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3:49 AM
How can we approximate $f(x+a)$ by a series involving only $f(x)$ and some constants?
I know by Taylor series we can do something like that, but that involves derivatives of $f$ and we are not sure if it differentialble.
4:01 AM
@Knight I'm aware of but don't know a lot about en.wikipedia.org/wiki/Puiseux_series
This is usually for smooth nice functions but it can handle some similar issues en.wikipedia.org/wiki/Asymptotic_expansion
You can also truncate the Fourier series, or truncate various "smoothed" versions of the Fourier series
Thank you
@Knight anything specific in mind, ie whats the domain of $f$, and you want approximation in what norm?
Calvin, if $f$ and $g$ are positive functions then I’m unable to find any counter-example of $$avg \left(\frac{f}{g}\right) = \frac{avg(f)}{avg(g)}$$ Where avg means the average of the function.
i suspect something like $f,g:[0,1]\to(0,\infty)$ defined by $f\equiv 1$, $g=x$ will work
in fact i suspect the equality will be hard to find
I meant $$avg \frac{f}{g} \geq \frac{avg(f)}{avg(g)} $$
I’m unable to find any counter-example of the inequality above, keeping in mind that f and g are positive functions.
4:17 AM
this one looks true
oh no
What happened ?
i thought it was soemthing else lol
It is true but I’m unable to confirm it
really? fair
then why were u looking for a counterexample lol
I couldn’t find any counter-example so I concluded that it is true
But I want to prove it.
4:20 AM
well im worried because its stronger than the bound $\int fg \le \int f \sup g$
at least when $\int := \int_0^1$
Yes, we are taking the interval to be [0,1]
I think we have to prove $$ \int_{0}^{1} \frac{f(x)}{g(x)} dx \geq \frac{\int_{0}^{1} f(x) dx}{\int_{0}^{1} g(x) dx}$$
4:44 AM
set $f=FG$ and $G=g$ then the inequality is the same as $\int FG \le \int G \int F$ which is why i stated it as the above without division
desmos.com/calculator/orbr7urfne if you wobble B_1 on this enough, it goes from over the black straight line to under it, which is the inequality
theres a nicer argument via scaling instead of semi-brute force search but cba atm lol
also the example on two points turns into an example on [0,1] by considering step functions
I really cannot understand your argument, sorry lol
basically, if you could prove something stronger than Holder's inequality
it would hold for all versions of Holder's inequality
like how there's for two vectors $\sum a_i b_i \le \sum_i a_i \sup_i b_i $
Holder’s inequality is beyond my scope :(
i just used the name its just pulling out the sup from the integral or sum
its just a name and i dont want the full power for all $p$, just the easy edge case which this is a part
5:06 AM
I’ll come back with a nicer argument
5:32 AM
@Knight,@CalvinKhor, can we show that an odd degree polynomial has atlest one real root?
It's ture, but I am having a hard time accepting it
How can we prove it?
@Arjun see limit x tends to infinity and -infinity
equation will be llike ax^n+..... where n is odd
at infinty the term ax^n will be dominated
@user69608 ,then?
I haven't done calculus for quiets while now
@Arjun both limits will be opposite if one is +infinty then other will be -infinity so there exist atleast one real root
@Arjun u can imagine just plugging infinity
Think about it like this Arjun. The limit as $x$ tends to infinity is equal to the sign of the coefficient of $x^n$, as the $x^n$ term dominates the other terms, and $x^n$ is positive
Then the limit as $x$ tends to negative infinity is equal to the negation of the sign of the coefficient of $x^n$, as $x^n$ is negative
So we have a function that ranges between negative infinity and infinity
And also it's a polynomial, so we know it's continuous
Then just appeal to the intermediate value theorem
@Drathora he wont be knowing as he has not done calculus but it is easy to see
5:46 AM
Oh, so instead of me saying "appeal to the intermediate value theorem", instead let me say:
A continuous function when drawn on a graph is a connected line right?
-infinity se +infinity udd ke to jayega nhi x axis ko toh cut krna hi padega
So if a connected line on a graph goes from negative infinity to infinity, obvously it must go past every single possible value of $y$ on its way
And on our graph that we've drawn, the corresponding "root" is just which $x$ is lined up with $y=0$ on the connected line that we draw
@user69608 ,I have done calculus, I haven't used it for over a year now ;-)
@Arjun lol
Thanks to @Drathora,and@user69608
I found it amazing tho😯
5:57 AM
10th mein padha kya calculus?
