V 2 = det(PP T ), i.e., the square of the volume is simply the determinant of P multiplied by its transpose. This is actually easy to prove. The first step is the same as Samadhi's first step: note that applying any elementary row operation E to the m vectors (considered as row vectors) multiplies the actual volume by det(E). We should check that this multiplies V 2 by det(E) 2 in my formula: det((EP)(EP) T ) = det(E(PP T )E T ) = det(E)det(PP T )det(E T ) = det(E) 2 det(PP T ), which is the correct behavior. Therefore, we only have to prove the formula for some "basic" configurations to which all others can be reduced via row operations. The second step is to decide what row-reduction we want to use. Unlike Samadhi's case, we can't just row-reduce to the identity matrix, because there's no mxn identity matrix if m < n. Instead, we opt for Gram-Schmidt orthogonalization, which produces a collection of orthogonal vectors via row operations. (If this needs to be explained, I can; it's easy.) So now we just have to prove that V 2 = det(PP T ) if our m vectors are all perpendicular to one another. In this case, the actual volume of the now rectangular parallelopiped is obvious: V = ||p 1 ||*...*||p m ||. On the other hand, what does PP T look like? We use the normal definition of matrix multiplication: the entry in row i, column j is the dot product of row i of P with column j of P T . But this is just

(using <,> to denote inner product), which, since the vectors are orthogonal, is 0 if i != j, and is ||p i || 2 otherwise. Thus, PP T is a diagonal matrix with diagonal entries ||p 1 || 2 , ..., ||p m || 2 . The determinant of this matrix is simply the product of these diagonal entries: det(PP T ) = ||p 1 || 2 *...*||p m || 2 , which as we know is V 2 . Therefore, the formula works. The only thing left to show is that my original formula from post #42 is equivalent to this (better) one. [edit: Sorry to hijack your school-help thread with this, Samadhi. If I think of anything else, I'll put it in a different thread.]

CzarJ
Hot babe

Posted: Mon Mar 13, 2006 10:06 am    Post subject: 50

On a somewhat related tangent:

I've always been fascinated by the branch of mathematics that deals with computability, but just now I've run into it like a brick wall.

I set out the other day to use Mathematica to create a program (well, define some variables and functions) to prove the cross-product-analog proposition (that's what I figure I'll call it) in a given number of dimensions. Now, I knew Mathematica couldn't prove it in general, and that plugging in an arbitrary number of dimensions wouldn't shed much light on the problem, but I just got Mathematica, and I was really excited, so I wanted to program something, and I couldn't think of anything better.

Here's how (after much trial-and-error style programming) I defined my variables:

 Code: basiscreate[d_, i_] := If[d == i, 1, 0] basis[d_, i_] := Table[basiscreate[k, i], {k, d}] signedbasis[d_, i_] := basis[d, i]*(-1)^(i + 1) Rowr[d_, n_, r_] := Table[x[r, If[Less[t, n], t, t + 1]], {t, d - 1}] Matrixn[d_, n_] := Table[Rowr[d, n, r], {r, d - 1}] CrossProduct[d_] := Sum[signedbasis[d, n]*Det[Matrixn[d, n]], {n, d}] DotProduct[d_] := Simplify[Table[(Table[x[m, n], {n, d}].CrossProduct[d]), {m, d - 1}]] == Table[0, {d - 1}]

Typing in "DotProduct[x]" will yield "True" if the hypothesis holds in x dimensions and "False' if it doesn't. I was pretty proud of this, but after getting to about six dimensions, it starting taking some serious time to compute. I got through 8, but it's been working on 9 for about the past 10 minutes with no sign of stopping (not that there normally is one, but, you know, it's an expression). Now, like I said, I don't expect it to shed a lot of light on the subject, but just out of curiosity (a commodity I've got plenty of), is there anything in my programming that could be significantly trimmed, or is this just one of those problems (whatever they're called) that just takes a ridiculous amount of time and there's nothing you can do about it?

