Mathematics - 2015-03-30 (page 3 of 3)

meer2kat

20:04

okay i got nothing. i'm in shape math mode right now. have you looked at paul's online notes?

example 4 might help

Owatch

I'll ask the professor.

evinda

@eleanora Hello!!! Is your name greek?

Owatch

I am doing another problem right now anyways.

Thank you for trying at least.

meer2kat

mhmm i'll be able to do this stuff again in the fall when i have to take real math again

meer2kat

20:16

if we only know that a quadrilateral has one pair of parallel and congruent opposite sides, what does it have to be? a parallelogram?

Owatch

Is there no way to write what x = in $x^{2}+1 = 0$?

meer2kat

i hate this geometry crap

@Owatch not in the real system

@Owatch if you want to go imaginary maybe

Owatch

$\sqrt{|-1|}$

meer2kat

@Owatch i

Owatch

But I'm not allowed to add that.

Abs

meer2kat

20:17

@Owatch ah

@Owatch no you can't do that.

Owatch

This is frustrating.

meer2kat

that would mean that it had an x intercept of 1, which is super not true

@Owatch same problem?

Owatch

Yes

I split it differently. . .

$\frac{A}{x^{2}+1} + \frac{B}{(x^{2}+1)^{2}}dx$

meer2kat

i found the way to do the partial fraction step, but it's cray. do you want to see it?

Owatch

$x^{2}+x+1 = A(x^{2}+1) + B$

meer2kat

20:21

it involves matrices

Owatch

And there is the issue. I cannot make A zero.

Or I cannot find X such that A = 0

meer2kat

sorry moderator, but this is a long solution

If you can make sense of this, @Owatch ? I can't. But that's apparently the solution

Owatch

Nevermind

Thanks but I will just ask.

It cannot be that hard

Mike Miller

the numerator is $(x^2+1) + x$

split into two fractions

Owatch

It's by partial fractions though (The exercise)

The numerator does not play a part in the issue as far as I know yet.

I would use it to solve. But first I must make A or B equal to zero. I could not solve for B, as it had nothing in terms of X. But I could try to solve for A, since it was $A(X^{2}+1)$. Except you cannot solve for X there.

meer2kat

20:35

that's what i tried to do @mike

@owatc that works but you'd have to do partial fractions twice if you split

Karim Mansour

@BalarkaSen

Owatch

Yes I solved the other one!

@meer2kat Are you familiar with Approximate Integration?

Ramanewbie

20:56

how can I get the area of the pink stuff ? gyazo.com/89539e2b53c9307b0c7e202728250886

Owatch

Area of entire triangle - area of two unshaded ones?

Ramanewbie

I need the area of the pink rectangle, but I can't manage to find $AP$ or $MN$

P minus N ? what do you mean ? @owatch

Owatch

P to N

meer2kat

@Owatch vaguely. is that what you did?

Owatch

No.

Just a question since that was an upcoming problem I wanted to try.

@Ramanewbie I take that back. Hold on

Ramanewbie

21:03

How do we know $PM$ = $MB$ ? @owatch

ok

Owatch

@Ramanewbie Because then it would not be a rectangle

So it must be

It's corners are right angles.

meer2kat

@Owatch ah okay. i've done it before but not recently

Ramanewbie

@owatch wait you said $PN = 14-x$ ?

Owatch

No I said I took that back.

Hold on

Ramanewbie

ok

Owatch

21:08

@Ramanewbie I believe that (P to N) and (A to M) = 14-x.

Because M-B does certainly = 14-x. And since the rectangle has right angles, P-N must too.

Ramanewbie

@owatch not possible, $PN = x$ because it's a rectangle

Owatch

What?

Well they just wrote X there.

For the side

Ramanewbie

$x = AM$, not $AB$

Owatch

Sigh.

Ok wait

I've got to go

Sorry Raman

Just know that you can find the long missing side of the triangle

Because it is a right triangle

Ramanewbie

21:26

ok see you @Owatch

is there anyone here ?

@quid

quid

22:00

@Ramanewbie yes?

Mike Miller

hello @quid

evinda

Hello @quid!!! How are you?:)

quid

Hello @MikeMiller and @evinda

I am alright. Thanks.

Valery Saharov

22:14

Hi folks

evinda

Hi @ValerySaharov

Valery Saharov

Chris's sis confused me with some weird integral property. Could someone help me pls?

why is this positive if f(x) is convex?

