Mathematics

@shintuku you should distinguish between the $x$ outside and the $x$ as the integrand. presumably they are different. also, without limits it is hard to say what it going on.

Bob

I believe there is a mistake in it

what is the best way for me to find the mistake?

thanks a lot, I'm working on formulating the question properly so the comments on the ambiguity are super appreciated

I'm trying to prove there exists a limit and determine what it is for $g(x)$ as $x \to \infty$. Wouldn't in this case distinguishing between the interior and the exterior $x$ require writing $g$ as a two variable function?

in other words, I'm working on: prove there exists an $L \in \mathbb{R}$ such that $\lim \limits_{x \to \infty} g(x) = L$

1:59 AM

whispering the secret is that $\int f \ dx = \int_a^x f \ dt$

oh! thank you so much shintuku!

you're welcome shintuku

Ted Shifrin

2:42 AM

@shintuku Disagree. The LHS is an arbitrary antiderivative. If you mean the RHS, then write it explicitly.

hm. then, $\int f \ dx = \int_a^x f \ dt$ iff for any $C \in \mathbb{R}$ there exists an $a$ s.t. $F(x) + C = F(x) - F(a)$

wait it's not right give me a second

shin: LHS is indefinite integral and may often be regarded as a set that is $\int f dx=$ set of all functions which derivative is equal to f under appropriate situations. Can you equal a function to a set? If yes, you should define that equality. Professor Ted may correct me here if I am wrong.

Yorch

Professor Ted is he wrong or correct?

$\int f \ dx$ is a set! that makes too much sense!

bless you Koro, that is an excellent idea and I will frame it in my bedroom

you’re welcome @shin

2:53 AM

$\int f(x) \ dx = \{F(x) + C:(F(x) + C)' = f(x) \}$

Avra

@TedShifrin @Bob. Yes!

or more exactly $\int f(x) \ dx = \{ F(x): F'(x) = f(x) \}$

Suppose that I have a vector space of 100-tuple that is $\mathbb R^{100} (\mathbb R)$ over field $\mathbb R$ and I am given two subspaces $V$ and $W$ such that $V=\{(x_1,x_2,...,x_{100}): 4|i\}$ and $W=\{(y_1,y_2,...,y_{100}): 5|i\}$, I want to find dimension of subspace $V\cap W$.

I say dim V=100-25=75 and dim W=100-20=80 so dim $(V\cap W)=100-45=55$

Is that correct?

@shintuku shin, I also said "under appropriate situations" that is, you should also state domain etc. also :)

3:08 AM

hm, I'll think of a formal statement for that

is it not sufficient for $F(x)$ to be an antiderivative, no matter the condition?

But for F' to make sense, F has to be differentiable, right? So it's better to state domain so that all symbols being used make sense :)

hm: if we write instead, not $=$, but $:=$, then if $F$ is not differentiable, it is not part of the set

right?

yeah I think. But suppose you give this definition (using :=) to someone, what will they make out of it? To make it more meaningful and complete, I think that you should state domain also.

makes sense, thanks!

something like this, $\int f dx:= \{F: A(\subset \mathbb R) \to \mathbb R: F'(x) =f(x) \text{for all $x\in A$}\}$ etc.

where $A$ is non empty subset of $\mathbb R$ , etc.

Any ideas here?

23 mins ago, by Koro

Suppose that I have a vector space of 100-tuple that is $\mathbb R^{100} (\mathbb R)$ over field $\mathbb R$ and I am given two subspaces $V$ and $W$ such that $V=\{(x_1,x_2,...,x_{100}): 4|i\}$ and $W=\{(y_1,y_2,...,y_{100}): 5|i\}$, I want to find dimension of subspace $V\cap W$.

3:30 AM

the notation is slightly ambiguous, but for an index to be divisible by both 4 and 5 is for it to be divisible by 20. i think you can write down an explicit basis for this space.

hi leslie :)

why do you think that notation is ambiguous.

there is no mention of i except in 4 divides i and 5 divides i. i (the person) have to infer what i (the index) is.

Ahh, I see. My bad. I meant $1\le i\le 100$.

this was implicit, but not explicit. i think i understood but was not sure. i did understand.

I was thinking what lies in $V\cap W$. The elements whose coordinates with indices of the form $4k$ and $5m$ , where $1\le k\le 25$ and $1\le m\le 20$ are zero.

