Mathematics - 2017-09-08 (page 1 of 8)

@anon I need to show that they both obey the same group structure, so f (xy) = f(x) f(y) thus homomorphic , but the map i found is not like f(x) = e^x or something, I constructed it from the finite elements, you see what I mean ?

10 Replies

That would explain some of my confusion.

Semiclassical

this one is trickier as a result.

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Would it be easier to do trig substitutions instead then? Or can I still use partial fraction?

Kasmir Khaan

@anon If I do them all case by case , it would not be a good proof imo, because if the size were larger , it would take forever ><

Semiclassical

12:12 AM

can you confirm which version it is?

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x^4 + 1

Semiclassical

that's a good question. i'm going to cheat

drat.

Kasmir Khaan

@anon I dont want an answer , I need to figure it out bymyself, just a hint or something on how to think right =p

Dodsy

drat is a good exclamation.

Semiclassical

okay, the answer comes out in terms of arctan.

10 Replies

12:13 AM

anon

@KasmirKhaan how many maps have you ever shown are a group homomorphism? stop avoiding doing the work and just do it. you'll learn more advanced modes of thinking later.

Semiclassical

...ow.

Kasmir Khaan

@anon okay thanks =p ill keep trying

Semiclassical

kinda funny that the first term there is the easier to integrate despite looking worse.

but yeah, you definitely want to do a trig sub.

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Yeah, not only did they give me a ridiculously hard problem... it's also combined with a deceivingly easy problem. I already got the first one, hence why I didn't ask about it to begin with.

Semiclassical

12:15 AM

in fact, $\int \frac{du}{1+u^2}$ basically is an inverse trig function

if you've got access to even a small table of integrals, you'll probably see that one

I guess the hint I'd give is that you want to choose $u$ so that $1+u^2$ becomes simpler

if you do $u=\cos\theta$, though, then that becomes $1+\cos^2\theta$ which isn't nice.

$u=\sin \theta$ is similarly bad.

so you'll want either $u=\sec\theta$ or $u=\tan \theta$.

one of those will work, the other won't

so pick one and see what happens.

Dodsy

night semi, ttyl

Semiclassical

night

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I'm not sure how to do trig substitution. I googled it, and I don't see any square roots in this problem

Semiclassical

doesn't mean you can't do it. the same principle holds

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What exactly am I trying to substitute?

Semiclassical

12:25 AM

well, i gave you two options

either $u=\sec\theta$, or $u=\tan\theta$

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Is this a different U than in intgral ( du/(1+u^2))?

Semiclassical

wait.

i'm wrong. sec(theta) isn't an option to try.

u=tan(theta) is the only one you'd want to attempt

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I don't know how to go about attempting, thats what I'm saying.

Semiclassical

it's the same u.

so here's the logic

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1 + tan^2 = sec^2?

Semiclassical

12:27 AM

right. that's why this substitution will help.

so if u=tan(theta), then what's du?

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but doesn't U = x^2?

Semiclassical

sure.

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x^2 doesn't = 1+ tan^2.... right?

Semiclassical

you could do $x^2=\tan\theta$ directly.

no, but nowhere have I suggested that it should be

Kasmir Khaan

@anon I came to conclusion that I have to write down all the product f(xy) = f(x) f(y),nothing else makes sense to show that the structure is obeyed

Semiclassical

12:29 AM

the point is that if x^2=tan(theta), then 1+x^4=1+tan(theta)^2=sec(theta)^2.

so at this point we treat u=tan(theta) with u being understood as a function of theta. what's du/d(theta)?

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sec^2(theta)

Semiclassical

right. so du=sec^2(theta)d(theta)

but as you noted earlier, 1+u^2=1+tan^2(theta)=sec^2(theta)

so du/(1+u^2)=d(theta)

hence the integral becomes $2\int \frac{du}{1+u^2}=2\int d\theta$

and that last integral is easy if you don't trick yourself into thinking its hard :)

skullpatrol

...and don't forget the + C

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Semiclassical

denied

skullpatrol

12:37 AM

drat

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The last integral is easy? Is it just theta?

Semiclassical

yuuup (and the 2)

and now you just need to back substitute to get back to x

you've got $u=x^2$ and $u=\tan\theta$, so $2\theta=...?$

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2arctan(u)

Semiclassical

sure, and u=x^2

so that's

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2 arctan(x^2)

Semiclassical

12:41 AM

ding

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Do I bring the 4 back in, or are we done with that?

