Mathematics - 2017-06-22 (page 1 of 7)

heh, that's something that matters in quantum mechanics @akiva

Akiva Weinberger

Here's a possibly easier one:

Namely, it's the basis for Bloch's theorem in solid state. That's huuuuuge

Akiva Weinberger

Prove that a linear operator that commutes with $E$ commutes with $E^{1/2}$.

Daminark

Hey @Dodsy, just saw this, and I'm alright, you? I'll check it out

Akiva Weinberger

12:02 AM

$E^{1/2}f(x)=f(x+\frac12)$

@Semiclassical Cool. He mentioned something about quantum mechanics I think but I didn't really understand it

hello

Bloch wave

A Bloch wave (also called Bloch state or Bloch function or Bloch wave function), named after Swiss physicist Felix Bloch, is a type of wavefunction for a particle in a periodically-repeating environment, most commonly an electron in a crystal. A wavefunction ψ is a Bloch wave if it has the form: ψ ( r ) = e i k ⋅ r ...

How do I find the roots of 3+4sin(2x)?

I diff it 8cos(2x)
I try to solve for zeros,
8cos(2x)=0
cos(2x)=0
arccos(cos(2x))=arccos(0)
2x=pi/2
x=pi/4?

More generally one talks about Floquet theory in ODEs: en.wikipedia.org/wiki/Floquet_theory

@Computer Zeros of the derivative don't tell you where the zeros of the function are. (There's some info you can glean, but not that much.)

I say its a min because when I diff it again, I have -16sin(2x) which is a negative number

12:05 AM

@Computer Not if $x=3\pi/4$.

Typhon

@Semiclassical me neither. I'm mostly just flabbergasted at this point. I thought me and amWhy were on good terms. Probably another dupe or something.

well basically what I am trying to do is "Find a lower bound and an upper bound for the area under the curve by finding the minimum and maximum values of the integrand on the given integral:" 0 to pi/2 of 3+4sin(2x)

Ah. I guess I don't understand why you labelled that as "find the roots of 'blah'"

Did you mean 'find the critical points'?

I say find zeroes from algebra but I guess they are critical points, yes

But those won't be zeros of the function :/

they're zeros of the derivative of the function

Anyways. What you've got boils down to finding solutions of $\cos(2x)=0$.

First, from what you know of a cosine function, when does it equal zero?

12:11 AM

for all pi/2

That's one solution.

But not the only one.

3pi/2

-pi/2

et cetera?

Right. More generally, pi/2 + pi*n for some integer n

yes

So you need $2x=\pi/2+n\pi$ here

So that's $x=\pi/4+n\pi/2$. (you saw nothing)

...in which case, yeah $x=\pi/4$ is the only root in the interval $(0,\pi/2)$.

12:13 AM

I have that x = pi/4

okay

What was the next step? (I want to check how you understood it.)

so, I am confounded. I have evaluated my minimum as pi/2 * f(pi/4) and it is wrong?

no, I was just making things more complicated than they ultimately had to be :/

You've indeed got a critical point at $x=\pi/4$. How did you figure out min vs. max?

I took the derivative of 8cos(2x) and it ended up as -16sin(2x) so I said it was a min

Careful. -16sin(2x) as written is an expression, not a value.

For instance, if I plug x=3pi/2 into there it would actually be positive.

12:19 AM

so -16sin(2x)=0
sin(2x)=0
arcsin(sin(2x))=arcsin(0)
2x=0
x=0?

no.

You'ore missing my point.

er

Where do I care about the value of the second derivative?

Do I care if the second derivative is positive at x=3pi/4?

no

Right. At what value of x do I care about it?

12:21 AM

pi/2+pi*n for all n in integers

?

That's where $\cos(x)=0$. You've got $\cos(2x)=0$.

oh 0

stare

cos(2*0) is not zero.

yeah, cos(pi/2)?

...okay, and what value of x is that?

12:24 AM

pi/4

Right. That's where the critical point is, and that's where you care about the second derivative.

So: what's the second derivative at that point?

4 sqrt(2)

Right. That's an actual value, and that is indeed positive.

so we have a maximum

Positive means concave up.

12:26 AM

oh minimum

Point being: When you say that it's a min, you need to be saying that the second derivative evaluated at the critical point is positive.

yes

Just saying "the second derivative is positive" is not enough, particularly when the second derivative is positive somewhere else.

makes sense

With that in mind, though, we can indeed conclude that the function has a local minimum at x=pi/4.

