« first day (1792 days earlier)      last day (3228 days later) » 

Huy
11:00 AM
@Rememberme: I think you're confusing something with something else here.
 
@Rememberme wot ??? Then Europe is full of Dictatorships LOL
 
Yes most probably @Huy Its been days since I have read political science
 
@Rememberme The people should vote if they agree with the terms of the agreement.
 
Okay might be .. As I said its been days since I studied political science
@Huy I have an algebra precalculus question for you (My sister gave me but I haven't been able to solve it yet).....
It says that $f(x)=ax^2+bx+c$ is a quadratic equation and given that f(2)=10 and f(-2)=-2 . Find the value of the coefficient of $x^2$@Huy
 
@Huy If the most people will say Yes, then since the actual president doesn't agree with the terms, we will have elections for a new president.
 
11:05 AM
The question is pretty weird...... I can solve for the coefficient of $x$ But for $x^2$ there seems to be no way...
I am surprised that this is a question for 6th graders....
 
@Rememberme There is not a unique solution
 
@Rememberme Yeah a parameter is missing. Infinitely many polynomials go through two points
 
you have two linear equations with three variables
 
But the answer is 1
How I have got no damn idea
 
@Rememberme No, one answer is $1$
any other number will also be an answer
 
11:08 AM
There can be many answers right?
Yes but she has her assignment tomorrow and she is scolding me saying "I keep on doing maths but I don't know any maths"
 
@Rememberme find two solutions that work then
 
But the answer is just given 1..... So should I just show 1 works?
Hello @Soham
 
@Rememberme No, you should show that the question cannot be answered in the given form because all numbers work
 
Okay I will do that.......
 
Huy
11:13 AM
@evinda: Who do you think is at fault for the current situation in Greece?
 
@Tobias Tell me which topology is better if equipped to a space Product topology or Box topology.... ?
 
@Rememberme Better for what?
 
As in a topology which comes more naturally or occurs more naturally in maths
 
@Huy I don't know. All the politicians of the past.
 
@Rememberme Hmm, which ever is the one that gives the product in the category of topological spaces (I think this is the product one)
 
11:16 AM
Any reason for choosing product (A more strong reason)?
 
@Rememberme that it actually being the product then? I don't see any reason that could be stronger
I mean, it does not get much more natural than that
Note that a product has some very strong properties
 
Huy
@evinda: Of the past? Does that mean not including Tsipras?
 
I like Box topology more .. It is more natural
$\prod{U_\alpha}$ where $U_\alpha$ is open in $X_\alpha$
 
@Huy I don't know. Maybe he is also at fault.
 
@Rememberme But as far as I recall, that does not atually give you the product in Top
 
Huy
11:19 AM
@evinda: Do you vote too?
 
I didn't get you @Tobias
 
@r9m I think I also found a brilliant solution to that integral. Very fast. :-)
 
Is it $2^9 \cdot 8 \cdot 50$? Or is it also wrong? @TobiasKildetoft
 
@MaryStar You have removed too many options in the first step now
 
@Huy I will think about it what to vote. I haven't decided yet. :/
 
11:20 AM
@TobiasKildetoft What do you mean?
 
@Rememberme There is a general notion of a product in a category, and in the category of topological spaces, this is given by the product of the sets with the product topology
 
Okay...ahh
 
@MaryStar You should only avoid one of the options in step 1 from before (the one where you chose nothing)
@MaryStar instead, you have avoided half of them
 
How do we calculate it? I haven't understood it... @TobiasKildetoft
 
@MaryStar what is $2^{10} - 1$?
 
r9m
11:22 AM
@Chris'ssistheartist well I often don't trust your 'fast' .. they are too often warp speed fast! :D
 
@r9m That one I kill it one line. :-)
 
r9m
@Chris'ssistheartist Nice Question! (+1) :D
 
Can someone explain quotient fields of polynomial rings to me please?
 
r9m
@Chris'ssistheartist awesome!! would you add it in your book?
 
@Icuttrees Possibly, if you have a concrete question
 
11:23 AM
@r9m yeah, there are some cool versions to study there! :D
 
@TobiasKildetoft So, we don't calculate one option, but not a specific one, right?
 
