Y'all are a bunch of wankers!

### hey.  Math question.

Can k^2 + 4 be reduced?  Can it be represented as the square of something?

I'm almost done with this dang problem.  Yes I should go to a math forum.  Yes, I prefer getting my ass kicked on CoT for no good reason.

Thank you.
sharkfish
September 28th, 2006 5:48pm
Maybe I mean "factored", not "reduced".
sharkfish
September 28th, 2006 5:49pm
Yes it can be factorized, depending on what kind of factors you allow. No it is not the square of something.

Since you hate me, feel free to consider this a hateful message.
bon vivant
September 28th, 2006 5:51pm
Memorize the solution to the quadratic equation:

ax^2+bx+c=0  has solutions

x= -b +- (b^2-4ac)^1/2
---------------------
2a

Or, w/out lousy monospaced notation:

You've got:

a=1, b=0, c=4

So you get imaginary roots, +2i and -2i
Ward
September 28th, 2006 5:52pm
Forgot the last line:

So it _can_ be factored as  (k+2i)(k-2i) = k^2+4  but usually that sort of question is looking for real factors only.
Ward
September 28th, 2006 5:56pm
Dammit ward, if you don't know you should say something.
Bot Berlin
September 28th, 2006 5:56pm
It's a case of memorizing common patterns.

In this case we have something in the form of a^2 + b^2, and then the standard pattern is:

a^2 + b^2 = (a + ib)(a - ib)

It's one of those things like memorizing your times tables. You can't really work them out, you just have to know them.
bon vivant
September 28th, 2006 5:57pm
According the Fundalmental Therom of Algebra, your equation has two roots.

More generally, a polynomial of n degree has n roots.

http://www.cut-the-knot.org/do_you_know/fundamental2.shtml
Rick Zeng/Tseng
September 28th, 2006 6:00pm
>>> a^2 + b^2,

AAAAAAAAAGGGGGHHH!  That's not the general pattern.  Memorize the damn solution to the quadratic and you're done.  Yes, you can see certain patterns w/out it  (e.g. x^2 - b^2 = (x+b)(x-b)) but just memorize the damn -b +- root(b^-4ac) all divided by 2a and that's all you need.
Ward
September 28th, 2006 6:06pm
Ward, with guys like you no wonder SoP complains about grades :)

No, the right way is to understand why. We have calculator for the formulas.
Rick Zeng/Tseng
September 28th, 2006 6:20pm
> Ward, with guys like you no wonder SoP complains about grades :)

I didn't say that. I thought that. But I didn't say that.
son of parnas
September 28th, 2006 6:21pm
"but usually that sort of question is looking for real factors only."

Yes. No imaginary numbers.  thanks.
sharkfish
September 28th, 2006 6:25pm
?? Sorry, Rick, you forget how old I am.  I'd have to find a book to refresh my memory now, but I learned the solution to the quadratic equation before there were calculators that could store it for you.

This is actually a counter-example to "learn processes, don't memorize facts."  It's turned out to be easier for me to have memorized the solution 25+ years ago than to remember how to re-derive it each time.
Ward
September 28th, 2006 6:26pm
>>> No, the right way is to understand why.

So, let's see your quick explanation of why an nth degree polynomial has n roots...
Ward
September 28th, 2006 6:28pm
But Ward, I was thinking more in terms of recognizing various general patterns that often come up, such as:

a^2 - b^2 = (a + b)(a - b)

a^2 + 2ab + b^2 = (a + b)^2

a^2 - 2ab + b^2 = (a - b)^2

a^3 + 3a^2b + 3ab^2 + b^3 = (a + b)^3

and so on. It's really helpful if you can learn to identify those and others on sight. The quadratic equation too of course, but is it really a good way to factorize all of the above?
bon vivant
September 28th, 2006 6:31pm
I'm ready to give up.  This is where the question comes from:

When I run into

n^3 = [n(n+1)/2]^2

I'm supposed to use mathematical induction to prove it.
That just means add k+1 to it and see if you can get it to look like the original, with the right side having n replaced by k+1, and the second n replaced by k+2. If I get the general shape of the original above with those values stuck in there, I'm all good.

my solution is

[k(k+1)/2]^2 + (k+1)

k^2(k+1)^2 / 4 + 4(k+1)/4

k^2(k^2 + 2K +1) + (4k +4)

k^2(k+1)^2 + 4(k+1)

(k+1)[(k+1)k^2 + 4(k+1)]

(k+1)^2(k^2+4) / 4

So now I'm supposed to take that last line and put it in the form of

[(k+1)(k+2) / 2 ]^2

But I can't get there.

sharkfish
September 28th, 2006 6:32pm
huh?

try n = 3. 3^3 = 27 on the left. (3*4/2)^2 = 36 on the right.
just me
September 28th, 2006 6:35pm
>>> It's really helpful if you can learn to identify those and others on sight. The quadratic equation too of course, but is it really a good way to factorize all of the above?

