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TuringTest, what now?

Mathematics
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Party time! :D
lol I dunno...
What can you teach me?

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Other answers:

a lot, there's a fair amount to algebra, but what's the next logical step?
Teach about polynomial.
I have already to an extent, but where to go next...
Ratios?
In maths the fundamentals concepts are limited but the problems are not, so how about practicing what you have already learned?
That's a good point FFM so lets do simplification that you know, but a little more intense: I think I need to make sure you can simplify bigger things:\[\frac{x^2y^5z}{xy^9z^3}\]
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awww, you're making me think of my cat in the hospital :(
http://a4.sphotos.ak.fbcdn.net/hphotos-ak-snc6/271016_253886774622476_100000034671401_1128907_4380462_n.jpg
How did it go with her?
broken leg... waiting for an operation
You beat you cat Turing?!!!
aww :(
turing sat on her leg. :(
she fell off the roof. how pessimistic
awww, sorry i was jking..
It's all good :P
Due to Turing's intense physics stress his cat decided to commit suicide :P
(x^2)*(y^5)*(z) ------------ = (x)*(y^5/9)*(z^1/3)? (x)*(y^9)*(z^3)
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Cats are cool.
why \( y^{5/9} \) ?
no doubt. ;)
you have inconsistent rules above inopeki\[\frac{x^a}{x^b}=x^{a-b}\]always...
http://i.imgur.com/48wOw.jpg
Oh right
(x^2)*(y^5)*(z) ------------ = (x)*(y^5-9)*(z^1-3)? (x)*(y^9)*(z^3)
yes, simplify...
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(x)*(y^5-9)*(z^1-3)=(x)*(y^-4)*(z^-2)?
http://i.imgur.com/6uzXf.jpg
cute^
yes, do you know another way to write\[x^{-a}\]???
No
http://i.imgur.com/eqTIb.jpg
\[x^{-a}=\frac{1}{x^a}\]so it's probably nicer to rewrite your expression with all positive exponents in this way.
Oh right, all negative exponents make the "total number" divided by one.
Btw who can explain why x^0 = 1?
(x)*(1/y^4)*(1/z^2)
Oh dear... I have my answers about x^0=1 (not true for x=0), but I'm sure FFM would not approve of them :/ @inopeki, yes now rewrite it as 1 fraction...
Turing that's a very important yet fundamental question.
another one is why \( (a^b)^c = a^{bc} \) ?
well I have a very simple proof of it and I can show why it is not true for x=0 what more do I need in your opinion?
the last rule is easier to explain, perhaps I should...
(x)*(1/y^4)*(1/z^2) (x)*(1/y^4*z^2)?
yes, now write it as a fraction what goes on the bottom? what goes on the top?
@FFMactually that's not so easy to explain now that I think about it.
Can you explain the second rule intuitively?
1 x* -------- y^4*z^2
no, I was thinking more of how easy it is to show x^ax^b=x^(a+b) intuitively, the other is tricky to me.
@ inopeki you can put the x on top, it means the same thing and looks nicer
x*1 -------- y^4*z^2
Can we say that all power function obeys the functional equation \( f(x)f(y)= f(x+y) \)?
Whats that f?
@Inopeki yes, no need to write the 1 though @FFM, I'd have to think about that :/
unspecified functions he's asking how far the rule about exponents can be extended..
x -------- Oh right y^4*z^2
Actually we can't but I believe that is true for exponential functions though.
Umm, ok?
it must be for simple ones\[2^x2^y=2^{x+y}\]of course
Tha's whats Inopeki is using right ?
basically Inopeki do you see the connection between what you are doing and the rule for multiplying exponents like\[x^ax^b\]?
Yeah, but does that mean that it becomes x -------- ? yz^6
no because y and z are different bases notice the rule above had both base x.