@TobiasKildetoft I just don't understand their definition. I have worked with $k[x]$ but not $k(x)$ before
 
@MaryStar I am not sure what you mean
 
r9m
@Chris'ssistheartist great! I wanna see too how different approaches work there ..
 
Is $k(x)=k[x]/(x)$?
 
11:25 AM
@Icuttrees Ahh, that sort of quotient field (more commonly called a fraction field or field of fractions to not confuse it with a quotient of the ring)
 
@r9m That one might not be that nice for a book.
 
r9m
@Chris'ssistheartist I read MNCE's comment to your previous question :D He's got that straight!!
 
@Icuttrees No, it is not a quotient of the polynomial ring at all
 
r9m
@Chris'ssistheartist :'( aww l.. why? I wanted to see nice solutions too!!!
 
It consists of all qotients of polynomials from the ring of polynomials?
 
11:25 AM
@Icuttrees It consists of all quotients of polynomials $f/g$ where $g\neq 0$
 
Can I have an example from you maybe?
 
@r9m Well, aesthetically speaking is not that nice. Besides that there is a competition amongst my problems to be added to the book. :-)
 
@Icuttrees example of what?
 
r9m
@Chris'ssistheartist Well!! Aesthetically speaking I find that result very beautiful!
 
@TobiasKildetoft of some k(x)'s elements
 
11:27 AM
@r9m I'll consider that! We'll see.:-)
 
r9m
@Chris'ssistheartist Yaaaay!! please do!! :D
 
@Icuttrees Well, we will have all the elements from $k[x]$ there, and then also for example $\frac{x^2+1}{x}$
 
Ahhh okay I have you
 
@Icuttrees The idea is to make the smallest ring we can which is a field and which includes $k[x]$ as a subring
(smallest can be made precise, but that is not important for now)
 
@r9m Well, it's a solution, but I don't see how is that faster. Maybe I miss something in the picture.
 
11:29 AM
Okay and this is called a fraction field, thanks very much
 
r9m
@Chris'ssistheartist Well he just reduced it to various component integrals that are discussed in Lewin's book .. page 159 onwards :-) So I'd call that the standard assassination plan!
 
@r9m Did you look at my solution? :-)
 
r9m
@Chris'ssistheartist No doubt your solution is unique!!
 
@r9m lol, you're funny now! :-)
 
@Icuttrees It is possible to more generally do this for arbitrary rings (well, integral domains)
 
11:31 AM
@r9m I mean I didn't understand why is that solution faster.
 
r9m
@Chris'ssistheartist I'm seriously impressed -_- don't make fun of my amazement!
 
@r9m Thanks! :D
 
@TobiasKildetoft I'll get to that :). For now I need to understand these a little better
 
r9m
@Chris'ssistheartist oh! I need your permission to share the series (special addendum) to my blog :-)
 
@r9m No need for my permission. Please do it! :-)
 
r9m
11:34 AM
@Chris'ssistheartist Thanks!! :D
 
@r9m :-)
@r9m One crucial question: Is that form you got the shortest form of the series? :-)
 
r9m
@Chris'ssistheartist yea! I don't have a different idea to do that atm .. writing it in terms of $H_n $ might look interesting but not helpful in solving it
 
@r9m I mean the closed form you got.
 
r9m
@Chris'ssistheartist yes .. is there a shorter way to derive it too?? :D
 
@r9m I refer at a different thing. Can you get a nicer closed form? :-)
 
r9m
11:39 AM
@Chris'ssistheartist I'm confused here then .. please clarify the context again (if you could)
 
@r9m Putting all the results of the 3 integrals, you get the closed form of the series, right?
 
r9m
@Chris'ssistheartist yes
 
@r9m What is the closed form you got by that?
 
r9m
@Chris'ssistheartist haven't calculated it yet (I can be that lazy) :P
 
@r9m when you calculate it then I tell you the rest. :-)
 
r9m
11:42 AM
@Chris'ssistheartist wait ..
 
@Hippalectryon Flajolet was really big in combinatorics, too, AFAIK.
@r9m, where do you study?
 
r9m
@SohamChowdhury on my table usually ..
 
aww, c'mon.
 
Huy
You sit on your table?
 
which uni?
 