It's helpful while you're taking a course that covers factoring quadratics, and after doing them (endlessly, it seemed at the time) in high school recognition of some of those patterns remains.

But you're generally better learning the general case (so to speak).  Quadratics come up w/ irrational and complex roots all over the place, so it's worth it to know the "complete" solution.
Ward
September 28th, 2006 6:35pm

n^3 is very definitely not equal to [n(n+1)/2]^2.

Something is missing here.
bon vivant
September 28th, 2006 6:37pm
Oops. I forgot to mention

this is a series proof.  So a series of n's, cubed, added together, equals the equation on the right.

So 1^3 + 2^3 + 3^3 = 36
sharkfish
September 28th, 2006 6:42pm
it looks like one of those combination thingies. but it's not.

n(n+1)/2 is what Gauss proved was the sum of natural numbers up to the number n. His teacher asked his students to add up all the numbers up to a 100. The teacher thought he could relax and play GameBoy or textmessage with his beau next door while all his pupils labored with their pencils and pads on the task.

But Gauss figured it out in like a minute - the general formula. He was 11 at the time.
just me
September 28th, 2006 6:43pm

[k(k+1)/2]^2 + (k+1)^3

NOT

[k(k+1)/2]^2 + (k+1)

Mathematical induction proofs, from my understanding, require you to figure out how to plug in the series and munge it back into the original frame of the equation.  You take the value plus 1, value plus 2, etc.  These (added in this case) in series work for the equation.

so

a series of integers N such that 1-cubed + 2-cubed + 3-cubed add up to something somebody deduced on the right side of the equation.

Prove it by replacing the right side unknowns with a representation of the series.
sharkfish
September 28th, 2006 6:46pm
i'm confused. we know that

sum(1 to k) of i^3 = [k(k+1)/2]^2

we want to add the next entry of the series, which is (k+1)^3, to each side to show that ...

sum(1 to k+1) of i^3 = [(k+1)(k+2)/2]^2 = [k(k+1)/2]^2 + (k+1)^3

correct?
just me
September 28th, 2006 6:47pm
in my book

1-cubed + 2-cubed + 3-cubed .... n^3 = [ n(n+1) / 2 ] ^2

if you plug 1 in on the right for n

[1(2) / 2 ] ^2 = 1

true.

plug in 2

[2(3) / 2 ] ^2 = 9 BECAUSE 1-cubed + 2-cubed = 9

...
sharkfish
September 28th, 2006 6:51pm
Start by assuming that

sum(1 to k) of i^3 = [k(k+1)/2]^2

as you have stated.

Then if you add one more cube to both sides, you should have

sum(1 to k+1) of i^3 = [(k+1)(k+2)/2]^2

Now on the left hand side you have just added (k+1)^3.

So you should have that [(k+1)(k+2)/2]^2 - [k(k+1)/2]^2 = (k+1)^3

If you can show that, you are almost there.
bon vivant
September 28th, 2006 6:52pm
THEREFORE

1.  n^3 = [ n(n+1) / 2 ] ^2

becomes

2.  n^3 = [ k(k+1) / 2] ^2

3.  n^3 + (k+1)^3 = [k(k+1) / 2] ^2 + (k+1) ^3

Now munge it so #3 looks rather like #1. I think I'm supposed to make it look like:

4.  n^3 = [(k+1)(k+2) / 2]^2
sharkfish
September 28th, 2006 6:55pm
bon vivant says I"m going for this

[(k+1)(k+2)/2]^2 - [k(k+1)/2]^2 = (k+1)^3

I say I'm supposed to go for

n^3 + (k+1)^3 = [k(k+1) / 2] ^2 + (k+1) ^3
sharkfish
September 28th, 2006 6:58pm
bon vivant and I are saying the same thing. the n^3 terminology is confusing. the n has been replaced by 'k'.

