Mathematics - 2014-08-28 (page 1 of 4)

Chris's sis

12:02 AM

Such a beautiful achievement, I think I'll sleep like a child this night!:-)

@robjohn I'm about to finish $$\sum_{n=1}^\infty\frac{H_{n}}{2^nn^4}$$ as well.

Mike Miller

5:44 AM

@blue do you have any thoughts on the question I linked above?

(I am willing to break my principles and ping people out of the blue for this one)

blue

hmm

GBeau

I'm being asked to express symbolically for every $y$ there exists an $x$ such that $x^2=y$

and we're letting $P(x,y)$ be the statement $x^2=y$

my question is mostly typographical, I guess...is there a normal way to separate this out so it looks better: $\forall y \exists x, P(x,y)$

like, how should I be formatting that?

blue

you can use \, for a small space

GBeau

$\forall y\, \exists x, P(x,y)$

blue

$\forall y\,\exists x \,P(x,y)$

GBeau

5:56 AM

my professor was switching between commas and colons today in his notes

Mike Miller

one can explicitly write down the order of GL_n(Z/mZ) so I've some hope this can be done with nothing more than some calculation

the non-abelian case will probably not fall to an attack like that

GBeau

@blue that looks much better, thanks

blue

at the extreme end of the spectrum, one could write a computer program that computes all G for which |Aut G|=n, and then find the minimum and maximum values of |Aut GxG|. this goes to show that we could theoretically say more than just 2|Aut G|^2 for a lower bound, but a more nuanced bound would depend sensitively on n. perhaps some arithmetic could be used. and then there's the upper bound to consider.

GBeau

and I'm not missing some subtlety in the way this is done, right?

blue

@GBeau ?

Mike Miller

6:00 AM

@blue I'm satisfied with the given lower bound. what I really, really want, is an upper bound that's polynomial in |Aut G|.

GBeau

the question seemed too simple, so I was worried if I'm missing something obvious

so now I have the negation of the statement is $\exists y\, \forall x, \neg P(x,y)$

In "solving" for that negation, I have this as an intermediate step: $\exists y\, \neg \exists x, P(x,y)$

is that the best way to format that typographically as well?

Chris's sis

Greetings @r9m @robjohn

robjohn

@Chris'ssis hey there... just get up?

Chris's sis

@robjohn Yeah (actually half an hour ago). My dogs are miraculously healed after praying to god for their healing. Yesterday they were about to die, now they are in a perfect shape.

robjohn

6:15 AM

@Chris'ssis close enough. Just curious, not spying :-)

@Chris'ssis what happened to them?

Chris's sis

@robjohn Some disease, but I don't know exactly what. At any rate they had some serious problems for some days. No eating at all.

Today everything is perfect!:-)

robjohn

@Chris'ssis Great!

Chris's sis

They eat, drink and play around like before. Here there were also other persons that had similar problems, all their dogs died in 4 days (from the beginning of the symptoms).

I'm very happy for that. For me they are like human beings. :-)

GBeau

6:32 AM

did I negate this properly: $\forall y\, \exists x, P(x,y)$ to $\exists y\, \forall x, \neg P(x,y)$ or should I keep looking at it

the negation seemed broader than I expected when applied to a case where P had meaning

Mike Miller

@GBeau yes, that's correct

Vrouvrou

Hello, someone speak french ?

Anthony

Alo.

Can someone answer a question I have about spherical harmonics?

Mike Miller

@Anthony what's a spherical harmonic

Anthony

<3

:(

Spherical solutions to Laplace's Equation, or something like that.

Mike Miller

6:56 AM

nope nope nope

skullpatrol

hope hope hope

Hello

can i ask a question?

GBeau

$\exists x\in\mathbb{R} (x^2\leq 0)$ is true as written because 0 as an example?

that's sufficient?

Hello

oh sorry. didn't know you were in the middle of one

Mike Miller

@Hello you're always free to ask, but be warned you may not get an answer

GBeau

7:03 AM

would I need to do anything else other than provide an example to prove this?

Hello

$\int e^{2x} \sqrt{e^x+1}dx$

i think its u-sub

Mike Miller

indeed

Hello

i made (e^x)+1 u, and end up with 2√u

so 2√(e^x)+1

not sure where to go from there

anyone here?

insert my name here?

GBeau

$\forall x\in \mathbb{R} (\exists y\in \mathbb{R} (x\leq y))$ I assume my professor only wants intuitive proofs of these statements given our length into the course...

