Mathematics - 2014-04-07 (page 2 of 3)

skullpatrol

05:18

miracrown was the "geek"

Sush

@usukidoll, please help me here

Why does Andre say "How many divisors, not necessarily positive, does 36 have?"

anon

x-6 times y-6 equals 36

a product of things is 36...

Mew

a product of two integers is 36

these two integers must be the divisors of 36

(36,1),(2,18),(3,12),(4,9) and 6

for the positive case

obviously if you'll have twice that for the negative case

but obviously we halve it again, since y>=x

so the answer is 10

Sush

05:35

@Mew, but my book says answer is $9$

Mew

yeah i meant 9

wait

10

let's think

the combinations are:

36*1

18*2

9*4

12*3

6*6

-1*-36

-18*-2

-3*-12

-9*-4

-6*-6

Sush

@Mew, see this

Mew

what about x = 12, y = 12

Mike

@anon i have to admit i liked link better

Sush

@Mew, I think Andre has something like $2^2*3^2=36$

anon

05:42

I got bored

Mew

i think they forgot x = 12, y = 12

Sush

So total $18$ divisors.

@Mew

And there are $18$ solutions, TOO!

eXtremiity

When you are given that something lies on the unit circle.

Sush

But, can't understand, Why?

@eXtremiity, HI

eXtremiity

Do you assume that this unit circle is centered at the origin?

Hey @Sush.

Sush

05:44

Are you answering me?

Mew

sush

they have 18 solutions

but do they have the constraint that x>=y that you have?

I have 18 solutions too if i excluded that contraint

Karl Kronenfeld

@eXtremiity I don't, if I read that statement out of context.

Sush

Ok, symmetry will do!

eXtremiity

@KarlKronenfeld. After reading that - what do you think?

Mew

but i think the answer is 10 not 9

because even though there are 18 solutions, you don't just halve for symmetry

Sush

05:46

@Mew, but how did we know those $18$ sols?

Mew

because both the 6*6 and the -6*-6 combinations still are satisfied by x>=y

easy

just list the ways to multiply to get 36, there are 18 ways

Sush

ok

Mew

36*1, 1*36, 2*18, 18*2, 3*12, 12*3, 9*4, 4*9, 6*6

that is half

which is 9

the other 9 are just the negatives of those 9

such as -36*-1

etc.

Sush

Yes.

Mew

giving 18

if we temporarily exclude the 6*6 and the -6*-6, we have 16 solutions

half these 16 to give 8 solutions that will satisfy the x>=y contraint

then add 2 for the 6*6 and -6*-6 we excluded, gives 10 solutions

Sush

05:49

And how to solve this without listing all solutions?

Mew

by thinking how many factors 36 has

excluding 6*6, there are 4 positive pairs of integers that multiply to give 36

Karl Kronenfeld

@eXtremiity It probably doesn't matter, to be honest. If it does turn out to matter, then, and only then, assume it's centered at 0

Sush

@Mew, thank you!☺

eXtremiity

Thanks @KarlKronenfeld.

Mew

doulbe it for the negative cases = 8

then add 2 for the 6*6 and -6*-6

= 10

np

0

Q: Randomized Block Design for repeated trials - Please help

Suppose I measure reaction time of 4 students, initially before any treatment, then after treatment 1, and finally after treatment 2. (Note all treatments will eventually be applied to all students) Also because of potential error with reaction time measurements, suppose 3 trials are done for ea...

statistics

please return the favor by answering the above ^

Sush

05:53

@Mew, m not brilliant☻

Mew

06:06

@Sush, i think the answer is 9

Sush

@Mew, ok, why?

Mew

nvm

answer is 10

just miscounted

Sush

@Mew, How can we count factors of 36 without listing them?

Mew

in our head

well

you count the basic 4

Sush

that is 2*2*3*3, right?

Mew

06:14

(36,1)(18,2)(12,3)(9,4)

no

i mean as above

i mean the "pairs"

so there are 4 pairs

Sush

ok,

Mew

now these pairs can be reversed

that makes 8

Sush

and 1 is 6,6

so 9. i can see that.