say replace it with either sum(1 to n) of i^3, or the more generic "fun(n)"

you know (ie, assume) ...

fun(k) = [k(k+1)/2]^2

we know fun(k+1) - fun(k) = (k+1)^3 ... that's the def of fun()!

so we want to show,

fun(k+1) - fun(k) =  (k+1)^3
=  [(k+1)(k+2)/2]^2 - [k(k+1)/2]^2 =

that is,

(k+1)^3      =  [(k+1)(k+2)/2]^2 - [k(k+1)/2]^2

That's nice and clean and doens't have the fun(n) or "n^3" in there which was confusing.
just me
September 28th, 2006 7:08pm
fun(k+1) - fun(k) = (k+1)^3

How did you come to that conclusion?

That's what I'm missing here.
sharkfish
September 28th, 2006 7:11pm
the minus is what is getting me.

I thought I am supposed to just add (k+1)^3
sharkfish
September 28th, 2006 7:12pm
OK, what you are looking for is this:

[k(k+1)/2]^2 + (k+1)^3 = [(k+1)(k+2)/2]^2

This is because you are adding the (k+1) cube to the series and then showing that the formula still holds. You will see it is the same as before, but swapped around.

Now you can try to rearrange the left side to make it look like the right side, but IMHO that is hard. However, anything you do that shows both sides are equal is going to be satisfactory.
bon vivant
September 28th, 2006 7:15pm
or,

fun(k) + (k+1)^3 = fun(k+1)

that's the definition of fun(k+1). that it, take the previous sum (fun(k)), and add the cube of the next number (k+1) to it.
just me
September 28th, 2006 7:16pm
The point of mathematics induction is to show that if we assume something is true for k, it's also true for k + 1.

Thus, paraphrasing just me,

If we assume that
sum(1 to k) of i^3 = [k(k+1)/2]^2

we can show
sum(1 to k + 1) of i^3 = [(k + 1)(k + 2)/2]^2
=>
[k(k+1)/2]^2 + k^3 = [(k + 1)(k + 2)/2]^2
Rick Zeng/Tseng
September 28th, 2006 7:19pm
Right.

And now I have to make it look equal somehow.

I hate myself.
sharkfish
September 28th, 2006 7:19pm
Hate the evil math, not yourself.
Aaron F Stanton
September 28th, 2006 7:20pm
Reaarrange it so the cubed term is to the left and the sqyared terms are to the right.

You have then
(k+1)^3 = [(k+1)(k+2)/2]^2 - [k(k+1)/2]^2

now a^2-b^2 = (a+b)(a-b)

and on you go.
trollop
September 28th, 2006 7:21pm
What I have so far

http://cherinet.brinkster.net/images/problem.gif
sharkfish
September 28th, 2006 7:22pm
Good lord have you guys been helpful.
sharkfish
September 28th, 2006 7:23pm
trollop: No no no no no! Ward says we have to plug [(k+1)(k+2)/2]^2 - [k(k+1)/2]^2 into the quadratic equation formula and find the roots. :-)
bon vivant
September 28th, 2006 7:24pm
I see trollop. It will take me a few to apply that...
sharkfish
September 28th, 2006 7:24pm
Sorry, I should have said:

Let assume that
sum(1 to k) of i^3 = [k(k+1)/2]^2

sum(1 to k + 1) of i^3
=[k(k+1)/2]^2 + k^3

...

-----
=[(k + 1)(k + 2)/2]^2
Rick Zeng/Tseng
September 28th, 2006 7:27pm
"And now I have to make it look equal somehow."

No, you have to show it is equal, somehow.
Rick Zeng/Tseng
September 28th, 2006 7:29pm
I don't see the difference.
sharkfish
September 28th, 2006 7:31pm
You mean my rephrasing of the problem or my rephrasing of your sentence?
Rick Zeng/Tseng
September 28th, 2006 7:33pm
hmm

(k+1)^3 = [(k+1)(k+2)/2]^2 - [k(k+1)/2]^2

= ((k+1)(k+2)/2  +  k(k+1)/2) ( ((k+1)(k+2)/2  -  k(k+1)/2 )

= ((k^2+3k+2)/2 + (k^2+k)/2) (((k^2+3k+2)/2 - (k^2+k)/2)

= ((2k^2+4k+2)/2) ((2k+2)/2)

= (k^2 + 2k +1) (k+1)

= (k+1)^2 (k+1)

= (k+1)^3

qed
trollop
September 28th, 2006 7:35pm
Superfluity of parentheses in there btw.

