$$\lim_{n\to\infty} 2^n\sum_{k=2^n}^{2^{n+1}-1}\frac{1}{k^2}$$

(actually, this is a very funny question)

Then, another one is $$\lim_{n\to\infty} \sum_{k=2^n}^{2^{n+1}-1}\frac{1}{k}$$

@Chris'ssis Ooh, ooh! pick me!

@robjohn Hello @robjohn! I didn't see you in the last period of time!

@Chris'ssis I was not here.

@Chris'ssis why is that funny?

@Chris'ssis it suffices to bound it by a suitable integral, right?

@IanMateus if the integrals you use to bound it converge to a limit, yes

@robjohn That series is to me like a person that uses a broken mask to steal its real identity, but it's a poor mask, a broken one, and you may see his/her face. :-)

It's strange, but these things are like living creatures to me, each one having its own personality ...

@BalarkaSen Hi, what's the solution to that transformation problem?

@ParthKohli Which?

@BalarkaSen $x^3 + \alpha x + \beta$ to $x^3 - ax^2 - b$

@Ian I am tired of explaining to people that Abel-Ruffini is not the same as what Galois proved. Sigh. People can be so idiotic time to time.

I figured that $\beta = -b$

That leaves us with $\alpha x = -ax^2$

@ParthKohli Wrong.

@BalarkaSen How can the product of roots change?

$a + b = c + d$ does not mean $a = b, c = d$

@Chris'ssis well, you can use $\frac1{n-1}-\frac1n\ge\frac1{n^2}\ge\frac1n-\frac1{n+1}$

Both the polynomials are equal right?

@ParthKohli Yes, they are both $0$.

But the roots are not really equal.

They have a rational relation, though,

Ah. OK, that makes sense.

I am going to the General room now, there I will explain.

@BalarkaSen hello, I would like to ask you something about a ENT question :-)

@robjohn Yeah.

Not in the right mood, @Ian. Ask @Chris'ssis.

any ideas on attacking $\displaystyle \lim\limits_{n \to \infty} \int_0^1 \cdots \int_0^1 f\left(\frac{n}{\sum\limits_{i=1}^n \frac{1}{x_i}}\right)\,dx_1\cdots \,dx_n$ .. where $f:[0,1] \rightarrow \mathbb{R}$ is a continuous function ?

@Chris'ssis ok. today I discovered that the power series of log(1+x) yields two different series expansion of log(2): $$\sum_{k=1}^\infty \frac{(-1)^{k+1}}{k}$$ and $$\sum_{k=1}^\infty \frac{1}{k2^k}$$ . The last converges really fast

@GabrielR. Indeed.

@GabrielR. There are even better and faster series for those.

Heya

@r9m Weierstrass approximation theorem?

@GabrielR. Have you seen the 2 limits I posted above?

@Chris'ssis I can't evaluate $\displaystyle \lim\limits_{n \to \infty} \int_0^1 \cdots \int_0^1 \left(\frac{n}{\sum\limits_{i=1}^n \frac{1}{x_i}}\right)\,dx_1\cdots \,dx_n$ to begin with ..

@Chris'ssis not yet

@Chris'ssis Seems to be log 2

@N3buchadnezzar Indeed.

@Chris'ssis Something like let $2^n$ tend to infinity instead of n, and then you get the same sum as Gabriel posted above

@r9m I might think is $\displaystyle f(1/4)$.

someone can help me about a recursion question? bounty will vanish soon. I'm tryng to find the link betwen recursion definition of function and other fields like category theory even if the question is mainly about the existence of a special terminology.

50 repps. bounty math.stackexchange.com/questions/752079/…

@Chris'ssis Have you seen this?

@r9m This should be $0$.

@r9m and this one should be $f(0)$.

@ParthKohli Why did you block me?

@r9m I think I can finish this one, but I need some time. It probably works by integrals manipulation, too.

Apples!

@Chris'ssis awesome :)

please tell me how to do it .. once you are done :)

@r9m I'm done.

