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NV-US
NV-US
9:07 PM
What is the difference between a characteristic value of a linear functional and a characteristic value of a linear opearator.
 
Ted Shifrin
Ted Shifrin
@NV-US: I've never heard of the first one.
 
NV-US
NV-US
Ok. Thank u
 
Ted Shifrin
Ted Shifrin
Have you really seen that?
 
NV-US
NV-US
No, sorry, was confused
I have a question, typing......
 
Ted Shifrin
Ted Shifrin
Heya @Ali!!
 
NV-US
NV-US
9:13 PM
Given a linear operator T on the space of all cts. real valued functions given by, T(f) = integral (from 0 to x) f(t)dt. Show that T has no characteristic values. I solved T(f) = pf and got, f(x) = Ae^(x/p), where A is constant
How to proceed now?
Sorry about, no latex. I am on my phone, it hangs
@TedShifrin
 
Ted Shifrin
Ted Shifrin
OK, so let's see. You get $pf'(x)=f(x)$, so $f'(x)/f(x) = 1/p$.
I agree with your solution.
Hmm ... So I'm confused.
 
NV-US
NV-US
How can i show that T has no char. values?
 
Ted Shifrin
Ted Shifrin
It seems like you showed every $p\ne 0$ is a characteristic value.
 
NV-US
NV-US
For all i know, the char. function i got, does belong in the space for p not equal to 0
 
Ted Shifrin
Ted Shifrin
Ohhhhh ...
 
NV-US
NV-US
9:17 PM
Yeah, i think so too
 
Ted Shifrin
Ted Shifrin
But we're being silly.
What is your $A$?
 
NV-US
NV-US
Constant
 
Ted Shifrin
Ted Shifrin
Is there some value of $x$ where you know what $T(f)(x)$ must be?
 
NV-US
NV-US
I dont understand
 
Ted Shifrin
Ted Shifrin
Look at $T(f)$.
Do you know its value, for sure, at some value of $x$?
 
NV-US
NV-US
9:19 PM
Like a constant value?
 
Ted Shifrin
Ted Shifrin
Yes.
Look at $\int_0^x f(t)dt$.
 
NV-US
NV-US
f(t)=5
T(x)=5x
f(t)=t gives T(f)=x^2/2
 
Ted Shifrin
Ted Shifrin
No, no. We don't know anything about $f(x)$. No matter what continuous function $f$ you look at, what value of $x$ can you tell me the value of $\int_0^x f(t)dt$?
 
NV-US
NV-US
At x=0
 
Ted Shifrin
Ted Shifrin
BINGO!!!
And the value is .... ?
 
NV-US
NV-US
9:23 PM
0
 
Ted Shifrin
Ted Shifrin
So now tell me what you know about your constant $A$?
 
NV-US
NV-US
Thinking
A is 0?
 
Ted Shifrin
Ted Shifrin
Yup. Now can you answer the question? :)
 
Balarka Sen
Balarka Sen
Hi @Ted
 
Ted Shifrin
Ted Shifrin
Oh oh, @Balarka, you're breaking your sleep cycle again? :)
 
Balarka Sen
Balarka Sen
9:29 PM
I have to go to school tomorrow so probably I'll go to bed in half an hour
 
Ted Shifrin
Ted Shifrin
Wow, you actually go to school?
 
Balarka Sen
Balarka Sen
Heh
 
Ted Shifrin
Ted Shifrin
BTW, with regard to that $S^{\omega + 1}$ question, I realized that if you glue two contractible things along something contractible, van Kampen isn't very interesting. :P
 
Ali Caglayan
Ali Caglayan
@TedShifrin hello Ted :)
 
Ted Shifrin
Ted Shifrin
I said hi cuz it said you were here earlier :)
 
Daminark
Daminark
9:32 PM
Hey @Ted, @Ali, and @Balarka!
 