Try p 158 in Gullberg for a good summary of the trivial "sum the integers" problem.

also here:

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

Rats, now I have NO excuse for not researching estimating systems ...
trollop
September 28th, 2006 7:43pm
= ((k^2+3k+2)/2 + (k^2+k)/2) (((k^2+3k+2)/2 - (k^2+k)/2)

= ((2k^2+4k+2)/2) ((2k+2)/2)

How did you jumpt from that to that?
sharkfish
September 28th, 2006 7:44pm
This is stupid.  I give up.

NO more math.  What's the point.

I am so fucking frustrated.
sharkfish
September 28th, 2006 7:44pm
oy, vey. can't we just factor out (k+1)^2 from both terms on the very first line of trollop's QED?

then we're left showing what Gauss (or that wikipedia page above), showed.
just me
September 28th, 2006 7:47pm
Yup. Factoring out k + 1 or (k + 1)^2 is easiest.
Rick Zeng/Tseng
September 28th, 2006 7:49pm
Sharkfish,

It'll be easier if we enjoy the process and learn the material rather than relying on math to get to the material and psychological goal in real life.
Rick Zeng/Tseng
September 28th, 2006 7:52pm
= ((k^2+3k+2)/2 + (k^2+k)/2) (((k^2+3k+2)/2 - (k^2+k)/2)
should have been
= ((k^2+3k+2)/2 + (k^2+k)/2) . ((k^2+3k+2)/2 - (k^2+k)/2)

so summing like-powered terms within the outer parens we get

= ((k^2 + k^2  + 3k + k  + 2)/2) . (k^2 - k^2  + 3k - k  + 2)/2)

which is

= ((2k^2+4k+2)/2) ((2k+2)/2)

...
trollop
September 28th, 2006 7:52pm
Here's how the teacher did it.  Ratface mofo JUST NOW posted it.

http://cherinet.brinkster.net/images/hisProblem.gif
sharkfish
September 28th, 2006 7:55pm
So the idea isn't to SOLVE for it, but get the final resolution in the shape of the original.

I'm pissed and I told him so.  I wasted all day on this and haven't gotten to any of the other problems and this is due tonight at midnight (at least four more of these).
sharkfish
September 28th, 2006 7:57pm
you can take (k+1) out right at the start to save a lot of drill.

Just me has a good eye there.
trollop
September 28th, 2006 7:59pm
Hmmm. Taking out the (k+1) is exactly why the teacher made it look so easy. You really do need an eye for that kind of thing.
bon vivant
September 28th, 2006 8:02pm
I agree with about the needing an eye thing.

You get stuck and waste time.

I told this fucking teacher I want all the advantages the class has and one more day to hand in this assignment since he didn't do a relevant example in class and posted an answer for the class who saw it this morning meanwhile I'm working it out here with you guys and getting angry.

Goddammit.
sharkfish
September 28th, 2006 8:03pm
Notice that the sane thing to do was to prove them equal.

Notice the prof didn't bother with that difficulty.

trollop's solution is beautiful but I'm not going to be able to do that under test pressure without a lot more training.

I screaming angry right now because this same prof has the nerve to make comments about the online students when if you are going to post answers to the homework, THEY have all fucking day and they have an actual example to follow, while WE struggle to figure out WTF he was thinking.

The kids in class just go to class and whine about the difficult ones and get him to post the answer.
sharkfish
September 28th, 2006 8:10pm
Sharkfish,

I suggest use the form I wrote down and do as much as you can.

Assume
funcA(k) = funcB(k)

then
funcA(k + 1)

...
func(funcA(k),k,k + 1)
func(funcB(k),k,k + 1)

...

----
funcB(k + 1)
Rick Zeng/Tseng
September 28th, 2006 8:13pm
Notice that the prof did prove them equal, only he did it in the "elegant" way.

Take the starting point from earlier in the thread:

[k(k+1)/2]^2 + (k+1)^3 = [(k+1)(k+2)/2]^2

This is the same starting point the prof had.

Now he went like this:

LHS = [k(k+1)/2]^2 + (k+1)^3

= (k+1)^2 [k^2/4 + (k+1)]

= (k+1)^2 [k^2 + 4k + 4] / 4

= (k+1)^2 (k+2)^2 / 4

= [(k+1)(k+2)/2]^2 = RHS

I thought this was hard to do, and it was hard to do unless you have done it before and remember the way. You especially had to notice about the (k+1)^2 factor. This kind of thing really can be painful.
bon vivant
September 28th, 2006 8:25pm
It's kind of hard.