@user69608 , ha 12th pura kiya tha
6:07 AM
@Knight I have a proof by a scaling argument that a counterexample exists but its easier just to look at this explicit counterexample, where I have done the conversion to step functions for you desmos.com/calculator/q5vb6vq8rj
a wobbly blue line at the bottom appears iff the parameters form a counterexample for $\int ab \le \int a \int b$
the proof is fiddly because of the setup of the problem (it would be nice if it was just $\mathbb R$ instead of $[0,1]$, and if we allowed the function to take the value 0) but these are things that can be worked around
@CalvinKhor Means it is not true in general that $$avg \left(\frac{f}{g}\right) \geq \frac{avg(f)}{avg(g)} $$ ??
= is very easy to disprove, and its not true that ≥ always holds
Okay, thank you so much friend !
you can guess this because Holder's inequality (sorry lol) is basically optimal
6:15 AM
np, tell me if you dont get the graph i think its kinda self explanatory if you read the function definitions
find the number of 7 digits numbers the sum of whose digit is even. i want to do this multinomial theorem. by $(x+x^2....x^9)(1+x+x^2....x^9)^6$ how to do further??
this will be $a1x+a2x^2....a63x^63 $
how to do now?
@CalvinKhor This example and Holder etc is way too complicated. Just take $f = g$ to be characteristic functions for the interval $[0, 1/2]$. $fg$ is still the same function as $f$, whose integral is $1/2$. $1/2$ is not less than $1/4$ last I heard.
Knight wants to divide so modify this by making it $\varepsilon$ instead of $0$ outside the interval $[0, 1/2]$
much better thanks
though i never used holder lol
Yeah I'm just saying
im in the middle of something but i was getting there, i had already set $f=g$
thx for the simple example
6:26 AM
Ah OK maybe I should have looked at your example a bit more carefully then. I concede.
no no i mean i was in the middle of trying to get a simple example when i had to do something else
please tell my problem
@CalvinKhor nvm its done
7:13 AM
@user69608 sry also i dont really like to do combinatorics lol
@CalvinKhor ok
1 hour later…
8:26 AM
If we have $$ \int_{0}^{1} g(x) dx \geq m $$ then is it necessary that we have $$g(x) \geq m$$ ??
@Knight imo it depends on value of x.
@Knight so not always
@user69608 Can you please say a little more?
It is true that $\sup g \ge m$. For if the opposite inequality held it would give $\int g < m$
Means the only necessary condition is $$sup~g \geq m$$ ??
this means there exists $x$ such that $g(x) \ge m$
8:35 AM
All right
no, the necessary condition is the integral $\int_0^1 g \ge m$
what is sup?
@CalvinKhor lol, I was asking what’s the necessary condition for that integral to be true
8:36 AM
oh, i misunderstood.
@user69608 The maximum value of a function is denoted by $sup~ f$ but remember that sup may not be the actual value of the function $f$
What I said is, $\int g \ge m$ implies $\sup g \ge m$ implies $\exists x : g(x) \ge m$
$g(x) \ge m$ is sufficient but not necessary to imply $\int g \ge m$
if by $g(x) \ge m$ you mean $\inf g \ge m$
Yes, I meant for all $x$ lol
But now I can see your reasoning, the graph of $g$ only has to go above the line $m$ for a small interval and we will be okay
8:39 AM
sorry wrong image
@CalvinKhor So, if we somehow can prove that $sup ~g \geq m$ then can I say $$\int_{0}^{1} g dx \geq m $$ ??
no, you need a condition on the infimum
(I just doubt myself so much that’s why I keep on confirming things)
8:40 AM
@CalvinKhor What?
if $\inf g \ge m$ then $\int_0^1 g \ge \inf g \int_0^1 1 \ge m$
Okay Cal
if $\sup g \le m$ then $\int_0^1 g \le \sup g \int_0^1 1 \le m$
Got ya
8:42 AM
You’re are very toućhe
my internet is very slow so ill take a long time to view your picture
@CalvinKhor ok
18. If $\int(\log x)[x](\operatorname{sgn} x) d(x-[\sin x])$
$=x \log (f(x))-g(x),$ where $[\cdot]$ denotes greatest integer function, then
(1) $f(x)=x$
(2) $g(x)=x$
(3) $\lim _{x \rightarrow 0} x f(x)=0$
(4) $\lim _{x \rightarrow 0} x f(x)=1$
that's your picture
i prefer questions typed
Hi guyz, if R and I represent real and imaginary number sets respectively, is n(R)=n(I)?
