Mathematics

9:34 AM

@N3buchadnezzar

9:49 AM

Greetings

@Chris'ssis hey there

@robjohn How is it going? :-)

@Chris'ssis pretty well... It is hot and muggy and it rained some earlier tonight.

10:05 AM

@robjohn I wanna show you a nice problem.

@robjohn $$\sum_{n=1}^{\infty} (-1)^{n+1} n \left(\arctan(x)-x+\frac{x^3}{3}-\cdots +(-1)^{n+1} \frac{x^{2n+1}}{2n+1}\right)$$

@Chris'ssis Remainder in Taylor series.

@BalarkaSen well, it's not hard. :-)

@Chris'ssis I am not saying anything about it's simplicity. I am just saying that the initial idea should be to write down the Taylor series remainder for $\arctan$

Alyosha

@Nimza No. What is mechmat?

@BalarkaSen Yeah, I got your point.

10:10 AM

@Chris'ssis I'll look at its derivative when I get back in a few minutes.

@robjohn No hurry with it. When you have time.

@Alyosha Hello again.

Alyosha

Indeed. I am beginning Alekseev now

Coolomundo!

Have fun.

@Chris'ssis Probably the integral form of the remainder would do the trick.

@BalarkaSen hmmm ...

10:19 AM

I am just sprouting ideas, not attempting to solve anything in any way! Probably having the integral form would allow you to interchange the summation and the integral, thus my suggestion. =)

@BalarkaSen The result is not that friendly (by the way).

You mean the final closed form expression?

Or do you mean the result of subing the integral form of remainder?

@N3buchadnezzar Comrades, let us bring radical!

hi

@BalarkaSen if you really want to, I can give you a very nice series to make you fall in love with calculus ...

wait

@Chris'ssis It's not that I hate calculus, only that I hate series and integral manipulations.

10:32 AM

@BalarkaSen $$\sum_{n=1}^{\infty} \arctan\left( \frac{ n^2+n-1}{ (n^2+n+1)(n^2+n+2)} \right) = \arctan \left( \frac{1}{2}\right)$$

2

I have seen those. They telescope.

@BalarkaSen I'd like to learn how it telescopes, really.

If it doesn't, I have honestly no idea.

@Chris'ssis For most of the arctan series, one just looks for $\frac{a+b}{1+ab}$ type expressions to use the addition formula and make it telescope.

If it doesn't work here, I don't know what would

@BalarkaSen I didn't say it doesn't work. It might work that way, but how?

I will star it and think about it when I get a little more time.

10:39 AM

@BalarkaSen OK

Byes for now.

@BalarkaSen bye

Q: Finding the expression of a projection

Can somebody help to find such exercises in this post

0

Suppose that $\mathbb{R}^3=K\oplus L$ as $K=Vect(k)$ and $L = Vect(l_1,l_2)$ and $k=(1,2,1)$, $l_1=(1,0,-1)$, $l_2=(-2,1,1)$. And Supposing that $q$ is the projection on $K$ in parallel to $L$. The question is what is the expression of the projection $q$ and how can I find it?

A: Finding the expression of a projection

11:06 AM

what do you think about this answer

0

Let $$P=\begin{pmatrix}1&1&-2\\2&0&1\\1&-1&1\end{pmatrix}$$ the transition matrix from the standard basis to the basis $\mathcal B=(k,l_1,l_2)$ and let $x=(x_1,x_2,x_3)^T\in \Bbb R^3$ a vector in the standard basis so its components in $\mathcal B$ are $$(x'_1,x'_2,x'_3)^T=P^{-1}x$$ and final...

china math

11:23 AM

Hello everyone,

@Chris'ssis

@chinamath Yes?

china math

Find a example function $f$,such $f\in C^2(R),and such $$\left(\sup_{x\in R}|f'(x)|\right)^2=2\sup_{x\in R}|f(x)|\cdot\sup_{x\in R}|f''(x)|$$

@chinamath I'm only interested in integrals, series and limits until I publish my book.

china math

Thank you all the same

@chinamath You might like my series in the right panel, the arctan series. Please do not post anymore my problems on main.

Julius

11:41 AM

Hi. Let $X$ be a discrete r.v. and $Y$ a continuous r.v. with bivariate density function f(x,y). I would like to compute a mean of $X$ when I know what is the value of $Y$.. Actually, I need to use $\frac{\sum_{i=1}^M i f(i,y)}{\sum_{i=1}^M f(i,y)}$, but what is it?

