induction questions that are hard to do in tha wee hours of tha mornin (1 Viewer)

[KeypadSDM]

Originally posted by GuardiaN
Assume true for n = k
ie (k^1 - 1) = M * ( x - 1 )

I think you're a little mixed up with that step.

Assuming true for k gives you:

x^k - 1 = M(x - 1)
not
k^1 - 1 = M(x - 1)

 

[ND]

Originally posted by GuardiaN


He is on a higher plane of maths existence ( ) than me, so I thought he just thinks a bit too complicated.

People good at maths make complex things simple (i.e. elegance), not the other way around.

 

[Grey Council]

mm? Usually thats true, but sometimes its like i said. heheh, trust me, it happens.

I think you're a little mixed up with that step.

Assuming true for k gives you:

x^k - 1 = M(x - 1)
not
k^1 - 1 = M(x - 1)

whatever, i got confused, i did it in my head. hehe

But the solution is right regardless. innit?

 

[Grey Council]

Ok, its been edited. The solution is now correct, or i think so anyway.

 

[KeypadSDM]

Originally posted by ND
People good at maths make complex things simple (i.e. elegance), not the other way around.

Nonetheless, I really can't see how I could simplify my solution ...

 

[ND]

I'm not saying that it could be done simpler. I'm saying that Guardian's assumption that because the solution is by you, it would be complex, is not a valid one.

 

[Grey Council]

Originally posted by ND
I'm not saying that it could be done simpler. I'm saying that Guardian's assumption that because the solution is by you, it would be complex, is not a valid one.

Hold your horses, i didn't say that! lol, i just said that his method, which is VERY similar to my one, has all this unnecessary stuff, that to me looks very complicated:

"(Where Z(x) is a polynomial in x where the leading co-efficient is x^(k-1))"

Thats what i mean, I don't quite understand that. But then i quite clearly said:

Im prolly wrong, but thats cause i didnt quite get what the hell he was saying in his proof. hehe, my bad.

I didnt say that BECAUSE the solution is by Keypad, it would be complex. When'd i say that?

And anyway, I think my solution is a BIT simpler than Keypads. Not much, but it is a bit more straightforward. I think.

 

[ND]

Originally posted by GuardiaN

I didnt say that BECAUSE the solution is by Keypad, it would be complex. When'd i say that?

It was implied when you said "He is on a higher plane of maths existence than me, so I thought he just thinks a bit too complicated." If you didn't intend it to mean that then that's ok.

And anyway, I think my solution is a BIT simpler than Keypads. Not much, but it is a bit more straightforward. I think.

Just a note: something more straight forward is not necessarily something simpler. I'll give you an example:

http://www.boredofstudies.org/community/showthread.php?s=&threadid=5144

Here's a question posted by spice girl at the beginning of last year. My first solution to it was straight forward, but hardly simple (all i can say is yuck). My 2nd solution isn't as straight forward, but far simpler. Spice girl's solution is even more simple (and elegant).

 

[Grey Council]

damn

*has been outmaneuvered by ND at every turn*

I give up

*sigh*

 

[Grey Council]

whoa!

*takes a step back*

ND is a big gun? Didn't know that. I thought you were doing your HSC with me. Now you tell me your a pro.

 

[mazza_728]

meh so lost

 

[ND]

Originally posted by GuardiaN
whoa!

*takes a step back*

ND is a big gun? Didn't know that. I thought you were doing your HSC with me. Now you tell me your a pro.

Heh. Pro? Hardly, it's just that i've already done my HSC. Remember that that question was done in march, so i was still a couple of months ahead of where you are now.

 

[Grey Council]

lol, fair enough

 

[KeypadSDM]

"(Where Z(x) is a polynomial in x where the leading co-efficient is x^(k-1))"

Whoops ... It's not actually co-efficient is it ....
Lol

 

[mazza_728]

KeypadSDM = i think u use very sophisitcated jargon to scare us away!

 

[KeypadSDM]

I was just trying to clarify. If you remove that line the whole thing makes just as much sense.

 

[Grey Council]

yeah, but WITH that line it confused the shit outta me. Or more like, it scared the hell outta me.

I was like, wtf does that mean, f(x) is a polynomial in x with leading coeffecient blah blah blah. the heck, man, calm down.

 

[KeypadSDM]

Where Z(x) is a polynomial in x...
We all know what a polynomial is, and we usually deal with polynomials in x, so there's no problem there.

with the leading term being x^(k-1)
Just means that the greatest power of x is is x^(k-1)

It's like saying a cubic has a leading term of x^3

(I know 'term' isn't the right word, but I can't remember it for the life of me. So 'term' will have to suffice.)

Edit: (Actually, I think it could be leading term. There's nothing like 8 hours of office duties to kill of your wits)

 

[ND]

Sure it's not 'leading term'? I think that that's how i've always known it.

 

[evilc]

polynomial in x of degree (k-1)?