Semiclassical

well, let's check to see by differentiating

if that's our antiderivative, we should be able to differentiate w/r/t x and get back 4x/(1+x^4)

best way to do that is to use the chain rule with u=x^2 and remember how differentiating arctan works

...which, not going to lie, seems painful.

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I should not have waited 2 years between calc 1 and 2, I don't remember any of the derivative or integrals of trig functions or log and ln and whatever.

Semiclassical

the thing is: no, you don't bring the four back in. the 4 has been there the entire time; we split the initial numerator up as 2*2x

the 2x went with the differential (since du/dx=2x) and the other 2 stuck around after

so the answer is complete.

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yay

skullpatrol

12:45 AM

Why the two year gap?

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Think the gap is about 1.5

Took calc I in 9th grade, then I took linear algebra in 10th and then 11th I have no more math at my highschool, so I walk over to the university nearby for calc II

skullpatrol

Hmm...

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I took AP calc, which didn't help because I learned just enough calc to pass a test, but, not enough for me to remember a significant portion of it.

I got a 5 on AB and 4 on BC though, so it worked temporarily lol.

skullpatrol

...what textbook are you using?

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essential calculus: early trancendentals

by James Stuart

Only 15 problems left :/ I did 7 in far more time than I would like to admit

Will you guys hate me if I choose to never take another math class after calc II? I'm ready to be done with this whole math thing

skullpatrol

1:04 AM

::puts 10 replies on ignore::

::changes username to 9.999... requests::

:-D

Kasmir Khaan

hi chat

Who can help me proving isomorphism of groups?

Semiclassical

1:30 AM

@KasmirKhaan I'm not confident this will work, but

there's only two groups of order 6. one of them is nonabelian and isomorphic to S3, the other is abelian and isomorphic to Z/6.

Kasmir Khaan

I moved on to another exercice, Z/7 * isomorphic to Z/6

Semiclassical

ah. that should be a lot easier.

Kasmir Khaan

@Semiclassical I dont think we are allowed to use such ideas because we did not prove them =p

let me tell you what I got so far

Semiclassical

you'd have to prove more than what I just said, yeah.

Kasmir Khaan

Z/7* = <3> = <5>

Z/n is cyclic for all n

Z/6 is generated by {1,5}

Semiclassical

1:33 AM

1) prove that GL2(Z/2) is a group of order 6. 2) prove that there are only two groups of order 6. 3) prove that it's not Z/6. 4) conclude that it's S3.

Kasmir Khaan

Its smart but we are not allowed to use something we did not go thru =p

so I will make a table of both groups

Semiclassical

yeah, i figured.

Kasmir Khaan

and show that structure is the same =p

it is the same as in the second example

Semiclassical

you may be able to make a few shortcuts.

Kasmir Khaan

I think he picked small groups for that reason

Semiclassical

1:35 AM

for instance, $\begin{pmatrix}1 & 0 \\ 0 & 1\end{pmatrix}$ definitely acts as the identity under matrix multiplication

Kasmir Khaan

so we do them by incpection =p

Yes I got all the products to work , without using any fancy stuff i dont think I can do better than that

Semiclassical

If we label that as $f(e)$, then we definitely have $f(e)f(a)=I_2 f(a)=f(a)=f(ea)$

so that establishes the $f(e)$ row of the operation table, and similarly for the column

Kasmir Khaan

How to make table on texstudio?

Semiclassical

but this only does so much.

there's a few ways, I think.

none of which I remember too well

Kasmir Khaan

haha =p

Semiclassical

1:38 AM

\begin{array}{|c|c|} \hline \\ e&a \\ \hline \\ a& e \\ \hline \end{array}

Kasmir Khaan

gave error

Semiclassical

well. that would be right except that it's too tall :/

tbh you're best off googling it

Kasmir Khaan

I think Ill just hand write them if it will be too much to write ><

he does not like that but

btw semi

on the second example

if a map 3 to 1 , 5 to 5 , 2 to 2

the first 2 are generators and the order of 2 is 3

2^3 = 8 = 1 mod 7

and 2+2+2 =6 =0 mod 6

1 to 0

idenity maps allways to identity

Semiclassical

one way to describe that simply

write each element 1 through 6 as 3^x

and then map 3^x -> x

Kasmir Khaan

Oh nice :D

it will be also easiar to prove hom, surjective and injective that way

Semiclassical

1:45 AM

this preserves the multiplication structure in (Z/7)*, since 3^x 3^y = 3^(x+y)\mapsto x+y=(x+y)

and it's also easy enough to establish the bijection: 3^0 = 1, 3^1=3, 3^2=9=2, etc.

