Jyrki Lahtonen
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 10h awarded Nice Answer 18h awarded Nice Question 18h revised what is the field over “ $K^0$ ”? edited tags; edited tags 1d comment Galois group of splitting field It is a quartic, so the question can be settled by more elementary means. IIRC an algorithm for identifying the Galois group of an irreducible quartic is outline in Jacobson's Basic Algebra I. 1d comment Galois group of splitting field I suspect this has been asked and answered on our site before. I couldn't find it right away though. Did you search? 1d comment Galois group of splitting field What tools have you got in your bag? If you know of Dedekind's theorem (See Qiaochu Yuan's post for an on-site account)), then it is easy. Modulo $p=2$ your polynomial is irreducible, so the Galois group (when viewed as a subgroup of permutations of the zeros) contains a 4-cycle. Modulo $p=3$ the polynomial is a product of a linear factor and an irreducible cubic, so Dedekind tells us that the group also contains a 3-cycle. The only subgroup of $S_4$ that contains both a 4-cycle and a 3-cycle is clearly $S_4$ itself. 1d comment Number of ways to write $n \in \Bbb N$ as a sum of $k$ positive integers? I dare guess that the downvote is from someone who thinks this is a homework problem you put next to zero effort into yourself - I disagree to the extent that my guess is that there is no clean formula (and hence this is not homework). Anyway, the number of ways of writing $n$ as a sum of $k$ positive integers is equal to the number of ways of writing $n-k$ as a sum of at most $k$ non-negative integers. I would 1) open a number theory book (Hardy&Wright comes to mind), 2) look for a chapter titled Partitions, and 3) try and absorb what they say about partitions of such restriced type. 2d comment Alternative Proof: if $n$ is an integer, prove that $\frac{n ( n^4 - 1)}{5}$ is an integer A good idea to use that factorization. If you want to avoid splitting into cases you can further observe that $$n^2+1=n^2+5n+6-(5n+5)=(n+2)(n+3)-(5n+5),$$ so $$\frac{n(n^4-1)}5=\frac{(n-1)n(n+1)(n+2)(n+3)}5-n(n-1)(n+1)^2.$$ The first numerator is a product of five consecutive integers, so it is always divisible by five. Apr 29 comment What is the value of $k^2$ Because $g(2x)=4f(\pi/2)^2f'(x)^2\ge0$ no matter what $f(x)$ is, the integral is clearly $\ge\int_1^31\,dx=2$. This lower bound can be achieve simply by using an $f$ such that $f(\pi/2)=0$ when $g(2x)=0$ for all $x$. Therefore $K=2$ and $K^2=4$. I really suspect something is wrong, because this is too trivial. Apr 29 comment What is the value of $k^2$ @almagest: Did you perhaps miss that $g(2x)\ge0$ for all $x$ trivially? The derivative $f'(x/2)$ is squared! Apr 29 comment Explaining the Ackermann function as A: $\mathbb N \times \mathbb N \rightarrow \mathbb N$ I would guess that you are expected to show that the function is defined for all valid inputs $(n,m)$ in such a way that the calculation terminates after finitely many recursive steps. A suitable induction seems to be called for (based on the lexicographical ordering of inputs). Apr 29 revised Explaining the Ackermann function as A: $\mathbb N \times \mathbb N \rightarrow \mathbb N$ added 1 character in body Apr 29 reviewed Approve Explaining the Ackermann function as A: $\mathbb N \times \mathbb N \rightarrow \mathbb N$ Apr 29 comment Existence of Type I Self Dual Codes Would the span of $110000\ldots, 001100\ldots, 000011\ldots,\ldots,00\ldots0011$ be one such? Apr 29 answered Angular momentum operators Apr 29 comment What is the value of $k^2$ Make $f(\pi/2)=0$ ?!!? IOW I guess we should have $g(x)=4 f(x/2)^2f'(x/2)^2$. Otherwise it's trivial. Apr 29 revised Uses of step functions deleted 358 characters in body Apr 29 comment Distance of bch code @Evinda Other users have flagged your comments like the one above (asking somebody to take a look at another question). Many view of them as mildly rude. Apr 28 comment Distance of bch code Yes, @Evinda. That's roughly the idea. But we usually need to permute many positions (such as translate the entire finite field, which is what I did in my answer). The upshot is that using an automorphism we can move one of the $1$s of an even weight word to the extended position, and get a word of the unextended code of weight one less. The other $1$s go God knows where, but we don't really care. Apr 28 comment Any math competitions dedicated to calculations by hand (college level math)? I'm not aware of any math competitions that allow the use of calculators. Can you point me at some, @SimpleArt (so that I know to steer clear - the calculators are an abomination in competitions IMHO :-)