Mathematics

user19161

6:00 PM

@jordan Do you know that a(b+c)=ab+ac ?

Jordan

yes

user19161

(a+b)c=ac+bc ?

Jordan

but (a+b)(c + d) is different

ac + ad + bc + bd

user19161

(a+b)(c+d)=a(c+d)+b(c+d)

Jordan

why is that split of the terms valid

user19161

6:01 PM

So you expand twice.

user19161

Once using left distributive, once using right distributive

user19161

You can use either one first

Jordan

oh I have never seen anyone not use FOIL

user19161

WTF is foil

Jordan

first outer inner last

user58512

6:02 PM

FOIL is a technique used to confuse children by shielding them from mathematics

Jordan

multiply the first terms, the outer terms, inner terms and the last terms

user19161

The problem with all these acronyms is that people may not know how things work

Jordan

mathematics is intended to be memorized not understood :P

user19161

FOIL just means do the outer brackets first, that is all.

Jordan

I was never taught math, just told to memorize it for maximized test scores

user19161

6:02 PM

It has little to do with what is here.

user19161

Here, there are no outer brackets and inner ones.

user19161

Actually, I am not even sure now what foil means, haha.

Jordan

anyways when I am as a part like $x^2 -2x -15(x-2)$ is it valid to say that I am doing $x^2(x-2) - 2x(x-2) - 15(x-2)$?

user58512

@Jordan, by x^2 -2x -15(x-2) did you mean (x^2 -2x -15)(x-2)

Jordan

the first terms, the outer terms, the inner terms and the last terms

user19161

6:04 PM

@Jordan Why would they be equal?

Jordan

@user58512 yes

@JacobBlack Becuase that is what I got

user58512

(x^2 -2x -15)(x-2) = x^2(x-2) - 2x(x-2) - 15(x-2)

that is correct

because

Jordan

that gives me an incorrect answer I think

user58512

(x^2 -2x -15)(x-2) = ((x^2 -2x) -15)(x-2) = (x^2 -2x)(x-2) -15(x-2) = x^2(x-2) -2x(x-2) -15(x-2)

user19161

First you need to understand left and right distributive.

Jordan

6:05 PM

I have never even heard those terms

user19161

The (a+b)(c+d) example is good enough.

Jordan

If I have (x^2 -2x -15)(x-2) is it wrong to FOIl it?

Like multiply everything by x and the everything by -2?

user19161

I don't even know what foil means.

user19161

Is foil a common term in the US?

Jordan

yeah, everyone learns it

user19161

6:07 PM

@Jordan So that is equal to (x^2-2x-15)x+(x^2-2x-15)(-2)

user19161

See?

Jordan

no....

I was doing it differently

I messed it up I guess

I was trying to multiply everything by x and then by -2

user19161

Can you expand a(b+c+d)?

Jordan

which gave me -2x^3

ab + ca + da

user19161

Also, expand (a+b+c)d

Jordan

6:08 PM

everytime I come here I basically learn that I have 20 years of math to learn

in two months

user19161

You'll catch up soon.

user58512

this is primary school level stuff

user19161

@jordan expand the above.

Jordan

ad + db + cd

I have no idea what primary school is

user19161

Good, why do you write bd as db?

Henning Makholm

6:10 PM

@JacobBlack It's a didactic acronym that helps you remember the correct order of the terms in (a+b)(c+d)=ab+ad+bc+cd.

Jordan

I dont know

user19161

@Jordan It is strange because you have intentionally inverted the order.

Jordan

I have to take calc 2 over again in the summer, I am screwed, it is a five week class here

It id easier for me to type, db

user58512

lol

@Jordan, did you do calc 1 ?

Charlie

@nimza Alexeyyyyyyyy! hahah

Nimza

6:13 PM

Hi, is there a name for a pair $(X,\to)$, where $X$ is a set, $\to$ is a convergence on $X$?

@Charlie param-pam-pam

Jordan

@user58512 I took calc 1 twice

user58512

you should be fine then

Henning Makholm

@Nimza What is a "convergence"?

Charlie

@Nimza ;P

user19161

So @jordan can you expand out now?

user19161

6:14 PM

8 mins ago, by Jordan

If I have (x^2 -2x -15)(x-2) is it wrong to FOIl it?

Nimza

@HenningMakholm a subset of $X^{\mathbb{N}} \times X$ with 3 axioms, should I write them?

user19161

It is 3 terms on the left and 2 on the right factor.

Jordan

@JacobBlack Yeah I see how it works, I just never looked past FOIL

user19161

So that is like (A+B+C)(D+E)

user19161

Rather than remember acronyms, one should understand how the algebra works. QED.

3

Jordan

6:15 PM

so AD + BD + CD + AE + BE + CE?

user19161

@Jordan Yes.

user19161

It is not hard to try have a feel of the distributive laws.

user19161

I will give you an example to help you.

Jordan

I think I have a better understanding

Henning Makholm

Rather than remember acronyms, one should understand how the algebra works. QED. QFT.

4

user19161

6:16 PM

Consider (a+b)(c+d)

user19161

Think of a rectangle with sides a+b and c+d

user19161

The product (a+b)(c+d) is then its area

user19161

Now this rectangle can be split up into 4 small rectangles.

user19161

Now the sum of the areas of the 4 rectangles is the area of the big rectangle.

Nimza

@HenningMakholm if we denote $(\{ x_n \}_{n=1}^{\infty}, x) \in \to$ by $x_n \to x$ then axioms are: 1) $x_n \to x \implies x_{n_{k}} \to x$, 2) $x_n \equiv x \to x$, 3) $x_n \nrightarrow x \implies \exists x_{n_k} \colon \forall x_{n_{k_l}} \nrightarrow x$

user19161

6:19 PM

@jordan The four small rectangles have areas ac, ad, bc and bd. See?

Jordan

Isnt the formula for area of a rectangle 1/2 base * height?

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$Mathematics$ Mathematics