YEah but then i dont see the connection..
just that dividing and multiplying are inverse operations x^a=x*x*x*...*x (a times) x^b=x*x*x*...*x (b times) so if we have a=3 b=2 we get x^a --- x^b x*x*x =-----= x x*x which is x^(a-b) if we have (x^a)(x^b) we get (x*x*x)(x*x)=x*x*x*x*x=x^(a+b) so the rules for exponents here come directly from their definitions. You can count the x and see that this relationship holds.
I know about that, like x^2*x^8=x^10
x^10/x^5=x^5
Basic
right, I want you to see how that and the rule for division are inverses of each other for a reason...
Turing you have a heck of patience :D
eh it's easy FFM, doesn't require too much concentration. good practice too I learn by teaching ;) ok factor \[3x^3y^3+6xy+9x^2y\]
:-)
The GCF of 3,6,9 is 3 The GCF of x,x^2,x^3 is x The GCF of y,y^2,y^3 is y 3xy(3x^3*y^3/3xy)+3xy(6xy/3xy)+3xy(9x^2*y/3xy) ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ Middle step, right?
yes exactly
3xy(3x^3*y^3/3xy)+3xy(6xy/3xy)+3xy(9x^2*y/3xy)=3xy(x^2*y^2+2+3x)?
yep :) great!
Really? :D
Note to Inopeki: Turing is a great resource, learn new things from him and practice as much as you can on your own :-)
Yes I've stressed the importance of studying on your own as well and yes Inopeki really, FFM would have noticed a mistake I'm sure ok quick, foil (a-b)(a+b)
When hes not here i try to go on purplemath and khan :)
a^2+ab-ba-b^2
simplify
Books books books!! online learning has it's own limitation :-)
a^2-b^2?
it doesnt @ffm i disagree
Foolformath, im having trouble finding books here in sweden
and what is the name of that form? remember?
The fundamental theorem of algebra? Or want that the one with the multiplex?
Multiplex? no I think you're thinking of 'multiplicity' but this is the 'difference of squares' \[a^2-b^2=(a-b)(a+b)\]
with ocw.mit , khan, purple math and openstudy and people like turing there is no limit to online education ^
Thanks but you get what you put in, I think is true with all this stuff.
Oh right
The fundamental theorem of algebra: "the sum of the multiplicity of the roots of a polynomial is equal to its order" difference of squares\[a^2-b^2=(a-b)(a+b)\]both important, but very different
so remember that we can run this backwards, and I can say 'factor'\[x^2-4\]using difference of squares wanna try?
run the FOIL backwards I meant*
x(x-4)?
Oh
no look at the form\[a^2-b^2=(a-b)(a+b)\]so what is a and b in\[x^2-4\]???
(x-2)(x+2)?
there ya go :)
how about\[x^4-y^4\](this is a favorite question on OS)
(x^2-y^2)(x^2+y^2)
Why?
good, now is that all we can do with it though? (I've seen a lot of tutors stop here too ;-)
hint:look at the first set of parentheses
Ummm
what is the first sett of parentheses?
(x^2-y^2)?
yes, and can you factor that?
GCF of x^2 is x GCF of y^2 is y xy(x^2/xy)-xy(y^2/xy)?
no you're over-thinking x^2-y^2 is difference of squares again!
\[a^2-b^2=(a-b)(a+b)\]
So (x-y)(x+y)?
yes! so now factor\[p^4-q^4\]completely!!
you will apply difference of squares twice
(p^2-q^2)(p^2+q^2) (p-q)(p+q)(p+q)+(p+q)?
Without the +
between the parenthesis
(p^2-q^2)(p^2+q^2) stop here we cannot factor the second set of parentheses, it is not a /difference/ of squares, it is a /sum/ of squares, and we have no formula for that. check that (a+b)(a+b) is not a^2+b^2 so only factor the one that is a /difference/ of squares
(p-q)(p+q)(p^2+q^2) You said i had to do difference in squares 2 times? Btw, lets move to a new thread, this one is starting to lag a little bit
That is above correct, you did difference of squares twice: p^4-q^4 (p^2-q^2)(p^2+q^2) <-once (p-q)(p+q)(p^2+q^2) <-twice ok new thread...
Ohhh

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