Mathematics - 2014-09-10 (page 3 of 4)

5:16 PM

back

Look at it, admire it and if you want it then worship it $$\int_0^1 x^{\large x^2} \ dx=1-\frac{1}{3^2}+\frac{1}{5^3}-\frac{1}{7^4}+\frac{1}{9^5}-\cdots$$
:-))))))))

Hippalectryon

worships

2

Chris's sis

@PedroTamaroff see above ...

rehband

@Chris'ssis Incredible

Khallil

Isn't stuff like that already known, @Chris'ssis?

Chris's sis

5:18 PM

@Khallil No

Khallil

I thought it was elementary. I derived that one in high school.

2

Will Hunting

@Chris'ssis I worship you.

Hippalectryon

@Chris'ssis And $\int\int_0^1x^{y^{2}}\mathrm{d}x\mathrm{d}y$ ?

Chris's sis

@Hippalectryon You meant $$\int_0^1 \int_0^1x^{y^{2}}\mathrm{d}x\mathrm{d}y$$?

$$\pi/4$$

Hippalectryon

Uh

How do you get that ?

Chris's sis

5:21 PM

@Hippalectryon Elementary integration. No need for pen and paper.

Hippalectryon

Show me :c

Chris's sis

@Hippalectryon There is nothing to show. What exactly to show you? Where are you stuck?

Khallil

It's just $\displaystyle \int_{0}^{1} \dfrac{1}{1+y^2} \text{ d}y$, @Hippa.

Hippalectryon

@Khallil How ?

Chris's sis

@Khallil True.

Hippalectryon

5:23 PM

Oh wait

Of course

-_____-

Ice Boy

@Chris'ssis don't provoke him please

Khallil

Think of it like this, @Hippa. $$ \int_{0}^{1} \left( \int_{0}^{1} x^{y^2} \text{ d}x \right) \text{ d}y $$

Oh, I see you got it.

Hippalectryon

:)

Khallil

Yea, it's pretty cool!

Hippalectryon

However

What's $\displaystyle \int \dfrac{1}{1+y^2} \text{ d}y$ ?

Khallil

5:25 PM

$y = \tan \theta$

Hippalectryon

It doesn't ring any bell to me

Oh

Chris's sis

@Hippalectryon I don't believe you.

Hippalectryon

$tan'=1+tan^2$ ?

@Chris'ssis Uh ?

@Chris'ssis I rarely use trig substitution

rehband

@Hippalectryon $\arctan(y)$ :)

Hippalectryon

Oh yes

I remember now

Khallil

5:27 PM

$y = \tan \theta \implies \text{d}y = \sec^2 \theta \text{ d}\theta$ $$ \begin{aligned} \therefore \int_{0}^{1} \dfrac{1}{1+y^2} \text{ d}y \ \ & \overset{y=\tan\theta}= \int_{0}^{\frac{\pi}{4}} \dfrac{\sec^2 \theta}{1+\tan^2 \theta} \text{ d}\theta \\ & = \ ... \end{aligned} $$

I feel like I'm always too late with these.

-______-

Hippalectryon

xD

Khallil

^_^

Huy

@Khallil: How's your room?

Khallil

I've not even cleaned through half of my papers, @Huy. Then there's the rest ... T_T

Chris's sis

@IceBoy let it go ... :-)

Huy

5:36 PM

T_T

Balarka Sen

@Chris'ssis It's not hard.

A variant of that dream thingy

Huy

I just made some dinner. Bread with jam. I should become a cook.

Chris's sis

@BalarkaSen Be the first that post the solution in chat. :-)

Balarka Sen

Moi eating. Wait a few minutes.

Chris's sis

@BalarkaSen It's not hard, that's true.

Balarka Sen

5:47 PM

Back. let me write it up.

$$\int_0^1 x^{x^2} \, \mathrm{d}x = \int_0^1 \exp(x^2 \log(x)) \, \mathrm{d}x = \int_0^1 \sum_{n = 0}^\infty \frac{x^{2n} \log(x)^n}{n!} \, \mathrm{d}x = \sum_{n = 0}^\infty \frac1{n!} \int_0^1 x^{2n} \log(x)^n \, \mathrm{d}x$$

Chris's sis

@BalarkaSen I only hope it's your work. :-)

Balarka Sen

@Chris'ssis It is, although it can be hardly said so as it closely resembles the Sophomore's dream.