But highschool algebra is about factoring, expanding and grouping anyway.
Rick Zeng/Tseng
September 28th, 2006 8:33pm
It's been way too long since fourth form maths ... but we did so much drill it sems to have stuck. (Also helped to have had one of the world's most inpirational maths teachers. Made all the difference.)
trollop
September 28th, 2006 9:08pm
"have you considered starting the assignment _before_ the day before it is due?

you know, like on the day it is given to you, or something.."

Look I"m not a complete fucking idiot.  We just handed in a paper on Tuesday, plus homework problems.

I'm pissed at the prof for posting everybody's grade.
Pissed because he didn't provide a decent example and made me waste my precious time.  Pissed because he posted the answer to the question, a fucking giveaway for the rest of the class, that I broke my head on all fucking day.  Yes, it is now a giveaway and everybody has the right answer.  What the fuck.  I'm double-pissed that he makes a quip, in class, about how the online students aren't doing as well and then doesn't post his fucking giveaway until late in the day.

Fuck that.  I have demanded an extra day for the homework.
sharkfish
September 28th, 2006 9:55pm
>>> But highschool algebra is about factoring, expanding and grouping

Which is why I've stayed out of it since the thread switched to the details of doing the proof by induction... It's been too long since I had to do that much algebra.  I did "participate" by nuking one comment, though...
Ward
September 28th, 2006 9:56pm
Oh, and all those students are using their TI calculator to do the factoring. WTF is that.

See, I get no points for doing the problem.  I get no points for figuring it out because guess what, nobody after this class will give a fuck because even though I UNDERSTAND the gist of the lesson, what I am being graded on is the ANSWER.

I'm too angry to explain this shit point by point to your idiocy, wSV.
sharkfish
September 28th, 2006 9:57pm
"Which is why I've stayed out of it since the thread switched to the details of doing the proof by induction... It's been too long since I had to do that much algebra. "

Yeah well, I haven't done it in 20 years.
sharkfish
September 28th, 2006 9:58pm
>>> Oh, and all those students are using their TI calculator to do the factoring. WTF is that.

Are the TIs fast enough for this to be an issue during a test?  ...assuming they're allowed at all - for this type of course, they shouldn't be, if there's any point to math it's to teach you symbol manipulation.  I guess for homework, the calculators are fast enough, though.

>>> See, I get no points for doing the problem.  I get no points for figuring it out because guess what, nobody after this class will give a fuck because even though I UNDERSTAND the gist of the lesson, what I am being graded on is the ANSWER.

Does this belong over in the "grades" thread?  That's the way the system works - ideally, you're supposed to "understand the gist of the lesson" but in fact the profs don't care much.  They have to teach, they have to give grades, they don't have to strive for the grades to accurately reflect how much you understand.

I remember one test where the exam was 90% based on material not covered in the course, it was all stuff that was one step beyond what he'd done.
Ward
September 28th, 2006 10:02pm
TI's are allowed for the test.

And grades are fine.  I just don't want to be compared to other students, in a derisive tone, as though I'm not doing my part.  If I'm off campus, not using a calculator, and the homework review isn't available to me until much later in the day than the other students, I am at a distinct disadvantage and I want this to be acknowledged at least by giving me an additional day to complete the problems, or weighting the homework appropriately such that I don't have to hear his bullshit offhand comments about online student performance.
sharkfish
September 28th, 2006 10:07pm
He gave me the extension to turn in the homework tomorrow.

I'm sure he's wondering WTF is this student a psycho or what.

Won't be the first time.  And yes, I am psycho.
sharkfish
September 28th, 2006 10:29pm
You should not be pissed. You should take some satisfaction that you may have learned the subject better than those other students who got given the answer without having to work it out. Using TI calculators is the same thing. What are they learning if they don't work things out on paper?

If I may suggest, you came into this thread with every appearance that you had not understood the real meaning of proof by induction. When you wrote,

"fun(k+1) - fun(k) = (k+1)^3

How did you come to that conclusion?

That's what I'm missing here.

the minus is what is getting me.

I thought I am supposed to just add (k+1)^3"

that was really showing that you hadn't grasped the gist of the lesson.

