Episode 2: Math equation

Oct 9, 2012 10:17 AM

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Feb 2010

2171

sorry, can't resist lol:
anyway,

ab(a-b)+bc(b-c)+ca(c-a) = -(a-b)(b-c)(c-a)

^dafuq happened?

Should be:
a^2 b-a^2 c-a b^2+a c^2+b^2 c-b c^2
(expanded form), right?

"Your taste is shit cause you like what I hate. Believe me I have 1000 cartoons that I rated with less than 5."

Oct 9, 2012 10:45 AM

Kirimoshi

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Mar 2010

Just memorize it. ©
This equation is surprisingly correct no matter how odd it may look to you. :)

Oct 9, 2012 1:00 PM

Sayan_

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Jul 2010

1027

Easier to see if you work backwards.

-(a-b)(b-c)(c-a)

= - (a-b) (c-a)(b-c)……. Step not needed. Rearranged to make it easier to see.
= - (ac -a^2 –bc +ab)(b-c)…. Expanded the first two terms
= - (abc -a^2b -b^2c +ab^2 -ac^2 +a^2c +bc^2 -abc).. Expand with remaining term.
= - abc +a^2b +b^2c -ab^2 +ac^2 -a^2c -bc^2 +abc….. Multiply by (-) sign
= a^2b +b^2c -ab^2 +ac^2 -a^2c -bc^2… Step not needed. Rearranged to make it easier to see
= a^2b -ab^2 +b^2c -bc^2 - ac^2 -a^2....Partial Factor
= ab(a-b)+bc(b-c)+ca(c-a)

Oct 9, 2012 1:36 PM

lehka

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Apr 2011

Haha, I have no idea what you're talking about. Lol /cry

Oct 11, 2012 10:55 AM

Cashdax

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Feb 2010

2171

Sayan12 said:
Easier to see if you work backwards.

I have no idea how that worked out lol
can you show us the the other formula? (not backwards)

EDIT: yeah, tried it. It's really heck easier when backwards.

"Your taste is shit cause you like what I hate. Believe me I have 1000 cartoons that I rated with less than 5."

Oct 11, 2012 2:44 PM

muntasir123

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Aug 2012

1056

The factorization is relatively simple, here is a proper solution without bash that would generally be solved on an actual proof. I made this longer simply for explanation once you practice you can do this very fast. Also whoever said to memorize this is completely wrong, you should never memorize anything in math other than basic operation proofs are essential whether it be simple formulas or not.

Consider the multivariable polynomial
f(a,b,c) = ab(a-b)+bc(b-c)+ca(c-a)

We want to "force" it to factor
It is well known that there will exist a value in which there will be a zero if it does factor

Suppose a = b

Therefore a-b is a root

Plug it in and we get 0

Suppose b = c

We get zero

b-c is a root

Suppose a = c

a-c is a root

we get zero

There we know at the least there exists a function with constant k ( determines whether the function is opening)

k(a-b)(b-c)(c-a) = ab(a-b)+bc(b-c)+ca(c-a)

We can plug in any real value for a b c to solve for k ( pick a pair 1,0,1 or anything)

then solve for k like a normal equation and you will get -1

Motivation

When we want to force a factorization let the variables equal each other or set them to simple integers such as zero or one, so we can see if we get zero, if so it is a root!!

Sayan12's approach

This works however the proof is not correct the original proposal was to prove the left equals right not backwards however we can go backwards then present our proof frontwards given the data we collected from going backwards.

muntasir123Oct 11, 2012 2:48 PM

"届どいた....手紙?" - スワ(オレンジ)

Oct 14, 2012 9:03 PM

Kirito-Dragneel

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Sep 2012

519

dafuq did I just click on @_@

Oct 14, 2012 9:46 PM

muntasir123

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Aug 2012

1056

ikr who notices this in anime XD

"届どいた....手紙?" - スワ(オレンジ)

Oct 14, 2012 10:27 PM

hentai_proxy

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Oct 2010

2269

Edit: modified for clarity.

muntasir, your answer is correct, but your claim that setting a=b and getting f identically zero implies (a-b) is a factor needs justification. A novice may easily be trapped. To see the subtleties of what you claim, consider what lies beneath: a=b are exactly the solutions of the equation a-b=0, i.e. the roots of the polynomial a-b. Your claim is that if f(a,b,c) is identically zero on the roots of the polynomial a-b, then that polynomial is a factor of f(a,b,c). But let's do the same thing for the polynomial (a-b)^2. The solutions of (a-b)^2=0 are exactly pairs (a,a), i.e. whenever a=b. f(a,b,c) is identically zero whenever (a-b)^2=0. Does this mean that (a-b)^2 is a factor of f(a,b,c)? Definitely not. This ties in with taking the radical in the Nullstellensatz.

Of course, you don't need the strength of the Nullstellensatz to justify your claim in the case of a-b, you can do things like that by hand. But think about this: suppose a polynomial f(a,b,c) is zero whenever a^2+b-3bc+ab=0. Can you then conclude that (a^2+b-3bc+ab) is a factor of f(a,b,c)? How can you prove that? These things are much more difficult and intricate for polynomials of more than one variable (one variable polynomials have a Euclidean division algorithm, so proofs are extremely simple there).

hentai_proxyOct 15, 2012 12:39 AM

Oct 14, 2012 10:38 PM

#10

GrownUp

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Aug 2012

542

Jesus Christ... As if you would notice this stuff.

