Mathematics - 2013-06-17 (page 1 of 2)

Will Jagy

00:02

people do not seem to have liked this one:
http://math.stackexchange.com/questions/422322/how-much-wood-could-a-woodchuck-chuck

Peter Tamaroff

@WillJagy Hello. I am bad at counting.

Peter Tamaroff

00:14

@WillJagy How are you?

@anon Will this do?

pourjour

00:29

does anybody know a good book for functional analysis with a lot of problems

Peter Tamaroff

01:07

@pourjour You're taking a functional analysis course?

AlanH

01:33

$H \leq G$ s.t. $H$ is not normal. Show there exists cosets $aH$ and $bH$ such that $(aH)(bH)$ is not a coset:

Let $b = a^{-1}$. Since $H$ is not normal, not all products of $aHa^{-1}$ are in $H$. By definition of coset, $xH$ is a coset for all $x \in G$. So $(aH)(bH)$ is not necessarily a coset.

Is this correct?

Peter Tamaroff

01:56

@AlanH If $H$ is not normal, there exists $g\in G$ and $h\in H$ such that $ghg^{-1} \notin H$. You ought to work with that.

In what you write, what is $b$? What is $a^{-1}$?

AlanH

@PeterTamaroff I meant to write $aHa^{-1}H$ somewhere in there. $b$ is the inverse of $a$

uh okay let me think about it again

robjohn

@PeterTamaroff It's neat, but on the border of obvious. It takes a minute to figure out what is going on.

Peter Tamaroff

@robjohn Heh, true.

@robjohn Now I have to prove every natural $n$ can be written in $2^{n-1}-1$ ways as a sum of positive naturals less than $n$.

robjohn

02:12

The basic idea is $(1+x)(1+x^2)(1+x^4)(1+x^8)(1+x^{16})\dots=\dfrac1{1-x}$

@PeterTamaroff Case it on the number of $0$s on the right in binary

no $0$s on the right, there are just as many ways as $n-1$

Peter Tamaroff

@robjohn Come again?

robjohn

Sorry, I was thinking of something else with a similar solution. Case on the smallest number in the sum...

Peter Tamaroff

@robjohn "Case on..."?

robjohn

Break up the sums based on the smallest number in the sum and equate that with the number of ways to make up a previous number.

AlanH

@PeterTamaroff So if $H$ is not normal, there exists some$ g $and$ h$ such that $ghg^{-1}$ is not in $H$. Then since $(ghg^{-1})H \in (gH)(g^{-1}H)$, so $(ghg^{-1})H$ cannot be a left coset?

Peter Tamaroff

02:24

@AlanH Could you $\TeX$ that up?

AlanH

@PeterTamaroff Nevermind. Don't bother with it

Peter Tamaroff

@AlanH Wait. What you want to show is that the equivalence relation fails to be a congruence.

Ah, you call $mH$ a right coset, yes?

OK.

AlanH

@PeterTamaroff I think I got it. Thanks.

Peter Tamaroff

@AlanH Good. Byes.

AlanH

@PeterTamaroff See ya. Thanks for your help again today

Vincent Tjeng

03:01

In 2-dimensional geometry, is it correct to say that the perimeter of the convex hull of a set of points is the curve with the minimum perimeter bounding all those points?

What's the name of a curve of minimum area bounding all points in 2-dimensions?

MRS1367

04:49

Hi all

Bageer

Hello

MRS1367

Hell @Bageer

I need some help

I've need to a formula for achieve to this:

Bageer

Well you can ask and someone might help.

MRS1367

My question is for calculating place of an object in an computer app

Alexander Gruber

wtf lol

Bageer

04:55

Looks like someone is swimming in rep

MRS1367

1. Maximum mode = when one of the objects is placed in 4792.7 px

my other objects must place in 2266.7 px

2. Middle mode = when one of the objects is placed in 1826.496 px
my other objects must place in 685.85 px

Alexander Gruber

@Bageer i leave MSE alone for half a day and look what happens.

this was not exactly what i imagined my highest voted answer would look like.