11:43 AM
lol
 
r9m
@SohamChowdhury well obviously the one that I got myself enlisted into
 
hahaha
 
duh.
which one is that?
 
Should I tell@r9m
 
r9m
@SohamChowdhury I don't know how else to say this .. but the one that has me is the one I'm studying in
@Rememberme =P
 
11:45 AM
OKay I am disclosing it.....
 
r9m
@SohamChowdhury cmi
 
@Soham Going to TIFR... :p
 
I figured that out. Google told me "UG 3rd year in an anonymous college in Chennai".
So . . . CMI is pretty logical from there.
Apr 14 at 19:12, by Sawarnik
@r9m Btw, I suppose its your last semester at CMI now? :D :D
 
Okay gtg and solve some topology problems
 
where are you?
in Munkres?
 
r9m
11:56 AM
@Chris'ssistheartist aghh! there are cubes and stuff in that ,,, don't feel like it (plum face)
 
12:11 PM
Does anyoe have an electronic copy of Cohomological Induction and Unitary Representations by Knapp and Vogan? I am at home so I can't go get a copy at the university library and I need to check some things in it
2
 
MGA
12:30 PM
I happily went through my undergraduate engineering degree thinking of vectors as n-tuples. Now I'm studying linear algebra more rigorously using Axler's book (LA Done Right), and now realize that a vector is a much more general concept (e.g. the set of functions R -> R is a vector space). But I have a small question I'd like answered before I continue reading: since LA is concerned with finite-dimensional vector spaces, do these turn out to still be n-tuples after all?
 
@MGA Yes, up to isomorhism at least
 
MGA
Fantastic, thank you!
@TobiasKildetoft Another small question: Am I correct that the study of infinite-dimensional vector spaces is done in functional analysis?
 
@MGA Not only there, but that is certainly one of the places it is studied
though very little study is usually done purely of infinite-dimensional vector spaces (not much more to them than the finite-dimensional ones)
usually one puts some extra structure on them
 
@r9m $$\frac{1}{12} \left(3 \text{Li}_3\left(\frac{1}{4}\right)+6\log (2) \text{Li}_2\left(\frac{1}{4}\right) +4 \log ^3(2)-4 \zeta (3)\right)$$
that I also added it to my post.
 
12:45 PM
My iPad keeps trying to change all occurrences of the word "invertible" into "infertile."
 
mini?
you gotta train it :-)
 
1:44 PM
@Fargle Do you mind if I ask you a question in topology?
 
@Rememberme Feel free.
 
Okay I am trying to learn product topology by thinking of some examples.....
But I have this problems in the definition...
It says that In the definition of product topology of $X=\prod_{(i \in I)}X_i$, where I is any set, the open sets are the unions of subsets $\prod_{(i \in I)}U_i$, where $U_i$ is an open subset of $X_i$ with the additional condition that $U_i=X_i$ for all but finitely many indices $i$

I don't get the additional condition @Fargle
So I want to know that is the set
$\Bbb{N} \times \Bbb{R} \times \Bbb{R} \times \cdots$
An open set in the product topology@Fargle
 
@Rememberme The additional condition comes from the fact that the product topology is actually defined so that, if $\pi_i : X \rightarrow X_i$ is the projection map onto the i-th coordinate, the sets $\{\pi_i^{-1}(U_i)\}$ form a subbase.
 
Okay what does it mean $U_i=X_i$ for all but finitely many indices ..
Does that mean that all the elements in $U_i$ should be equal to $X_i$ ? or there has to be few elements but how many??@Fargle
 
@Rememberme What topology are you putting on $\Bbb N$?
 