Oct 14, 2012 11:08 PM

#11

phoenixalia

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Mar 2010

3041

I don't understand a thing lol. XD

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Oct 15, 2012 3:25 AM

#12

removed-user

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Jul 2018

564533

Cmon people. 8th grade Algebra ain't rocket science

Oct 15, 2012 1:09 PM

#13

IntroverTurtle

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Nov 2010

26413

edwd2 said:
Cmon people. 8th grade Algebra ain't rocket science

I forgot everything about 8th grade algebra as soon as 8th grade ended.

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Oct 15, 2012 2:53 PM

#14

Florete

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Jun 2008

2216

This topic is awesome.

Oct 15, 2012 3:16 PM

#15

Karhu

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May 2011

2420

Huh? You should be able to see that it is correct just by looking at its form.

I definitely have superpowers. I can feel it in my balls.

Oct 15, 2012 3:23 PM

#16

FluffyDango

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Apr 2008

2146

IntroverTurtle said:

edwd2 said:
Cmon people. 8th grade Algebra ain't rocket science

I forgot everything about 8th grade algebra as soon as 8th grade ended.

Same :P

Plus, I only watch this anime for the romance and comedy, not to do math :P

Oct 15, 2012 4:30 PM

#17

wara56

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Jun 2011

Math rocks and so does this thread :)
Btw: http://myanimelist.net/manga/27505/Suugaku_Girl

if you like math-related topics

Oct 15, 2012 4:39 PM

#18

ClementIV

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Nov 2008

1739

I'm glad I was not the only one that noticed and it's actually interested in how that came about.

But, math is not my best or favorite subject, still I found it interesting that they added it into the episode.

Oct 15, 2012 4:52 PM

#19

Kurogashi

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Jul 2009

2641

Yes, my math majors, flex that frontal lobe!

Oct 15, 2012 5:12 PM

#20

shanimebib

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9157

I love you guys for discussing this! Brings back memories. Those Wonderful High School Days :')

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Oct 15, 2012 10:29 PM

#21

Atikal

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Oct 2010

848

F*ck this sh*t!

Oct 15, 2012 10:40 PM

#22

hiromee

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Feb 2010

357

I love this thread

Oct 15, 2012 10:43 PM

#23

Roloko

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Sep 2010

6759

*brain explodes*

Math was always my worst subject. xP

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Oct 27, 2012 3:53 AM

#24

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Jan 2008

1254

Haru has good reason to ask Natsume to remember instead of explaining, replies in this thread is good proof. ><

Oct 27, 2012 2:57 PM

#25

Amutofan123

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Nov 2009

230

Yup. I have no clue what any of you are talking about. Once I finish a math class, everything I learned in it disappears from my head. God, I hate math. Although, we are doing trig in my math class right now and... I kind of actually enjoy it.

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Oct 29, 2012 5:56 PM

#26

Omniknight

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Oct 2008

930

The key thing is that you have to add a +abc and -abc to be able to factor it forwards.

Oct 29, 2012 10:37 PM

#27

Liila

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Jul 2011

....why? o.o

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Oct 30, 2012 7:39 AM

#28

Masnoon

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Jun 2010

246

I have the weirdest boner now.

Oct 30, 2012 8:08 AM

#29

modderfokker

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Jun 2012

Liila said:
....why? o.o

I suck at math but I know you can add +abc and -abc because the result of the two is 0 and therefore it doesn't change the original equation. Likewise, you can add as many +0's in your equation as you like but the end result will always be the same.

Albert Einstein said:
Education is what remains after one has forgotten everything he learned in school.

Oct 30, 2012 9:37 AM

#30

EvL-666-DanZ

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Jul 2012

I have no idea about this but this an awesome thread XD

Oct 30, 2012 12:22 PM

#31

itsKatsudon

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Apr 2009

264

holy shit

if we die we'll meet again in valhalla...

Nov 11, 2012 3:29 AM

#32

lionheart04

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Oct 2012

807

probably the most intelligent thread that you will ever see in mal forums

Nov 12, 2012 9:56 AM

#33

ataraxial

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Mar 2012

1255

This is awesome. I paused and checked it in my head when I saw it.

At first I was like, wait, 6 terms and 8 terms not possibly equal.
And then I was like, oh lol I'm dumb abc and -abc cancel.

Jan 5, 2014 4:05 PM

#34

krylios

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Jan 2014

Well as some naive highschool student who just turned 16 and has no life here is 30 minutes worth of work

ab(a-b)+bc(b-c)+ca(c-a)

Solving 1st and 3rd bracket 1st...
a^2b - ab^2 + c^2a - ca^2 + bc(b-c)

Re-arranging
a^2b - ca^2 - ab^2 + c^2a + bc(b-c)

a^2(b-c) - a(b^2 - c^2) + bc(b-c)

a^2(b-c) - a(b - c)(b + c) + bc(b-c)

Taking (b-c) common
(b - c) [a^2 -a(b + c) + bc ]

(b - c) [a^2 - ab - ac + bc]

(b - c) [a (a - b) - c (a - b)]

(b - c) [ (a - b) (a - c) ]

(a - b) (b - c) (a - c)

Edit: I was curious too XD

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Episode 2: Math equation

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