Bageer

Anyone know if there is some sort of and operation for tags, for example suppose I don't have a problem with looking at questions that have homework or calculus but not both, so I would have in my ignore list something like calculusANDhomework.

@AlexanderGruber Yah, in a way it is sort of sad.

MRS1367

3. Minimum mode = when one of the objects is placed in 219.15 px
my other objects must place in -167.6 px

Alexander Gruber

@Bageer i don't think this exists, but you should ask on meta (maybe a feature request? i'd use that.)

Bageer

05:00

@AlexanderGruber I guess I will do that.

MRS1367

totally, I must say that when the first object moves, the second object must change its position

I check many formula for find a formula for do that in the coorect method

but I couldn't find a good method

I can't find a similar formula for doing that for changing the second object position

I check for finding a same multiple for doing that

but I can't

I check for doing that with some other formulas

but I can't find anything

Dominic Michaelis

05:50

Does someone here have Mathematica?

anon

....

Vincent Tjeng

@DominicMichaelis actually, me ... want me to run some code?

Dominic Michaelis

@VincentTjeng something like that. This `FindDistributionParameters[
Table[PDF[BinomialDistribution[10000, 0.2], x], {x, 0, 10000, 1}],
NormalDistribution[\[Mu], \[Sigma]]]` gives an unexpected result and I don't know why :/

FindDistributionParameters[
 Table[PDF[BinomialDistribution[10000, 0.2], x], {x, 0, 10000, 1}],
 NormalDistribution[\[Mu], \[Sigma]]]

Vincent Tjeng

okay, I run mma 9.0 on windows

Dominic Michaelis

Same here

Vincent Tjeng

05:53

running...

{[Mu] -> 0.00009999, [Sigma] -> 0.000833768}

Dominic Michaelis

The result should be mu -> 2000, and sigma -> 40 in my opinion

but I get the same result as you

Vincent Tjeng

hmm

Dominic Michaelis

I mean the mean of a binomial distribution with n=10000 and p=0.2 is clearly 2000 isn't it ?

Vincent Tjeng

yep. reading through the documentation right now

@DominicMichaelis - now I'm confused ... what does this mean?

FindDistributionParameters[
Table[PDF[NormalDistribution[100, 0.2], x], {x, 0, 200, 1}],
NormalDistribution[\[Mu], \[Sigma]]]

I get {[Mu] -> 0.00992401, [Sigma] -> 0.140346}

Dominic Michaelis

Ah I see my input makes no sense. I try to make a distribution over the given propabilities not over the the data

Vincent Tjeng

05:59

I see.

Dominic Michaelis

I thought I just give the propabilities of the old distribution to estimate another one ...

FindDistributionParameters[
 RandomVariate[BinomialDistribution[10000, 0.2], 10^5],
 NormalDistribution[\[Mu], \[Sigma]]]

this would be a correct input

Orangutango

Hello, I have an off topic problem I would like to introduce. It is a question I posed yesterday which has not gotten a response.

I do not wish to violate chat etiquette, so stop me if I am.

Here is the link to my problem.

Dominic Michaelis

06:19

your problem is not well defined

you sum up to $f(n)$ but $n$ isn't in the Domain of $f$

Orangutango

ah thats a typo, thank you.

Alexander Gruber

@Orangutango this does violate chat ettiquette (see point 3 here), but at least you're being polite about it, so i'm ok with it this time

Orangutango

but is the question otherwise clear? i am searching for an algorithm to maximize the sum.

Bageer

@Orangutango Does it not maximize if your order smallest^0 +largest + second largest^2+...+smallest^n-1?

Dominic Michaelis

@Bageer it does

Bageer

06:24

Maybe I am just being naive

or not...

Orangutango

The optimal way to sum $\{ \frac{1}{10},\frac{1}{2},\frac{4}{7},\frac{3}{5},\frac{2}{3}\}$ does not follow that algorithm.

user1

I almost had an answer to this question, using a bit of abstract algebra + analysis.