1:53 PM
The subspace topology..@Fargle
 
"all but finitely many" means true for all, except for finitely many cases where it's false
like, all but finitely many naturals are bigger than 10
because there are only finitely many that are not bigger than 10
 
@Rememberme Then an example of an open set in that space would be $\{1\} \times (0,1) \times \Bbb R \times (0,1) \times \Bbb R \times \Bbb R \times \cdots$.
@Rememberme math.stackexchange.com/questions/871610/… also addresses this better than I can. (I just woke up)
 
@Fargle {1} is not open
oh, it is in $\Bbb N$, nvm
 
Yeah, I was gonna say lol
 
I want to know why
$(0,1)\times\Bbb R\times\Bbb R\times\cdots$ is open in product topology...?
It says this since,
only one of the factors is not the whole space
 
1:57 PM
exactly
all of them are the whole space, except for one
one is finitely many
 
one is finitely many?
 
one is finite, yes
 
No no
 
but is it many?
 
don't be a spoonwood skull
 
1:58 PM
I am like what do you mean when you say one is finitely many...
Do you mean its finite
 
yes
 
He means "one set that's not the whole space, is finitely many sets that aren't the whole space"
 
if you only have one thing out of infinitely many things you could have had, then you have finitely many
you have a finite number of things out of the total
 
Okay...
Now I am slowly getting a nack of this...
 
the basic open sets of the product topology are direct products of open sets, finitely many of which are proper, and the rest being the whole thing
(according to urban dictionary, a "basic open set" is an open set that thinks she's the shit but really ain't)
 
2:01 PM
So lets say now I have two factors which are not the whole set .... that would still imply the set is open right?@anon
 
@anon I must confess it's a difference I do not get. @Rememberme don't worry, just nail it for finite cases
 
@Rememberme assuming your two factors are open
 
Yes
 
like $A\times B\times X_3\times X_4\times\cdots$ is open in $X_1\times X_2\times X_3\times X_4\times\cdots$ assuming $A\subseteq X_1,B\subseteq X_2$ are open
 
Yes that is what I mean
 
2:03 PM
@Rememberme Similarly, if you have $U_1 \times \cdots \times U_n \times X_{n+1} \times X_{n+2} \times \cdots$ in $X_1 \times X_2 \times \cdots$, this is an open set if $U_i$ is open in $X_i$ for $1 \leq i \leq n$, for any choice of $n$
 
but $U_1\times X_2\times U_3\times X_4\times U_5\times X_6\times\cdots$ is not open if $U_1,U_3,U_5,\cdots$ are proper in $X_1,X_3,X_5,\cdots$ (respectively)
since $\{1,3,5,\cdots\}$ is not finitely many of the natural numbers!
 
Okay....
 
questions?
 
Lets say I got to prove that $(X_1\times X_2 \times X_3\times \cdots X_{n-1})\times X_n$ is homeomorphic to $(X_1\times X_2 \times X_3\times \cdots X_n)$

I just define the function as
$f((x_1,x_2,\cdots x_{n-1}),x_n)=(x_1,x_2\cdots x_n)$ .... THen I can prove it right @anon
 
I don't know - can you prove it? :)
try!
 
2:09 PM
By taking some open set $V \in (X_1\times X_2 \times X_3\times \cdots X_n)$
 
If $f:X\to Y$ is a map, to prove that it's continuous it suffices to prove the preimages of $Y$'s basic open sets are open. Do you know why?
 
@anon Yes you can prove this from the open ball definition of continuity
 
are you working in metric spaces?
topological spaces don't always have balls
also you mean $V\subseteq $ not $V\in$
 
No when I was working They had given me this exercise to make a definition of continuity which doesn't involve metric spaces .., So I proved it
@anon What I just got to do is give the form of a basic open set in the range right and then apply the function on it and prove the preimage is open....
 
if you mean apply the inverse function to get the preimage, then yes
 
Huy
2:31 PM
@anon: Speaking of balls, to get a Riemannian manifold I endow some differentiable manifold with a Riemannian metric. Is there anything I can say about the topology induced by that metric and the original topology of the differentiable manifold?
 
well, they should be the same topology no?
 
hi @anon
 
hello
 
Huy
@anon: Are they?
If they are, does that mean that every Hausdorff and second countable space is metrizable?
 
every hausdorff and second countable space is metrizable, yes
 
Huy
2:34 PM
That's pretty surprising.
 
you have to assume the manifold is connected, but yeah same topology
 
yep
 
Huy
Very cool.
 
it's called Uryshon metrizability theory
 
Huy
I guess I should have paid more attention in topology class.
 
