TuringTest, what now?

- anonymous

TuringTest, what now?

- Stacey Warren - Expert brainly.com

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- schrodinger

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- saifoo.khan

Party time! :D

- TuringTest

lol
I dunno...

- anonymous

What can you teach me?

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## More answers

- TuringTest

a lot, there's a fair amount to algebra, but what's the next logical step?

- anonymous

Teach about polynomial.

- TuringTest

I have already to an extent, but where to go next...

- anonymous

Ratios?

- anonymous

In maths the fundamentals concepts are limited but the problems are not, so how about practicing what you have already learned?

- TuringTest

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:\[\frac{x^2y^5z}{xy^9z^3}\]

- saifoo.khan

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- TuringTest

awww, you're making me think of my cat in the hospital :(

- anonymous

http://a4.sphotos.ak.fbcdn.net/hphotos-ak-snc6/271016_253886774622476_100000034671401_1128907_4380462_n.jpg

- anonymous

How did it go with her?

- TuringTest

broken leg...
waiting for an operation

- anonymous

You beat you cat Turing?!!!

- anonymous

aww :(

- saifoo.khan

turing sat on her leg. :(

- TuringTest

she fell off the roof.
how pessimistic

- saifoo.khan

awww, sorry i was jking..

- TuringTest

It's all good :P

- anonymous

Due to Turing's intense physics stress his cat decided to commit suicide :P

- anonymous

(x^2)*(y^5)*(z)
------------ = (x)*(y^5/9)*(z^1/3)?
(x)*(y^9)*(z^3)

- saifoo.khan

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- anonymous

Cats are cool.

- anonymous

why \( y^{5/9} \) ?

- saifoo.khan

no doubt. ;)

- TuringTest

you have inconsistent rules above inopeki\[\frac{x^a}{x^b}=x^{a-b}\]always...

- anonymous

http://i.imgur.com/48wOw.jpg

- anonymous

Oh right

- anonymous

(x^2)*(y^5)*(z)
------------ = (x)*(y^5-9)*(z^1-3)?
(x)*(y^9)*(z^3)

- TuringTest

yes, simplify...

- saifoo.khan

##### 1 Attachment

- anonymous

(x)*(y^5-9)*(z^1-3)=(x)*(y^-4)*(z^-2)?

- anonymous

http://i.imgur.com/6uzXf.jpg

- saifoo.khan

cute^

- TuringTest

yes, do you know another way to write\[x^{-a}\]???

- anonymous

No

- anonymous

http://i.imgur.com/eqTIb.jpg

- TuringTest

\[x^{-a}=\frac{1}{x^a}\]so it's probably nicer to rewrite your expression with all positive exponents in this way.

- anonymous

Oh right, all negative exponents make the "total number" divided by one.

- anonymous

Btw who can explain why x^0 = 1?

- anonymous

(x)*(1/y^4)*(1/z^2)

- TuringTest

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...

- anonymous

Turing that's a very important yet fundamental question.

- anonymous

another one is why \( (a^b)^c = a^{bc} \) ?

- TuringTest

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?

- TuringTest

the last rule is easier to explain, perhaps I should...

- anonymous

(x)*(1/y^4)*(1/z^2)
(x)*(1/y^4*z^2)?

- TuringTest

yes, now write it as a fraction
what goes on the bottom?
what goes on the top?

- TuringTest

@FFMactually that's not so easy to explain now that I think about it.

- anonymous

Can you explain the second rule intuitively?

- anonymous

1
x* --------
y^4*z^2

- TuringTest

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.

- TuringTest

@ inopeki
you can put the x on top, it means the same thing and looks nicer

- anonymous

x*1
--------
y^4*z^2

- anonymous

Can we say that all power function obeys the functional equation \( f(x)f(y)= f(x+y) \)?

- anonymous

Whats that f?

- TuringTest

@Inopeki yes, no need to write the 1 though
@FFM, I'd have to think about that :/

- TuringTest

unspecified functions
he's asking how far the rule about exponents can be extended..

- anonymous

x
-------- Oh right
y^4*z^2

- anonymous

Actually we can't but I believe that is true for exponential functions though.