Chris's sis

@BalarkaSen Good.

Balarka Sen

Your problem is just a plain mimicry of that, so nothing serious and nothing to worship either.

Chris's sis

@BalarkaSen There is no mimicry. And the second point is that I asked you in the past a question (some days ago) that you posted on a site and received a solution that you showed here and you said it is yours.

Balarka Sen

5:51 PM

$$\int_0^1 x^{2n} \log(x)^n \, \mathrm{d}x = n! (-1)^n (1+2n)^{-n-1}$$

@Chris'ssis I never said that.

Look at the dates.

Chris's sis

@BalarkaSen Well, that's less important. Anyway ... :-)

Balarka Sen

The problem was posted on the site after I produced the solution, acutally.

Hippalectryon

Acutally indeed :)

Balarka Sen

@Chris'ssis No, no it's very important if you are to accuse people that they copy-paste solutions.

I can accuse you to be copy-pasting problems and solutions taken from other sites, actually, @Chris'ssis

Be respectful to people. Don't be arrogant.

4

Hippalectryon

He's not arrogant -__-

she*

Balarka Sen

5:54 PM

She kind of is.

Hippalectryon

And fix your typo >:c

Chris's sis

@BalarkaSen What? Show me only one such a case! I create, you know ...

Balarka Sen

@Chris'ssis No, you do that. You are accusing people baselessly.

I too create.

Chris's sis

@BalarkaSen I didn't accuse you, I only told you the reality.

Pedro Tamaroff

zips soda

munchs on popcorn

Balarka Sen

5:57 PM

@Chris'ssis Compare the dates please : mathhelpboards.com/challenge-questions-puzzles-28/… and chat.stackexchange.com/messages/17406898/history

Hippalectryon

@PedroTamaroff Uh ? zipped soda ?

Balarka Sen

@Chris'ssis The solution of mine was presented on August 29, while I posted it as a challenge on August 30.

It was an accusation.

A false one and as well as offensive.

Yes, a mimicry of this

And I solved it.

Hippalectryon

@BalarkaSen It's not a mimicry if you haven't heard of the original before ...

Balarka Sen

FYI, everyone, it was not me who flagged it.

Was it, @MichaelHampton?

mathh

Hey everyone. Is there a way of seeing all the chat messages I've ever posted here?

Michael Hampton

6:04 PM

@BalarkaSen Don't bring me into this!

Balarka Sen

@MichaelHampton I am just asking you to confirm whether it was me or not.

Pedro Tamaroff

@BalarkaSen Who cares?

2

Balarka Sen

@PedroTamaroff Well, she should know before coming back here after 1/2 an hour later and offending me with her silly comments!

I'm sorry guys, but this chat is way too aggressive for me.

Pedro Tamaroff

@BalarkaSen Poor thing!

Hippalectryon

This room is turning into an arena :c

Hippalectryon

6:29 PM

chat.stackexchange.com/rooms/17052/…

For those who are tired of fights here -__-

Huy

Never.

Chris's sis

Rules are the rules, they must be respected.

(I thought I didn't break anyone, I used a star for replacing a letter in my word that was considered inappropriate)

At the same time, I noticed that @PedroTamaroff told me mad (more times, without any star in the word) here and nothing happened. That's weird.

Will Hunting

@Chris'ssis Don't worry, say whatever you want here, you will at most be suspended from chat for 30 min or more.

@Chris'ssis What did you say? You called him mad?

Chris's sis

@WillHunting No, he called me like that for a few times.

Will Hunting

Well, he is a bit mad, yes.

Daniel Fischer

6:38 PM

@Chris'ssis a) one needs a flag for a chat-suspension, and if nobody flags, nothing happens. b) "mad" isn't such a bad word. I don't know which word you used, but if it resulted in a suspension, it was probably more serious.

Chris's sis

@DanielFischer I wouldn't say that since it's a very big difference if you call someone "mad", and on the other hand you only refer to a text like being a bu****it.

Will Hunting

@DanielFischer I got suspended for saying "I just had a great shit"

Huy

@WillHunting: Then stop sharing so much personal information.

2

Will Hunting

@Huy I don't care. I will share as much as I want.

Huy

If you don't care then don't complain.

Will Hunting

6:41 PM

Wrong, I can complain.