It is not a conditional mean since P(Y=y)=0, is it then a restricted mean of $X$ given a known value of $Y$? Does not seem so either since I think that $E[X; Y=y]=\sum_{i=1}^M i P(X=i, Y=y)=0$ and $E[X; Y\in dy]=\sum_{i=1}^M i f(i,y)$.

12:17 PM

@Chris'ssis Such math, shiba shiba

@N3buchadnezzar What do you mean?

dunno

@Chris'ssis What are you doing?

@N3buchadnezzar dunno

=(

@N3buchadnezzar :-)

@N3buchadnezzar I create some very nice question ... and I ponder over life things ...

12:26 PM

@Chris'ssis I am doing serious physics.

@N3buchadnezzar Do you like it?

A problem asks me to integrate over a sphere and then in the solution they write the integral as $$ \int_0^{2\pi} \int_0^{\pi} r^2 \sin^2\theta \, \mathrm{d}\theta \,\mathrm{d}\phi$$

I mean then they are not integrating over the whole sphere just the upper half (becomes angry)

Should be just a simple sine, not squared sorry

12:54 PM

can someone tell if this is correct: $vect(l_1,l_2)=\alpha l_1+\lambda l_2$

I really shouldn't mess with the cranks who post mathematical nonsense. I guess I'm just mean.

@ThomasAndrews that's what I'm asking for is the expression above make sense or not?

No, I wasn't referring to that, sorry, but a question on the site. I have no idea what you are talking about - the context or the question at hand. @pourjour

@ThomasAndrews Sorry hhhh

1:20 PM

is $\forall n \in \Bbb N : \exist k \in \Bbb N: 5^m +1 = k^3 $ the correct way of expressing the negation of the following : For each natural number m the number $5^m +1$ is not a cube of a natural number?

Sorry for my bad grammar

.

Ah, the person I am tangling with is the infamous Thierno M. SOW, who has claimed to have solved the ABC Conjecture, the Beal Conjecture, the Goldbach Conjecture, the Riemann Hypothesis and the Twin Primes Infinity, all in a 12-page paper.

Seems unlikely

Should be just a simple sine, not squared sorry

Given the levels of wrongness in his "question' in SE, it seems even more unlikely. @Matthew :)

1:49 PM

2:05 PM

I have a question that I know the answer too but I don't no how to express the answer

with words?

I mean express it the right manner

*in

2:31 PM

@Chris'ssis how about $$\frac14\left[\frac{x}{1-x^2}+ \log\left(\frac{1-x}{1+x}\right)+\arctan(x)\right]$$

Sorry, I had a nap, and now I have to go to the park.

OK. I got that in terms of arctanh.

Here is my work $$
\begin{align}
f'(x)
&=\sum_{n=1}^\infty(-1)^{n+1}n\left[(-1)^{n+1}\left(\frac{x^{2n+3}}{2n+3}-\frac{x^{2n+5}}{2n+5}+\dots\right)\right]'\\
&=\sum_{n=1}^\infty\frac{nx^{2n+2}}{1+x^2}\\
&=\frac{x^4}{(1-x^2)^2(1+x^2)}
\end{align}
$$
$$
\begin{align}
\int\frac{x^4}{(1-x^2)^2(1+x^2)}\mathrm{d}x
&=\int\frac18\left(\frac1{(1-x)^2}+\frac1{(1+x)^2}-\frac2{1-x}-\frac2{1+x}+\frac2{1+x^2}\right)\,\mathrm{d}x\\
&=\frac18\left[\frac1{1-x}-\frac1{1+x}+2\log(1-x)-2\log(1+x)+2\arctan(x)\right]\\

BBL

Alizter

Hello

I am sitting in a room that is 40 degrees

@robjohn Yeah, proceeding like that is the way to choose.

Alizter

2:47 PM

@Chris'ssis Hello

@Alizter Hi :-)

Freddy

Hello everyone

Alizter

@Chris'ssis I was talking with my cousin earlier and she was talking about some difficult diff equations and integrals she got in engineering. I showed her one of those fractional part integrals and she stopped complaining. :)

2

Random Variable

@robjohn Did you actually need to define $\log \left( \frac{z^{n}}{1-z^{n}} \right)$ like you did to answer Mark's question? Is it not enough to notice that $f(z)$ doesn't have a branch point at infinity?