GBeau

7:24 AM

True, because for any given x there's always a y equivalent, clearly?

Hello

1

Q: Calculus (integration [multiplication])

Determine $\int e^{2x} \sqrt{e^x+1}dx$ Is there a multiplication rule for integration or something?

GBeau

stating merely the equals statement is sufficient here, correct?

Hello

@GBeau where is everyone

blue

@Hello what don't you get in the answers?

they've explained things pretty clearly methinks

have you ever done or seen substitution before?

Hello

Yeah i got 2√u when i did it and i'm not sure where to go from there

i have the answer but i'm not sure how to get to it

blue

7:31 AM

if u=e^x + 1 then (e^2x)√(e^x + 1) becomes what?

2√u like i said

no

what is it then?

what does e^(2x) become?

Hello

when integrating?

blue

7:32 AM

write e^(2x) in terms of u, where u=e^x + 1

Hello

let me just show you how i did u-sub

because i'm not too sure what you're getting at

u = (e^2x) + 2

du/dx = e^(2x)

blue

wait, you're doing u=e^(2x)+2 ??????

jesus

Hello

1**

blue

tell me again, what's your u?

Hello

dx = du/e^(2x)

e^(2x) + 1

blue

7:34 AM

but the thing underneath the √ sign is e^x + 1, not e^(2x) +1, isn't it?

Hello

lol. that's what i meant. my bad

u = e^x + 1

good

final^

so continue

dx = du/e^x

7:35 AM

yes

Hello

the e^x cancels and the e^(2x) becomes a 2

blue

huh?

Hello

hard to explain in text

blue

e^(2x) divided by e^x is not 2

Hello

my mistake is at that step then

r9m

7:36 AM

@Chris'ssis hi :)

Chris's sis

@r9m Hello! :-)

blue

@Hello what is e^(2x) divided by e^x ?

Hello

e^x?

blue

yes

Hello

ok

ill continue and get back to you

blue

7:38 AM

so you have (e^x) √(e^x + 1) du, how do you write that in terms of just u?

r9m

@Chris'ssis I have too many classes today, over that no sleep last night 'cos of construction noise :(

blue

first write e^x in terms of u, then write √(e^x + 1) in terms of u

Hello

im not sure what you mean in terms of u

Chris's sis

@r9m I usually sleep like 5, 6 hours per night. Sometimes only 4 hours since I have much work to do here.

blue

@Hello you've never seen the words "write blah in terms of blah"?

Hello

7:39 AM

well yes. but im not sure how to write it in terms of u

blue

rewrite e^x as an expression involving the variable u but not the variable x

well, u=e^x + 1, so what is e^x in terms of u?

GBeau

I'm being asked to find the complement of $\{ x : x < -1\}\cup \{x : x \geq 3\}$ inside $\mathbb{R}$ and express the answer is set builder notation.

Hello

u-1√u or something?

r9m

@Chris'ssis I need atleast 6-7 hrs o/w I feel like a junkie vagabond while the classes go on =P

GBeau

proper set builder notation would be like: $\{ x\in\mathbb{R} : x \geq -1\wedge x<3\}$

Chris's sis

7:39 AM

@r9m :-)))))))) I'm still thinking you're some kind of professor ... but, well, ... :-)

GBeau

right?

blue

@Hello yes, the final integrand will be (u-1)√u du (remember your parentheses!)

@GBeau sure, or just $-1\le x<3$ for compactness

Hello

do i replace all the u's now?

wait

lol nevermind

blue

to integrate (u-1)√u du, distribute (u-1)√u and write it as (u^power - u^power)du, then you can use the power rule

r9m

@Chris'ssis well if I were to be even near qualified as a professor .. :P that'd really reflect a poor image of our country education system .. but wait our education system on an average is poor :P .. so who knows =P

Hello

7:42 AM

u^(3/2) - u^(1/2)?

blue

@Hello right

Hello

now i can integrate them individually

and replace the u's?

blue

@Hello after you finish integrating with respect to u, you'll go back by writing u=e^x + 1

Chris's sis

@r9m :D

r9m

@Chris'ssis lulz .. I have never seen a 20 yr old prof anywhere in my country =P

Chris's sis

7:44 AM

@r9m really?

Hello

@blue thanks for your help!

Chris's sis

@r9m Still you have had great mathematicians during the time. By the way, I'd like to have all the books of the professor Srivastava ...

blue

np

GBeau

@blue let's say I was writing these on one line and then the next...would it be more proper to say those two lines were $=$ or $\equiv$?