Huy

I'm proud of you.

Will Hunting

I had accounts of 20k on each of 2 sites. I think I have the moral right to say "shit" in this chat.

Huy

What exactly does one have to do with the other?

Will Hunting

I think I will ignore you.

Huy

Go ahead.

Will Hunting

6:43 PM

And I don't think Chris Sis is arrogant

Hippalectryon

For those who want to talk about math, and not the meaning of life of whatever other totally unrelated discussion being held here -__-

chat.stackexchange.com/rooms/17052/…

Will Hunting

This chat is crazy. I got flagged for saying "I am on bananas".

This room is becoming the Islam room where every small shit is flagged.

Huy

@WillHunting: I think you should read up on what it means to "not care". Because clearly you don't "don't care".

Will Hunting

I think you are very irritating.

Chris's sis

The pity is that these discussions often push me in a zone I don't wanna be, I waste a lost of precious time, I wanna talk about mathematics only, but I'm often attracted into these traps ...

Hippalectryon

6:46 PM

@Chris'ssis Then come to my room until they're finished and let's do awesome series :D

Chris's sis

@Hippalectryon I posted this one on main math.stackexchange.com/questions/926627/…

Hippalectryon

or so I'd like to say, I have chem to finish

Chris's sis

@Hippalectryon Did you upvote?

Hippalectryon

@Chris'ssis I just did

Chris's sis

@Hippalectryon Good (it'll be more attractive)

Hippalectryon

6:48 PM

And I faved it :) I wanna know the answer

ryagami

Hi, everyone

Chris's sis

@Hippalectryon lol, I was also downvoted ...

Hippalectryon

Hello @ryagami

Huy

Good evening, @ryagami.

Hippalectryon

@Chris'ssis -__-

Will Hunting

6:49 PM

@Chris'ssis How often do you get downvotes?

Balarka Sen

1) I am in leagues with @Huy, @Will. 2) @Hippa you don't do series, you just say "it's so awesome" and nothing else.

Hippalectryon

@BalarkaSen Nope, I store them and try to do them actually

Chris's sis

@WillHunting There were some days in which I had downvotes almost every day. Now it's a bit better.

Hippalectryon

And to me it IS awesome

@BalarkaSen i'm not nearly as good in that field as you

Will Hunting

@Chris'ssis Once they downvoted me on 3 consecutive days.

I come to this room to worship @Chris'ssis series, lol.

Chris's sis

6:50 PM

@WillHunting lol :-)))))

Balarka Sen

@Hippalectryon The question is not being good. The problem is that @Chris'ssis's policy is to get praises from people and accuse people of copy-pasting solutions if they post a solution. I mean, she is a genius alright, but she should have some respect for people out there!

Hippalectryon

@BalarkaSen I'm praising her work because I want to, not because she's asking me to ...

'accuse people of copy-pasting solutions if they post a solution' It's not a generality. Everyone makes mistakes ...

ryagami

@Chris'ssis the function {1/x} looks super weird

Balarka Sen

@Hippa I have nothing against what you do. I am against her attitude towards amateurs.

Hippalectryon

I mean, I'm an amateur, but I haven't has any problem with her.

Will Hunting

6:53 PM

@BalarkaSen I think you are just blowing things out of proportion

Balarka Sen

@WillHunting I am not. Why did she (falsely) accuse me of copy-pasting solutions, then?

Chris's sis

@BalarkaSen Well, not really, you don't have to take literally all I write. Of course, I like my people enjoy my questions and solutions, but I don't go that far as suggested. Well, you say a lot of things, and then I also say a lot of things. The best thing is to drop it.

Will Hunting

We must treat children with care in this room.

Balarka Sen

ignores @Will

Chris's sis

@BalarkaSen These discussions push me away from my objectives. I only wanna focus on math.

Will Hunting

6:55 PM

I think every SE chat should be like the Eng chat!

Balarka Sen

@Chris'ssis It apparently seems like you focus on praises. just sayin'.

nvm, i'll drop it.

Chris's sis

@BalarkaSen Apparently you never talk about mathematics, you apparently seem to be always attracted to these discussions that have nothing in common with math.

rlemon

1 invalid flag + 900 chat rooms = 5 new users joining

am I doing this math thing right?

