Mathematics

3:00 PM

Hello!!! Happy New Year!!!

When we have a differentiable function $f(x) \geq 0, \forall x \in \mathbb{R}$, does it stand that $f'(x) \geq 0, \forall x \in \mathbb{R}$ ??

Huy

@evinda: Don't you know the function $f(x)$ with $f'(x) = \frac{1}{\sqrt{1-x^2}}$ by heart?

@MaryStar: $f(x) = e^{-x^2}$.

evinda

@Huy Is it $arc \sin x$? But how could I continue to calculate it?

Huy

@MaryStar: Or even simpler, look at $f(x) = e^{-x}$.

@evinda: What is $\frac{x}{|x|}$?

evinda

@Huy $\frac{x}{|x|}=\pm 1$ right?

Huy

@evinda: Did you already come across the signum function?

Mary Star

3:05 PM

@Huy We have that $f(x)=g(x)-h(x)$, so to give a counterexample do we set $g(x)=e^{-x}$ and $h(x)=0$ ??

Huy

@MaryStar: Sure.

evinda

@Huy Yes, so is it like that? $\int \frac{\cos y}{|\cos y|} dy=\int \frac{\cos y}{\cos y sgn(y)} dy=\int \frac{1}{sgn(y)}dy$? If so, how can we continue?

Mary Star

@Huy Ok!! Thank you very much!! :-)

evinda

@Huy Do you have an idea?

Huy

@evinda: $\frac{1}{\operatorname{sgn}(x)} = \operatorname{sgn}(x)$ and $(|x|)' = \operatorname{sgn}(x)$.

evinda

3:19 PM

$\int \frac{\cos y}{|\cos y|} dy=\int \frac{\cos y}{\cos y sgn(y)} dy=\int \frac{1}{sgn(y)}dy=\int \sgn(y)dy=|y|+c=|\arcsin y|+c$
So is it like that @Huy ?

There shouldn't be the absolute value, right @Huy ? So have I done something wrong?

amWhy

Hi @Mary,

Hi @evinda!

You might have learned that my real name is Amy ;-)

And of course, hi @Huy. Certainly didn't intend to leave you out!

Mary Star

3:35 PM

@amWhy Hi Amy!! Happy New Year!! :-)

evinda

Hello @amWhy!!! Beatiful name!!!

amWhy

Happy New Year, @Mary. Yes, I thought my username was a bit clever ;-)

evinda

Yes it is @amWhy

Do you teach at a university @amWhy ?

amWhy

Yes I do. Mainly undergrads, but at least one grad class a year.

Oops...My "late" breakfast is ready. I'll check in later. It's been awhile since I've been in chat!

evinda

Which is your field? @amWhy

@amWhy A ok :)

Hello @JorgeFernández

Happy new year!!!

Chris's sis

3:44 PM

@robjohn I worked a lot for going this way. The point is that without publishing some articles before, it's hard to publish a book, I mean it's about your activity as a mathematician until you publish it. With my latest work I'm able to write a very nice book. Sometimes it seems unreal to be that I managed to go that far, but with much passion and extremely hard work there is no limit.

Jorge Fernández

@evinda Thank you,I wish you the best in 2015.

evinda

@JorgeFernández Thank you :)

Chris's sis

Extremely hard work - for succeeding in anything .

Chris's sis

4:02 PM

This one looks pretty cool

8

Q: A fractional part integral giving $\frac{F_{n-1}}{F_n}-\frac{(-1)^n}{F_n^2}\ln\left(\!\frac{F_{n+2}-F_n\gamma}{F_{n+1}-F_n\gamma}\right)$

I've been asked to elaborate on the following evaluation $$ \begin{align}\\ \displaystyle \int_0^{1} \!\cfrac 1 {1 + \cfrac 1 {1 + \cfrac 1 {\ddots + \cfrac 1 { 1 + \psi (\left\{1/x\right\}+1)}}}} \mathrm{d}x & = \dfrac{F_{n-1}}{F_{n}} - \dfrac{(-1)^{n}}{F_{n}^2} \ln \!\left(\!\dfrac{F_...

Chris's sis

4:17 PM

Basically all is done elementarily (when decomposing the work) although at first sight it seems like a very complicated job.

Mary Star

How could I show that $\sin^2(x)=\frac{1}{2}(1-\cos(2x)), \forall x \in \mathbb{R}$ using the Cauchy-product??

Huy

@MaryStar: $\sin(x) = \sum_{k=0}^\infty (-1)^k \frac{x^{2k+1}}{(2k+1)!}$.