Hello

@blue can i ask you another?

r9m

7:46 AM

@Chris'ssis I don't have a list of all his books :| .. are they available online ?! idk :o

blue

@GBeau equality of sets is expressed with =

@Hello sure

Hello

here's the question (the pictures are there so it's easier to look at): http://www.purplemath.com/learning/viewtopic.php?f=14&t=3946
i have no idea what that 3 means

GBeau

@blue thank you

Hello

the first 3 before the f(x)

wait i just got it

Chris's sis

@r9m math.uvic.ca/faculty/harimsri There you find what he published during the time.

Hello

7:47 AM

i'm just used to seeing y instead of f(x)

blue

the 3 is just a 3

r9m

@Chris'ssis ic :) thanks :D

Hello

yeah i got confused. answer is 5

Chris's sis

@r9m Welcome ;)

r9m

@Chris'ssis mother of god .. that list is gigantic :O

Chris's sis

7:50 AM

@r9m Yeah, there is much stuff to learn. :D

r9m

:)) .. gtg now .. classes after lunch :) bbl

Chris's sis

@r9m Have fun! :D

Hello

@blue how would i integrate 4/(1+4t^2)dt?

blue

do you know the antiderivative of 1/(1+t^2) ?

Hello

no. probably something i haven't learned. i just finished precalc and im doing an ap calc exam

nevermind that question

blue

7:53 AM

it's a trig function

Hello

same thing for 1/((√4-t^2))dt?

blue

more trig

Hello

ok

how would i find the maximum possible area for a rectangle?

a rectangle has one side on the x-axis and the upper two vertices on the graph of y=e^(-2x^2)

blue

let the left and right edges be line segments whose points have x-coordinates a and b respectively

compute the area of the resulting rectangle in terms of a and b

that's how you start presumably

Hello

0.6065

lol

well that's the answer but i'm not sure how to get to it

blue

8:01 AM

suppose the lower left corner of the rectangle is (a,0). what is the upper left corner then?

Hello

(a, something)

would it be b?

blue

I haven't introduced b yet

remember your own words

> the upper two vertices on the graph of y=e^(-2x^2)

Hello

(a,1)?

blue

if x=a then what does your formula say y is?

Hello

e^(-2a^2)

blue

8:04 AM

yes, the point is (a,e^(-2a^2))

the other point is (b,e^(-2b^2))

Hello

those are the two top points?

blue

convince yourself that in order for these two points to be connected by a horizontal line (the top edge of the rectangle), we must have a=b. either look at the graph to see this, or argue e^(-2a^2)=e^(-2b^2) implies a=+/-b

@Hello yes

Hello

i'm looking at the graph. it looks like a hill

how do numbers come into play to find the maximum area?

blue

what is the area of the rectangle in terms of a?

you compute that, then maximize it if you can

Hello

(e^(-2a^2))(a+b)?

blue

8:08 AM

no

the height is indeed e^(-2a^2), but the length is 2a

Hello

2a?

blue

since it goes from (-a,0) to (a,0) (the lower corners)

Hello

how do i "maximize" it?

blue

the area is 2a e^(-2a^2), do you know how to maximize or minimize a function (of one variable)?

Hello

maybe if i hear it, but not at the top of my head

blue

8:10 AM

usually you learn how to do a problem before the problem is asked of you...

Hello

well i just finished precalc and i was assigned an ap calc exam

so maybe this isn't one of those problems

blue

one doesn't usually take an ap calc exam right after a pre-calc class

Hello

it's my summer assignment for ap calc

blue

okay

Hello

if it's something i didn't learn i have no problem with not doing this question

blue

8:12 AM

maximizing/minimizing a function means finding the extreme values, and plugging them in

extreme values of a function occur at the endpoints of its domain (including +/- infinity) and arguments for which its derivative vanishes

Hello

der=0?

blue

f'=0, yeah

Hello

f' = e^(-2x^2) * -4x?

blue

f(a) = 2a e^(-2a^2)

can you find the derivative of that?