Hippalectryon

HERE A TABLE FOR YOU (╯°□°）╯︵ ┻━┻ A TABLE FOR YOU (ノಠ益ಠ)ノ彡┻━┻ TABLES FOR EVERYONE ┻━┻ ︵ヽ(`Д´)ﾉ︵ ┻━┻ NOW EVERYONE IS HAPPY, LET'S GO BACK TO MATHEMATICS. THANK YOU

5

Balarka Sen

@Hippalectryon HAHAH

Hippalectryon

6:58 PM

I'm serious though

Chris's sis

@Hippalectryon :D

rlemon

┻━┻ ︵ヽ(Д´)ﾉ︵ ┻━┻` -- how have I never seen this one before

Will Hunting

@Hippalectryon You should make an inversion of my blue square.

rlemon

and why did that formatting fail so badly.

Hippalectryon

@WillHunting Uh no then I would abandon Chris's sis's avatar

Will Hunting

6:59 PM

@Hippalectryon Are you in love with Chris Sis? lol

Hippalectryon

Uh ... awkward eww

I love her integrals :D

@Chris'ssis Any hint on showing that $\left(\prod_{k=1}^n(a+kb)\right)^{1/n}\sim b(n!)^{1/n}$ ?

$a,b>0$

Chris's sis

@Hippalectryon Sorry, no :-(

Balarka Sen

$$\begin{align}\int_0^1 x^{x^2} \, \mathrm{d}x = \int_0^1 \exp(x^2 \log(x)) \, \mathrm{d}x = \int_0^1 \sum_{n = 0}^\infty \frac{x^{2n} \log(x)^n}{n!} \, \mathrm{d}x &= \sum_{n = 0}^\infty \frac1{n!} \int_0^1 x^{2n} \log(x)^n \, \mathrm{d}x \\ &= \sum_{n = 0}^\infty \frac1{n!} \cdot n!(-1)^n (1+2n)^{-n-1} \\ &= \sum_{n = 0}^\infty \frac{(-1)^n}{(1+2n)^{n+1}}\end{align}$$

Huy

@Hippalectryon: Asymptotic wrt.?

Hippalectryon

wrt ?

Balarka Sen

7:03 PM

There, a better write up.

Huy

with respect to?

as $n \to \infty$?

Hippalectryon

Yes

Huy

isn't it clear since $\prod_{k=1}^n k = n!$?

Balarka Sen

the lower order terms are dominating, yes

Hippalectryon

@Huy It seems clear intuitively, but writing it formally isn't that easy for me

Balarka Sen

7:06 PM

@Hippa formalities are usually forgotten while thinking about asymptotics.

expand $\prod^n (a + kb)$

Hippalectryon

And idk if $d\sim c\Rightarrow d^{1/n}\sim c^{1/n}$

Balarka Sen

@Hippalectryon it is.

Hippalectryon

@BalarkaSen How do I show that ?

Balarka Sen

$d \sim c$ means $d = c + o(c)$

Hippalectryon

Oh right

True

Balarka Sen

7:07 PM

i am surprised by the mood-shifting i can do while doing math

O_o

Hippalectryon

@BalarkaSen WOn't expanding it bring awful coefficients ?

Huy

@Hippalectryon: You don't need to explicitly expand it because the coefficients are just constants. The exponent is what matters.

Balarka Sen

@Hippalectryon yes, and they will be dominatory against $n!$.

oh coefficients are useless. all $O(1)$.

it's the order of $n$ you want

Hippalectryon

But why isn't the $b$ on the RHS affected by the $1/n$ power ?

Chris's sis

@BalarkaSen I posted a new question on main - math.stackexchange.com/questions/926627/…

Balarka Sen

7:08 PM

@Hippalectryon what $b$?

Huy

@Hippalectryon: Because $(b^n)^{1/n} = b$.

Hippalectryon

$\left(\prod_{k=1}^n(a+kb)\right)^{1/n}\sim b(n!)^{1/n}$

Balarka Sen

@Hippalectryon nah.

that ^

you are multiplying something $n$ times and then taking $n$-th root =P

Hippalectryon

I still have some issues with writing a formal proof

Balarka Sen

@Hippalectryon State everything in big O and small o

formal enough

that's the fun of finitary analysis =D

Hippalectryon

7:10 PM

Uh can you give an example ? I'm not sure I get it

I mean, the first coefficients aren't dominated at all

Balarka Sen

@Hippalectryon you want to prove$\prod^n (a + kb) \sim b^n n!$. expanding the product one gets $n! b^n + O((n-1)!)$

as $n \to \infty$, the error term of $O((n-1)!)$ is $o(n!)$.

my bad.

thus, $\prod^n (a + kb) = b^n n! + o(n!) $ hence by definition, the result follows.

