André Nicolas
Reputation
976/1000 score
 1d comment Number of subsets in set with sum greater than 10 For sum equal to $10$, or $n$, look for information under partitions into distinct parts. Big literature, dating back to Euler. No nice closed form formula, but pleasant enough generating function, and recurrences. 1d comment Two cards are drawn without replacement from an ordinary deck of 52 cards. What is the probability that the cards are of the same suit? Your second is not quite right. For the bottom and top, we need to use "combinations" in both, or "permutations" in both. Your $\binom{13}{1}\binom{13}{1}$ counts the number of hands that have a diamond and a spade. So the denominator should be $\binom{52}{2}$. If you want permutations at the bottom, for the top you need to count diamond then spade, and spade then diamond, $2\binom{13}{1}^2$. 1d comment Two cards are drawn without replacement from an ordinary deck of 52 cards. What is the probability that the cards are of the same suit? Here is a simpler way. Imagine the cards are drawn one at a time. Whatever card was drawn first, the probability the next one matches it in suit is $12/51$, which if you like (I don't) simplifies to $4/17$. 1d answered Calculate the average acceleration and average speed of a particle 1d comment Proving a sequence converges to 1 Note that $b^n\lt a^n+b^n\lt 2b^n$, and therefore $b\lt (a^n+b^n)^{1/n}\lt 2^{1/n} b.$ Use the fact that $2^{1/n}\to 1$ as $n\to\infty$, and Squeezing, to finish. 1d comment Let $n,k \in Z^+$. Show using Euclid's algorithm that $\gcd(n, nk+1) = 1$ You are welcome. A better way I think is to observe that if $d$ is a positive integer that divides $n$ and $nk+1$, then it divides $nk$ and $nk+1$, and therefore it divides their difference $1$. Since $d$ divides $1$, we conclude that $d=1$. That brings out part of the logic behind the Euclidean algorithm. 1d comment Let $n,k \in Z^+$. Show using Euclid's algorithm that $\gcd(n, nk+1) = 1$ Apply the Euclidean Algorithm, First step, $nk+1=(k)(n)_1$. Now the algorithm is over! 1d comment How to prove that $X_n\equiv 1\pmod 3$ Use strong induction. The induction step is to show that if our assertion is true for all $j\lt n$ it is true at $n$. That will be very straightforward in this case. 1d comment Smallest n such that there's a polynomial in $\Bbb Z_n [X]$ of degree 4 that has 8 roots in $\Bbb Z_n$ Yes. But my $4x^3(x+1)$, though "legal." is kind of cheating, since the $4$ gives a huge boost toward zeroness. That's why I added the less cheating $x(x+1)(x+2)(x+3)$. 1d comment Smallest n such that there's a polynomial in $\Bbb Z_n [X]$ of degree 4 that has 8 roots in $\Bbb Z_n$ Yes, pretty easy, of any two consecutive evens, one is divisible by $4$. 1d comment Smallest n such that there's a polynomial in $\Bbb Z_n [X]$ of degree 4 that has 8 roots in $\Bbb Z_n$ I am literally minded! 1d revised Smallest n such that there's a polynomial in $\Bbb Z_n [X]$ of degree 4 that has 8 roots in $\Bbb Z_n$ added 2 characters in body 1d answered Smallest n such that there's a polynomial in $\Bbb Z_n [X]$ of degree 4 that has 8 roots in $\Bbb Z_n$ 1d comment Arclength of parametric curve You are welcome, This kind of sign issue (not quite "taking out" a term tight from under a square root sign) comes up quite often. 1d comment Calculus 2, need some help on this question please! Integrate the density from $0$ to $1$. Do you have trouble evaluating the integral? If so, note that $x(1-x)=x-x^2$. (I am assuming that the density is $(0.04)(x)(1-x)$, but maybe the first $x$ in your density is intended to be a multiplication sign. 1d comment Arclength of parametric curve When $\sin t$ is negative, we have $\sqrt{4\sin^2 t\cos^2 t+\sin^2 t}=-\sin t\sqrt{4\cos^2 t+1}$. Better than breaking up the integral, use symmetry. 1d comment Is this sequence bounded? Certainly bounded (that is, both bounded above and bounded below). Every term is $\le 17$. And every term is $\ge 0$. We can of course find tighter upper and lower bounds. 1d comment Is this a correct proof of why $\gcd(a,b) = \gcd(b, a- b)$? It is fine. The amount of detail to give is not well-defined. I would say it is obvious that any integer that divides $a$ and $b$ diviides $b$ and $a-b$. Also, any integer that divides $b$ and $a-b$ divides $a$ and $b$. So the set of common divisors of $\dots$. Two points about your new proof: (i) It is understandable and (ii) it is right. 1d comment Differentiation of a Summation You are welcome. A formal write up can be done by separating the two sums. In the one that has the $n$ on top, cancel with the $n$ in $n!$, getting $\frac{1}{n-1}!$. The replace everywhere in that sum $n-1$ by $j$, and then $j$ by $n$. Now observe the cancellation with terms of the second sum. I did not feel like writing it up since it is tedious TeX-wise. 1d comment Differentiation of a Summation Imagine $k$ large, like $12$. You have a difference of sums. Note that for example the term in the first sum corresponding to $n=4$ cancels the term in the second sum corresponding to $n=3$. And so on for almost all terms.