Orangutango

I verified this in Mathematica.

In my post, the last sum is the optimal one.

I can post the Mathematica code to verify if anyone doubts this.

Dominic Michaelis

I am honestly surprised

Orangutango

thats why this problem caught my attention! my intuition was totally wrong, and I still don't know exactly why

Bageer

06:30

That is cool

Dominic Michaelis

But post the Mathematica Code :)

Bageer

was the optimal one the second one you posted

?

Vincent Tjeng

@DominicMichaelis someList = {1/10, 1/2, 4/7, 3/5, 2/3};
Last@SortBy[{N[Total[#^(Range[Length@someList] - 1)]], #} & /@
Permutations[someList], First]

:)

Orangutango

beat me to it

Dominic Michaelis

@VincentTjeng thats a Neat code :)

Orangutango

06:34

and very slickly done

cleaner than my code XD

@Bageer yes, it is

Vincent Tjeng

thanks :)

anyway @Orangutango - I think part of the reason why your question didn't receive so much attention is that some of us didn't understand it! :)

Orangutango

well at least I'm not alone then! it's been troubling me for almost a week now.

Maisam Hedyelloo

@MRS1367: hi

Orangutango

well, i'll let you all toy with it as you will. it's very late my time and i've been thinking about this problem for far too long today. thanks for hearing me out, i hope one of you gains some insight on the problem.

Vincent Tjeng

07:09

@Orangutango - doing a study for sets of 5 random reals between 0 to 1...

data = Table[someList = RandomReal[{0, 1}, 5];
Ordering@
Ordering@(Last@
SortBy[{N[Total[#^(Range[Length@someList] - 1)]], #} & /@
Permutations[someList], First])[[2]], {i, 1, 10000}];

Tally[data]

I got {{{1, 3, 4, 5, 2}, 3605}, {{1, 4, 5, 3, 2}, 2664}, {{1, 4, 3, 5, 2},
452}, {{1, 5, 4, 3, 2}, 659}, {{1, 2, 3, 4, 5},
1834}, {{1, 3, 4, 2, 5}, 450}, {{1, 3, 2, 4, 5},
112}, {{1, 4, 3, 2, 5}, 59}, {{1, 2, 4, 3, 5},
48}, {{1, 3, 5, 4, 2}, 85}, {{1, 2, 3, 5, 4}, 11}, {{1, 2, 4, 5, 3},
21}} for 10000 runs

Dominic Michaelis

That doesn't look good :/

Vincent Tjeng

strangely, only 12 possible orderings pop out ... not 1, 4, 2, x, x for instance

not sure whether i coded it wrongly or what

Dominic Michaelis

Show[
 Plot[
  Evaluate@
   Table[
    x^j - x^k,
    {k, 1, Length@someList},
    {j, 1, k}],
  {x, 0, 1}]]

Ethan

how can you select 'random reals' between [0,1]

Dominic Michaelis

RandomReal[{0,1}]

Ethan

07:14

lol.

Whats your definition of random?

Dominic Michaelis

Pseudorandom ;)

Bageer

Is there an easy way to figure out the parity of the permutation?

Ethan

why do you ask

Dominic Michaelis

@VincentTjeng well we can say for sure that the first one is fixed

the smallest number is always the first

Vincent Tjeng

oh well... i run 100k times and look what pops out

{{{1, 3, 4, 5, 2}, 35695}, {{1, 4, 5, 3, 2}, 26819}, {{1, 3, 4, 2, 5},
4708}, {{1, 2, 3, 4, 5}, 18862}, {{1, 5, 4, 3, 2},
6257}, {{1, 3, 5, 4, 2}, 959}, {{1, 4, 3, 5, 2},
4388}, {{1, 3, 2, 4, 5}, 1011}, {{1, 4, 3, 2, 5},
623}, {{1, 2, 4, 5, 3}, 213}, {{1, 2, 3, 5, 4},
58}, {{1, 2, 4, 3, 5}, 403}, {{1, 2, 5, 4, 3}, 4}}

@bageer what do you mean by parity? I think it can be done

Bageer

07:22

The permutation sign (whether the permuations is odd or even)

I have test, almost all the ones you put up and so far they have all been even.