Balarka Sen
Balarka Sen
Sure, $\pi_1$ is not hard to calculate :P
 
Ted Shifrin
Ted Shifrin
Hi Demonark
 
Ali Caglayan
Ali Caglayan
Hi @BalarkaSen
 
Balarka Sen
Balarka Sen
Gluing two contractible things along something contractible actually gives you a contractible object though
 
Ted Shifrin
Ted Shifrin
Right :)
 
NV-US
NV-US
9:32 PM
So T has no char values because we cannot find a non zero function such that T(f) = pf
 
Ted Shifrin
Ted Shifrin
Perfect, @NV-US.
 
Ali Caglayan
Ali Caglayan
van kampen is quite nice
 
Balarka Sen
Balarka Sen
Hi @Ali
 
NV-US
NV-US
Thank you for the help :)
 
Ted Shifrin
Ted Shifrin
You're most welcome.
 
NV-US
NV-US
9:33 PM
But how did u get that we were doing something wrong earlier?
 
Ted Shifrin
Ted Shifrin
We didn't do anything wrong. We just didn't figure out $A$ ... And I saw it instantaneously.
You have to use some initial value information when you have it!
 
NV-US
NV-US
Ok, i will remember that
 
Balarka Sen
Balarka Sen
@Ted What does it mean to take connection of tensors? I suspect $\nabla_X T(X_1, \cdots, X_n) = -X T(X_1, \cdots, X_n) + \sum_i T(X_1, \cdots, \nabla_X X_i, \cdots, X_n)$
 
Ted Shifrin
Ted Shifrin
Good :)
 
Balarka Sen
Balarka Sen
Oh also hi @Daminark
 
Ted Shifrin
Ted Shifrin
9:35 PM
@Balarka: A connection on vector fields induces a connection on tensors of type $(k,\ell)$.
It is a derivation that commutes with contraction.
 
Balarka Sen
Balarka Sen
For (k, 0) tensors is it by the formula I wrote?
Ah
 
Ali Caglayan
Ali Caglayan
If you consider a covering of X with included finite intersections then you can make this into a category O with morphisms the usual inclusions. Then all van kampen says is the fundamental groupoid of X is the colimit of fundamental groupoids of objects of O
 
Balarka Sen
Balarka Sen
i don't really care about that point of view
 
Ted Shifrin
Ted Shifrin
Are you talking to me or to Ali?
 
Balarka Sen
Balarka Sen
That was to Ali
 
Ali Caglayan
Ali Caglayan
9:36 PM
probably me... :(
 
Ted Shifrin
Ted Shifrin
If $T$ is an $(n,0)$ tensor then $T(X_1,\dots,X_n)$ is a scalar function and it's just differentiation in the direction of $X$. But it equals the sum when you use the derivation property. So your formula isn't right.
Categorical language is a sure way for me to ignore you, Ali, but that's OK. :P
Particularly today. I had a tooth extracted, and I'm bleeding and in pain :P
 
Balarka Sen
Balarka Sen
@TedShifrin Hm
I thought there had to be a minus sign somewhere because metric connections satisfy $\nabla g = 0$
 
Daminark
Daminark
@Ted sorry, I hope you feel better soon
 
Ted Shifrin
Ted Shifrin
Give me 6-8 months, Demonark.
The issue is that what you wrote is ambiguous, Balarka. You mean $(\nabla_X T)(X_1,\dots,X_n)$. :P
 
Daminark
Daminark
Wait is it gonna take that long? Damn
 
Balarka Sen
Balarka Sen
9:40 PM
@Ted Oh, I thought it was apparent?
 
Ted Shifrin
Ted Shifrin
It wasn't to me until you bitched.
 
Balarka Sen
Balarka Sen
Sorry lol
 
Ali Caglayan
Ali Caglayan
@TedShifrin is torture standard procedure now?
 
