lmfaoa

Whats a good way to look at completing the square

I know it's in the form a(x+d)+e

Completing the square ?

Ok, done with chemistry.

oh man, now you've got my brain thinking legendre transforms

@Dodsy A small correction: a(x+d)^2+e

oh right

What do you mean by a good way to look at it?

@Dodsy usual way I look at completing the square of $ax^2+bx+c$ is to write it as $a(x^2+(b/a)x)+c$

and then focus momentarily on the $x^2+px$ part (p=b/a).

So $y=-2x^2+24x-63$

That's the simplest case of completing the square, and you do it by recognizing it as being almost of the form $(x+p/2)^2=x^2+px+p^2/4$.

Sure, an example will do well here.

you've got $-2(x^2-12x)-63=0.$

Can you see how to complete the square of $x^2-12x$?

uhhh

$(x-6)^2$

no

Close

You want to subtract off a constant

Almost. If you expand that out, how does it differ from $x^2-12x$?

it would have +36

Right. So $x^2-12 x = (x-6)^2-36$.

right.

So therefore you've got $-2((x-6)^2-36)-63=0.$

Test: $\fbox{(x-6)^2-36}$

test?

If I expand the outer parens and regroup, that's $-2(x-6)^2+9=0.$

Test: $\boxed{(x-6)^2-36}$

he's testing mathjax stuff

@Dodsy I was trying to see what the LaTeX command does

@Semiclassical sorry what did you do there?

fbox apparently reads it as text, boxed reads it as math

$$-2((x-6)^2-36)-63 = -2(x-6)^2-2(-36)-63 = -2(x-6)^2+72-63=-2(x-6)^2+9$$

Oh right

it was in the brackets

Right.

that makes sense.

So now you've got $-2(x-6)^2+9=0$. Can you see how to solve that?

$-2(x-6)^2 = -9$,$\sqrt\frac{-9}{-2} + 6 = x$

but it doesn't work...

Correct. Interesting, however; why did you write it as (stuff) = x instead of doing step-wise?

Also, you want a $\pm$ in front of the square root

hm.

What would step wise entail?

To say it a little differently, $x=\sqrt{\frac{-9}{-2}}+6=\frac{3}{\sqrt{2}}+6$ is a root of your quadratic.

It's just not the only one

Like, divide by -2 both sides to get $(x-6)^2 = \frac{-9}{-2} = \frac{9}{2}$, then taking square root both sides to get $x - 6 = \cdots$, and then adding $6$ both sides to get $x = \cdots$

I mostly want to know why you wrote it as (...) = x

And not x = (...)

Did you "revert" the functions in some mechanical process instead of unwrapping it step-by-step?

@Secret Might be better to represent that in a matrix form.

doesn't really help in root solving , though

Sorry my laptop died

RIP in peace

You might be right there.

@Secret what do you want to even do

@BalarkaSen I don't usually do the math on paper

I just do it in my head and then write it down

@Dodsy Ah, alright

$x^5+x+1$ is reducible @Secret

You can factor it into a quadratic and a cubic

$x^5 + x + 1 = (x^2 + x + 1)(x^3 - x^2 + 1)$ I think

But if it would serve me bettter to write it out, then maybe I should start doing that

Yeah

I am not sure what exactly I want to do with this, I just felt like completing the penteract, then tesseract, then cube then square

You might look at cyclotomic polynomials.

@BalarkaSen In fact, it's easy to see that the first of those is a factor, since $\omega$ (the cube root of unity) is a root of $x^5+x+1$

I think that's an example of one.

@Semiclassical what did you mean "it's not the only one"

@Dodsy If you do it correctly in your head, do that!

There's two real solutions to that equation :)

I am bad at head-calculating myself

@Secret Yeah, that kinda fails for cubics and up, doesn't it?

(I think they're both real)

@Semiclassical but it's the vertex, is it not?

@AkivaW Right.

No.

$x=6$ is where the vertex would be.