Hippalectryon

But then we have $o(n!^{1/n})$... is that a known form ?

Balarka Sen

@Hippalectryon what?

where do you get that from?

Hippalectryon

I mean, what happens to the $o(n!)$ after we apply the $1/n$ power ?

Balarka Sen

you don't have to. prove independently that $f(n) \sim g(n)$ if and only if $f(n)^{1/n} \sim g(n)^{1/n}$

you can set $f(n)^{1/n} = F(n)$ and $g(n)^{1/n} = G(n)$ which translates your problem into $F(n) \sim G(n)$ iff $F(n)^n \sim G(n)^n$. Be binomial.

Chris's sis

7:17 PM

@robjohn I posted a very nice question on main.

math.stackexchange.com/questions/926627/…

Hippalectryon

@BalarkaSen That gives me $f^n=g^n+\sum_{k=1}^n \binom{n}{k}g^ko(g)^{n-k}$

Balarka Sen

just eliminate the dominated terms.

better, divide by $g^n$.

Chris's sis

@BalarkaSen let me give you a real hint: @Hippalectryon knows a lot of stuff.

:-)

Hippalectryon

@Chris'ssis Not a lot :c

Balarka Sen

@Hippa all the terms you have are $g^k o(g)^{n - k}$ so that cancels a lot of stuffs, right?

in particular, prove that the sum there is $o(g)$

that should be enough hints for you.

Hippalectryon

7:24 PM

@BalarkaSen Thanks.

I have another really basic problem but I'm just stuck on it

Balarka Sen

fire away

Hippalectryon

show that $1+\frac{x}{1+\sqrt{x}}-\sqrt{1+x}>0$ for $x>0$

Just a stupid function

Balarka Sen

meh meh inequalities. but you can just minimize it.

Hippalectryon

I tried derivating and stuff but didn't succeed :/

Balarka Sen

differentiate and look for extremas.

Hippalectryon

7:26 PM

Oh wait

Balarka Sen

@Hippa well, what did you find after diffing?

Hippalectryon

I might be stupid

So we look at $1+\sqrt{x}+x-\sqrt{1+x}-\sqrt{x}\sqrt{1+x}$

We differentiate

$1/2\sqrt{x}+1-1/2\sqrt{1+x}-\sqrt{x}/2\sqrt{1+x}-\sqrt{1+x}/2\sqrt{x}$

Balarka Sen

yes, you are right, @Hippa

does that match yours?

Huy

@BalarkaSen: I think $x=3$ is the only zero of its derivative.

Balarka Sen

oh oops made a mistake

Hippalectryon

7:28 PM

Uh i differentiated the original multiplied by $1+\sqrt{x}$ for simpler calculs

Ok now we need to find a zero

Balarka Sen

yes.

Hippalectryon

How do I do that ?

Balarka Sen

well wait. i am getting $0 + \frac1{1+\sqrt{x}} -\frac1{2\sqrt{1+x}} - \frac{\sqrt{x}}{2(1+\sqrt{x})^2}$

@Hippalectryon simplfy and turn into a polynomial.

Hippalectryon

I tried that

Simplified :

Balarka Sen

@Huy you're correct. $x = 3$ is the only solution.

@Hippalectryon how did you compute that derivative? it seems wrong.

Hippalectryon

7:33 PM

Well

1'=0

Balarka Sen

oh you computed the derivative of that multiplied by $1 + \sqrt{x}$ but that won't give you the correct answer.

Hippalectryon

$\sqrt{x}'=1/2\sqrt{x}$

@BalarkaSen Why ? We're looking at the sign

Balarka Sen

@Hippalectryon what sign? it would change you're equation completely, arriving at different roots!

Hippalectryon

We're trying to prove an inequality

So multiplying by >0 should not change anything

Balarka Sen

you're finding a root. roots of $f'(x) = 0$ and $f'(x)g(x) + g'(x)f(x) = 0 $ are vastly different.