If you have understood now what you didn't understand then, then you have learned something real. If you had been in the class with the other students, you would have lost out because you would even now have not understood.
bon vivant
September 28th, 2006 10:57pm
"And grades are fine.  I just don't want to be compared to other students, in a derisive tone, as though I'm not doing my part."

Agree 100%; there is no point in comparing.
Rick Zeng/Tseng
September 28th, 2006 11:02pm

If you look at his lecture note, you'll see his approach isn't in a balanced equation so much as fretting with it until it looks right.

To get to the beginning of the work...in this case it is a series of numbers being added, in this case, cubes of a series.  So you represent these sequential numbers/series as k+1^3 and that is the next number in the series.  You just tack that on to the initial supposition.

Putting in a minus sign for the _true_ equation this represents isn't part of the lecture, so therefore, isn't part of my understanding.

In my opinion, induction proofs for a series is just replacing the unknowns with the next value in the series and munging the equation as in his note.

No where in his notes is he equating (k+1)^3 to anything.

He is mushing it into the existing equation.

We know the goal:

[(k+1)(k+2)/2]^2

can you make the original supposition look just like that after adding (k+1)^3?

I have simplified this into baby speak because that's what I understand.

Sometimes, people go overboard and make it more complicated than what it really is.

If I'm wrong, I'm wrong.  But that's my understanding.  I know WHY the other guys built the equation.  But their way isn't going to make my life easier.  It makes me understand, as you say, the way the big people do, but I won't pass the test with big people ways, under pressure.

Just go with it, bon vivant.

I know.  It's like when I try to explain computing concepts to users.  I have a talent for simplifying things.  Some say I oversimplify. I say if it gets the user to the same point, I and they are better off at that moment.
sharkfish
September 28th, 2006 11:10pm
To be fair, it could be the case that the teacher did not demonstrate the lesson.
Rick Zeng/Tseng
September 28th, 2006 11:11pm
P.S.  My boss really hates when I explain some of his work to non-computer programmers.  Here is an example:

"Shark, I just wrote a routine that takes an invoice, recalculates the hourly rate depending on whether the rate is based on activity or staff rate or flat rate, and applies it across the board without regard to time-frame.  There's no history because storing the values would mean extremely lengthy lookups and the SQL is daunting."

Me to the customer:

"If you click that link, it recalculates the rate such that there is no historical evidence of the previous rate."

This is a nonsensical scenario (we do have rates histories etc.).  But the boss cringes if he stands there and tells me something like this and then I "translate".

I really do simplify everything this way, and it makes mathematics extremely difficult for me.

The help in this thread is much appreciated, and I will never forget this one.

:)
sharkfish
September 28th, 2006 11:16pm
"To be fair, it could be the case that the teacher did not demonstrate the lesson."

I might have ot agree with you there, Rick.  This thread is much more informative than that lesson.  Seriously.
sharkfish
September 28th, 2006 11:17pm
If the prof wanted to be complete, he could have written it out this way

http://www.crazyontap.com/topic.php?TopicId=11101#137933

But he didn't.  I listen to the audio lecture, and I view the notes as in the link.  I see munging, not any equality. I don't even know to ask the question because he doesn't provide enough examples for me to see the pattern and guess at what he is really doing.

The page above the link I gave to his solution doesn't have the balanced equation, either.

He just sticks the equals sign on the left side of the page, and starts with the equation as you see there.
sharkfish
September 28th, 2006 11:35pm
A bunch of posts jumped in while I was composing this post. But just one last thing.

It would be really good for you to understand that "fretting with it until it looks right" is *exactly the same* as putting it in a balanced equation.

He wrote the top line, and then he wrote "aside, what do we need to find out?", and then he wrote the other half of the balanced equation underneath. You need to understand why the squiggly line there is exactly the same as an equals sign, or you have not understood the gist of proof by induction.
bon vivant
September 28th, 2006 11:35pm
He didn't write the top line, I'm telling you.
sharkfish
September 28th, 2006 11:36pm
YOU need to understand. Damn you bon vivant. you are an irritating twit, even if you are good at math.

Toastmasters for you, sucker.
sharkfish
September 28th, 2006 11:37pm
Sorry, cross posting. I'm just going by the link you posted to his notes. I guess I'm not good at writing anyhow. I always want to change what I wrote after I have posted it.
bon vivant
September 28th, 2006 11:39pm
page 1 of profs explanation

http://cherinet.brinkster.net/images/page1HisProblem.gif

page 2

http://cherinet.brinkster.net/images/problem.gif

And the audio lecture doesn't mention the important point, either.
sharkfish
September 28th, 2006 11:42pm
dammit.