@Chris'ssis how did you show $\displaystyle \lim\limits_{n \to \infty} \int_0^1 \cdots \int_0^1 \left(\frac{n}{\sum\limits_{i=1}^n \frac{1}{x_i}}\right)\,dx_1\cdots \,dx_n = 0$ ?

then I can have $\displaystyle \lim\limits_{n \to \infty} \int_0^1 \cdots \int_0^1 \left(\frac{n}{\sum\limits_{i=1}^n \frac{1}{x_i}}\right)^k\,dx_1\cdots \,dx_n \le \displaystyle \lim\limits_{n \to \infty} \int_0^1 \cdots \int_0^1 \left(\frac{n}{\sum\limits_{i=1}^n \frac{1}{x_i}}\right)\,dx_1\cdots \,dx_n$

@r9m Use this $$\frac{1}{\sum\limits_{i=1}^n \frac{1}{x_i}}=\int_0^{\infty} e^{\large -s\sum\limits_{i=1}^n \frac{1}{x_i}} \ ds$$

@Chris'ssis That was AWESOME !!

Friends, how do I disprove the statement that when a function f is one-to-one, then it is strictly increasing?

And Happy Easter :D

@DrJonesYu Happy Easter. :)

@ParthKohli @DrJonesYu Happy Easter.

@Sawarn Marry Easter

@DrJonesYu Happy Halloween!

You racist.

Huh.

-.-

Sorry discrete math has got me done :c

DOWN*

@ParthKohli Its not halloween idiot :P

WOAH

Dem strong words....

@robjohn @N3buchadnezzar @GabrielR. those limits above can be seen as Riemann sums, and we're done.

How to calculate $\large{\int\limits_0^4 \frac{dx}{4+2^x}}$ ?

@Chris'ssis SuperSis please help.

@Sawarnik This one is elementary ...You should think of it for a while.

wait...

@Sawarnik Use this $$\frac{1}{4+2^x}=\frac{1}{4}\frac{(4+2^x)-2^x}{4+2^x}$$

@Chris'ssis Oh thanks! Yup, it was easy.

@Sawarnik You see!!! You need to think of it a bit! Never give up! :-)

@Sawarnik one idea is to use $\int_a^b f(x)\,dx = \int_a^b f(a+b - x)\,dx$ :) .. that way you won't have to integrate anything except integrating a constant :)

@r9m I love the graphical representation of that form of integration by substitution!

The symmetry is awesome!

@Shisui its the symmetry that makes it look awesome :P :D

@r9m Have you got any other cool integral tricks?

@Shisui Its difficult for me to start listing them .. when I see a problem .. depending on the circumstances they pop up in my head :)

$$\int_0^{\sqrt{2}} \left(\int_0^{\sqrt{2}} e^{a b} \, da\right) \, db$$

@r9m Fair enough! I'm the same!

@r9m maybe you didn't note, but that trick above allows you to finish the integral immediately. (referring to what I showed to @Sawarnik)

@r9m Thanks genius :)

@r9m $$\int \frac{2^x}{4+2^x} \ dx=\frac{1}{\log(2)}\int \frac{(4+2^x)'}{4+2^x} \ dx=\frac{\log(4+2^x)}{\log(2)}$$

@MatsGranvik I got $$ \int_{0}^{\sqrt{2}} \left( \int_{0}^{\sqrt{2}} e^{ab} \text{ d}a \right) \text{ d}b = \dfrac{e^{4}-1}{b\sqrt{2}} - \sqrt{2} $$

@MatsGranvik I thought you'd need to evaluate the inner integral w.r.t $a$, then the result of the inner integral with respect to $b$ ...

@Shisui Yes you are right. I did not notice that you did the first step only.