Kasmir Khaan
Kasmir Khaan
yelloha
 
Balarka Sen
Balarka Sen
Yeah you're right
 
Ted Shifrin
Ted Shifrin
9:40 PM
Demonark: A bunch of time to heal, then implant, then a few months, then crown ...
So eating will be interesting for a while, even after a few days :P
 
Kasmir Khaan
Kasmir Khaan
@TedShifrin I gotta prove that GL_2 (Z/2) is isomorphic to S_3
 
Balarka Sen
Balarka Sen
Also I hope you get better.
 
Kasmir Khaan
Kasmir Khaan
I dont need help but just tips
 
Ted Shifrin
Ted Shifrin
Thanks, Balarka.
 
Kasmir Khaan
Kasmir Khaan
I have the 6 elements and studiying them by order
 
Ted Shifrin
Ted Shifrin
9:41 PM
Just explicitly write out $GL_2(\Bbb Z_2)$, Kasmir. Write down the 6 matrices.
Ohhh.
 
Kasmir Khaan
Kasmir Khaan
-.-
 
Balarka Sen
Balarka Sen
Tooth problems are the absolute worst. Unfortunately I don't brush so maybe my teeth will fall off in my twenties
 
Ted Shifrin
Ted Shifrin
That's really bad news, Balarka, plus you stink.
4
 
Kasmir Khaan
Kasmir Khaan
I just need some tips :D
Ehm well
 
Ted Shifrin
Ted Shifrin
Make the multiplication table, Kasmir.
 
Kasmir Khaan
Kasmir Khaan
9:42 PM
the order of them are 1, 2, 3
 
Ali Caglayan
Ali Caglayan
hint how many groups of order 6
 
Ted Shifrin
Ted Shifrin
He doesn't know that yet, @Ali, I bet.
They're trying to get him to write down the isomorphism explicitly, I bet.
 
Kasmir Khaan
Kasmir Khaan
2 of order 3 how to distingusih them in a fast way ?
 
Ted Shifrin
Ted Shifrin
I have such an exercise in my algebra book!
 
Kasmir Khaan
Kasmir Khaan
you got a book ?
on algebra too ?
 
Ted Shifrin
Ted Shifrin
9:43 PM
Yup.
 
Balarka Sen
Balarka Sen
@Ted "Stink" is the mainstream counterculture style that comes after "punk"
 
Kasmir Khaan
Kasmir Khaan
OMGGGGGGG
 
Balarka Sen
Balarka Sen
I embrace it
 
Kasmir Khaan
Kasmir Khaan
where can I buy it?
 
Ted Shifrin
Ted Shifrin
LOL, Balarka.
 
Kasmir Khaan
Kasmir Khaan
9:43 PM
Ted pls give me link :D
 
Ali Caglayan
Ali Caglayan
type in Ted Shifrin anywhere
 
Ted Shifrin
Ted Shifrin
You can find it linked on my university homepage, which is linked on the homepage here
 
Kasmir Khaan
Kasmir Khaan
I thought you did multivarible calculus and geometry only =p
you told me this in fact i think :D
 
Daminark
Daminark
Abstract geometry: An algebraic approach
 
Ted Shifrin
Ted Shifrin
LOL, Demonark. Right.
 
Kasmir Khaan
Kasmir Khaan
9:44 PM
Okay searching now :D
 
Balarka Sen
Balarka Sen
Smacking Demonark: A Geometric Approach
 
Ted Shifrin
Ted Shifrin
Actually, my very first coauthor complained that I was too much of an algebraist because I loved to use differential forms. ... We never wrote a second paper together :P
 
Kasmir Khaan
Kasmir Khaan
this ?
 
Ted Shifrin
Ted Shifrin
Anyhow, @Kasmir, who cares? The multiplication tables will match up regardless. You just have to do it.
 
Balarka Sen
Balarka Sen
@TedShifrin hahah
 
Ted Shifrin
Ted Shifrin
9:45 PM
Yes, yes. Stop making it look like I am running a business here!
 