Mew

also these pairs can have their signs switched to -ves

that makes 16

then add (6,6) and (-6,-6), that makes 18

Sush

yes and 6,6 and-6,-6 is 18.

Ye.

Mew

06:15

the reason i left the (6,6) until last is because you can't reverse thsi pair

since (6,6) is the same as the reverse

Sush

ok.

Mew

so you don't really have to list all 18 factors to know there is 18

and by factors i mean "pairs"

which are also the factors

actualy so doesn't matter

Sush

So, i can count just by noting $2^2*3^2$ and then 3*3 for positives, right?

@Mew

Mew

order does matter

Sush

Ya!

Mew

06:18

because (36,1) corresponds to x = 42, and y = 7

but (1,36) corresponds to x = 7, y = 42

which are different solutions

skullpatrol

Sorry :(

Mew

except we know that x>=y so

we only keep one solution per pair

np

Mike

@KarlKronenfeld I think I'm going to quit math

Mew

as punishment you must answer this question:

math.stackexchange.com/questions/743116/…

Karl Kronenfeld

@Mike oh no

Mew

06:19

0

Q: Randomized Block Design for repeated trials - Please help

Suppose I measure reaction time of 4 students, initially before any treatment, then after treatment 1, and finally after treatment 2. (Note all treatments will eventually be applied to all students) Also because of potential error with reaction time measurements, suppose 3 trials are done for ea...

statistics

Mike

become a doctor instead

seems easier

Karl Kronenfeld

lol yeah

Mew

that's what i did

Mike

skullpatrol

You are becoming a numbers doctor :-)

Mike

i am stuck on the stuff for both of my classes

lol

Mew

06:21

na

medical

but i still have statistics troubles

what r u stuck on mike

skullpatrol

@mew are you an MD?

Mew

gonna be

Mike

math, @Mew

Mew

still studying

skullpatrol

What year of med school are you in?

Mew

06:23

2nd

skullpatrol

nice

Mew

it's all about memorization in med

easier than maths conceptually, but still much harder overall

skullpatrol

Do you have a "photographic" memory?

Mew

lol na, otherwise med would be easy

skullpatrol

imo all of education would be

I once read that one of the major evolutionary reasons man's brain is larger than apes is that we had to remember where we kept our food. Just some food for thought :-)

Sush

06:35

@Mew, In am exam, the score in each of the 4 subjects A,B,C and D - can be integer between 0 and 10. Then number of ways in which a student can secure a total of 21 is 880. I can find that with finding coefficient of $x^{21}$ in $(1+x+\dots+x^9+x^{10})^4$

But is there an easier way? As the question had only 2 marks!

@skullpatrol, i kept my food in refrigerator! lol:-)

skullpatrol

:D

Sush

professor Robert Langdon has Photo memory, I think!

eXtremiity

@Mike. Just emailed my lecturer about the metric we talked about. His response:
"Yes, I had intended that you use the d_\infty metric on C(T)...I should
have been clearer on that."

usukidoll

kutsuwa eraser kit ^^'

Sush

@eXtremiity, can you please suggest an easier way?

For this

eXtremiity

06:41

Hmmmm.

Combinatorics?

Sush

Ya

eXtremiity

I usually resort to the axiom of choice for these types of questiosn.

Hmmmm

Sush

OK!

You tried to explain me the axiom of choice, but :\

eXtremiity

It's quite simple. I'll give you an example. Say you have 5 marbles and you want to see how many combinations of 4 you can make.

Well the answer is 5 x 4 x 3 x 2.

You have 5 choices for the first marble, 4 for the second, 3 for the 3rd and so on.

It's quite powerful as you can see.

Mike

@eXtremiity :D

eXtremiity

06:44

Now, this question hmmm..

I guess I'm going to have to think of a mark as a marble.

And Im asking myself how many combinations of 4 will give me a final score of 880.

Maybe . . .

