7,570 reputation
31146
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location Besançon, France
age 34
visits member for 2 years, 1 month
seen 1 hour ago

Mathematics is a hobby for me, and MSE helps me not to forget too much what I learned in university (MS in applied mathematics at Lyon 1). I like especially real and complex analysis, trigonometry, numerical analysis, finite groups theory, number theory and combinatorics. I am not allergic to other fields.

I like experimenting with a computer, but by hand as well, with a good old slide rule or a table of logarithms for computations.

Contact: arbautjc at gmail dot com


15h
comment Help with understanding the polynomial long division algorithm
For the (1), it means it's a relation that is true after each loop (a loop invariant, if you prefer). Suzu Hirose already answered for the (2).
1d
comment A Ferris wheel has a radius of 20 m. Passengers get on halfway up on the right side.
@BMWurm Actually, I have since found this Wikipedia article about Colossus in Missouri, and according to this article, its speed is $16 km/h$. In the example above, the speed would be $13 km/h$. Hence, even if I still think it's very fast, it's not unheard of.
1d
comment bessels equation
I tried to translate to LaTeX, but it has to be checked.
2d
comment Symbols for “odd” and “even”
I wasn't aware of these, thank you, and +1 then
Nov
22
comment Parametric equation of a cone
In the case $z=\sqrt{4x^2+y^2}$, the base of the cone is not circular but elliptic. Notice that a cone is not limited to circular or elliptic bases, see the Wikipedia article on cone.
Nov
19
comment Alternating Series , why start at n = 1?
@Henry Agreed, but if the question is about why we start at $1$ in general, it's not because $b_0$ is never defined (because someteimes it is). Actually, I find more interesting to notice that you can start anywhere, even at $b_{100}$ if necessary, and sometimes it is: if your sequence is decreasing only after a number of "randomly behaving" terms. Of course you don't start on undefined/nonexistent terms.
Nov
19
comment Alternating Series , why start at n = 1?
@Henry But it's just because $b_0$ is not even defined. For defined terms, you can start anywhere you wish.
Nov
19
comment Alternating Series , why start at n = 1?
You can start at any index $n_0$, the behaviour of terms before $n_0$ has no impact on convergence, provided you still have $b_{n+1}\leq b_n$ for $n\geq n_0$ and $b_n\to0$.
Nov
16
comment Right ascension / Declination of a satellite
After downloading this and a database from space-track, I have no clear answer: elevation and azimuth are correctly related to the observer (for instance, the green oval line is where elevation=0, which is the "horizon", from the satellite), but I can't explain why RA and DEC vary when you change the observer. Maybe it would be better to ask directly to Alexander Lapshin (heavensat@mail.ru).
Nov
16
comment Right ascension / Declination of a satellite
What about the second part of my comment?
Nov
16
comment Right ascension / Declination of a satellite
Right ascension and declination refer to the equatorial coordinates system, which does not depend on the observer. However, I see on the images that time is local, and local time depends obviously on the location of the observer. Does the problem disappear if you switch to UTC?
Nov
16
comment Nasty integration?
What is divergent? I'm not sure one can compute the integrals in "closed form", but they look convergent to me.
Nov
15
comment Why $2^{3^n}=-1 \mod 3^{n+1}?$
Ah, ok! Thank you.
Nov
14
comment Why does the google calculator give tan 90 degrees = 1.6331779e+16?
You can switch between radians and degrees. So, in degree, you get an error for $\tan 90$, and in rdians, you get an error for $\tan \pi/2$. However, if you ask "tan 90 degrees" directly in Google, you get the value shown by the OP. Same if you ask directly "tan (pi/2)".
Nov
13
comment Give a supremum of any two interger?
@askuyue Ok, you want the inf, ask for the sup, and you are not satisfied that yo got the sup? Is this just some troll?
Nov
13
comment Give a supremum of any two interger?
@askuyue What did you expect?
Nov
13
comment $A\in M_2(\mathbb C)$ and $A $ is nilpotent then $A^2=0$.How to prove this?
@learningmaths The trace is also the sum of eivenvalues (notice that the trace is invariant by transformation $A\to P^{-1}AP$, see here‌​). And since a nilpotent matrix $A$ satisfies $A^n=0$ for some $n$, all eigenvalues must be zero.
Nov
13
comment $A\in M_2(\mathbb C)$ and $A $ is nilpotent then $A^2=0$.How to prove this?
@I didn't downvote, but your equation should be $A^2-tr(A)A+det(A)I_2=0$.
Nov
13
comment this is a conjecture or a result? every arithmetic progression contains a sequence of $k$ “consecutive” primes for possibly all natural numbers $k$?
@quit You are probably right, but the OP should clarify this I think. Then, only the case $a=1$ is interesting, since for $a>1$ he already showed $k$ must be bounded. Thank you for your comments, I realize I didn't understand the question.
Nov
13
comment this is a conjecture or a result? every arithmetic progression contains a sequence of $k$ “consecutive” primes for possibly all natural numbers $k$?
While your question is open, the OP's one has a crystal clear answer.