Dominic Michaelis

{1,3,4,5,2}-> {1,2,4,5,3}-> {1,2,3,5,4}-> {1,2,3,4,5} is odd isn't it ?

Vincent Tjeng

@DominicMichaelis do help me check through my code... I don't want us to be trying to find reasons for wrong results :)

Bageer

Damn, I have been putting in the cycles into wolfram alpha... its late, okay!

Vincent Tjeng

{Signature[#[[1]]], #[[2]]} & /@ Tally[data]

Bageer

(cycles as in the list)

Vincent Tjeng

07:27

{{-1, 35695}, {-1, 26819}, {1, 4708}, {1, 18862}, {1, 6257}, {1,
959}, {1, 4388}, {-1, 1011}, {-1, 623}, {1, 213}, {-1, 58}, {-1,
403}, {-1, 4}}

does anyone find it curious that 99% of the lists have either the 2nd or the 5th smallest as the last term?

Bageer

Oh no particularly good pattern. I have to say that his problem is surprising me.

Dominic Michaelis

no that is expectable

Look at the plot

Show[
 Plot[
  Evaluate@
   Table[
    x^j - x^k,
    {k, 1, Length@someList},
    {j, 1, k}],
  {x, 0, 1}, PlotLegends -> Automatic]]

if your number is very small or very big raising it's power don't hurt the sum

MRS1367

Hi @MaisamHedyelloo

Vincent Tjeng

oh, i see.

Dominic Michaelis

and by the plot the second smallest is most likely to become the last

which is with our data at over 70 % of the cases true

@VincentTjeng Why do you use two orderungs?

Vincent Tjeng

07:42

@DominicMichaelis Ordering@{1, 5, 2, 3, 4} gives you {1, 3, 4, 5, 2}

applying Ordering to that again gives you the original order. a bit difficult to explain but i think its correct from the documentation

Dominic Michaelis

I am running for 300 k

in your code you make at first a list of 5 randomreals

Then you make a list of its permutations and on the permutationslist you apply a list of functions namely the sum we are trying to minimize and the identiy

this list you sortby the sum,

but this should give the smallest as the first ?

Vincent Tjeng

yes

Dominic Michaelis

ah right so the largest is in the end and you take the last one

Vincent Tjeng

mm thats why i take the last

quick and dirty code

Dominic Michaelis

afterward you take the second part which gives you the permuted list of randomreals

your code seems to work

i mean that it looks correct oto me

Vincent Tjeng

08:04

hmm ok

Dominic Michaelis

@VincentTjeng I gonna write an answer with our intermediate results

Vincent Tjeng

thanks! interested to see the details from a big run :)

Dominic Michaelis

10:07

hi

I posted some things

Vincent Tjeng

10:27

@DominicMichaelis nice work. hope it inspires some further thinking from other people

Dominic Michaelis

i will try with longer permutations

for longer permutations we need another code

Vincent Tjeng

i think we can force the first one to be smallest

not sure wha other simplifications we can make

if you can you should extract some examples for the rare cases like {1,2,5,4,3}

Dominic Michaelis

yeah we need to do more analysis before improving the code

if we take your command for a list of length 9 it takes 100 seconds at me

Vincent Tjeng

same.

Dominic Michaelis

length 10 *

Vincent Tjeng

10:39

around there

Dominic Michaelis

we could reduce it by faktor 10 with forcing the smallest to be first

in the best case

this sounds totally like a project euler problem to me :D

maybe there is something like that on project euler

Vincent Tjeng

perhaps :)

Chris's wise sister

10:58

Greetings all.