No, @Hippa, we're trying to find a minima

Hippalectryon

7:37 PM

Urm ok let's re-derivate

indeed, $\frac1{1+\sqrt{x}} -\frac1{2\sqrt{1+x}} - \frac{\sqrt{x}}{2(1+\sqrt{x})^2}$ it is

Chris's sis

@Hippalectryon When you're done with that one, try this one $$x^x \le x^2-x+1, \space x\in (0,1]$$

Balarka Sen

yes

Hippalectryon

How do I simplify that ?

@Chris'ssis Ok

Balarka Sen

@Hippalectryon first multiply by $2(1+\sqrt{x})^2$.

Hippalectryon

$2(1+\sqrt{x})-\frac{1+\sqrt{x}}{\sqrt{1+x}}-\sqrt{x}=0$

Balarka Sen

7:40 PM

remove all the denominators.

@Huy Are you an inequality-enthusiast?

Hippalectryon

$2(1+\sqrt{x})(\sqrt{1+x})-(1+\sqrt{x})-\sqrt{x}\sqrt{1+x}=0$

Huy

@BalarkaSen: Not really. I'm thinking about an easy way to solve it and almost got there but made a sign mistake which messed all up.

Khallil

I feel like I'm missing out on so much good stuff by cleaning my room.

T_T

Hippalectryon

$2\sqrt{1+x}+\sqrt{x}\sqrt{1+x}-(1+\sqrt{x})=0$

Balarka Sen

@Hippa ermmm

that's not right, no.

you have made some mistake before.

Hippalectryon

7:43 PM

Where did I make a mistake ?

Balarka Sen

@Hippa that's yours to figure out -__-

@Huy I initially thought about giving this to r9m but he is absent for some time : prove that $x^y + y^x > 1$ for all positive reals $x, y$.

Huy

@Hippalectryon: mathb.in/19933

Chris's sis

Prove by high school knowledge that $$\int_0^1 x^{ x^{\large 2}} \, \mathrm{d}x \le C$$ where $C$ is Catalan's constant

Balarka Sen

I have no idea though

Huy

@Hippalectryon: Check if I made any mistakes. I'm tired, so it's likely.

Hippalectryon

7:45 PM

@Huy Wow that's a beautiful proof that $3 > 2$

xD

Thanks anyway

Huy

@Hippalectryon: Just turn your monitor around by $\pi$.

Balarka Sen

@Chris'ssis doesn't that easily follow from my calculations?

Chris's sis

@BalarkaSen It's a different requirement there, it's high school knowledge.

Balarka Sen

$$\begin{align}\int_0^1 x^{x^2} \, \mathrm{d}x = \int_0^1 \exp(x^2 \log(x)) \, \mathrm{d}x = \int_0^1 \sum_{n = 0}^\infty \frac{x^{2n} \log(x)^n}{n!} \, \mathrm{d}x &= \sum_{n = 0}^\infty \frac1{n!} \int_0^1 x^{2n} \log(x)^n \, \mathrm{d}x \\ &= \sum_{n = 0}^\infty \frac1{n!} \cdot n!(-1)^n (1+2n)^{-n-1} \\ &= \sum_{n = 0}^\infty \frac{(-1)^n}{(1+2n)^{n+1}} \\ & \leq \sum_{n = 0}^\infty \frac{(-1)^n}{(1+2n)^2}\end{align}$$

@Chris'ssis I thought I used no special functions? Wouldn't that be considered high school?

Huy

@Hippalectryon: Is it correct?

Chris's sis

7:50 PM

@BalarkaSen It depends on the high school. Can we find a solution without using Taylor series?

Balarka Sen

@Chris'ssis Hmm, nice catch! Taylor is not high school.

Interesting, interesting, hmm.

Chris's sis

:D

Hippalectryon

@Huy I think so

Khallil

I've got a quick question. Can anybody here find an integral which is equal to $\Gamma(x) \zeta(x)$?

Balarka Sen

@Khallil yes.

Khallil

7:52 PM

Is it as simple as differentiating $\int f(x) \text{ d}x = \Gamma(x)\zeta(x)$, @Balarka?

Chris's sis

@Khallil see $(14)$ here mathworld.wolfram.com/GammaFunction.html

Balarka Sen

@Khallil no, absolutely not.

Khallil

I was hoping for a hint or two (or twelve), @Chris'ssis, but thanks for the link.