@Chris'ssis YAS .. I understood that :) .. I tried $\int_0^4 \frac{1}{4+2^x}\,dx = \int_0^4 \frac{1}{4+2^{4-x}}\,dx = \int_0^4 \frac{4+2^x - 4}{4.2^x+2^4}\,dx = \frac14 \int_0^4\,dx - \int_0^4 \frac{1}{4+2^x}\,dx$

@MatsGranvik My mistake! $$ \begin{aligned} \int_{0}^{\sqrt{2}} e^{ab} \text{ d}a & = \dfrac{1}{b} \left( e^{ab} \right) \Big|_{0}^{\sqrt{2}} \\ & = \dfrac{1}{b} \left( e^{b\sqrt{2}} - 1 \right) \end{aligned}$$ I integrated the outer integral incorrectly. I took $b$ out as a constant ... facepalm

@Chris'ssis my attempt gives a decrease on the order of $n^{-1/2}$

@robjohn I am still unable to find some upper bound of $e^{-s/x}$ , or there a different way ?

Happy Easter all

I ate chocolate for breakfast

@user127001 Happy Easter!

@user127001 I'm eating chocolate right now!

@robjohn Sorry, I don't answer the questions since I have some problems here with my pc.

@robjohn That should be true.

@Chris'ssis: I am at lunch on phone. I will compute and post when back.

Is there a general form for $$\sum_{k \ge 1}\left(\frac{k}{r}\right)\frac{1}{k}$$, where $\left(\frac{k}{r}\right)$ is the Legendre symbol?

For instance, for $r=2$ it's $\frac{1}{\sqrt{2}} \log(1+\sqrt{2})$, and for $r=5$ it's $\frac{2}{\sqrt{5}} \log(\frac12(1+\sqrt{5}))$.

This guy must be a physicist.

@Alyosha yes, see the very end of this section on wikipedia

actually that's only with r prime

Have you ever felt time flies with a very high speed? This is what happens to me for some time. When I believe there were 10 minutes passed I realize it was about 3 hours passed or something like that.

(I begin to hate the clocks)

Mmm

@Chris'ssis You checked the link I showed you?

@N3buchadnezzar Which one?

Integral

@N3buchadnezzar No

@Chris'ssis Here.

@N3buchadnezzar Thanks.

Seen this one before?

@N3buchadnezzar I don't remember right now.

I am proving that for all f,g in vector space of functions $h(f,g) = \int_{0}^{1}f(t)g(t)dt$ is an inner product... How to prove it's positive definite?

@Chris'ssis Okay =) I can not remember seeing it before and it looks interesting =)

@N3buchadnezzar Indeed.

Of course, vector space of functions defined and continuous on (0,1) :)

In other words, why should $q(f) = \int_{0}^{1}{f^{2}(t)dt} \geq 0$?

well, if the integrand is non-negative everywhere, then the integral is non-negative everywhere.... or you want to prove that thing itself?

@JayeshBadwaik Yes!

@mirgee you can derive this from the basics of riemann sums for sure.... apart from that, let me see.....

ping @Mike

@JayeshBadwaik That's exactly what I want to do! Can you just give me some hints what to use, where to start?

hello @FernandoMartin

@mirgee Write down the definition of the riemann integral in terms of riemann sum (Rudin does it as the convergence of the sums of the infimums and the sums of the supremums over partition) you just have to worry about the sum of the infimums, show that the infimum of the sum is greater than or equal to 0, and hence, the integral is non-negative for sure....

@Mike: do you read any math blogs?

i look at tao's and gowers' occasionally, but i don't actively read any

y

@seaturtles Thanks.

@JayeshBadwaik Thank you, I will take look at it :)

@robjohn After a while I reach this point $$\lim_{n\to\infty} n \int_0^{\infty} (1-x \log(1+1/x))^n \ dx =\int_0^{\infty} \lim_{n\to\infty} n (1-x \log(1+1/x))^n \ dx \rightarrow 0$$

@robjohn well, if you ask me about this way then I wanna tell you it can be improved. :-)

@Chris'ssis I am back from lunch and grocery shopping.

@robjohn OK. I wrote you above some of my thoughts on the last steps of my way.

@robjohn If I let above $x\mapsto 1/x$ things become much more clear.

apologies if this is the wrong place to ask. I am learning graph theory in uni, in a computing course. Can anyone explain a scenario that I might find that useful in a job? In other words should I focus on graph theory or just try to pass the unit?

are you a cs major?

I dont know the term major. its a bachellor degree.