@robjohn are you around?

I'm looking for some nice example of limits where $\lim_{n\to\infty} n(b_{n+1}-b_{n})=l>0$, $\lim_{n\to\infty} b_n=b>0$

Ethan

$\ln(n)=b_{n}$ works for the first one..

@Chris'swisesister I don't think any exist

Chris's wise sister

11:14

@Ethan are you sure?

Ethan

@Chris'swisesister If $n(b_{n+1}-b_{n})$ is bounded

Then $b_{n+1}-b_{n}$ over $\frac{1}{n}$ is bounded

so that $\frac{c_1}{n}<b_{n+1}-b_{n}<\frac{c_2}{n}$

Sum both sides from $n=1$ to $n=m-1$

$c_1\ln(n)<b_{n}-b_{1}<c_2\ln(n)$

For sufficiently large $n$

Sense $\sum_{n=1}^m\frac{1}{n}=\ln(n)+O(1)$

Chris's wise sister

That seems right.

Ethan

Assuming $b_1$ is defined and bounded, if its not we can always start the sum at a different number and use the same argument

in which case regaurdless we get $c_1\ln(n)<b_{n}<c_2\ln(n)$

for constants fixed constants $c_1$, $c_2$, and for large enough $n$

and it follows $\lim_{n\to\infty}b_n$ is not bounded

Though if we lift the second constraint

The only $b_{n}'s$ that work for the first one, are non zero multiples of the natural logarithm I think

Chris's wise sister

I felt it's something wrong with this question ...

skullpatrol

11:31

@Chris'swisesister Pardon me asking, but what motivated you to look for such limits?

Chris's wise sister

@skullpatrol I was doing some research on limits.

robjohn

12:30

@Chris'swisesister yes

blob

13:11

Hello, if we throw a dice 3 times what is the probability that the first is an odd number if we know that we got 1 odd number?

Chris's wise sister

@robjohn it was about the problem I discussed with Ethan. Now it's clear that point.

EwokNightmares

13:46

Hi, I have a question. What would I classify a 2D surface that is closed but has a hole in it?

like 2 concentric circles defining the boundaries of a closed surface

Dominic Michaelis

torus ?

annulus ?

EwokNightmares

a torus can be 2D?

Dominic Michaelis

the surface of a torus is 2 dimensional

Annulus (mathematics)

In mathematics, an annulus (the Latin word for "little ring", with plural annuli) is a ring-shaped object, especially a region bounded by two concentric circles. The adjectival form is annular (as in annular eclipse). The open annulus is topologically equivalent to both the open cylinder and the punctured plane. The area of an annulus is the difference in the areas of the larger circle of radius and the smaller one of radius : :A = \pi R^2 - \pi r^2 = \pi(R^2 - r^2)\,. The area of an annulus can be obtained from the length of the longest interval that can lie completely inside the ann...

EwokNightmares

an annulus sounds like the exact definition of the 2nd description I gave

So a torus would be a more generic term for 2D closed surfaces with holes in them?

while annulus is specifically circles?

Dominic Michaelis

i missunterstood you

you want the whole thing t obe 2 d not only it's surface

EwokNightmares

13:49

yes it is just a 2D shape to consider

it has a boundary on the outside everywhere, but basically just has a hole in it anywhere

and the shape can be squigly

and an annulus would fit into this category

Babak S.

14:07

@robjohn: Hello! ;-)

pourjour

14:18

@PeterTamaroff I want just to start learning them

robjohn

@Ethan $\log(\color{#C00000}{m})+\gamma+O(1/n)$

@BabakS. hey there! good day?

Babak S.

Good day @robjohn our rescue here. ;-)

@robjohn: I have a question. We know that there is not a general formula ruling the primes numbers. But for many finitely prime numbers $p$ we see that $2p+1=4k+3$ wherein $k$ is some positive integers. What can we do with this case. I am trying to generalize a presentation of a group and in it the above equality is being used frequently again and again intuitively. May I ask you any hint? Thanks

robjohn

14:38

If $p$ is odd, then $p=2k+1$ for some $k$, so $2p+1=4k+3$. Is that what you mean?