Mew

i think it is much harder extremiity

eXtremiity

It is.

Mew

because it involves the partition of no. 21

eXtremiity

Oh I have missed something. I can only score between 0 and 9 for each subject.

Mew

what about 10?

eXtremiity

06:46

Oh sorry and 10.

This is tricky.

Mew

what exam was this for sush?

Sush

@Mew, Entrance exam, for B.Stat.

We got 2 hrs, 30 questions.

Mew

bachelor of stats?

so you will soon be able to answer my statistics question?

Sush

The biggest problem I face is expanding that BINOMIAL

No, bro, no@Mew

Mew

hi jessica

eXtremiity

06:53

@Sush. Question has got me :( .

Mew

are yo good at statistics?

eXtremiity

Expand $(1+x+\dots+x^9+x^{10})^4$?

Sush

not quite!

I am economics undergrad

tried several exams!

(ENTRANCE)

Mew

what is B.stat?

Sush

@eXtremiity, ya!

it is bachelors of statistics

@Mew, as you mentioned.

Mew

06:55

so you want to do a second degree?

eXtremiity

Well first, it's not a binomial. Binomials look like this $(a+b)^n$ :).

Bi = Two.

Sush

no, i just gave exam!

usukidoll

polynomial

Sush

Yup@eXtremiity

eXtremiity

Now, that exam. Ahhh, what term are you looking for?

The coeffecient of x^21?

Sush

06:56

yes.

eXtremiity

Well that is an easier challenge.

Sush

i used wolfram :\ to calculate.

ok,

eXtremiity

$(1+x+\dots+x^9+x^{10})\times (1+x+\dots+x^9+x^{10}) \times (1+x+\dots+x^9+x^{10}) \times (1+x+\dots+x^9+x^{10}) $

JessicaK

Is there a way to contact a moderator without flagging an arbitrary question? I can't find a way to directly contact one otherwise

Mew

you can bring your issue to me jessica

skullpatrol

06:57

Just call one here

Sush

@JessicaK, ask @robjohn

he is moderator.

@eXtremiity, ok.

ok!

JessicaK

Basically, someone deleted their post shortly after I answered, then rewrote the entire answer when they did not understand the first time

eXtremiity

So, $(1+x+\dots+x^9+x^{10}) \times (1+x+\dots+x^9+x^{10})$ will not generate a coeffecient for $x^{21}$, right?

The highest power you're going to get from that expansion is $x^{20}$

skullpatrol

Stuff like that @jessica happens all the time

Karl Kronenfeld

@JessicaK See answer and comment thread on this meta post: meta.math.stackexchange.com/a/8529/67848

JessicaK

07:03

There was a metapost that suggested it was worth bringing up to a moderator, otherwise I wouldn't have mentioned it, but I'll just drop the matter. Thanks

skullpatrol

Drop it like its hot :-)

As Mariano says it can be a form of cheating.

Mew

Jessica, are you any good at statistics?

usukidoll

I would scream if that happened to me

skullpatrol

@mew use "@"

JessicaK

I've studied statistics, stochastic calculus and other similar areas long enough for me to realize I know nothing

Mew

07:07

@JessicaK, if you can answer this question, I will not delete your answer and repost it: math.stackexchange.com/questions/743116/…

0

Q: Randomized Block Design for repeated trials - Please help

Suppose I measure reaction time of 4 students, initially before any treatment, then after treatment 1, and finally after treatment 2. (Note all treatments will eventually be applied to all students) Also because of potential error with reaction time measurements, suppose 3 trials are done for ea...

statistics

usukidoll

I love stuffed jalapenos

skullpatrol

Stuffed with?