Kasmir Khaan
Kasmir Khaan
HAHAHA
 
Daminark
Daminark
Lol I want to see that author talk to Nori...
 
Kasmir Khaan
Kasmir Khaan
omg I really did not know you have a book on algebra
That is the best news :D
 
Ted Shifrin
Ted Shifrin
It has a bunch of projective geometry in the last chapter, which makes it a bit unique.
 
Kasmir Khaan
Kasmir Khaan
becaue you explain things in a good way
 
Ted Shifrin
Ted Shifrin
9:45 PM
Well, I do groups way after everything, so it won't match your course.
I did integers, polynomials, commutative rings first because they're easier than groups.
So, as I say, it won't match your course.
 
Kasmir Khaan
Kasmir Khaan
It is good anyway to pick few chapters
we gonan do rings and polys anyway
what did you mean by they will match ?
there are 3 elements of order 2
 
Ted Shifrin
Ted Shifrin
I think normal subgroups and quotient groups are harder than ideals and quotient rings, but I guess it's readable without the earlier chapters.
 
Kasmir Khaan
Kasmir Khaan
so how do I know whitch is mapped to which
 
Ted Shifrin
Ted Shifrin
You just need to make the multiplication tables consistent.
You don't want to think just about order.
 
Kasmir Khaan
Kasmir Khaan
Yes how the elements "multiplication" do to each one
but I ll do it my way, and then tell me if there is a better systematic way ( if i did not find it )
=P
 
Ted Shifrin
Ted Shifrin
9:49 PM
Where one element of order 2 goes and where one element of order 3 goes will determine everything, Kasmir. Check it out.
 
Kasmir Khaan
Kasmir Khaan
okay =p
 
Ted Shifrin
Ted Shifrin
@Balarka: I think you're off by a minus sign with your formula!
 
Balarka Sen
Balarka Sen
@TedShifrin Actually maybe the minus should have been on the second term instead of the first.
 
Ted Shifrin
Ted Shifrin
Right.
 
Balarka Sen
Balarka Sen
I was writing it all out and confuzzling myself with that. Thanks
 
Irregular User
Irregular User
10:09 PM
Can anyone think of ANY function g such that $\lim_{x \rightarrow 1} \frac{g(x)}{x-1} = 4$ but $\lim_{x \rightarrow 1} g(x) \neq 0$?
Tried a bunch of things but I keep getting = 0
ANY function is fine, need not be continuous
 
Wildcard
Wildcard
0
Q: Can a cube be cut according to these rules?

WildcardI'm reading Concrete Mathematics: A Foundation for Computer Science (for my own amusement), and after working out the recurrence for the number of three-dimensional regions that can be defined by $n$ different planes, I got interested in another question, more geometrical than combinatoric in nat...

recurrence-relations volume solid-geometry
 
orlp
orlp
@Wildcard for the record, even though intuitively the process is 'smooth' (you could balance the lines out by slowly moving them) I think with larger line (or plane) counts you start running into the problem where the sides of the square/cube are very large compared to the combinatorial explosion of tiny regions you get in the center of the cube where all lines (or planes) overlap
 
Daminark
Daminark
10:26 PM
@IrregularUser it isn't possible, x-1 tends to 0 so if g(x) doesn't tend to 0, the quotient blows up
 
orlp
orlp
@Wildcard $L(n) = \frac{1}{2}n(n+1) + 1$, $P(n) = \frac{1}{6}n^3 + \frac{5}{6}n + 1$
@Wildcard even stronger statement that I can't disprove is the following
given a set of lines on a square, can you smoothly translate and/or rotate the lines such that all regions have equal area, without ever reducing or increasing the amount of regions?
 