Lord_Farin

@BabakS. It says nothing but that $p$ is odd.

Babak S.

Ohh yes. Thanks I got where I was mistaking.

robjohn

@BabakS. misstaking a vampire can be hazardous to your health.

Babak S.

@robjohn: Yes :-( but sometimes I lost the very basic facts. I think I am getting Alzheimer at 41.

EwokNightmares

Are you math professors/PhD grads?

Babak S.

14:45

@EwokNightmares: I am a lecturer.

EwokNightmares

I admire math experts so much

Shayna

15:03

Good morning! :D

blob

15:49

good evening

keri

16:01

Hey1

Shayna

17:13

Anyone here good at scientific notation?

I've started the problem [(5600000)(0.000000081)]/[(900)(0.000000028)] and gotten to (5.6x8.1x10^6x10^-8)/9x2.8x10^2x10^-8)... Then to [(5.6X8.1)/(9X2.8)]X10^? And I'm stuck LOL. How would I determine that exponent?

Based on that my answer would be 1.8x10^? So confused on how to get that exponent though. :/

Vrouvrou

17:37

Hello, can someone help me please :math.stackexchange.com/questions/422940/…

Bandeira Gustavo

17:47

@amWhy WTF is the problem of the finite groups guy?

amWhy

@BandeiraGustavo I don't have a clue...but the attitude is surely a turn-off!

AlanH

Is the following a typo? If $a \equiv b \pmod{m}$, then for some scalar $c>0$, $ac \equiv bc \pmod{mc}$

Lord_Farin

@AlanH Such is likely. It holds for all suitable scalars $c > 0$.

Hello people.

Peter Tamaroff

18:20

@Lord_Farin $\sup$?

Lord_Farin

@PeterTamaroff $\inf$ :(.

Peter Tamaroff

@Lord_Farin Oh, man. I saw that one coming. Why?

Lord_Farin

@PeterTamaroff Progress on thesis is too slow.

And how are you doing?

anon

Chatroom manners | ChatJax install (meta post)

13

Peter Tamaroff

@Lord_Farin I'm hungry. And I have a olympic runner of a nose.

Lord_Farin

18:36

@PeterTamaroff Interesting; since yesterday, such has presented itself to me as well.

I've used my handkerchief as much today as I did all of last week together.

Peter Tamaroff

@Lord_Farin Heh. I just answered. I will get to 800 answers this month, I will. =)

Lord_Farin

@PeterTamaroff These numbers mean nothing of course; yet I felt somehow satisfied for passing 200 a few days ago :).

Peter Tamaroff

@Lord_Farin Good!

Lord_Farin

(Can't upvote by the way; today, I've burnt through my votes exceptionally fast.)

Peter Tamaroff

@Lord_Farin This is my latest answer

Some days ago I posted something erroneous.

Lord_Farin

18:48

@PeterTamaroff I suspect several instances of such unfortunate events in my answers list as well.

Peter Tamaroff

$$\prod_{k\geqslant 1}(1-q^kz)=\sum_{k\geqslant 0}\frac{q^{}}{q-1}\frac{q^2}{q^2-1}\cdots \frac{q^k}{q^k-1}z^k$$

The first term is (obviously) $=1$.

On the other hand $$\prod_{k\geqslant 1}(1-q^kz)^{-1}=\sum_{k\geqslant 0}q^k (1-q)(1-q^2)\cdots (1-q^k)z^k$$

If I am not remembering wrongly. I lost the page. =(

@anon Do you know an explicit formula for $\mu_n=\text{ the number of } 1 \text{s in the binary expression of } n$?