(Edit: Ope, it's true for 9 dimensions. I'm gonna plug in 10 and go to bed. =P)
Bicho the Inhaler
Daedalian Member

Posted: Tue Mar 14, 2006 9:40 am    Post subject: 51

CzarJ, kevinatilusa's proof actually generalizes to any number of dimensions. If A is the nxn matrix whose entry in row i, column j is a_ij, then det(A) can be evaluated as

a_11*det(B_11) - a_12*det(B_12) + ... + (-1)^n*det(B_1n),

where B_ij is the (n-1)x(n-1) matrix you get by deleting row i and column j from matrix A. That formula, called "cofactor expansion" (we expanded on the first row, but there's a similar formula for every row and column) is an essential theorem about determinants. Therefore, the generalized cross product is orthogonal to all of the factors by kevinatilusa's argument.

About your program:
• It is slightly repetitive, since you end up computing CrossProduct[d] d-1 times, while the answer never changes.
• CrossProduct[d] is expensive to compute, if only because the answer is so complicated. To get a feel for this, try the following:
 Code: SL[d_]:=StringLength[ToString[CrossProduct[d]]]
and look at SL[d] for d up through 7. You'll notice that it grows roughly as a factorial function (!) (heh...no pun intended).
• Simplify is by nature an expensive operation. Simplify on a huge expression is a hugely expensive operation.
Samadhi
+1

Posted: Tue Mar 14, 2006 1:46 pm    Post subject: 52

 Quote: Therefore, the generalized cross product is orthogonal to all of the factors by kevinatilusa's argument.
Which I'd previously argued.
_________________
And he lived happily ever after. Except for the dieing at the end and the heartbreak in between.
CzarJ
Hot babe

Posted: Wed Mar 15, 2006 10:09 am    Post subject: 53

Bicho--I'll have to look over your post when I'm more awake--it's not really sinking in now. I think if I just go back and stare at kevinatilusa's post until it clicks I'll be fine, though. =P

 Bicho the Inhaler wrote: You'll notice that it grows roughly as a factorial function

I do understand that part. I had thought that might be the case, in fact, since CrossProduct has to essentially evaluate (d!)/2 2x2 determinants (that's assuming it evaluates determinants by cofactor expansion--I don't know any more efficient way to do it, but maybe there is one).

Anyway, Samadhi, you can have your thread back; I'm done. =P
Samadhi
+1

Posted: Sun Mar 19, 2006 3:03 am    Post subject: 54

Well I'm having trouble compiling my program. I'm supposed to create a class with various methods and then run a program to test that it works. The class program compiles fine but I get an error trying to assign it.

Here's the Counter program:
 Code: import java.util.*; public class Counter {    private int value;    public void setZero()    {       value = 0;    }    public void increment()    {       value = value + 1;    }    public void decrement()    {       value = value -1;       if (value < 0)          value = 0;    }    public int countValue()    {       return value;    }    public void printCounter()    {       System.out.print(value);    } }

Here's the program to test it
 Code: import java.util.*; public class CounterTest {    public static void main(String[] args)    {       Counter count1 = new Counter();       int value;       int n;       count1.setZero();       value = count1.countValue();       System.out.println("After setZero(), counter value = " + value);       for (n=1; n <= 3; n++)       {          count1.increment();          value = count1.countValue();          System.out.println("After increment() in loop " + n + " using countValue(), counter value = " + value);       }       for (n=1; n <= 4; n++)             {                count1.decrement();                System.out.println("After decrement() in loop " + n + " using printValue(), counter value = ");                count1.printValue;       }    } }

Here's the error I get using textpad to compile:
 Code: C:\Documents and Settings\Sam\My Documents\CounterTest.java:7: cannot find symbol symbol  : class Counter location: class CounterTest       Counter count1 = new Counter();                 ^ C:\Documents and Settings\Sam\My Documents\CounterTest.java:7: cannot find symbol symbol  : class Counter location: class CounterTest       Counter count1 = new Counter();

I even put counter.java and counter.class in my java path and it doesn't work. Anyone help?
Samadhi
+1

 Posted: Sun Mar 19, 2006 3:43 am    Post subject: 55 I'm guessing I need to put it somewhere so that it will import or I need to change the first line so it will import it...but the book doesn't say.
Death Mage
Raving Lunatic

Posted: Mon Mar 20, 2006 1:42 am    Post subject: 56

The line in question is:
 Code: Counter count1 = new Counter();

Right at the top.