2:35 PM
ok, apparently a lot of topology has been discussed. what did I miss?
 
hey @Fargle
 
@BalarkaSen It's not Hausdorff + second-countable, it's regular + second-countable
 
@BalarkaSen I just dropped in for product topologies
anyway, I must shower
 
Hiya @Cbjork
 
oh right, Hausdorff second countable regular
anyway, that's irrelevant, as there are no topological spaces except metric spaces /joke
@anon ah, I see. thinking about anything interesting?
 
2:38 PM
@BalarkaSen I once posted a completely wrong answer about visualizing braid groups as fundamental groups of configuration spaces. I plan to go back and fix it today or tomorrow. And also include details about visualizing them as mapping class groups too to make up for it.
 
I never grokked why $\pi_1$ of trefoil knot is the braid group on three strands.
 
@Fargle fun challenge: W/o using L'Hôpital's rule, can you calculate $\lim_{n \rightarrow \infty} n\log(1+\frac{1}{n})$ ?
 
there seems to be a lot of literature on that out there. modular lattices, etc, etc.
 
@BalarkaSen I will think about that too then :)
 
it seems that the trefoil knot complement can be thought off as the complement of the curve $\{(x, y) \in \Bbb C^2 : x^2 = y^3\}$ in $\Bbb C^2$
and that's where eisenstein forms jump in (recall the identity about $g_2$ and $g_3$?)
@anon let me know if you find anything interesting
 
2:41 PM
@BalarkaSen I thought g2 and g3 were independent? (we're talking about mod forms and elliptic curves right?)
 
can someone discuss this question with me?
 
that question is banned from the internet
 
@KartikWatwani ?
 
(I jest, go ahead)
 
question:ecpress the ratio of the sum of squares of the odd number of terms of a GP to the sum of those terms as a polynomial of the common ratio of the GP.
*express
 
2:43 PM
I think you mean the odd-numbered terms, not the odd number of terms
 
@KartikWatwani (by the way I don't if you've seen, but I solve the problem in the other chatroom yesterday)
 
@anon well, yes, i suppose they are.
 
@Cbjork Uh, probably not. I've not taken any upper-level analysis/calculus >_>
 
but how is that relevant?
 
you asked me if I knew of some identity involving g2 and g3
y'all keeping me from my shower yo
 
2:45 PM
yeah, I mean, recall that the discriminant is g_2^3 - 27g_3^2
 
oh, yeah
 
@anon i mean odd number of terms eg,5 terms,11 terms
 
so your ell curve is well defined iff g_2^3 \neq 27g_3^2
i.e., when you're away from that curve
 
@h
@Hippalectryon how can i see the answer to the question
@hi
 
@KartikWatwani Well, just go back to the chatroom we used yesterday, the conversation is still there
 
2:47 PM
@Hippalectryon that was some other question
 
@KartikWatwani I know, but since you never replied at the end, I wasn't sure if you got my final answer or not
 
@Hippalectryon Let me see
@Hippalectryon i am not getting link to that page ,plz provide here
 
@Hippalectryon i am checking ,plz wait
@Hippalectryon in this ,i think we dont have to $aq^{k_1}\in\mathbb{Z},aq^{k_1}+1=aq^{k_2},aq^{k_2}+1=aq^{k_3}+1$
@Hippalectryon $aq^{k_1}\in\mathbb{Z},aq^{k_1}+1=aq^{k_2},aq^{k_2}+1=aq^{k_3}+$
 
2:56 PM
@KartikWatwani Why ? That's just the direct translation of the problem into mathematical terms
 
@Hippalectryon $aq^{k_1}\in\mathbb{Z},aq^{k_1}+1=aq^{k_2},aq^{k_2}+1=aq^{k_3}$
@Hippalectryon see this change
@Hippalectryon do u agree?
 