8:46 AM
so what is the meaning of $\int f dg$? is it $\int_{(0,x]} f dg$? I forget if it should be left open or right open
@TooFatManNoNeck what is $n(R)$ and what is a 'real number set'?
@CalvinKhor sorry i didnt get u
$n(R)$ the number of elements in the real number set R.
cardinality of R = n(R)
what is a 'real number set'? why did you say 'the' real number set?
@TooFatManNoNeck Believe me or not but your name has made my day! Thanks for the laugh that you gave me
@Knight lol
8:49 AM
OK. Thank you @all.
is it the set of all real numbers? then i guess yes, if $I = \{ ix : x\in R\}$....? its afterall indexed by $R$
@user69608 what is the meaning of $\int f dg$? I only understand things like $\int_A f dg$ where $A$ is some measurable set
@CalvinKhorL R is a set of all real numbers.
that would be "the" set of all real numbers
OK. I am not english native speaker.
and usually written \mathbb R $\mathbb R$ or \mathbf R $\mathbf R$
sorry, I'm not trying to annoy you with english class
saying "the" means unique
8:51 AM
I think all he wants to know is whether the cardinality of real numbers is equal to the cardinality of imaginary numbers
Yes. That is right
@Knight yeah, i did answer that interpretation
@CalvinKhor what does that mean that u are telling A... i am in high school
@TooFatManNoNeck But mind you, complex number is not the same as the imaginary number.
if you are in highschool, how did you find a question about stieljes integrals
8:53 AM
complex = real "+" imaginary
Calvin is in full form
@TooFatManNoNeck Yes
@CalvinKhor i am preparing for JEE exam
JEE sounds scary lol
idk stieljes
8:54 AM
well i can probably figure out the answer given enough time, but I have no idea how to explain it to you then
@user69608 Do you know how to pronounce it?
In a Venn diagram, then I can draw a circle, label it as the real number set, and divide them into equal areas, one represents THE imaginary number set and the other one represents THE real number set.
@Knight does that matter?
@user69608 I’m not mocking you brother, I was just having fun with ya coz I have hard time seeing the letter in that name
i missed a t lmao srry mr stieltjes
8:55 AM
@feynhat @Thorgott Help I'm being percolated
Any comments, please ping me. I have to leave for a couple of minutes.
@CalvinKhor may be it have a simple solution
@Knight i know
what are some easier questions of this type you can solve? @user69608
It is said like “Steel Jes”
8:58 AM
@CalvinKhor i am seeing such for first time. may be we have to work for x positive or negative .how do u simplify that d(x-floor(sinx))
of course you dont know what integral means yet but floor(sin(x)) is a very simple function
its constant on certain intervals; break the integral according to that info
@BalarkaSen but its an indefinite integral
ah i didnt notice that, that makes things way worse because the question is meaningless
you have to interpret $\int$ as $\int_0^x$
@BalarkaSen hence why i first asked for clarification
@BalarkaSen oh
9:01 AM
@CalvinKhor yeah this stuff is bad
typical engineering math horrors
Balarka are you a 4th year undergrad student?
its typeset in idk word equation editor or something
second year
lmao @Calvin
@BalarkaSen In which Year you completed your school?
uh, 2 years ago?
9:03 AM
@Knight lmao what did u expect?
@user69608 Actually he is Indian and I don’t know much about Indian system of education.
@Knight ok
@BalarkaSen Did your university took a entrance test for admission?