I thought that would be the roots

Note that it's of the form $y=-2(x-6)^2+9.$

(x-3) an (x+3)

@AkivaWeinberger That's why I don't really get polynomials. It's one of those things that are "nonlocal" i.e. you cannot even figure out easily how the graph changes by varying one coefficient of powers of x, yet this thing is supposed to behave like a vector

Without knowing the answer beforehand, I like the following process: $x^5 + x + 1 = x^5 + x^4 + x^3 + x^2 + x + 1 - x^4 - x^3 - x^2 = x^3(x^2+x+1) + 1(x^2 + x + 1) - x^2(x^2 + x + 1) = (x^2 + x + 1)(x^3 - x^2 + 1)$

oh you're right...

The first term is always negative unless $x=6$, in which case it's zero.

oops

So it'll start at $(x,y)=(6,9)$ and fall off to the left/right

So one root should be a bit to the right of 6, and the other a bit to the left.

right

so then $x = \pm\sqrt\frac{9}{2} + 6$

Right. You can use \pm for $\pm$.

completing the square is super useful because with that form, you can pin down where the vertex and other geometric features are. Sadly that fails for cubics and up

Right.

So how can you tell the vertex?

just because of what's in the brackets?

Well, suppose I've got it in the form $y=a(x-b)^2+c$.

right.

Then the first term is either strictly positive or negative (depending on the sign of a) unless x=b.

@Secret You can sort of solve for local extrema with $f'(x)=0$

So the vertex will be at $(x,y)=(b,c)$ and the parabola will curve up/down in both directions.

But, yeah, quadratics are much nicer

especially since all parabolas are similar

By contrast, if it was $y=a(x-b)^3+c$ then the curve curves up in one direction and down in the other.

e.g. $x^3$

@Semiclassical so given $y=-5(x+6)^2 + 9$

then the vertex is (-6,9)

Right.

and, given $y=5(x+6)^2+9$

No way I can do better than 9, since if I move away from x=-6 then y goes down.

then the vertex is (-6,9)

any good problems today? I don't have anything fun to do, just write

Right. In that case, though, the parabola curves up not down.

oh right

Hence the sign of a business.

So to find the roots, you solve for x

and to decide which way the parabola opens you look at the coefficient

This is also handy for optimization stuff, since it tells you immediately that the largest output you could ever have for $y=-5(x+6)^2+9$ is $9$. The first term can only ever make things worse.

Right.

Note that that also lets you see whether there are roots in the first place. For instance, if $y=(x-2)^2+1$ then the vertex is $(2,1)$ and the curve bends up.

How can we tell if a parabola opens to the right or left

It opens in the same direction on both sides.

Alright, I think I genuinely understand that now.

Mmkay.

@MikeMiller I found a nice description of a decomposition of R^3 into circles today.

Thanks @Semiclassical

#mypersonaltutor

I think I've seen this

it's not a foliation right?

I don't know a foliation though

One way to sum things up: If the vertex is at $(x_0,y_0)$ and $y_0>0$, then the only way to get (real) roots is if $a>0$.

Yeah

And if $y_0<0$, then you need $a<0$.

There's a paper by Vogt that constructs a foliation by circles.

I have to include that parenthetical because, after all, $y=x^2+1$ has complex roots $x=\pm \sqrt{-1}$.

It definitely doesn't have real roots, though, since the curve can't get below $y=1$ for any real $x$.

Blah blah blah

@Semiclassical so to go over completing the square, we take $-2x^2+24x-63$ and make it $-2(x^2 -12x) -63$ and then slightly factor it to $-2[(x - 6)^2 -36] -63$

Then simplify.

My take home final is due in 5 hours and I'm still trying to figure out this damn ALOHA protocol throughput analysis using markov chains :'(

Right.

Okay great :)

The description was that, throw out the z-axis. Then you can foliate it by annuli around z-axis of decreasing radii, and foliate the annuli by circles themselves. So R^3 - line admits a decomposition by circles; re-embed the line as a "U" shape so both the ends go the same way. Now embed another line of the "U" shape inside a tubular neighborhood of the 1st "U" (which is an R^3 itself!), then foliate the exterior of the 2nd "U" inside the 1st

and since the coefficient is negative, the graph opens downwards

Correct to the second.