Irregular User
Irregular User
@Daminark Ah, thanks for that
 
Wildcard
Wildcard
10:43 PM
@orlp I think that is the key.
@orlp I fixed the definition. Got interrupted but now it's done.
 
orlp
orlp
@Wildcard it's definitely a stronger statement though
 
Wildcard
Wildcard
Let me try again....
 
orlp
orlp
although I have to say I have 0 knowledge in advanced geometry proofs
@Wildcard \binom{a}{b} $ = \binom{a}{b}$
 
Wildcard
Wildcard
@orlp $L(n)={n\choose0}+{n\choose1}+{n\choose2}$
@orlp $P(n)={n\choose0}+{n\choose1}+{n\choose2}+{n\choose3}$
 
orlp
orlp
Is it safe to assume that the maximum number of regions in $d$ dimensions using $n$ cuts of $d-1$ dimensional dividers is $\displaystyle R_d(n) = \sum_{k=0}^d \binom{n}{k}$?
(there must be a proper term for '$d-1$ dimensional dividers' that form regions, right?)
 
Kasmir Khaan
Kasmir Khaan
10:56 PM
@anon hi
 
anon
anon
hi
 
Kasmir Khaan
Kasmir Khaan
am trying to show that GL_2 (Z/2) is isomorphic to S_3
using table ( we did not do fancy formulas yet )
Is there a systematic way of doing this?
I mean a way that is clever ( not trial and error)
I found the order of the elements
 
anon
anon
you can interpret the vector space as {0,a,b,c}, where adding a letter to itself gives 0 and adding two distinct letters gives the third. every permutation of {a,b,c} (fixing 0) is then a linear map (preserves this addition operation).
 
Kasmir Khaan
Kasmir Khaan
can you give an example of such method?
I mean how it works
 
anon
anon
I just did
if you interpret the addition the way I said to, the result becomes obvious
but if you're in an intro class, this problem is not too much to do by simply working out everything
 
Wildcard
Wildcard
11:00 PM
@orlp I haven't worked out why the pattern works the way it does.
I didn't work out that pretty closed form on my own, unfortunately.
 
Kasmir Khaan
Kasmir Khaan
@anon am in intro class =p
 
orlp
orlp
@Wildcard have you looked at the relevant OEIS sequences and their references?
 
Wildcard
Wildcard
The exercise in the book is to work out what the recurrence is, not to put it in closed form. I did that successfully. When I checked my answer, the neat patterned closed form was given.
I can see that the recurrence holds that:
 
Kasmir Khaan
Kasmir Khaan
@anon that is a fancy way :D thanks
But if I list the two tables
how much info can I get to find what element maps to what
 
Wildcard
Wildcard
$R_d(n)=R_d(n-1)+R_{d-1}(n-1)$ for all $d$ and all $n$, and that $R_d(0)=1$ for all $d$.
 
Kasmir Khaan
Kasmir Khaan
11:03 PM
I have three elements of order 2 , and two elements of order 3
 
anon
anon
do you know about group actions?
 
Kasmir Khaan
Kasmir Khaan
nope =p
 
Wildcard
Wildcard
@orlp So all that's left is to prove that the recurrence I've defined above is equivalent to this summation and you're all set.
 
Kasmir Khaan
Kasmir Khaan
We only had the definition of isomorphism
I need to find the map phi
after that I can work out 1-1 and onto
I think we have to do it the hard way on this exercice to appriate the fanciar methods
 
Wildcard
Wildcard
@orlp But my question is really about the specific case of cutting a cube with 4 planes. That's the first interesting 3 dimensional case.
 
anon
anon
11:05 PM
you don't know what the correspondence between the matrices and the permutations is?
 
Kasmir Khaan
Kasmir Khaan
exactly
 
anon
anon
label the three nonzero vectors 1,2,3. then for each matrix, figure out how it permutes those three vectors.
 
orlp
orlp
@Wildcard I don't know why you're jumping straight into cubes
try solving squares & lines first
(lines and points is trivially true)
 
Wildcard
Wildcard
@orlp If a square can be cut in 7 equal area portions by 3 straight lines then I think a cube can probably be cut into 15 equal volume portions with 4 planes.
@orlp because it's what I thought of first. ;)
 
Kasmir Khaan
Kasmir Khaan
@anon hmm can you please explain in more detail? what vectors?
 