I have been lurking internets and didn't find anything like that

I also can't figure out the proof

moreover I doubt it

evinda

Hey @Christopher

Christopher

hey evinda

evinda

@Christopher Are you a student?

Christopher

22:19

@evinda Yes

evinda

@Christopher Are you german?

Christopher

@evinda I can speak some of it. Why do you ask?

evinda

@Christopher I thought so because I thought that your name would be german

@quid Could I ask you something?
Suppose that we have: $f(t^{n+1},y^{n+1})=y^{n+1}+4 \pi \cos(4 \pi t^{n+1})y^{n+1}$

Does it then hold that:
$\frac{\partial{f(t^{n+1},y^{n+1})}}{\partial{y^{n+1}}}=1+4 \pi \cos(4 \pi t^{n+1})$ ? Or am I wrong?

quid

@evinda if there are no further dependencies then I think yes this is fine.

evinda

@quid Nice.. Thanks a lot :-)

quid

22:32

you are welcome @evinda

robjohn

There should be a FAQ regarding multiplication by a conjugate. We have a large number of questions about forms involving $\frac1{a+\sqrt{b}}$ etc. Does anyone know of an answer that is particularly good and deals with this topic?

@ValerySaharov Did you see my proof of this?

Jan 29 at 15:50, by robjohn

@Chris'ssis If $f$ is convex, then $$ \begin{align} f(\pi-x)&\le\frac\pi{2\pi-2x}f(x)+\frac{\pi-2x}{2\pi-2x}f(2\pi-x)\\ f(\pi+x)&\le\frac{\pi-2x}{2\pi-2x}f(x)+\frac\pi{2\pi-2x}f(2\pi-x)\\ f(\pi-x)+f(\pi+x)&\le f(x)+f(2\pi-x) \end{align} $$ Therefore, $$ \begin{align} \int_0^{2\pi}f(x)\cos(x)\,\mathrm{d}x &=\int_0^{\pi/2}[f(x)-f(\pi-x)-f(\pi+x)+f(2\pi-x)]\cos(x)\,\mathrm{d}x\\ &\ge0 \end{align} $$

0celo7

Does anyone here have experience with basic modular transformations?

columbus8myhw

Hey, how do you prove that $\lim_{x\to0}\dfrac{e^x-1}x$ exists? (This is equivalent to asking how to prove that $\lim_{x\to0}\dfrac{2^x-1}x$, since a simple transformation, $x\to\log_2x$, takes you between the two.)

Actually, I'm after a more specific question, which is similar:

How do you prove, rigorously, that there exists a constant $k$ such that $2^x\ge kx+1$ for all $x$? (Desmos tells us that $k\approx0.7$. Calc tells us that $k=\ln2$.)

0celo7

@columbus8myhw L'Hopital (for the first question)

columbus8myhw

@0celo7 Circular reasoning — you need to evaluate that limit in order to find the derivative of $e^x$ in the first place. Sorry.

robjohn

22:45

@columbus8myhw Use the inequality $e^x\ge1+x$

David Wheeler

@columbus8myhw That depends on how you define $e$

0celo7

^^

columbus8myhw

@DavidWheeler True enough.

0celo7

I define my $e$ such that $e^x$ is its own derivative.

:)

robjohn

@columbus8myhw Then substitute $u\log(2)=x$

columbus8myhw

22:46

@robjohn True, but...

...when I said this:

2 mins ago, by columbus8myhw

How do you prove, rigorously, that there exists a constant $k$ such that $2^x\ge kx+1$ for all $x$? (Desmos tells us that $k\approx0.7$. Calc tells us that $k=\ln2$.)

I meant, before you've already developed the tools of calculus. In particular, we haven't even proven yet that there exists an $e$ such that $e^x\ge x+1$ for all $x$, yet.

robjohn

@columbus8myhw You can show that $e^x\ge1+x$ using Bernoulli's inequality, which requires only induction.

columbus8myhw

@robjohn Elaborate, please? (I forgot what Bernoulli's inequality was...)

evinda

@robjohn @quid @DanielFischer I have to calculate approximations of the solution with the method $y^{n+1}=y^n+h \cdot [\rho \cdot f(t^n,y^n)+(1-\rho) \cdot f(t^{n+1},y^{n+1})] , n=0,....,N-1 \\ y^0=y0$ for various values of $\rho$ and the errors for uniform partitions with N=64, 128, ...., 4096,8192 subintervals. Determine the value of the parameter $\rho \in [0,1)$ such that the method has maximum order, let $\rho=\rho_m$.
PS: The order accuracy can be computed numerically as follows.