@evinda: I'm a bit busy and preparing to go out for dinner soon, did you solve it?

evinda

@Huy No, I didn't... :)

Huy

@evinda: I'll be back in 2-3 hours, if you're still around then. =)

Alec Teal

@PedroTamaroff need your attention real quick

math.stackexchange.com/questions/1087614/… There's a serious error in the answers to this question, I'm right (by luck) but the answer is 1 not 0.

Mary Star

4:31 PM

@Huy So, should it be as followed??

$$\sin^2(x) = \sum_{k=0}^\infty (-1)^k \frac{x^{2k+1}}{(2k+1)!}\cdot \sum_{k=0}^\infty (-1)^k \frac{x^{2k+1}}{(2k+1)!}=\sum_{k=0}^\infty \sum_{i+j=k}(-1)^i \frac{x^{2i+1}}{(2i+1)!}(-1)^j \frac{x^{2j+1}}{(2j+1)!}=\sum_{k=0}^\infty \sum_{i+j=k}(-1)^{i+j} \frac{x^{2(i+j)+2}}{(2i+1)!(2j+1)!}$$

Huy

@MaryStar: I thought you wanted to use the Cauchy product?

evinda

@Huy Of course I will :D Have a nice time!!!

Mary Star

@Huy Isn't the Cauchy product used at the second equality??

evinda

Let $a, b \in \mathbb{Z}, b \neq 0$. How could we show that:
a div b = ⌈a/b⌉ ,if b < 0
Could you give me a hint?

Jasper Loy

Hi.

evinda

4:45 PM

Hi @JasperLoy

@JasperLoy Happy new year

Jasper Loy

@evinda Happy new year. I am sad though. I need to figure out how to best live the rest of my life now.

evinda

What options are there? @JasperLoy

Jasper Loy

@evinda I don't know. I need to get well from my OCD first, which is what I have been trying to do. I have not been working for 7 years.

evinda

Aha @JasperLoy Do you want to find a job?

Jasper Loy

@evinda I might try to find one before going back to study more math. We'll see.

Each year I try to get better and get well enough, but each year bad things happen and I am back to square one. It is very miserable.

@Chris'ssis Hope your book is published this year.

Andrew Thompson

4:55 PM

I note that there are less than 100 questions tagged Type Theory; are there few people on this site who's familiar with it?

Chris's sis

@JasperLoy I hope that too. :-)

evinda

@JasperLoy :(

Jasper Loy

@AndrewThompson Yes. I do not know what type theory is.

@evinda :(

r9m

cloasing votes here please !!

@Chris'ssis Happy New Year :-)

Chris's sis

@r9m Happy New Year! How are you doing these days? :-)

r9m

5:10 PM

@Chris'ssis surviving resolutions ;)

Chris's sis

@r9m lol, OK :-)

user153330

5:26 PM

@Alizter you wrote a program that renders cayley diagrams chat.stackexchange.com/transcript/message/16194613#16194613 do you know where i can learn to do that?

Don Larynx

@Chris'ssis Programming isn't hard work yet I succeed in it.

wow I am hungover as a pootenany

user153330

LEL just scored 8/10 in individual.utoronto.ca/somody/quiz.html

Don Larynx

oh god @user153330

I got an 8/10 as well lulz

teadawg1337

@user153330 The answer to the ninth one is definitely Hobo

Don Larynx

@teadawg no person with a youth and beard like that is a prof. the quiz lied.

user153330

5:34 PM

Baby Pedro

teadawg1337

I took it seriously and got a 10/10

Mary Star

We have the polynomial $f(x)=x^3+6x-14 \in \mathbb{Q}[x]$. We have that $f(x)$ has exactly one positive real root $a$. That means that $f(x)$ can be written as followed:

$$f(x)=(x-a)(x^2+px+q)$$

Where do $p,q$ belong to?? Are they in $\mathbb{Q}, \mathbb{R}$ ??

mikeonly

@MaryStar Is $\mathbb{Q}$ a complex or rational numbers set?

Mary Star

@mikeonly $\mathbb{Q}$ is the set of rational numbers

mikeonly

5:57 PM

@MaryStar I think they are for sure in $\mathbb{R}$. For your example they are not in $\mathbb{Q}$.

Can somebody have a look on the proof books.google.com.ua/…

iwriteonbananas

@JasperLoy just step into radical action, it'll free you

mikeonly

It's The Nested Interval Lemma (or Cauchy–Cantor Principle).

iwriteonbananas

@user153330 lol

mikeonly

I have a question whether this proof can be simplified to just the axiom of completeness of $\mathbb{R}$?

If we have $$a_1\leq a_2\leq \dots \leq a_n \leq b_n\leq \dots\leq b_2\leq b_1$$ isn't there a point between $a_n$ and $b_n$ for sure due to the axiom of completeness?

user153330