Are you studying computer science rather than mathematics

yes

Then yes, you should absolutely learn the graph theory

Graph theory is essential and everpresent in the modern world

Because pretty much everything can be modelled as a graph

do you mean the analysis and design phase of programming?

I guess?

ok cheers

I'm not a computer programmer, but I'm aware that graph theory is essential to those who are

I will focus on it

we learn so much stuff outside of just programming, and I am anxious to get the pieces of the puzzle connected

@robjohn I think I can let it that way.

@Mike Also to Alexander.

sup @Pedro

@FernandoMartin Hey.

@PedroTamaroff Yes, discrete mathematicians go bonkers for it too.

@SalehenRahman There's the floor function.

@FernandoMartin What's the news?

not much

@PedroTamaroff interesting. I always thought that the floor function is to round down to the nearest integer.

@SalehenRahman Oh.

@SalehenRahman $\lfloor \log_2(n) \rfloor$

You floor the log.

I misunderstood.

What @Mike said.

i win

@PedroTamaroff @Mike thanks a lot!

That was such a simple solution... silly me.

no worries my man :)

@robjohn $$1-x \log(1+1/x) < \frac{1}{2x+1}, \space x>0$$

@robjohn I'd like to request that this question be locked but not deleted.

@robjohn it's actually $$1-x \log(1+1/x) < \frac{1}{x+1}, \space x>0$$

@robjohn tomorrow I'll write down my proof with all the details. It is pretty elementary. By the way, I also used the inequality $\displaystyle e^{-x} \le \frac{1}{1+x}$.

f: B -> N where f(S) is the number of 1 bits in S and B is all the possible bit strings.

Is that a function?

are bit strings finite?

I'm assuming right

There is no capped number of a length of bit strings

So I think its a function

i see no reason it shouldnt be

why do you think it might not be

I think it is

define 'function'

define 'define'

jackass

=o

A function is just a mapping from one set to another, right?

@DrJonesYu 'mapping'

Where every element in the first set is mapped to an element in the 2nd set

'mapped to'

@Pedro: ;__;

The ball bustin :(

Technically, a function is a bunch of pairs.

@FernandoMartin what's that? the bad guy from Pan's Labyrinth?

In your case, we'd have (S,# of 1 bits in S), @DrJonesYu

Yeah

And thats a function! :D

f: B -> N where f(S) = 0 if S is the empty bit string, and otherwise f(S) is the smallest integer i such that the ith bit of S is 1

Is that a function too? I think so

@DrJonesYu We don't have some magical pair space for these pairs to live; you have to ensure that the first part of each pair is in the domain and the second part is in the codomain. (That's why @Mike asked about whether the strings are finite)

I don't think the set theory definition of function matters except to set theorists and punks

@DrJonesYu Also no two distinct pairs can have the same first part. So that's technically what you're checking when asked to show well-definedness.

But since your teacher isn't a punk/set-theorist, you can work at a reasonable level of abstraction.

Yeah thanks, so I think both of those functions meet those rules

@DrJonesYu I think so too.

^_^ Well I am happy now Karl

Are you happy?

hm, yeah, I drew a nice commutative diagram, so I am happy

I did a lot of function proves

proofs

provves*

proofs*

provves

that's a new one.

=)

Pedro loves to bust balls

What's the opposite of busting balls

@Karl "But since your teacher isn't a punk/set-theorist..." haha

@karl so what have you been studying lately

@Mike I'm learning Galois Theory

Hello @PedroTamaroff, I am a bit confused. Let the partial sum $P_m$ of a power series be $\sum_{k=0}^{m}a_kz^k$. I want to show that if $\forall \varepsilon\gt 0\, \exists N >0 : m, n\gt N \implies |P_m-P_n|\lt 2\varepsilon$, then the series converges. How can I prove it? I think I could use the existence of a infimum $\inf\{P_k : k> N\}$ and a supremum $\sup\{P_k : k> N\}$, so $\lim \inf_{k\to \infty} P_k - \lim \sup_{k\to\infty} P_k =0$. Do you think this is fine?