@KartikWatwani That's just a typo from me. What I did after that follows the right version
There's no '+1' at the end indeed. Doesn't change what follows though hopefully
 
@Hippalectryon i am checking i just checked 1st line only
 
morning chat
 
@Semiclassical o/
 
3:03 PM
@Hippalectryon $q^{k_2-k_1}=\frac{1}a=q^{k_3-k_2}$
@Hippalectryon i think third part is not correct
@Hippalectryon how did u brought that relation
 
@KartikWatwani ? It should follow from the relations above $aq^{k_1}+1=aq^{k_2},aq^{k_2}+1=aq^{k_3}$
@KartikWatwani Oh wait I have apparently made a very stupid mistake
 
@Hippalectryon correct it and then we will see
till the time @Hippalectryon is correcting ,is someone there to discuss this question. question:express the ratio of the sum of squares of the odd number of terms of a GP to the sum of those terms as a polynomial of the common ratio of the GP.
 
hi
 
@KartikWatwani Sorry, even though it was a stupid mistake, it makes my following reasoing obsolete :(
 
any puzzle solvers interested in a hard problem? math.stackexchange.com/questions/1341400/…
 
3:16 PM
@KartikWatwani For your new problem, what do you mean by "the odd number of terms of a GP" ?
 
@Hippalectryon that is a doubt to me as well but i think it means 5 terms or 11 terms etc..
 
@KartikWatwani Sure, but which 5 (or 11) terms ?
 
@Hippalectryon any odd number of terms being general (2n+1) terms
 
Ok, and what do you call the "common ratio" of the GP ?
 
@Hippalectryon let there be a GP $a_1,a_2...a_n$ then the ratio $\frac{a_2}{a_1}$ is common ratio
 
3:21 PM
ok
So basically we've got a progression $aq^m$ and we want to express $\dfrac{(aq^{2k_1+1})^2+\dots+(aq^{2k_n+1})^2}{aq^{2k_1+1}+\dots+aq^{2k_n+1}}$ as $P(q)$
 
@Hippalectryon no
@Hippalectryon let me write
 
Oops wait not what I meant give me a sec
 
@Hippalectryon ok
 
$\dfrac{(aq^{k_1})^2+(aq^{k_2})^2+\dots+(aq^{2k+1})^2}{aq^{k_1}+aq^{k_2}+\dots+a‌​q^{2k+1}}=P(q)$
that's what I understood. Is that wrong ?
 
can someone please help me on my question: math.stackexchange.com/questions/1345924/…
 
3:26 PM
@Hippalectryon ya thats correct
 
@user168530 Formatting your question with proper mathematical expressions would most likely help us answer it.
 
^done
credits: Hirshy
0
Q: Limit of Convergent Sequence Property Proof Help

user168530I have a question about this property: Let $\lim\limits_{n\to\infty} a_n = a$, then $\lim\limits_{n\to\infty}(ca_n) = ca$ for all $c \in \mathbb R$ If we consider when $c$ doesnt equal $0$, my book's proof says: Consider $|ca_n - ca| < \varepsilon$, then we have $|c||a_n-a|<\varepsilon$. Si...

 
@user168530 See my answer, feel free to ask me for additional information if you need to.
@KartikWatwani May I ask where the exercise comes from ?
 
@Hippalectryon thank you very much
 
@Hippalectryon a book by A Das gupta
 
3:33 PM
So to drive the point home, @Hippalectryon in the proof, we are choosing a particular epsilon?
 
@user168530 yes, since the property holds for any $\epsilon>0$
@KartikWatwani Well, why not start with the easiest non trivial case ? Can you express $\dfrac{1+q^{2k_1}+q^{2k_2}}{1+q^{k_1}+q^{k_2}}$ as a polynomial of $q$ ?
 
@Hippalectryon can i talk to u after eating dinner?
 
@KartikWatwani sure
 
3:55 PM
@MikeMiller Hi !
 
@Hippalectryon i m back
 
@KartikWatwani Ok. So, can you do the case above ?
 
@Hippalectryon i am trying
@Hippalectryon no
 
@KartikWatwani I can't either. I actually don't think it's possible at all. Which is why your exercise seems most weird.
 
@Hippalectryon but answer has been provided in the book
 
4:04 PM
@KartikWatwani Doesn't mean that the exercise isn't flawed. Maybe they mean consecutive terms for instance. Do they just give an answer, or do they explain how they got it ?
 
@Hippalectryon they just give answer
 
Because if it's consecutive terms then one gets a polynomial in $q$
@KartikWatwani What's the general answer ?
 