Kasmir Khaan

Yepp it works perfectly :D

Semiclassical

this is a map from (Z/7)* to Z/6, to be clear, not the other way around

Kasmir Khaan

Yes we used that Z/7* is generated by 3

Semiclassical

right.

this same idea should work for any (Z/p)^* with p an odd prime, in fact. one just needs to find a generator.

and one can then show an isomorphism with Z/(p-1).

Kasmir Khaan

because both are cyclic right ?

Semiclassical

1:49 AM

right.

if it were (Z/n)^* with n composite then it'd be trickier

Kasmir Khaan

this stuff is intresting but a bit tricky =p

Semiclassical

I forget if there's an easy way to identify a generator of (Z/p)*, come to think of it.

Kasmir Khaan

i think by fermat theorem

a^p-1 = 1 mod(p)

or wait that cant be right ><

Semiclassical

it's a^(p-1)=1 mod p

but it doesn't really help here. 2^6=64=1 mod 7, but 2 isn't a generator.

Kasmir Khaan

yes >< I think one just have to find a generator by incepection

then once that found , the others will follow

<a^k> = < a^gcd(n,k) >

Semiclassical

1:54 AM

What I want to say is that if p is prime then (p-1)/2 generates the (Z/p)^*

Kasmir Khaan

where a is a generator and n is the order of the group

Semiclassical

but i'm not at all certain that's true.

=p

google time

haha

1:57 AM

right, this generator is known as a primitive root mod p

Kasmir Khaan

oh thanks :D

hmm am having trouble proving hom here

Semiclassical

even with primes it doesn't seem like there's a simple pattern to the primitive root. neat

Kasmir Khaan

Yeah when we did fermat in other classes , some numbers gives wrong answer

they are not prime and they "pass " as primes

Semiclassical

well, suppose you've got two elements of (Z/7)*. since 3 generates this group, these two elements must be of the form 3^x and 3^y where x,y are between 0 and 6.

what's the product of these two numbers in (Z/7)*?

Kasmir Khaan

3^(x+y)

But do we do f : 3^x --> x

or f :x -->3^x

its comfusing having 3^x first

Semiclassical

2:01 AM

either works.

if you find x->3^x clearer, go ahead

in that case you've got a map from Z/6 to (Z/7)*

Kasmir Khaan

lets do f: 3^x --> x

yes I know it should work both ways =p

Semiclassical

okay, so (Z/7)* to Z/6.

Kasmir Khaan

f(xy) = 3^xy right?

that is not what we want

Semiclassical

indeed not.

let's go the other direction, just to save some energy

Kasmir Khaan

okay ><

f:x ---> 3^x

grrr

same thing I get

Semiclassical

2:03 AM

the group operation in Z/6 is addition, and we want to prove that f preserves this operation

Kasmir Khaan

okay

Semiclassical

so what does that mean here? i.e. what does it mean for f to preserve addition in Z/6?

Kasmir Khaan

f(xy) = f(x) + f(y)

the first product mod 7

Semiclassical

careful.

Kasmir Khaan

the second sum is mod6

oh

got it backwards

Semiclassical

2:05 AM

right. the way you wrote it would work for 3^x |-> x.

Kasmir Khaan

f(x+y) = f(x) * f(y)

Semiclassical

right. that's what we want for x->3^x.

but what's f(x)*f(y), based on what we decided about f?

Kasmir Khaan

f(x+y) = 3^(x+y) = 3^x 3^y = f(x)* f(y)

Semiclassical

right.

the only thing one should be a bit careful about, I think, is that 3^(6)=1

so it's legitimate to think of 3^(x+y)=3^x 3^y with x+y taken mod 6.

Kasmir Khaan

it should be 1

oh

Semiclassical

2:07 AM

right.

one might want to be careful with the writing at that point.

okay, back another time

Kasmir Khaan

okay thanks for help ! @Semiclassical

J. M.

2:26 AM

It can be a bit annoying when relevant unupvoted comments get buried under petty upvoted ones.

Oh well.

Typhon

2:39 AM

indiedb.com/games/block-builder

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concept is cool... graphics are hideous

Faust

3:10 AM

morning

yhey typhon

@TedShifrin second day was harder graph theory is throwing me a curve ball also i didnt like my first geometry class hope it gets better also i got 1 analysis question for you next time i find you ^^