Try to either define it right away with some dummy data, or try [] instead of ().

Also, to call something from another file, it's filename.function or filename.variable

That's just my take after quickly glancing at it.
_________________
* These senseless ramblings brought to you by Insanity™. If you just can't figure the dang thing out, it must be Insanity™.
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Samadhi
+1

 Posted: Mon Mar 20, 2006 2:21 am    Post subject: 57 I know that's the problem. It's not recognizing the class. It should work just like any other class, such as if I wanted to create a new object of the Scanner class I would type Scanner keyboard; (class name, followed by the variable. This is exactly how you do it with primitive types too)
cloudRunner
Daedalian Member

Posted: Mon Mar 20, 2006 5:37 am    Post subject: 58

I think you might need a constructor.

Try definining a method in the Counter class that looks something like:
 Code: public Counter(){      value = 0;  // you can initialize it to zero here, as well as in the setZero() method }

just a guess...
_________________
Forever is such an unpleasant word
Samadhi
+1

 Posted: Mon Mar 20, 2006 6:17 am    Post subject: 59 Not a bad idea, as far as troubleshooting goes (when all else fails, try everything. And I did try it after you suggested ) However, from what I can tell initializing variables is not required when creating a class. And since I got the error when compiling CounterTest, the problem should logically be with CounterTest, not Counter (since it compiled OK). In fact it wouldn't accept initialization on a separate line, I had to add '= 0' to the end of the 'private int value' And that didn't change the error I got when attempting to compile CounterTest at all.
extropalopakettle
No offense, but....

 Posted: Mon Mar 20, 2006 9:33 am    Post subject: 60 Do you have the Counter class in files named Counter.java and Counter.class (and not counter.java and counter.class)? And you need to have the directory that contains Counter.class in your classpath (not Counter.class itself). You can also try compiling both in one shot (compile *.java) - but you'll still need to resolve the classpath issue to run it.
Samadhi
+1

 Posted: Tue Mar 21, 2006 4:31 am    Post subject: 61 My old programs don't run. I think I need to do a system restore.
Samadhi
+1

 Posted: Tue Mar 21, 2006 5:27 am    Post subject: 62 Somehow my JAVA_HOME environment variable was changed. I am so very annoyed.
Samadhi
+1

 Posted: Wed Mar 22, 2006 8:09 am    Post subject: 63 Took my first test in Calc III today. I was the third person to finish the test.....5 minutes before class ended. Blew the questions about tangent lines to curves. Just totally couldn't remember the formulas. Total brain fry. (I also totally forgot the formula for finding the distance from a point to a plane but luckily, the point was (1, 1, -5) and the plane was 12x + 13y + 5z = 0 ) I also missed whether (x 2 /(x 2 + y 2 )^ 1/2 ) is differentiable at 0,0. At 0,0 f(x) = 0, when it isn't 0, 0 it's the above formula. I did get whether it's continuous at 0,0 (part A) but trying to find the limits of the partials got sooooooo ugly I gave up. There was only a quarter inch of space provided on the test so I think I was doing it wrong.
CzarJ
Hot babe

Posted: Wed Mar 22, 2006 3:52 pm    Post subject: 64

I'm not really awake, and I couldn't tell you anything about the last part at least until I am (if then), and I'm sure someone else will beat me to it, but here's some thoughts on the other parts:

 Quote: Took my first test in Calc III today. I was the third person to finish the test.....5 minutes before class ended.