usukidoll

cream cheese

skullpatrol

icic

usukidoll

errr I don't want to take my midterm tomorrow

lol

Mew

07:09

it'll be easy

usukidoll

the only problems I'm confident in is the modular arthimetic

JessicaK

Sorry, that isn't within my expertise

usukidoll

dude issuing a midterm after spring break is a bad idea

Mew

what else do they assess

usukidoll

cuz during spring break all the students do is party and get drunk XD

so by the time we come back and this pops up, we're like omglolwut

skullpatrol

07:10

Those are the ones the profs don't want around

Mew

if anyone can answer my question, I will upvote all of their answers

usukidoll

skull you know something is wrong if everyone got a 3 out of 10 on the last 2 homeworks

Mew

for all of their other questions and answers

eXtremiity

Is $|\cdot|$ the symbol for usual metric?

usukidoll

and failed the first midterm

that just tells me that prof isn't doing his job ^^

skullpatrol

07:11

Does he care?

usukidoll

no

this is like last semester all over again

when everyone failed the second midterm and 3 or 4 people had 28/72

which one of those people was me

eXtremiity

Happens at my Uni too.

usukidoll

like anything can happen...I mean last semester no one did score high on homework and midterms yet everyone passed

WHY THE H((((( DOES ThAT HAPPEN ?

eXtremiity

One of my lecturers made terrible exams.

usukidoll

like seriously what if this is med school?

and the profs would be lol yeah you passed everything

so the student is still lacking training

Mew

07:13

dude everyone passes in medschool

usukidoll

once they have their license and make mistakes...I smell lawsuits

oh gawd D:

this is why our education system sucks

skullpatrol

That makes the law students rich

Mew

dude don't learn from the teachers

learn from the textbooks

skullpatrol

True dat^

usukidoll

won't work if the textbook itself is complicated

it's worse than Rosen's Discrete Math

or any of the Schaum outline books

Mew

07:16

maybe you don't spend enough time trying to read it

usukidoll

I do -_____-

eXtremiity

I have learn't to do a lot of self study. Use Stack a lot.

Mew

but otherwise you could always buy a different textbook

usukidoll

can't help it if the chapters are super short and don't go into detail

like Rosens

eXtremiity

And ask lecturers personally.

usukidoll

07:17

reminds me of Differential Equations by Bear. ewwwwwwww worse book ever

yet I passed that course last semester with an A- ^^

I won't recommend that book however...too many steps being skipped and has typos up the bo bo

Mew

what are you learning about this semester?

skullpatrol

Some people just use the textbooks available in the library.

Mike

who here is tired

i'm tired

usukidoll

math 321 is what me and the whole class is taking which is Intro to Advanced Mathematics

at least one good thing on this course is that I can do simple proofs that don't require a lot of definition and theorem mashups

skullpatrol

Me too

Mew

07:20

Mike, i'm not tired

Mike

well screw you, buddy

usukidoll

Mew? Pokemon? Gotta catch em all?

I slept at 3 am last night wonderful X_X

Mike

i gotta be up at 7

but i'm not ready to sleep

usukidoll

next semester I gotta be up at 5:45 am curse the distance and the early lecture

Mew

@Sush math.stackexchange.com/questions/743217/…

usukidoll

07:22

but the professor is good for that course so it's going to be worth getting up

just like last semester...there was this really cool professor who replaced someone else at the last minute. I don't know why the class didn't like him...he's knowledgeable and totally awesome...and... hehheehe ^^

Mike

maybe i shoul djust sleep and work in the morning

i'm quickly wearing out

skullpatrol

Sounds like a good plan

Don't wear yourself out so early in the week.

eXtremiity

08:06

Hey, does anyone know how to draw the arrow with the f(x) on top on lyx?

Or a convenient way to do this?

Studentmath

08:27

I think my prof. got my test wrong by 7 points..

Pretty sure I got this answer right.

Alex Youcis

09:09

@user91500 You're a grad student studying analytic number theory?

user91500

@Alex: Yes

Alex Youcis

@user91500 What institution are you at? :)

nerdy

so, i need to prove that for n>=2, and for any positive integer k, (n-1) | (n^k - 1 ) , i have proved doing induction on k ( starting with base k=1 ).. Do i need to do something else about n ? Do i need to do induction on n now with k fixed ? im kinda confused with two-variable inductions

Sawarnik

@r9m (removed) Need your help anyways, you here!

r9m

@Sawarnik ya ..