12:28 AM

but I have now a min at x=pi/4...

(In fact, -8sin(2x) is positive for all 0<x<pi/2. so it's actually true over the entire interval of integration)

i have some integrand 3+4sin(2x) on the bounds 0 to pi/2

Right.

Note, though, that when checking for min/max you also should check on the boundary.

s0 f

so f(0) and f(pi/20

Right. What values are those?

12:29 AM

f(0)= 3

f(pi/2) = 3

Right. And what was f(pi/4)?

7

...huh.

Something's fishy.

f(x)=3+4sin(2x)

f'(x)=8 cos(2x)

f''(x) = -16 sin(2x)

2*2^(1/2)+3

Oh, woops.

what's f''(pi/4) ?

12:32 AM

-16

Right. (You said 4sqrt(2) earlier)

That's negative, so the function is concave down at that point

hence, maximum not minimum.

which makes a lot more sense given than 7>3.

So you're saying what exactly?

pi/4 is my max?

Right.

pi/2 is my min bound

?

Careful.

That's where the min/max occur.

They're not what the bounds on f(x) will be.

And those are what will matter.

So what is the min/max for f(x)?

12:35 AM

0,pi/4?

Are those values of f(x) ?

pi/4 is a value of f(x)? no?

No. pi/4 is a value of x. f(pi/4) is a value of f(x).

so f(pi/4) = 7

on the upper bound

Right. That's the max value of f(x).

12:37 AM

on the lower bound f(0)= 3

3,7?

Correct.

Those are the upper and lower bounds.

whether it is f(0) or f(pi/2) its still 3

Yeah. In fact, the integrand is symmetric around the point x=pi/4.

So you could just as well integrate from 0 to pi/4 and double the result. In that case there's a unique max and unique min.

so those are my bounds, I still have the integrand of 3+4sin(2x) and the given bounds of 0 to pi/2 to evaluate maximum and minimum value.

Right.

12:39 AM

Unless you're saying those are my maximum and minimum areas?

no, not quite.

You'll use those to get the min/max areas, but they themselves aren't the min/max areas.

But, imagine you drew a graph of y=3+4 sin(2x) along with the horizontal lines y=3, y=7.

so pi/2*f(pi/4)= max area?

Right.

or is it pi/2*f(7)?

Again, is 7 a value of x or a value of f(x) ?

12:41 AM

oh pi/2*7

Right.

as my maximum area

and 3pi/2

as minimum

Visually: The curve y=3+4sin(2x) lies below the line y=7. So the integral is bounded above by 7*pi/2

and similarly 3*pi/2 below

In fact, I think the true value might be 5*pi/2.

true value for the total area?

Right.

12:43 AM

hm

thank you

that was very helpful

If I'm visualizing it right, the curve should be a sinusoidal oscillation around the line y=5

and that oscillation should spend as much time above as below the line, so it'll integrate to zero area.

I don't know what sinusoidal means

sine or cosine

nvm sounds like it has something to do with sine

or anything that has the same shape

12:44 AM

spring shape

Oh, I'm wrong actually. Lemme show the picture.

what is worthy of noting in this picture?

hang on, picture's silly

On there, you see very clearly that the red line is bounded above by 7 and below by 3

However, my final assumption was wrong: It spends a bit more time above y=5 than below

So the integral will actually end up being a bit bigger than 5pi/2

1:09 AM

What's the primary intuition behind Radius of Convergence

Radius of convergence

In mathematics, the radius of convergence of a power series is the radius of the largest disk in which the series converges. It is either a non-negative real number or ∞ {\displaystyle \infty } . When it is positive, the power series converges absolutely and uniformly on compact sets inside the open disk of radius equal to the radius of convergence, and it is the Taylor series of the analytic function to which it converges. == Definition == For a power series ƒ defined as: f ( z ) = ...

From intuition the Raidus of COnvergence can informally be described as the distance of convergence from the centerpoint of our disk to the boundary of our disk

^ is my intuition correct I feel like it's not

@Zophikel That's basically right, though I'd say it a little differently.

@Semiclassical how would you say it i'm just glad my intuition is correct

For any power series, there's the point around which you're expanding. (i.e. if it's got powers of $x-x_0$ it's an expansion about $x=x_0$.)

yeah all right

If $R$ is the radius of convergence of a series $\sum_{k=0}^\infty a_k(x-x_0)^k$, then that series will converge absolutely for any $x$ within $R$ of $x_0$.