From my experience so far, as well as what my brother and a couple of other people have told me, this is kind of the norm in math classes. I didn't finish my first Calc IV test (probably the second hardest test I've ever taken, after my first AP US History test) and got a 70. I was talking to the professor later and said, "I felt like I really knew my stuff, but I just couldn't do it on the test," then he said, "Actually you did quite well." So I think it's a safe bet in most of your classes from here on out that if you finish, you probably did fine. =P

 Quote: I also totally forgot the formula for finding the distance from a point to a plane but luckily, the point was (1, 1, -5) and the plane was 12x + 13y + 5z = 0

I hate you. =P We had to derive the formula on one of my Calc III tests. I guess I shouldn't complain--at least I'm not going to forget it now.
Bicho the Inhaler
Daedalian Member

Posted: Thu Mar 23, 2006 12:44 am    Post subject: 65

Samadhi - >>first let y = 0, so f(x, 0) = x^2/sqrt(x^2) = |x|. This is not differentiable at x = 0, so the function as a whole can't be differentiable at the point (x = 0, y = 0). You can rephrase this as saying f x doesn't exist at (0,0).<<
 CzarJ, on 3/15, wrote: I'll have to look over your post when I'm more awake...
 CzarJ, on 3/22, wrote: I'm not really awake...
I can get pretty lazy, but I've never slept for an entire week. :-P
CzarJ
Hot babe

 Posted: Thu Mar 23, 2006 2:59 am    Post subject: 66 I wish I was asleep that whole week. =P
Samadhi
+1

 Posted: Thu Mar 23, 2006 4:13 pm    Post subject: 67 Czaj: A 70 isn't 'fine' unless possibly if you're grading on the curve (assuming that's out of a 100). Math is my major, I don't think Cs are a good thing. As for the distance from the plane thing. If it makes you feel better I spent 20 minutes wracking my brain trying to refigure the formula to no avail. I literally laughed out loud (chuckled would maybe be better) when I realized the answer.
Antrax
ESL Student

 Posted: Thu Mar 23, 2006 4:54 pm    Post subject: 68 <-- is completely happy with some of his Cs. I think you should give it a couple of semesters, CS has some fiendishly difficult courses._________________After years of disappointment with get rich quick schemes, I know I'm gonna get rich with this scheme. And quick!
GH
Daedalian Member

 Posted: Thu Mar 23, 2006 5:27 pm    Post subject: 69 I spent a few semesters in college working for a professor who was the Chairman of the Education Department. He told me on more than one occasion that if a student scored 100 on any of his tests, he felt like that test was a failure, because it didn't give that student enough "room" to perform. I might compare it to an Olympic long jump competition in a pit that's only 27 feet long. You wouldn't be able to give a "real" score to anyone who was able to jump more than 27 feet. His personal policy, then, was to give tests on which no student could get a perfect score. He made his tests longer and more difficult than I've ever seen. He would then look for the "natural breaks" in the test scores and assign letter grades accordingly. This policy made him wildly unpopular with the students that didn't understand it.
Antrax
ESL Student

 Posted: Thu Mar 23, 2006 5:59 pm    Post subject: 70 And with the students who see a degree, and GPA in particular, as a way of impressing potential future employers, and who could care less about one professor's odd way of grading students._________________After years of disappointment with get rich quick schemes, I know I'm gonna get rich with this scheme. And quick!
Samadhi
+1

 Posted: Thu Mar 23, 2006 6:42 pm    Post subject: 71 Gh: Maybe he could have made it insanely difficult, but used a curve? Antrax: How dare they! (btw: where's the learning in either equation?)
GH
Daedalian Member

Posted: Thu Mar 23, 2006 7:05 pm    Post subject: 72

 Samadhi wrote: Gh: Maybe he could have made it insanely difficult, but used a curve?
Isn't that what this is?
 I wrote: He would then look for the "natural breaks" in the test scores and assign letter grades accordingly.