Keep doing this so you get it decomposed outside a decreasing intersection of "U" shaped solid cylinders

Once you expand that, you find as well that the overall +c is 72-63=9.

If you place it right, you'll push it off to infinity, so you have covered everything

If anyone wants to save me, please check this out cs.toronto.edu/~marbach/COURSES/CSC358_S17_1/multiaccess.pdf

right.

So you start above the y-axis and the parabola bends down.

From there it all hangs together.

@MikeMiller Huh, interesting.

Let me look it up

This is going to sound weird.

The main strategy there is to get rid of the annoying a and c in the original ax^2+bx+c form.

But I'm having trouble factoring $x(2x-5)-2(x^2-3x-4)$

and I know know why

Expand it out first?

Well, when you see something like that, I'd first check to see if both terms have a common factor.

But the only roots of the first term are x=0 and x=5/2, which don't seem like roots of the second equation. So they probably don't have a factor in common.

so then it would be $(2x^2 - 5x)-(2x^2 + 6x +8)$ = $(x^2 +x -8)$

Check thy algebra.

The distribution is right, but the subtraction isn't.

oh

eh, now your distribution is wrong :/

:/

$-2(x^2-3x-4)=-2x^2+6x+8=-(2x^2-6x-8)$.

The last term is what you had originally, and it was right.

Gosh darnit

So you've got $(2x^2-5x)-(2x^2-6x-8)$. That was correct.

What happens when you take the difference?

$x^2 + x +8$ ?

Look at the quadratic terms again

(the $x^2$ terms)

$(2x^2)-(2x^2)$

oh so that becomes 0?

Right.

so then it's just (x+8)

yup. So what's the root?

Yup

Oh okay that makes much more sense.

X = -8

not sure why I was having so much trouble with that.

So if I'm given a function and told to find the equation of the tangent line

and then told that it intersects the graph at another point

can I just equate them to each other?

Well, do you have an example?

for instance: $f(x)=2x^3 + 10x^2 -28x$ at $x=-3$, $y=-34x+18$

x = -3, sorry.

I don't even know how to start on this problem...

Well, depends on what you're doing.

Do they want you to find the tangent line, or is that given?

If it's given, you can just equate the two.

Ah. So they do want the tangent line itself.

So I found the tangent line to be -34x + 18

Ah, you've found that. Then yes, you can just equate the two.

However, there's something worth noting.

If you rearrange your tangent line a bit, it's equivalent to $y=-34(x+3)+120$.

Right, that was one of the forms I had

Right. That's a useful one here, because it focuses attention at $x=-3$.

The significance of this is that you know $f(-3)=120$.

when x = -3, y = 120

Or, more suggestively, that $f(x)-120=0$ has zero at $x=-3$.

hm.

Where I'm going is this. To solve the equation, we indeed should set them equal.

So that's $f(x)=y=-34(x+3)+120.$

If I move that $120$ over, though, I get $f(x)-120=-34(x+3)$.

yeah but then we have $2x^3 + 10x^2 -28x = -34x + 18$

which seems like it would be hard to solve...

oh so f(x) = 120

at x = -3

Right.

So let's remember: If $f(x)-120$ has a zero at $x=-3$, what does this tell us about the factors of $f(x)-120$ ?

that x = -3 is a factor

?

waves hai

Close enough: That $x+3$ is a factor.

yo

right.

So $f(x)-120 = (x+3)p(x)$ where $p(x)$ is some quadratic polynomial.

But now let's look at our equation: $f(x)-120 = (x+3)p(x) = -34(x+3)$.

So, we divide the function?

by the factor?

oh

Right. You're allowed to do this, because we're not interested in the root at $x=-3$.

We already know it's there

so then $p(x)$ = -34

Right. So now you just need to find $p(x)$ and solve the resulting quadratic equation.

So we definitely need long division here? or is there a way around it.