Page 2

http://cherinet.brinkster.net/images/hisproblem.gif
sharkfish
September 28th, 2006 11:43pm
If I were sitting in class, I might have asked the question.

But notice, it wouldn't matter because the homework is do the same day of his "explanation".

I think that's what pisses me off the most.  I'm supposed to be earning points.

It shouldn't be that you only get an A if you know the material already. I should be able to get a great grade with my current level of intelligence and intellectual curiousity. As it stands, the prof is making this very impossible.
sharkfish
September 28th, 2006 11:45pm
To avoid misunderstanding about which notes I am referring to, I mean this link that you posted earlier:

http://cherinet.brinkster.net/images/hisProblem.gif

The top line of that is

[k(k+1)/2]^2 + (k+1)^3

and then "aside, what do we need to find out?"

[(k+1)(k+2)/2]^2 = (k+1)^2(k+2)^2/4

When he says "what do we need to find out?" he means "what do we have to make the first thing equal to?"

That's all I'm trying to say.
bon vivant
September 28th, 2006 11:48pm
And ranting further, this is why I think school and education as it is now is a crock of shit.

He who begs the most has the best grades (assuming you don't already know the material before it is taught).  If the teacher isn't really teaching, you have to then state your case and beg for an extension. Which I did.  I now have until 1PM tomorrow.

sharkfish
September 28th, 2006 11:48pm
It is very important, though, to list the case for k as the 'assumption', and then try to use that to derive the case for k + 1.

For series, it's simple, just add the next term to the original 'conjecture', and that's why they are used as examples.

Backing that up with a base case (k = 1), we have, bingo, a solid proof.

Also, logic and maths are the only subjects you can have objective proof. Celebrate that :)
Rick Zeng/Tseng
September 28th, 2006 11:49pm
"When he says "what do we need to find out?" he means "what do we have to make the first thing equal to?" "

I said that way up thread, before you came to post here.

Furthermore, are you saying he wrote out all the steps?  I don't see that.

The "aha" isn't there.  It just isn't.  You would have to know what he's talking about in order to see it.  Because he didn't write it down, didn't speak of it in the lecture.
sharkfish
September 28th, 2006 11:50pm
And here's the thing: I KNOW what everyone else's grades are for every assignment.

He knows the grades, since he gave them. Knowing that people aren't getting it, why would he skip any step?

WTF.
sharkfish
September 28th, 2006 11:51pm
As I said in the grades thread, testing the same week of a lesson and then grading on it really truly is a crock of shit. I don't know anything to do about that. The world is just screwed up in so many ways, mainly because organizations act in stupid ways.
bon vivant
September 28th, 2006 11:52pm
And in my book, it has explanations that skip two or three steps!

You're just supposed to know that n(n+1)/2 + (n+1) is the same as (n+1)(n+2)                                                                              /2

That's an easy one.  What hard ones are they skipping?  I'm just not that stupid.  The books and the teachers make this stuff difficult.
sharkfish
September 28th, 2006 11:54pm
I am so tempted to send my prof this thread.  But I can't.
sharkfish
September 28th, 2006 11:56pm
Oh, the teacher *definitely* does not know how to teach.

assuming it's true for k + 1? No! You are trying to show it is true for k + 1 using the assumption that it is true for k!
Rick Zeng/Tseng
September 28th, 2006 11:57pm
Just to reinforce my last post, when I learned proof by induction for the first time, I didn't get it. I didn't get it, in fact, until weeks after the lesson. With a lot of the mathematics I learned, I sat in class feeling dazed and confused, and wondering why everyone else seemed to be understanding, and why was I so thick that I couldn't get it.

It turns out that the others in the class didn't get it either, but it took a while for me to realise that.
bon vivant
September 29th, 2006 12:00am
Yeah well he made it through PhD Nuclear Physics prgram.
sharkfish
September 29th, 2006 12:00am
That explains it.

Basically, he does not 'grok' maths, logic and CS.
Rick Tang
September 29th, 2006 12:07am
I think ricks got it.  if those are his notes in that image then he either doesn't understand proof by thingie himself, or he badly misspoke himself.
worldsSmallestViolin
September 29th, 2006 12:13am
Well f already, where is the suggestion we write a program that does this stuff for us.  Oh yeah, that's what the programmable calculator is for!