Because $$\prod_{k\geq 0}(1+qz^{2^k})=\sum_{k\geq 0}q^{\mu_k}z^k$$

anon

well, there is an explicit formula for the $n$th binary digit, and you can add these up over all $n$, but it will not be pretty

as long as you consider Mod(n,2^k) to be a reasonable function anyway

Peter Tamaroff

@anon Said sequence is pretty interesting. You can build it as follows. First, observe the first three terms are 1 1 2. Now, insert a 1 at the beginning, add 1 to the rest elements, and join this to the sequence you had. That is 1 1 2 1 2 2 3. Now, do the same, 1 1 2 1 2 2 3 1 2 2 3 2 3 3 4. This builds up the sequence.

So if you know the first $2^k-1$ terms, you know the next $2^k$ terms.

Peter Tamaroff

19:11

In particular $\mu_{2^k}=1$ for each $k$.

skullpatrol

Yo @Argon wazzup pal?

Vrouvrou

@Lord_Farin hello

Lord_Farin

@Vrouvrou Hello there. How are you doing?

Vrouvrou

i'm good

@Lord_Farin or @anon i need your help please for a derivative

can you help me please ?

Lord_Farin

@Vrouvrou I'm currently busy, sorry.

Vrouvrou

19:25

>_<

Peter Tamaroff

@Vrouvrou What is it?

Vrouvrou

{math.stackexchange.com/questions/422940/question-on-derivative}

Peter Tamaroff

@Vrouvrou No clue on that one, man.

Vrouvrou

>_< no one has an idea

Peter Tamaroff

@Vrouvrou Well, it is fairly advanced, isn't it?

Vrouvrou

19:32

yes but

i posted it on matoverflow i hope that someone could help me

Lord_Farin

@amWhy Just one instance of you happening to make a mistake does not imply that you should refrain from taking the liberty of editing my posts. In 95 out of 100 instances, your edits will correct genuine flaws.

I appreciate your effort of proofreading my answers. :)

amWhy

@Lord_Farin I do sincerely apologize. Thanks for commenting here. I rarely intervene in others posts (answerers), and doubt I'll need to with your posts...

Lord_Farin

@amWhy No offence taken. I'm notorious for skimming over "pathological" cases.

amWhy

I was still thinking "biconditionally" (given my answer), and jumped in too quickly.

@Lord_Farin Yes, you corrected precisely one example of a pathological post, authored by me, yesterday! :-) Thanks.

Lord_Farin

19:49

@amWhy Hm, I think there was a language barrier at work there. I meant to write "glossing" in place of "skimming".

skullpatrol

@Chris'swisesister Greetings :D

Chris's wise sister

@skullpatrol hi :)

skullpatrol

@Chris'swisesister How are you?

Lord_Farin

It pleases me to have found a use for my knowledge about Kripke structures (for intuitionistic prop. and pred. calculus).

Chris's wise sister

@skullpatrol in a bad mood. How about you? :-)

skullpatrol

19:56

@Chris'swisesister I'm Ok, thanks :-)

@Chris'swisesister Did working on those limits cause your mood change?

Lord_Farin

Please, more answers. :)

(Two were already deleted.)

Chris's wise sister

@skullpatrol In a way, yes. I'd like to be able to finish calculus problems without pen and paper ... but I still need to work to go there ... (I'm not serious enough in my work - - I need more discipline)

Lord_Farin

I'm out, going to try to do some useful stuff.

Bye all.

skullpatrol

later

Chris's wise sister

@Lord_Farin hi & bye

Lord_Farin

20:07

@Chris'swisesister Same. :)

skullpatrol

@Chris'swisesister Like any skill it takes time & effort to develop...

Chris's wise sister

@skullpatrol yeah ...

skullpatrol

@Chris'swisesister It's like the difference between the ability to do something and the ability to do something well.