Indians are exactly like Americans but eat less junk food and don't wear stupid hats
and more into IT i guess
9:05 AM
Or was it by an interview?
under the assumption that $\int$ means $\int_0^x$ so you cannot assume (2) holds cuz there can be an arbitrary constant, and because you can always change what $g$ is, you cannot assume what $f$ is either
@BalarkaSen which university?
there was an exam and an interview yes
so now you only need to care about the behavior near 0
@user69608 indian statistical institute
9:06 AM
@BalarkaSen oh
@BalarkaSen did u appear for JEE?
did you do this $JEE$ thing @BalarkaSen? what on earth is it, some exam for applied stuff?
lol sorry to doubleask
its like SAT but for engineers
markedly worse
@BalarkaSen which Year Thorgott (rhymes with Forgot) is in?
9:07 AM
same as me, i think
ask him
Okay @Thorgott Are you a second year undergrad?
actually 1,2,3,4 are all wrong, for the same reason
@BalarkaSen wdym
so either the question doesnt expect you to notice that you can treat $x \log f(x)$ as a remainder term, or its not $\int := \int_0^x$
if $\int = \int_0^x$ and you were to charitably force $f(x) = x$ to be true then for $1\gg x > 0$, [\sin x] =0 and. $[x]\sgn x = 1$ so the integral is just $x\log x$, and then (3) is also true
so thats what I'll go with, either none are true, or (1) and (3)
9:45 AM
@CalvinKhor what about g(x)
what about it?
yes, I am
10:01 AM
@Thorgott Did your university take entrance test for admission? And was it comptetive (I mean rate of acceptance was quite low)?
none whatsoever
@Knight tho The acceptance rate at IITs is merely 0.7 percent
@user69608 Yes it is very notorious
@Thorgott I want to study at same place.
10:41 AM
any 3d geometry SME?
10:59 AM
screenshot of 100% text
@CalvinKhor ?
idk man thats a lot of work
type it up, add your thoughts and post it as a question?
also I have no idea what it means to add two planes
and their definition of linearly dependent is not so correct
Here i'll do you a favour
Let $A$ is set of all possible planes passing through four vertices of given cube. Find number of ways of selecting four planes from set $A$, which are linearly dependent and one common point. (If planes $P_1 = 0$, $P_{2}=0, P_{3}=0$ and $P_{4}=0$ can be writen as $a P_{1}+b P_{2}+c P_{3}+d P_{4}=0$, where all $a, b, c, d$ are not equal to zero, then we say planes $P_{1}, P_{2}, P_{3}, P_{4}$ are linearly dependent planes).
Let OABC is a regular tetrahedron and $P$ is any point in space. If edge length of tetrahedron is 1 unit. find the least value of $2\left(\mathrm{PA}^{2}+\mathrm{PB}^{2}
I suggest you ignore everything in the brackets lol it just needs to be burned to the ground so that you can rebuild from scratch
11:22 AM
@CalvinKhor how so? how is it different from your definition?
well, i still have no clue what linear dependence of planes are, but if its anything like linear dependence of vectors, then $\{ v_1, v_2, v_2 \}$ can be a linearly dependent set even if the coefficients $a,b,c$ that witness this have $c=0$. What you need is that not all of them are zero (as opposed to all of them are not zero)
"Let $A$ is set"
oh actually i think the bracket makes sense. after you make the above change
@Thorgott i transcribed it fully faithfully
yeah, but why is that printed in a textbook?
or is it just some sheet
@Thorgott yes
11:36 AM
@Thorgott its probably just some exercise printed for preperation for some exam of user6numbers's apparently called JEE. Perhaps there's a .doc of this floating about or something
@CalvinKhor IT is a question from a coaching institute sheet for preparing jee
before I did UK-based IGCSEs my local exams were mostly like this. In fact even during my IGCSEs i think my teacher prepared some horrible .doc files though I didn't have the vocabulary to complain about typesetting
@Knight hålo
@CalvinKhor Hola
What is it there Calvin?
i think hello is globally recognised by now
even if it does tell them that you're a tourist
@CalvinKhor Hola is the Spanish form of Hello
11:42 AM
yeah but if I say hello to a spaniard they definitely would understand i think
spaniard the right word? idk
: (
I checked the equivalent of a dictionary apparently the proper way to spell it in malay is "helo" lol
@user69608 why the long face?
