Erick Wong
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 20h comment Find $r$ in the next formula Easiest, not horribly inefficient, way to find $r$ would be to use a binary search. I'm assuming that $r$ is an interest rate and the payments are all non-negative. 20h comment Prime Numbers and a Two-Player Game @Batominovski Ah I see the link with data going up to $10000$, interesting. 20h comment Prime Numbers and a Two-Player Game @Batominovski Does the data from OEIS go beyond 3 digits? Note that $\ln^2(500)$ is not markedly far from $50$, so I don't see how the data supports a faster growth rate. I haven't worked through this heuristic but it's in line with my expectations of $L$ being very sparse. $O((\ln n)^2)$ is in fact the conjectured maximum spacing between primes (but the best proven bound is exponentially worse). But positive natural density of $L$ is so far from what I'd expect that it might be in reach of a disproof. 1d comment Need my CRT work spot-checked What you have so far looks really good and complete, in that the final modular equation is exactly equivalent to the original. One remaining step is to reduce the system of equations so that the moduli are pairwise co-prime (either by breaking each $M$ into prime powers, or finding common factors between all the $M$s). But you might consider that to be part of the CRT process, so it isn't necessarily missing. 1d comment Why does this imply a congruence does not exist? @WillStein Ah, well here are some general thoughts for what it's worth. You need to prove "if $g \nmid a$ then there is no solution". A premise such as $g\nmid a$ is not very definite so it is often not practical to work with. This gives a clue that you might want to start from the contrapositive: "if there is a solution then $g \mid a$". Once you assume a solution, you have that $kx \equiv a \mod M$. At this point you can easily combine it with $gi=M$ and $gj=x$ to get $gkj \equiv a \mod gj$...From there you'd be pretty close to showing that $a$ is divisible by $g$, no? 1d comment Why does this imply a congruence does not exist? @WillStein It might help if you gave us some insight into what kind of things you tried during those hours of staring, suggest some more general problem-solving strategies you might have tried :). 2d comment The square of n+1-th prime is less than the product of the first n primes. This is Bertrand's postulate, which the OP specifically is trying to avoid using. 2d comment The square of n+1-th prime is less than the product of the first n primes. The prime divisors of $6$ are not all less than $\sqrt 6$. 2d comment Summation of prime multiples less than n @kaydee Thanks. To be clear, is $m$ the only fixed quantity in the sum? Are $p_i$ allowed to vary along with $x,y,z,w,a_i$, or are they fixed along with $m$? 2d comment Summation of prime multiples less than n @kaydee Are you sure $7$ is supposed to be included in both the $2,3,5,7$ part and a possible value for $p_i$? Definitely improved, but what terms count as distinct? If $a_1 = a_2 = a_3 = 0$ then you get the same number even when $p_1, p_2, p_3$ change. Are these counted in the sum multiple times or just once? 2d comment Summation of prime multiples less than n Could you be less vague about the summation terms? Is this a ham-fisted way of specifying all products of squarefree numbers with a $7$-smooth number? You don't say whether $p_i$ can be be $\le 7$ in which case the same integer has multiple representations. Is $m$ fixed? Basically, you haven't described your sum at all adequately. Jul27 comment How is $\frac{\sqrt{a}}{a+1} (0^{a+1}+1^{a+1})$ equal to $\frac{\sqrt{a}}{a+1} (-1)^a$ Care to explain how you arrived at the last equation from the second-last? You seem to be saying that $- (-1)^{a+1} = 1^{a+1}$. Why? Would you also simplify $-(-1)^2$ to $1^2$? Jul27 comment Find $f'(x)$ when $f(x) = x^2 \cos(\frac{1}{x})$ @Jellyfish Finding $f'(x)$ means finding the value of $f'$ (or determining that it does not exist) at all values of $x$. If you only compute $f'(x)$ for $x\ne 0$, that doesn't fully answer the question. Jul27 comment Find a plane with distance $3$ from $3x-y-z = 0$ @Guerlando OCs Your friend's solution makes no sense. If $P$ satisfies $-3x_0 + y_0 + z_0 = 0$, then so does $3P$, and $5P$. So it doesn't uniquely identify distance at all. Your solution sounds perfectly reasonable. Jul26 comment $6$ eigenvalues of a $4\times4$-matrix? The row operations you are using to reduce the matrix are NOT determinant-preserving. $r_1 \leftarrow r_1 - c r_2$ preserves the determinant, but $r_1 \leftarrow c r_1 - r_2$ does not. Jul26 comment Is there an efficient way to iterate vertex-transitive graphs? I don't have a good answer for your main question, but the graphs you're describing are called circulant graphs, which are a special case of Cayley graphs (which in turn are a special case of vertex-transitive graphs). The automorphism group of an arbitrary vertex-transitive graph might not contain a cycle (note: this is not the same as $\text{Aut}(G)$ being cyclic, only that it contains a subgroup acting cyclically on all vertices). Jul26 comment Does there exist an invertible matrix with 1,-1 and 0? A vector with all components equal is always in the kernel. The graph being connected is only relevant to the kernel having exact dimension $1$. Jul26 comment How can I tell that my matrix is nilpotent? @LebronJames Yes, the action of $L$ is linear; if it weren't, you couldn't represent it as a matrix in the first place. You might be thinking of the fact that the polynomials are not linear functions of $x$ and $y$, but that would be a red herring since the domain of $L$ Is not $x$ and $y$ but the vector space of polynomials. You are exactly right that all you need to show is that $L(f+g) = L(f)+L(g)$ and $L(af) = aL(f)$. Jul25 comment explanation of notation in programming problem @dagd1 The reason you shouldn't compute it naively from the definition is that factorials get really large very quickly (50 factorial has 65 digits), so you end up either a) overflowing your machine precision and getting inaccuracies or run-time errors, or b) spending needless effort dealing with high-precision large numbers when the final answer is often much smaller. Jul25 answered How can I tell that my matrix is nilpotent?