Sawarnik

09:16

@r9m A cyclic trapezoid has angles of 45, 45, 135, 135. What is the angle of intersection of the diagonals?

r9m

@Sawarnik whats a cyclic trap ?

LOL

Sawarnik

;)

r9m

what makes you think the angle of intersection of diags is constant ?

Sawarnik

cyclic trapezoid

@r9m It isn't?

r9m

draw a right angled isosceles triangle .. and draw lines parallel to the hypotenuse .. you will get cyclic trapezoids with angles of 45, 45, 135, 135 .. that has diagonal angle varying fron 90 to 180 degrees :)

Sawarnik

09:24

Ok, we have a length. top most parallel side = radius = 10.

@r9m ?

r9m

should be easy ,.,, did you use sine rule ?

basically its asking for the angle of intersection of medians (to non hypotenuse sides) of a right angled isocsalis triangle ..

Sawarnik

@r9m Wow! You are a genius.

r9m

just joking :P .. there is no need to complicate things :P

@Sawarnik what was that supposed to mean ? lol

Sawarnik

@r9m SE wants you to click on it.

joshworth.com/dev/pixelspace/pixelspace_solarsystem.html

Alec Teal

09:42

/me bumps math.stackexchange.com/questions/742913/…

Morning

user91500

10:13

@AlexYoucis See my profile :)

Chris's sis

10:46

Greetings.

Mats Granvik

11:26

Hi!

somaye

hi

Mats Granvik

If I can calculate the Mertens function from any number sequence as input, can I say anything about the Mertens function then?

somaye

@skullpatrol how are you?

nerdy

Statement : gcd( gcd(a,b) , b) = gcd(a,b) . I understand this perfectly by intution, and i can also prove by showing that every integer d that divides gcd(a,b) and b , must divide a and b, also the converse. My problem is by proving that with Bezout identity.

How can i make the fact that i can find x,y s.t ax + by = d, imply into the fact that there exists xo.yo such that d.xo + byo = (ax + by).xo + byo = a(x.xo) + b(yo + yxo) = d ? How do i prove that if x,y are minimal for a,b so is x.xo and yo + yxo ?

Mats Granvik

Is there an easy way to check if a graph is symmetric?

I know now. I have plotted the graph for a table: If n>=k then 1 else 0.

ccorn

11:56

@nerdy As to the minimality: Since d divides b, d also divides every xd+yb.

And 1d+0b=d. By the above, nothing between 0 and d can be achieved.

nerdy

12:12

hmm thats nice, so the outline is to prove that we know bezout coeficients (1,0) for c = xo(gcd(a,b) ) + yo.b are minimal because if they weren't minimal, say leading to c > gcd(a,b) then we would have that gcd(a,b) wouldn't divide c, which is absurd

thanks for the input again ccorn :)

nerdy

12:24

my problem is really about logic, to prove gcd(gcd(a,b),b) = gcd(a,b), dont we need to prove that gcd(gcd(a,b),b) = d implies gcd(a,b) = d and gcd(a,b) = d implies gcd(gcd(a,b),b) = d ?

what was the logical outline of our proof ccorn ?

meer2kat

12:41

@matsgranvik what was that message about yesterday...?

ccorn

@nerdy If you want to prove gcd(gcd(a,b),b)=gcd(a,b), an equivalent statement is: Given d=gcd(a,b), prove that gcd(d,b)=d. The latter is easy since d divides b, as shown above.

There are so many complex-analysis subscribers on this site. It would be nice if someone could properly answer this question, as the 2 answers given so far fail to achieve the goal. I could adapt Apostol's proof, but that is not something I would like to do.

nerdy

12:58

I guess its just some logic gap i have, i posted here math.stackexchange.com/questions/743478/…. By the way, your avatar is the graph of a complex function, isnt it ? So beautiful :D

Transcript for

$Mathematics$ Mathematics