$a[1-r+r^2-r^3+...+r^{2n}]$
 
Yep, so they mean consecutive terms
For instance $\dfrac{q^{2k}+q^{2k+2}+q^{2k+4}}{q^{k}+q^{k+1}+q^{k+2}}=(q^2-q+1)q^k$
 
@Hippalectryon how did u got that?
@Hippalectryon ya i got it ,lets move further
 
4:13 PM
:-) if you got that one you can get the general result right ?
 
@Hippalectryon but how did u got to know it contains consequtive terms
 
@KartikWatwani Because it seemed weird to me that it would work for randomly chosen terms, so I made a wild guess that they were consecutive. Turns out I was right.
 
@Hippalectryon they were not randomly chosen terms,they were 1st,3rd,5th,7th.....(2n+1) terms
 
2 hours ago, by anon
I think you mean the odd-numbered terms, not the odd number of terms
2 hours ago, by Kartik Watwani
@anon i mean odd number of terms eg,5 terms,11 terms
@KartikWatwani In what you said it's an even number of terms, not the even-numbered terms .... ?
In the exercise you wrote here, it's an even number of (any) terms of the GP.
I had to assume that those terms were consecutive otherwise the result doesn't hold
 
@Hippalectryon ohhhhhh.that was a mistake which caused 2 hours
@Hippalectryon now how would we do this one
 
4:24 PM
So, what's the correct version, to be sure ?
 
@Hippalectryon we have to consider 1st,3rd,5th,7th.....(2n+1)th terms
 
@KartikWatwani And we still want the ratio of the square on the normal in terms of $q$ ?
 
@KartikWatwani Ok, what have you tried ?
 
@Hippalectryon i am doing ,,,wil not take more than 3 min
no i am not gettin the result
 
4:32 PM
What did you try ? At simple approach leads me to $aq^3\dfrac{1+q^{2n}}{1+q^2}$
 
i have first let a GP
$r,r^2,r^3,....,r^{2n}$
then i have chosen odd terms and created a GP
$r,r^3,r^5,....r^{2n}$
 
$r^{2n+1}$
 
?? $r^{2n}$ isn't an odd term
 
because $(2n+1) $ th terms is r^{2n}
 
4:36 PM
@TedShifrin I just realized my mistake. I am so slow.
Lo siento señor :D
 
@Hippalectryon ya it is $r^{2n+1}$
 
Ok, go on
 
@Hippalectryon then the squared terms are $r^2,r^6,r^10....r^{4n+2}$
@Hippalectryon the squared terms have common ratio r^4
 
then ?
 
@Hippalectryon then find the ratio of the squared odd terms and unsquared odd terms but didnt got result
 
4:44 PM
@KartikWatwani I started the same way to get my result though. So, we have $\dfrac{\sum_{k=1}^nr^{4k+2}}{\sum_{k=1}^nr^{2k+1}}=r\dfrac{\sum_{k=1}^n({r^{4}}‌​)^k}{\sum_{k=1}^n({r^{2}})^k}$. What can you do from here ?
 
@Hippalectryon $r[\frac{r^{2(n+1)} +1}{r^2+1}]$
this is what i can get
 
Don't you mean $r[\frac{r^{2n} +1}{r^2+1}]$ ?
There are only $n$ terms in each sum
 
i think there are n+1 terms
 
Well it's a sum from $1$ to $n$ so ...
 
there are n even terms but n+1 odd terms
that is why it is $2n+1$
 
4:51 PM
No. I took the first $n$ odd terms. $\{2k+1\mid1\le k\le n\}$ gives the first $n$ odd integers.
Waiiit
We're both right haha
I'm right because the sums I wrote above start at $k=1$, but you're right since they should actually start at $k=0$
 
okayyy
 
@KartikWatwani Ok great, we have the same result then
 
now what to do
 
Let me check once more to be sure though
 
4:58 PM
@KartikWatwani Don't we have $r\dfrac{\sum_{k=0}^n({r^{4}}‌​)^k}{\sum_{k=0}^n({r^{2}})^k}= r\dfrac{\frac{1-r^{4n+1}} {1-r^4}}{\frac{1-r^{2n+1}}{1-r^2}}=\frac{r}{1+r^ 2}\dfrac{1-r^{4n+1}}{1-r^{2 n+1}}$ ?
 
is ur latex write
because i am translating it
 

« first day (1792 days earlier)      last day (3228 days later) »