- anonymous

Umm, ok?

- TuringTest

it must be for simple ones\[2^x2^y=2^{x+y}\]of course

- anonymous

Tha's whats Inopeki is using right ?

- TuringTest

basically
Inopeki do you see the connection between what you are doing and the rule for multiplying exponents like\[x^ax^b\]?

- anonymous

Yeah, but does that mean that it becomes
x
-------- ?
yz^6

- TuringTest

no because y and z are different bases
notice the rule above had both base x.

- anonymous

YEah but then i dont see the connection..

- TuringTest

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.

- anonymous

I know about that, like x^2*x^8=x^10

- anonymous

x^10/x^5=x^5

- anonymous

Basic

- TuringTest

right, I want you to see how that and the rule for division are inverses of each other for a reason...

- anonymous

Turing you have a heck of patience :D

- TuringTest

eh it's easy FFM, doesn't require too much concentration.
good practice too I learn by teaching ;)
ok factor
\[3x^3y^3+6xy+9x^2y\]

- anonymous

:-)

- anonymous

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?

- TuringTest

yes exactly

- anonymous

3xy(3x^3*y^3/3xy)+3xy(6xy/3xy)+3xy(9x^2*y/3xy)=3xy(x^2*y^2+2+3x)?

- TuringTest

yep :)
great!

- anonymous

Really? :D

- anonymous

Note to Inopeki: Turing is a great resource, learn new things from him and practice as much as you can on your own :-)

- TuringTest

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)

- anonymous

When hes not here i try to go on purplemath and khan :)

- anonymous

a^2+ab-ba-b^2

- TuringTest

simplify

- anonymous

Books books books!! online learning has it's own limitation :-)

- anonymous

a^2-b^2?

- Akshay_Budhkar

it doesnt @ffm i disagree

- anonymous

Foolformath, im having trouble finding books here in sweden

- TuringTest

and what is the name of that form? remember?

- anonymous

The fundamental theorem of algebra? Or want that the one with the multiplex?

- TuringTest

Multiplex?
no I think you're thinking of 'multiplicity'
but this is the 'difference of squares'
\[a^2-b^2=(a-b)(a+b)\]

- Akshay_Budhkar

with ocw.mit , khan, purple math and openstudy and people like turing there is no limit to online education ^

- TuringTest

Thanks but you get what you put in, I think is true with all this stuff.

- anonymous

Oh right

- TuringTest

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

- TuringTest

so remember that we can run this backwards, and I can say
'factor'\[x^2-4\]using difference of squares
wanna try?

- TuringTest

run the FOIL backwards I meant*

- anonymous

x(x-4)?

- anonymous

Oh

- TuringTest

no look at the form\[a^2-b^2=(a-b)(a+b)\]so what is a and b in\[x^2-4\]???

- anonymous

(x-2)(x+2)?

- TuringTest

there ya go :)

- TuringTest

how about\[x^4-y^4\](this is a favorite question on OS)

- anonymous

(x^2-y^2)(x^2+y^2)

- anonymous

Why?

- TuringTest

good, now is that all we can do with it though?
(I've seen a lot of tutors stop here too ;-)

- TuringTest

hint:look at the first set of parentheses

- anonymous

Ummm

- TuringTest

what is the first sett of parentheses?

- anonymous

(x^2-y^2)?

- TuringTest

yes, and can you factor that?

- anonymous

GCF of x^2 is x
GCF of y^2 is y
xy(x^2/xy)-xy(y^2/xy)?

- TuringTest

no you're over-thinking
x^2-y^2 is difference of squares again!

- TuringTest

\[a^2-b^2=(a-b)(a+b)\]

- anonymous

So (x-y)(x+y)?

- TuringTest

yes!
so now factor\[p^4-q^4\]completely!!

- TuringTest

you will apply difference of squares twice

- anonymous

(p^2-q^2)(p^2+q^2)
(p-q)(p+q)(p+q)+(p+q)?

- anonymous

Without the +

- anonymous

between the parenthesis

- TuringTest

(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

- anonymous

(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

- TuringTest

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...

- anonymous

Ohhh

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