Hello

i need the derivative of the derivative?

blue

8:16 AM

no

Hello

that's the derivative^

blue

the function you're trying to maximize is f(a) = 2a e^(-2a^2)

Hello

oh yeah

blue

you need to compute its derivative f'(a)

Hello

i was looking at the original

no i don't know how to

blue

8:19 AM

product rule

and chain rule

GBeau

woah uhh

$(A\setminus B)\cup (B\setminus A) = (A\cup B)\setminus (A\cap B)$

I'm being asked to prove this by translating to logic statement for all x

I've written $\{x: (x\in A\wedge x \notin B) \vee (x\in B\wedge x \notin A)\}$

Hello

@blue 2e^(-2x^2)-8x+2

blue

@Hello no

GBeau

can you prod me in the right direction here? o.o

blue

@GBeau can you write the second one in logical form too?

burn the candle from both ends

GBeau

8:22 AM

I'll work on that

Hello

@blue -16x?

blue

haha, no

Hello

(2)(e^(-2x^2))(-4x) + (e^(-2x^2))(2)

first term (2x)*

blue

yes

so (2-8x) e^(-2x^2), when is that 0?

Hello

take away the +2 and replace the -8x with -16x

blue

8:25 AM

where are you getting 16?

Hello

nevermind

forgot about the + sign

why didn't you add the e terms

?

areyou multiplying those two?

blue

multiplying what two?

Hello

2-8x and e

the terms

up there

blue

the first term is (-8x) (exponential), the second term is 2(exponential), adding them together is (2-8x)(exponential)

Hello

yes

blue

8:28 AM

when you write two things next to each other without any symbol between them, it means multiplication

Hello

what about the two e terms

add them?

blue

did you read what I said?

both terms have (exponential), so you factor it out

factoring out a common term is the distributive property in reverse

Hello

so it's not e^(-2x^2) + e^(-2x^2)?

GBeau

@blue I've written out so far $\{x\in A: x \notin B\}\cup\{x\in B: x \notin A\} = \{x: x\in A\vee x \in B\}\setminus \{x: x\in A\wedge x \in B\}$

representing $(A\setminus B)\cup (B\setminus A) = (A\cup B)\setminus (A\cap B)$

blue

@Hello it's (-8x)e^(-2x^2) plus 2e^(-2x^2), why are you erasing things from the expression?

you're not allowed to erase things you don't like from expressions

Hello

8:30 AM

yes that's what i have

where did you go from there?

GBeau

or uh, I shouldn't have changed the left

blue

@Hello which equals (2-8x) e^(-2x^2), by factoring out the common term e^(-2x^2)

Hello

gotcha

GBeau

$\{x: x\in A\wedge x \notin B\}\cup\{x: x\in B\wedge x \notin A\} = \{x: x\in A\vee x \in B\}\setminus \{x: x\in A\wedge x \in B\}$

blue

being able to factor common terms out of sums of terms is an important skill you need to have

Hello

8:31 AM

i knew how, i just didn't know you were doing that. hard in text

blue

@GBeau you don't want to show equality using set operations, you want to show two logical statements are logically equivalent

namely, you want to show $(x\in A \wedge x\not \in B)\vee(x\not\in A\wedge x\in B)$ is logically equivalent to $(x\in A\vee x\in B)\wedge \neg (x\in A\wedge x\in B)$

Hello

so x=1/4 when it's set equal to 0?

blue

you could compactify the notation by writing $(P\wedge \neg Q)\vee(\neg P\wedge Q)\iff (P\vee Q)\wedge \neg(P\wedge Q)$

@Hello correct, so you plug 1/4 back into the area formula

GBeau

@blue so $\setminus$ can go to $\wedge \neg$

blue

@GBeau right

GBeau

8:34 AM

I guess, I see this logically

blue

@GBeau if you wanted you could just write truth tables

methinks

Hello

.4413?

GBeau

@blue my sole concern here was not deviating from whatever my professor wants us to write in this homework

at least he's super lax about help; his policy was "get help from everybody you can and work together, but write out your own solution"

he goes off on a lot of tangents so we sometimes barely mention topics that end up on homeworks :s

blue

@Hello apparently

Hello

.4412 i meant

apparently?

blue

8:38 AM

@Hello you plug 1/4 in and that's what you get, and we seem to have done everything right, so that looks to be the answer (even though you originally stated the answer was like 0.6)

Hello

yeah the answer is .6065

blue

ah, actually you did compute the derivative wrong

(2a e^(-2a^2))' = (2a)' exp(-2a^2) + (2a) exp(-2a^2) (-2a^2)' = 2e^(-2a^2)-8a^2 e^(-2a^2) = (2-8a^2) e^(-2a^2)

so a=1/2, not 1/4

Hello

oof i see

GBeau

@blue I must be close $\forall x, (x\in A\wedge x \notin B) \vee (x\in B\wedge x \notin A) \equiv \forall x, (x\in A\vee x \in B) \wedge (x\notin A\vee x\notin B)$