What do you mean "where's the learning"? I was just pointing out that I knew a professor who felt like measuring each student's full potential on every test was more important that stroking the egos of students by testing only to minimum standards. And I don't think it's odd at all. The only thing uncommon about it is the desire to make his tests "un-ace-able."
Samadhi
+1

 Posted: Thu Mar 23, 2006 7:13 pm    Post subject: 73 This is a math teacher? I thought they liked to be objective.
Samadhi
+1

 Posted: Thu Mar 23, 2006 7:22 pm    Post subject: 74 And "un-aceable" is a ridiculous notion. Are you going to be able to write an arithemtic quiz that I couldn't ace? (go ahead and try) Hell, I'll even give you a chance to stump me at algebra or trig! No? Ok. How about Calc I? Granted, I think I'd have to revisit, but I could ace Calc I after some prep. It's math. It's defined. I don't see how you can make a test that is un-aceable when it comes to math. And the notion that you (the professor) must make some question that is impossible to beat is pure hubris.
worm
unregistered

Posted: Thu Mar 23, 2006 7:38 pm    Post subject: 75

 GH wrote: His personal policy, then, was to give tests on which no student could get a perfect score. He made his tests longer and more difficult than I've ever seen. He would then look for the "natural breaks" in the test scores and assign letter grades accordingly. This policy made him wildly unpopular with the students that didn't understand it.

my biggest problem with this method is the "natural breaks" part...seems very arbitrary. one prof. i worked for was pretty notorious for the difficulty of his class. his philosophy was fairly similar, but he used a set curve (75-100=A, 60-75=B, 50-60=C, 35-50=D). that may seem like a pretty big range for an A, but an A in his class meant something.

btw, he wasn't popular, either, but i think the students who took his class appreciated that their grades depended on what they knew and not what their peers knew.
worm
unregistered

 Posted: Thu Mar 23, 2006 7:43 pm    Post subject: 76 oh, and his test weren't "un-aceable". i don't specifically remember any 100's, but there were 2 students who were always in the upper 90's. funny thing, one was an english major and one was a history major and the course was o-chem.
GH
Daedalian Member

Posted: Thu Mar 23, 2006 8:24 pm    Post subject: 77

 Samadhi wrote: And the notion that you (the professor) must make some question that is impossible to beat is pure hubris.

I said longer and more difficult. The professor wouldn't have to have a question that you can't answer, s/he would just have to have enough test that you couldn't finish it in time.

The professor that I worked for was, as stated above, the Chairman of the Education Department, not the Math Department. He was teaching (prospective) teachers.

worm, don't you think that our perceived competence at something is almost always determined in realtion to our peers? Why should school be different?
Antrax
ESL Student

 Posted: Thu Mar 23, 2006 10:42 pm    Post subject: 78 Funny, the professor who taught us computability last semester did just that (give us a test that you couldn't finish on time). The result was a test that favoured quick thinkers in a way that had nothing to do with the subject matter (I aced the test, while much stronger students scored in the low 70s), and an incredibly low test average. Hurrah._________________After years of disappointment with get rich quick schemes, I know I'm gonna get rich with this scheme. And quick!
CzarJ
Hot babe

 Posted: Thu Mar 23, 2006 11:12 pm    Post subject: 79 The test isn't graded on a curve per se, but the numerical grade is pretty meaningless, because the way he does the semester grades is by looking at the percentage grades and just kind of deciding where A, B, C, etc. should go. I'm pretty happy with my 70, although I'm determined to do better on the next one. Point is, to have the majority of the class not finish a test is not uncommon. College tests are crazy, that's just the way they work. When I was in high school, teachers expected students to be able to make a hundred on a test, but that's just not how it goes in college. That's not to say it won't happen, but it's no reason to worry or be surprised if it doesn't. One of my favorite teachers in high school was telling me when he was in law school, he took a test on tax law and made a 23% and it was the highest grade in the class. But he passed the class (with flying colors, I imagine). Anyway, that's all. The end.
worm
unregistered

Posted: Fri Mar 24, 2006 12:00 am    Post subject: 80

 GH wrote: worm, don't you think that our perceived competence at something is almost always determined in realtion to our peers? Why should school be different?

sure, relative competence is important, and it shouldn't be any different in school. to me, tests are the most basic evaluation of relative competence in a subject. why is it necessary to massage the data so it fits a curve?

also, i think a curve-free class gives better grade consistency. that is, an A today requires the same level of competence as it did 2 years ago.
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