Let $E(N_1)$ (respectively $E(N_2)$) the error of the numerical method for $N_1$ ( respectively $N_2$) su…

robjohn

@columbus8myhw I prove the integer case here

and there is a link to an answer showing the rational extension

columbus8myhw

@robjohn OK, I looked up what it is now.

@robjohn Interestante.

In any case, I was working on proving stuff without calculus.

In particular:

If $k$ is the unique number such that $2^x\ge kx+1$ for all $x$, then
$$k-\frac1n<1-\frac12+\frac13-\dotsb\frac1n<k+\frac1n$$
Thus, using the squeeze theorem (which is slightly calculus-y, but I'll allow it), we have proven the following:
$$k=1-\frac12+\frac13-\dotsb$$

So, in a sense, I showed that $\ln2=1-\frac12+\frac13-\dotsb$ without calculus! (Except that I called it $k$, and never showed that it's $\ln2$. Of course, if I started from $e^x\ge x+1$, then I could show that $k=\ln2$, so I really could show that $\ln2=1-\frac12+\dotsb$. Without calculus. It was a fun puzzle.)

I'd go into the details, but I have a test to study for. In any case, here's two hints. Hint 1: $1-\frac12+\frac13-\frac14+\frac15-\frac16=\frac14+\frac15+\frac16$. Why? Hint 2: Try plugging in $x=\frac1{nk}$ to the original inequality, for various $n\in\mathbb Z$.

David Wheeler

22:58

Using Bernoulli's inequality, one can show that $e^x > \left(1 + \dfrac{x}{n}\right)^n \geq 1 + n(x/n) = 1 + x$.

The trick is the "middle term": first you have to show it is a strictly increasing sequence for $n$, and that:

columbus8myhw

In any case, proving this — without calculus — is my puzzle from me to you. Enjoy!

David Wheeler

$$\lim_{n \to \infty}\left[ \left (1 + \dfrac{1}{n}\right)^n\right]^x = \lim_{n\to \infty} \left(1 + \dfrac{x}{n}\right)^n$$

columbus8myhw

(Oh, small edit: The statement you start from should be, "$2^x\ge kx+1$, with equality iff $x=0$." That last bit isn't essential, but you'd need it if you want strict bounds on the partial sum, I think.)

@DavidWheeler You can just use a substitution on that, right?

$n\to\frac nx$?

David Wheeler

Pretty much

columbus8myhw

all right, bye

David Wheeler

23:03

Proving $\{a_n\}$ is increasing, where $a_n = \left(1 + \dfrac{1}{n}\right)^n$ is the fun part.

When you start expanding the terms, using the binomial theorem, you start to understand where the Taylor series comes from, too. It's kinda fun.

columbus8myhw

@DavidWheeler I'll think about that.

Gridley Quayle

23:25

Could someone help me with the last part of an axiomatic proof where I have to show that the bisector of an isosceles triangle meets the base at a right angle.

I have a triangle base BC with apex A, I first had to prove that if we have another isosceles triangle base BC apex A', then AA'M bisects the angle at A & A' (M is the intersection of the line extended through AA' and BC). Then I had to show that BM=CM. I did both these by proving SAS congruency then using that.

I'm just stuck on the last part, I have shown that the angle AMB=-AMC but how do I prove that they are both right angles?

Howcan

Hi, is anyone well versed in statistics here (or has some background in multivariate regression?). Cross validated chat is a bit slower and I think my question is minor.

KaliMa

hello, does anyone know basics of matrix multiplication and vectors

David Wheeler

I know a little bit

Howcan

23:40

Are you referring to me or KaliMa?

David Wheeler

:20824781

Howcan

:c, statistics will be the end of me

KaliMa

Is there an efficient way to compute $A^k b$, $A$ is a large $n$ by $n$ matrix, $b$ is an $n$ by $1$ vector (the long vertical kind), and I am computing this for each $k=0$ upward.

right now best I can do is $O(n^2)$ per $k$, is it possible to do better?

Mike Miller

if you could diagonalize $A$, ignoring that step, I think it should be $O(n)$ per $k$

KaliMa

how do you mean

Mike Miller

23:53

mm, no longer convinced

ignore me

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$Mathematics$ Mathematics