Balarka Sen

I think you need to expand $1/(e^z-1)$ in Taylor.

That'd involve Bernoullis

Lots of 'em.

Huy

@BalarkaSen: I tried looking for the zeroes of the gradient but I guess there must be a simpler approach?

Khallil

7:55 PM

I've no knowledge of Bernoulli numbers, but I'll keep it in mind for when I look at them, @Balarka.

Balarka Sen

@Huy yeah, finding 2D minima seems the only approach but I want something for lesser mortals ={

@Khallil expand $1/(e^z-1)$ in Taylor if you want to know.

Huy

@BalarkaSen: I'm too tired to do that explicitly, which is why I stopped. I'll try to think of something easier. :P

Chris's sis

@Khallil How about an integral which equals to $$\Gamma(x)\eta(x)$$ ?

Balarka Sen

math is not about knowing. it's about rediscovering again and again.

Khallil

What's $\eta(x)$, @Chris'ssis?

Chris's sis

7:56 PM

@Khallil Dirichlet eta function

Balarka Sen

@Chris'ssis Might as well : How about an integral which equals to $$\frac{\zeta(s)}{\Gamma(1-s)}$$?

=D

Chris's sis

@BalarkaSen good, good ... :-)

Balarka Sen

PS : It's not a real integral. I found it in Titchmarsh, I think it involved the Hankel contour or something.

@Huy If you do, be sure to post it here.

Huy

@BalarkaSen: No, I'll just tell you if I do. cba getting on yet another forum.

Balarka Sen

haha, OK.

I have no idea on that ineq, @Chris'ssis. Can you show me the solution? I'm interested.

Chris's sis

8:06 PM

@BalarkaSen Use the fact that $$x^x \le x^2-x+1, \space x\in (0,1]$$ You must prove it first.

Balarka Sen

@Chris'ssis Oh! So those weren't unrelated!

Let me see, let me see.

rubs hand

I didn't even try, @Huy.

sigh

Huy

There was a mistake, I think.

Balarka Sen

I stand corrected. trying again

$x^x - x\leq x^2 - 2x + 1 = (x - 1)^2$

Hmm.

ineq looks super-tight

i doubt if there is any other way than extremum approach.

@Chris'ssis It's much too hard for me. I don't know.

Huy

There is. I'm just busy with something else. Don't spoil the solution, @Chris'sis, please. :D

Balarka Sen

eh?

Chris's sis

8:17 PM

@Huy No, I won't say a word for some days, weeks. :-)

I can only tell you to encourage you that it can be proved in more ways.

Balarka Sen

ah.

then i can have a sleep in peace and look at it tomorrow.

Chris's sis

@BalarkaSen Take your time.

Balarka Sen

@Huy ping me when you post your solution.

Huy

@Chris'ssis: I most definitely won't find an elegant one but it looks solvable so I'll give it a try. :D

Chris's sis

@Huy Good! :D

Huy

8:22 PM

@Chris'ssis: Not gonna lie, I actually enjoyed solving the earlier problem with the n-th derivative. Been quite a while since I spend time with problems like these. However I couldn't spend as much time as you do on them, but I'm glad you enjoy them that much. :)

Chris's sis

@Huy :D

@Huy Once you get this $$\sum_{n=0}^{\infty} \frac{f^{(2kn)}(1)}{(2kn)!}=2^{k-1} \frac{2^k-\cos(k\pi/2)+\sin(k \pi/2)}{2^{2k}-2^{k+1} \cos(k \pi/2)+1}$$ then you're immediately done.

@ryagami Indeed. I noticed pretty late you wrote me. Sorry.

ryagami

@Chris'ssis fast response

:D

mick

Hey balarka ! How is your understanding of the prime gaps evolving ?

Huy

8:43 PM

@Chris'ssis: I took the logarithm and thus wanted to show
$$0 \leq \log(x^2-x+1) - x \log x.$$
For $x \to 0$ and $x=1$, the RHS vanishes. Since the derivative of the RHS only changes its sign once in $(0,1]$, the RHS is either non-negative or non-positive and by plugging in any value, we get the statement. Would that be correct? @BalarkaSen

Chris's sis

@Huy That's a good start.

Huy

@Chris'ssis: What am I missing?

Chris's sis

@Huy Some details like the derivative. The idea behind seems very good.