Chris's wise sister

@skullpatrol there is also a trick here. Since we're smart beings, it's also possible we try to explain somehow why we're in a bad mood (since we try to explain everything), but it's also possible that sometimes there is about some neurotransmitter imbalances due to excessive work on things, stress.

amWhy

Hello, all, hello, @skull

Chris's wise sister

20:20

@amWhy hi

amWhy

@Chris'swisesister :D

Chris's wise sister

:D

skullpatrol

Hi @amWhy

amWhy

@skullpatrol I just get frustrated, that's all...but some things are just part of life I guess.

@skullpatrol :D

skullpatrol

@amWhy :D

Shayna

20:23

Hello :)

skullpatrol

Hi

Shayna

How are ya this lovely afternoon?

skullpatrol

Fine thanks, how are you?

Shayna

Well, could be better. Trying to finish up some math homework with the worst textbook ever once again! Thank goodness it's the last assignment LOL.

skullpatrol

Orly? who wrote it?

Shayna

20:26

Do you happen to know if there is a good thread on simplifying exponential expressions with variables only containing positive exponents (assuming all variables represent nonzero real numbers)?

Pearson custom math text- I'd assume our professor picked the content he wanted in the text. sigh

I'm terrible at math other than statistics LOL... And I'm trying to do a math for teachers course :/ Already did a BA majoring in both Psychology and Sociology... Yet this math stumps me biiiig time LOL!

Like if I tried to do 4(5a^4)(2a)^-2 would it just be 40a^2 ?

skullpatrol

Hmmm... I would suggest you focus your professors notes.

Shayna

There is none- it's a distance course. We just have the text. :/

There's nothing with negative exponents in that section LOL

Denis Rodman

@Shayna Real multiplication is commutive, yes, but 4*(2a)^-2=a^-2.

skullpatrol

@Shayna Have you tried the Khan Academy on YouTube?

Shayna

I did try the Khan Academy. Problem is I just don't really know what I'm looking for most of the time. :/

skullpatrol

20:33

Polynomials would be a good start.

Shayna

@DenisRodman Thank you! In the text it says to just multiply 4X5X2... Has me confused LOL it doesn't even say what to do with the exponents

skullpatrol

Try "Rules of Exponents"

Shayna

@skullpatrol Here's the thing. I can do polynomials. I'm just unsure where to start when it has the 2 sets of brackets and numbers outside of those brackets :/

@skullpatrol Thank you! :D Hopefully I can find something!

skullpatrol

Start from the inner-most brackets.

Shayna

So the 5a^4 and 2a?

If I were to follow FOIL wouldn't it be 4(5a^4) and (2a)^-2?

skullpatrol

20:38

What does "FOIL" stand for?

I meant that as a rhetorical question(!)

Shayna

first inside outside last

XD

but then I wasn't sure if I should be following FOIL or BEDMAS :/

skullpatrol

Yes, you need BInomials to multiply to use it :-)

Shayna

Hahaha ohhh man I feel sorry for the kids I'm going to have to teach math to! I hope I end up teaching like grade 4 or lower!

skullpatrol

The FOIL method is for multiplying binomials.

BEDMAS is for order of operations.

Shayna

So I do bedmas right? LOL

so basically work left to right on 4(5a^4)(2a)^-2 ?

skullpatrol

20:43

You need to know the "Rules of Exponents." :-)

Shayna

Yeah I'm looking through the videos on Khan Academy... I just can't find anything with more than one part to the exponents questions :/ I know how to do the easy ones but when they end up compounded I get confused.

skullpatrol

@Shayna Try this to start with...

...math builds on itself.

Shayna

Yeah, I just watched that one before you linked it. I know how to do questions like those ones. It's just when they get more complicated I get lost unless I have pretty much an exact example to work from. :/

Like... I can do exponents well enough that I got 98% in Statistics LOL

skullpatrol

@Shayna 4(5a^4)(2a)^-2 = 5a^2

Shayna

Okay I see where the a^2 came from... But how did we get 5?

skullpatrol

20:56

4(5a^4)(2a)^-2

=4(5a^4)(2^-2)(a^-2)

...
(ab)^m = (a^m)*(b^m)

Transcript for

$Mathematics$ Mathematics