Wildcard
Wildcard
11:07 PM
If it is possible, I'd like a puzzle consisting of 15 pieces of different colors, all the same volume, that can be assembled into a cube.
 
anon
anon
@KasmirKhaan "label the three nonzero vectors"
 
Wildcard
Wildcard
@orlp yes, hmmm.
 
Kasmir Khaan
Kasmir Khaan
do you consider (123) a vector?
 
anon
anon
no. do you know what GL(2,Z/2Z) means?
 
Kasmir Khaan
Kasmir Khaan
yes its the general linear group of 2x2 matrices in mod 2
 
anon
anon
11:09 PM
2x2 matrices with entries from Z/2Z. what do we do with 2x2 matrices? we apply them to column vectors. column vectors with entries from Z/2Z in this case.
 
Kasmir Khaan
Kasmir Khaan
yes I have written all the 6 such matrices
 
anon
anon
if we write down row vectors, the finite vector space (Z/2Z)^2 has four elements: {(0,0), (0,1), (1,0) and (1,1)}
 
Kasmir Khaan
Kasmir Khaan
okat
okay*
 
anon
anon
the three nonzero vectors are (0,1), (1,0), (1,1)
 
Kasmir Khaan
Kasmir Khaan
okay so far so good
 
anon
anon
11:10 PM
call them a=(0,1), b=(1,0), c=(1,1)
 
Kasmir Khaan
Kasmir Khaan
okay
 
anon
anon
for each matrix, figure out how the corresponding matrix transformation permutes a,b,c
for example, the matrix with top row (0,1) and bottom row (1,0) swaps coordinates, so it permutes a and b while fixing c, so the permutation in one-line notation is (ab)(c).
 
Kasmir Khaan
Kasmir Khaan
genious :D
@anon Thanks alot ! :D first time i see something like this =P
 
anon
anon
mmhmm
 
Kasmir Khaan
Kasmir Khaan
@anon are you a proffesor of math olympiades?
 
anon
anon
11:13 PM
no
 
Kasmir Khaan
Kasmir Khaan
Okay =p
Ill keep working using this method and see where things go :D
@anon the top row (1 1) and bottom row (0 1 ) , does this correspond to (bc) ?
 
anon
anon
no, that matrix fixes b=(1,0)^T doesn't it?
 
Kasmir Khaan
Kasmir Khaan
do we interpet them as colums or rows?
 
anon
anon
the default is to apply a matrix to a column vector from the left
problem is writing columns in the chatroom is annoying and I don't know if you have chatjax
 
Kasmir Khaan
Kasmir Khaan
Hmm i see that you got a = ( 0,1 ) , i put it as (1,0)
as a colum , that was the mistake
 
anon
anon
11:23 PM
sure you can do that
 
Kasmir Khaan
Kasmir Khaan
okay
using a = (1 ,0 ) , b = (0,1) , c = (1, 1 ) as colums
the matrix first row (1,1) second row ( 0,1)
 
anon
anon
then yeah (a)(bc)
 
Kasmir Khaan
Kasmir Khaan
exactly a stayed fixe
fixed*
and c went to b
make sense that b went to c :D
okay one more try to see if I got the hang of this
first row ( 1, 1) second row (1,0) is (abc)
@anon sorry if am annoying you , but found this method quite cool =p
 
anon
anon
yep. it's a group action.
 
skullpatrol
skullpatrol
...are you going to write the Putnam again? @anon
 
anon
anon
11:30 PM
?
 