1:19 AM

ahhh ok

Moreover, for it to be the radius of convergence we also want that to be strict i.e. I can't make it any bigger.

aahh all right @Semiclassical

For instance, if $y=1+x+x^2+\cdots$ then the radius of convergence is 1.

If $-1<x<1$ then that converges to $1/(1-x)$ without issue.

If $x\to -1$, though, that's the series $1-1+1-1+\cdots$

The terms don't go to zero, so that's not a convergent series.

@Semiclassical and that's because power series will diverge or converge for specific values yes

And more obviously the series at $x\to 1$ would be $1+1+\cdots$.

In both cases that's not something convergent.

A somewhat weirder example is $1-x^2+x^4+\cdots$

In that case, both of the series at $x=\pm 1$ are alternating, so they're not so obviously divergent. But yeah, they in fact do diverge

1:24 AM

all right so what makes dertainming convergence of a power series is the values can alternate

Not really. For $1+x+x^2+\cdots$ it'll be $1+1+\cdots$ at $x\to 1$.

There's not any hope of that converging.

The example I really should be doing, though, is $x+x^2/2+x^3/3+\cdots$.

ahh ok @Semiclassical usually the way I check for convergence is by using tests

Yeah, and this is a good one for that.

@Semiclassical without using formal tests i'd say the series diverges

But one thing you can notice right off is that for $x=1$ you've got $1+1/2+1/3+\cdots$

1:26 AM

yeah

so yeah, harmonic series---diverges.

On the other hand, if $x\to 0$ then the series straightforwardly vanishes. So it's not clear based on that what the radius of convergence could be.

At most it could be 1.

@Semiclassical the easy way to deal with series of the form is to check if the $\lim$ exists and also depends on what the radius of convergence is defined to be

One thing that is suggestive, though, is what happens for $x=-1$:

$1-1/2+1/3+\cdots$

So for power series diverging or converging completly depends on the Radius of Convergence

Right. It characterizes it entirely.

You can use the alternating series test to see that the above series indeed converges.

1:29 AM

Wow thanks @Semiclassical :-) I owe you a math textbook

However, it doesn't converge absolutely

For that you'd need $\sum_n |a_n|=\sum_{n=1}^\infty 1/n$ to exist not just $\sum_n a_n$.

ahhh ok

and that's obviously not going to happen.

yes true

So you have that the series, which is centered at $x=0$, diverges at $x=1$ and is conditionally convergent at $x=-1$.

1:30 AM

@Semiclassical what about for a Double Series would it be the same case

No idea.

:( that should be an interesting question for MSE :)

And you should find that it has radius = 1. (I almost want to say that this follows from the first two facts, but I really don't know.)

Makes me wonder, in fact.

Suppose $a_n>0$ and $\sum_n (-1)^n a_n$ is conditionally but not absolutely convergent. Must the radius of convergence be 1?

Hmmmmmmm

Back later

@Semiclassical i'll save this for later

Soham Chowdhury

1:57 AM

@BalarkaSen starting again now that I'm less unsure about college.

what's with all the flagging nonsense? I leave y'all for a week and ...

2:08 AM

@Semiclassical I was going to write it as $(3-x)^{-2}-(1-x)^{-2} > \frac{1}{2}$

then simplify

not sure if that's right

probably not the same thing

so I guess first we square the $\frac{1}{2}$

oh I see my mistake

$(3-x)^{\frac{1}{2}}-(1-x)^{\frac{1}{2}} > \frac{1}{2}$

still doesn't relly do anything.

I should probably learn more math before I start trying these problems.

@Daminark I found this thing from Cambridge that they expect all incoming math grads to do in the summer before beginning their degrees maths.cam.ac.uk/sites/www.maths.cam.ac.uk/files/pre2014/…

anyways i don't know any of it

2:35 AM

@Dodsy First rule of thumb is move one of the square roots to the other side and then square.

The one thing to take care about is that you're in a situation where both sides are positive. (you wouldn't want to do this with -2<2 since you'd get 4=4.) But if I move the -sqrt term to the other side that's not a problem

2:51 AM

ah I see

So where does that put you?

hmmm.

$x-3-\sqrt{1-x}>\frac{1}{4}$ ?

That'd be $a-b>c\implies a^2-b^2>c^2$.