@CalvinKhor Yep
well, or anyone else who speaks spanish, they're not all from spain
Yes, example Rocky Martin
12:26 PM
@CalvinKhor it is from a exercise, they don't tell us from
@user69608 don't tell us what? sorry I don't follow
thank you korea for kimchi
@CalvinKhor it is last exercise of sheet which is tough(atleast for us) which are not discussed in institute.
except that all exercises are discussed
i think i can do the exercise but its a lot of work, i have some stuff i'm trying to do too, how about we meet halfway and you tell us what you've tried for your exercise. Do you know what this set $A$ of all planes is?
or ask it as a question so that you can find someone who wants the karma or has more free time
12:44 PM
@CalvinKhor i thought it would be quick here
maybe its quick if you've practiced this a lot, but I'm not super quick with geometry :) anyone who wants to take it up is welcome to chat...
@CalvinKhor second question is already there on site,found it
if you havent tried to solve them yet, you will learn more (and hurt more) if you try very hard for a long time, before you get the solution
1:11 PM
Expand all summands a la $(x-y)^2=x^2-2xy+y^2$ to obtain
$$ \vec a^2+\vec b^2+\vec c^2-2(\vec a+\vec b+\vec c)\vec p+4\vec p^2,$$
which is
$$ \left(2\vec p-\frac{\vec a+\vec b+\vec c}2\right)^2 + \text{something}$$
after this what will be summation a.b
1:24 PM
floor (floor(a/x)/x) = floor (a/x^2) ??
Not really sure what you're asking @user69608
@Re60K try x=-1 and lo and behold
@RE60K for counterexample with $x>0$, if $a=1,x=0.6$, then RHS = 2 while LHS = 1
1:41 PM
Sanity check: Consider $(\mathbb{Z}/2\mathbb{Z})^2$. As a finite ring, this is Noetherian. The zero is $(0,0)$. The only unit is $(1,1)$. The non-zero, non-unit elements $(1,0),(0,1)$ are both idempotent, hence reducible, so there are no irreducible elements in this and no non-zero, non-unit element factors into irreducibles. Does this make sense?
Every Euclidean ring is a principal ring I think
And product of local rings is always a Euclidean ring
Maybe the last implication is dubious (Edited out)
I can't remember
every Euclidean ring is a PID, yes
every PID is a UFD
Euclidean ring need not be a domain
Just have a Euclidean size function
urgh, it does to me
otherwise the implication is probs wrong
whatever, does my specific example make sense?
Euclidean ring is definitely principal I think
1:48 PM
it's just 4 elements, I only want to confirm I'm not missing something insanely stupid
I'm trying to remember the general thing lmao
your example is fine
ok, thanks
so now I can be smartass by pointing out this exercise asking me to prove every non-zero, non-unit element in a Noetherian ring factors into irreducibles is wrong and then be doubly smartass by pointing out it's true for Noetherian domains
lol yeah
of course they meant domains
it doesn't say so in the exercise
so I had to make sure I'm not being stoopid
Boolean rings are always good examples
to trick people that is
1:52 PM
ye, Boolean rings are weird
what is $(\Bbb R_{\ge 0} \cup \{\infty\}) \otimes_\Bbb N \Bbb Z$ as a semiring?
both operands are semirings
I think unique factorization ring is when any non unit non zero divisor factors into irreducibles
who cares about things that aren't domains
I needed to think about Euclidean rings once to prove that for any PID $R$, $\text{SL}_n(R) \to \text{SL}_n(R/I)$ is surjective for any proper ideal $I$ in $R$
Cuz $R/I$ is then a Euclidean ring and Euclidean rings are elementarily generated so you pullback elementary matrices
oh and btw $0 \times \infty = 0$ if you're wondering how multiplication is defined
who cares about rings, semirings ftw
1:56 PM
for domains, every non-zero non-unit factors into irreducibles iff the ring satisfy the ACC on principal ideals and, in a domain where every non-zero non-unit factors into irreducibles, such a factorization is unique iff all irreducibles are prime
yeah FD = ACC on principal ideals
I think that's how the story for domains goes
the first part is certainly true, I'd have to rethink the second to be sure, but I don't care to do so
Why are you guys doing ugly algebra again
yeah its correct
this is fairly nice
1:58 PM
Agree with Thorgott
I still have the ugliest exercise left
Euclidean rings are important
though it's only ugly cause I'm too stupid for it
do you know about the Dedekind Hasse norm Thor
I guess a more fundamental question is whether you can tensor semirings together
1:59 PM
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