So we're talking 100 -25-20= 55 non zero elements in a vector. But I am not sure.

Also the correct answer seems to be 66

:'(

Since indices divisible by 20 are double counted, I should take that into account as well but even then I get 100-25-20+5=60.

Ahh I got it. I am right :)

dim $(V\cap W)=60$.

I misread 60 as 66.

:)

3:46 AM

@Koro here I think we'd have a complete description: For a continuous $f: I \subset A \to B$ and $F: I \to B$, we have $\int f := \{ F \ | \ F' = f \}$

yes, I think. It also tells (secretly) by FTC, that $F$ exists and is differentiable also :)

I think you should also say that $f$ is bounded. Otherwise good luck integrating 1/x on $(0, 1]$ for example. :)

or better yet, $I$ is a closed and bounded set (let's consider $\mathbb R$ only) so that I is compact hence continuity of f ensures boundedness of $f$ also.

i just thought of this, since we can say a statement like "$\int f$ does not exist", isn't it wrong to define it with these conditions?

otherwise the statement "$\int f$ does not exist" would necessarily be a contradiction

I think you mean the situation $\int f=\emptyset$. But, how is it interesting?

oh, right. if $f$ has no antiderivatives, then we can say the set $\int f$ is empty

4:12 AM

right, well, then we can say $\int f(x) \ dx = \int_a^x f(t) \ dt$ if and only if $F(a) = 0$ where $F'(x) = f(x)$

Hello. How do I search for a webpage with posts authored at least by persons ABC and XYZ? Neither "user:ABC,XYZ" nor "user"ABC user:XYZ" does this. Thanks!

correction: nor "user:ABC user:XYZ"

@shintuku shin, I am not sure what you did in this but I'll suggest to have a look at a chapter in Spivak that deals with this. I think that will help you understand these things more.

my goal is to contact (thank) a particular user who has previously answered my questions

hm you're right. I'll go read it hehe

I think it's dealt with in chapter on derivatives or on integrals, though I don't remember right now which chapter exactly :'(

@shin

4:21 AM

oh! wait: actually, the proper statement is: $\int_a^x f(t) \ dt \in \int f(x) \ dx$!

exactly!

woohooo

thanks a lot Koro!

you're welcome @shin

4:48 AM

i am looking away

why

don't tell me the fruit of my strenuous labour is a false statement

Ted Shifrin

@RyanG what do you mean webpage?

5:13 AM

i mean MSE question-answer page containing posts by at least users ABC and XYZ . in this case i'm ABC, and i'd like to know which of my previous questions XYZ has answered.

5:26 AM

@shintuku i am not a fan of indefinite integrals.

i spent a few hours in a state of confusion about indefinite integrals, and am not a fan either

Martin Sleziak

@RyanG You can check the queries which I posted on MathOverflow Meta: Is it possible to search for posts/questions of an user $X$ commented on/answered by the user $Y$?

If you use some of those queries, do not forget to switch the site to Mathematics.

Hello, if f,g are continuous functions onto the unit sphere, then why is $\|tf(x)+(1-t)g(x)\|=0$ only if t=1-t? Any help is appreciated!

Can anyone give a hint please? I'm super stuck!

5:44 AM

the norm being zero means the points add to zero. there may be some g--m-tr- there.

Does it imply that f(x) must be equal to -g(x)? @leslietownes

Since they must be antipodal to add to zero?

the scalar thing is throwing me off but i do think they need to point in antipodal directions or it won't work.

it's not clear to me that you wouldn't be able to even infer information about t.

i'm still thinking it over.

It is from this by the way :math.stackexchange.com/a/3143172/810585

6:10 AM

@pritchard you must be missing something, that is certainly not enough to conclude $t=1-t$.

@copper.hat I am trying to figure out the reasoning behind this question math.stackexchange.com/questions/3142878/…

Specifically how Connor Mallin shows why the assumption holds true

Geometrically, it seems to me that the points must be antipodal for norm to be zero; is that true?

6:40 AM

@MartinSleziak Thanks, thank worked.

*that

russian bot

9:17 AM

privyet

person told me that each and every branch of math is boring

10:32 AM

the second derivative test is inconclusive

what do i do, i wasn't trained for this

10:47 AM

you use thought, shintuku, the power of thought! i summoned a directional derivative which was equal to zero, showing said point was not an extremum

wait that doesn't show anything

the multivariable definition of local extrema is either greater/lesser or equal