Kasmir Khaan
Kasmir Khaan
@anon did you mean that when we study the chapter group actions we will learn such methods? :D
 
anon
anon
what you did is use a group action
 
Kasmir Khaan
Kasmir Khaan
All right =p I dont know what that is yet so =p
it sounds like a fun topic
 
Dodsy
Dodsy
11:46 PM
whats up guys!
 
Semiclassical
Semiclassical
hi-yo
 
Dodsy
Dodsy
how's it going sir!
 
skullpatrol
skullpatrol
yo-hi
 
Kasmir Khaan
Kasmir Khaan
since the order of both groups are the same , to show isomorphism , is enuf to show injective right?
 
Semiclassical
Semiclassical
not bad
 
Dodsy
Dodsy
11:50 PM
I'm so excited for my first classes tomorrow!
feels like time has slowed down
 
Semiclassical
Semiclassical
cool
what's the schedule?
 
Dodsy
Dodsy
8:30- enriched calc 1
9:30 - enriched physics 1
10:30- Lin Alg
 
10 Replies
10 Replies
linear algebra is fun
loved that class
 
Semiclassical
Semiclassical
neat
feel free to ask about physics
 
10 Replies
10 Replies
I'm back at attempting to integrate 4x/(x^4+1)
I have the U substitution, but, you guys told me to do partial fraction inside of u substitution and I am confuzzled
Because now I need to solve for A and B, but, I have DU and a U
 
Semiclassical
Semiclassical
11:53 PM
to clarify what we had earlier
you can do that integration in two ways. one way is to do partial fractions immediately, and integrate each of the resulting fractions
the other is to do a u-substitution first, and only then do partial fractions and integrate the resulting fractions
 
10 Replies
10 Replies
Ok, I'll would like to figure out both, but, the second one is currently stumping me.
 
Dodsy
Dodsy
@semi
omg i keep doing that :(
I will certainly ask about physics
I love our math chats.
 
Semiclassical
Semiclassical
heh, cool
 
Dodsy
Dodsy
you really helped me study for calc :3
 
Semiclassical
Semiclassical
@10Replies kk. do you know what u-sub you want to do?
 
10 Replies
10 Replies
11:56 PM
U = x^2
 
Semiclassical
Semiclassical
having a bit of calc knowledge should help with physics.
 
10 Replies
10 Replies
then I get du/(u^2 + 1)
 
Kasmir Khaan
Kasmir Khaan
semi
if we have 2 groups of same order
 
Dodsy
Dodsy
here you have to take physics and calc together
 
Kasmir Khaan
Kasmir Khaan
injective hom is iso right?
 
Dodsy
Dodsy
11:56 PM
since physics is calc based
 
Semiclassical
Semiclassical
i mean, kinematics is basically "so here's what having constant second derivative implies"
 
Kasmir Khaan
Kasmir Khaan
@Semiclassical.
 
Dodsy
Dodsy
yeah
 
Semiclassical
Semiclassical
@KasmirKhaan not sure, tbh.
 
Dodsy
Dodsy
well the first assignment is all calc it looks like
some derivative stuff
 
Semiclassical
Semiclassical
11:57 PM
right
 
Kasmir Khaan
Kasmir Khaan
@Semiclassical okay thanks anyway =p
 
10 Replies
10 Replies
If you want an interesting thing to ask your teacher, ask them what the integral of position means
 
Semiclassical
Semiclassical
it's when you do forces that it stops being just calculus implications
@10Replies should still be a factor of 2 overall. (the u-sub gets rid of a factor of 2, but you start with a factor of 4)
 
skullpatrol
skullpatrol
not to mention energy :^)
 
Dodsy
Dodsy
Hopefully nobody tries to help me answer this.
because I want to try on my own first
 
Semiclassical
Semiclassical
11:59 PM
so you've got $\dfrac{2}{1+u^2}du$
 
Dodsy
Dodsy
and haven't even had first class
but this is a sampling of my assignment
 
Semiclassical
Semiclassical
"oh man that's so easy the trick is [REDACTED] and the answer is 42."
 
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