It sounds like any other art, the computer can do it better, but just for the personal growth (hopefully there is some and some professional growth to boot) you do it anyway.
LinuxOrBust
September 29th, 2006 12:15am
I always found it interesting when math teachers would say "massage" this into this, and I am thinking does this problem have a backache?

It's like puzzles and who knows, maybe even the teacher just studies the answerbook!?
LinuxOrBust
September 29th, 2006 12:18am
I knew a guy I worked with, he got an EET at DeVRY and then got a job analyzing rocket-launch data.  hehe.

Do YOU Sharkfish want to study rocket-launch data, math stuff for hours on end?  Of course, at my age it seems more interesting since distractions like women are far from beating down my door.  ;-P

It's alright, plenty other things than math I find extremely interesting.
LinuxOrBust
September 29th, 2006 12:37am
This brings it all home:

http://community.monster.com/Forums/ShowPost.aspx?PostID=173195

That's right, when you are done with your math degree, you can learn something really useful like PHP, setting up a server, and C# express edition.  hahahahaha.
LinuxOrBust
September 29th, 2006 12:47am
Math induction is very much like recursiveness in CS. You pull yourself up your bootstraps.

Though in CS you try to make each subsequent calculation simpler (-1 in some way), because your computer's time is not infinite. While in math induction each subsequent tends to be more complex (+1), because you're trying to cover an infinite set of possibilities.

The prof's classroom instruction (especially as he asked his students to finish it off at home) seems fine. Going down from your assumptions and coming up from your goal is a very realistic way to work - it's a process of discovery. Throwing things back and forth across the equals sign, too. It reminds me a bit of designing libraries and APIs. Sometimes instructions are on one side of the library/API function boundary, sometimes on the other. Often code is refactored to look more appealing, long after the initial coding process. Written out math proofs too are different from the sketch pad hacking.

I had a prof who gave us a take-home final exam (number theory, in fact). It had two extra credit questions on it. The last of these was an unproven conjecture - ie, even the professor didn't know how to solve. Frigging aye - I didn't know it was unproven, he never told us. I stayed up a good 48 hours trying to prove it, missing two of the other, normal, questions. Those were the days.

By the way, the more generic formula for a sum of powers,

sum (1 to n) of n^(2p+1) = [n(n+1)/2]^(2p)

doesn't work.
just me
September 29th, 2006 12:49am
"It's alright, plenty other things than math I find extremely interesting."

Studying math is akin to studying music.  I find it painful, but satisfying in some ways.  When I get to the parts that hurt, I think of it as a puzzle rather than work.

Yesterday I had a hard time because of time pressure and other responsibilities.  Take that away, and this whole thread would be a pleasant mental exercise instead of me ranting :)

It isn't for everyone.  You can write great software and know nothing about mathematics.  By the way, I'm not really trying to get a BS in Math, althought that would be nice.  I'm really trying to qualify for more selective grad school programs, possibly in computer science or computer engineering or maybe even math education.

I have to plan for my second life, after 45, when if you aren't in management, the corporate world thinks you are an idiot and even more expendable because of your health insurance costs.

You have reminded me that life is a journey.  I'm trying to enjoy the ride, and part of it is making myself a more disciplined person.  Completing math courses is HARD for me.

If I can do this, I can remove a substantial mental block that says "I'm not good enough."  I really do want to quit.  But I'm not.

I won't.  Even if I get a C in the course, I'm going to continue. I have decided I don't care.  Just because the professor is moving on, it doesn't mean I can't go back to the algebra that I need to practice, but this weekend rather than during the week when other things are pressuring me.

If that means a C, it means a C.
sharkfish
September 29th, 2006 10:30am
Gyad.  I'm much more reasonable when I'm not under pressure and not on sinus medication/allergy meds like yesterday. I can see things a lot more clearly this morning.

Good grief.
sharkfish
September 29th, 2006 10:32am
Just one step at a time.

The neat thing about Prove this = that problems is that the goal is shown already so you know it is attainable if you make the right moves (some of those moves can be a bit obscure or would grace a high wire artist).

I'm NOT in favour of rote learning but getting the basics down pat lets you see and do things more rapidly.

It's a bit like music. Practice helps.
trollop
September 29th, 2006 11:35am