50,950 reputation
bio website users.utu.fi/lahtonen/…
location Rusko, Finland
age 50
visits member for 3 years, 3 months
seen 21 mins ago

General non-sense:

Mostly I teach here. I want to encourage beginning students to think for themselves, so I use a lot of hints and comments. More advanced questions I often just answer.

However, I don't do 1-on-1 chat. Don't ask. My answers and comments are for all to see, so I don't like to hide them. Also I feel that entering a chatroom carries with it an obligation to continue an on-line discussion, and then I would not be using my time on my own terms.

Relevant personal history:

PhD from Notre Dame in '90.

Drifted from representation theory of algebraic groups to applications of algebra into telecommunications, mostly coding theory, and lately mostly teaching at college level.

3 graduate students with awarded PhDs

I have mostly worked at our local University at Turku, Finland. At one point I tried working for Nokia Research Center. It was ok, but an old dog didn't learn all the tricks, and then they downsized, so I returned to the Uni as a tenured lecturer.

revised How many solutions are there for the congruence $x^{14}+x^7+1 \equiv 0 \; (\text{mod } 343)$?
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comment How many solutions are there for the congruence $x^{14}+x^7+1 \equiv 0 \; (\text{mod } 343)$?
For the sake of variety I wanted to do this without Hensel lifting. Granted, Hensel is waaayyyy more general than an ad hoc method like the one above.
answered How many solutions are there for the congruence $x^{14}+x^7+1 \equiv 0 \; (\text{mod } 343)$?
revised class number of pure cubic fields and elliptic curves
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comment How find all positive real $\beta$ such A finite number of $\left|\frac{p}{q}-\sqrt{2}\right|<\frac{\beta}{q^2}$
I would guess that excluding the $a_n$:s means that $|p^2-2q^2|\ge2$. If correct proving it should not be too difficult (basically you need to show that all the solutions of the Pell equation have $q=a_n$ for some $n$). Using the theory of units of rings of integers of real quadratic extensions of $\Bbb{Q}$ leads to it, but that is probably not an allowed piece of theory. I would try "infinite descent" (or induction): if $p_n^2-2q_n^2=1$, then multiplying $(p_n-q_n\sqrt2)$ by $(\sqrt2 +1)$ gives a "smaller" solution. I'm off air, so cannot pursue this now.
comment How to compute homotopy groups of torus?
Homotopy groups of a product of two spaces are the direct products of the respective homotopy groups of the factors. Torus is a product of two circles. If you know the homotopy groups of a circle you know those of a torus.
comment How find the value $\beta$ such $\left|\frac{p}{q}-\sqrt{2}\right|<\frac{\beta}{q^2}$
@Winther (+1 to you was already there). That formula gives other solutions to a Pell equation when you have already found one solution. It does not work that simply always. For $k=6$ we have $\lfloor\sqrt6\rfloor=2$, and $$(\sqrt6-2)(\sqrt6+2)=6-2^2=2$$ instead of $1$. But $\sqrt6\approx 5/2$ and, indeed, $$(2\sqrt6-5)(2\sqrt6+5)=-1.$$ Therefore for all the pairs $(p_n,q_n)$ such that $p_n-q_n\sqrt6=(2\sqrt6-5)^n$ we have $$p_n^2-6q_n^2=(-1)^n.$$
comment Sum of a geometric series whose common ratio might be 1
In cases where $|r-1|$ is very small (and given separately) and $n$ large, you can try $$r^n=e^{n\ln r}$$ in the numerator. Together with the Taylor series $$\ln(1+x)=x-\frac{x^2}2+\frac{x^3}3-\cdots$$ When $r\approx 1$, you have $x\approx 0$, so the series converges quickly. You need to also take $n$ into account to decide how many terms are needed, so ... piecewise.
comment Sum of a geometric series whose common ratio might be 1
Adam, the problem is apparently numerical instability of the formula near $r=1$. It may be better for the program to be fed $r-1$ separately to avoid loss of precision in the denominator. For very high precision a piecewise treatment is likely necessary. I mean, it's not like there are any other formulas :-)
answered How find the value $\beta$ such $\left|\frac{p}{q}-\sqrt{2}\right|<\frac{\beta}{q^2}$
revised How find the value $\beta$ such $\left|\frac{p}{q}-\sqrt{2}\right|<\frac{\beta}{q^2}$
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comment Subgroups of Symmetric groups isomorphic to dihedral group
Elements of $D_n$ are faithfully represented by their action on the vertices of the regular $n$-gon, so... Yes!
revised Making Y chocolates from X chocolates
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comment Ternary Golay codes and correction probability
Just redo the derivation of that formula for $P_{corr}(C)$ in qSC. I'm not sure what happens, but an educated guess would be that $1-p$ always stays the same, and $p/2$ becomes $p/(q-1)$. Frankly, I think trying to memorize such things is pointless. Memorize/understand what those terms represent, and everything else comes naturally.
revised Ternary Golay codes and correction probability
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comment Ternary Golay codes and correction probability
Oh, now I see! You correctly used powers of $p/2$ multiplied by $\alpha_i$:s, because the error in a ternary symbol can take two distinct values, and in a symmetric channel they are equally likely. But that $1-p$ should stay, because it is the probability of no error in that symbol.
answered Ternary Golay codes and correction probability
comment Ternary Golay codes and correction probability
The weights of the extended ternary Golay code are multiples of three, because it is self-dual.
answered Extended Golay Code - Vectors of Odd Weight
comment Write down a linear operator $f:\mathbb{R}^4\to\mathbb{R}^4$ whose minimal polynomial is $m_f(t)=t^3-t^2$
A general recipe for finding a matrix with given characteristic/minimal polynomials involving higher degree factors is described by Arturo Magidin here.