I haven't looked through your working, but here's a much easier method: If $a^2+a+1=0$, then $(a-1)(a^2+a+1)=0$, that is, $a^3-1=0$, ie, $a^3=1$. On the other hand, $1+a+\dots+a^{2017}=\frac{a^{2018}... View answer Accepted answer 4 votes$C_{12}$is the cyclic group of order 12. So it's not$D_{12}$(aka$D_{24}$) which has order 24. So, looking at the two elements you've identified:$e$, or "do nothing". That doesn't have ... View answer 4 votes By rationalising the denominator,$x_{n+2}$simplifies to $$\frac{\sqrt{x_n^2+1}\sqrt{x_{n+1}^2+1}+x_nx_{n+1}-1}{x_n+x_{n+1}}.$$ Letting $$x_n=\frac12\left(p_n-\frac1{p_n}\right),$$ that is (say) $$... View answer 4 votes Let's represent the coins with a vector in Z_2^n, with 0 representing heads, 1 representing tails. Let G be the group acting on Z_2^n generated by a cyclic permutation of the basis vectors. ... View answer 4 votes The difference is in the first scenario, all you're told is that one of the coins is heads, but you don't know which one. This only rules out TT, so you still have three possibilities: HH, HT or TH. ... View answer Accepted answer 4 votes Are you given that a=d? Because that neither implies, nor is implied by, the fact that A has a repeated eigenvalue. Rather, your characteristic polynomial will be (x-\lambda)^2, so you know (A-\... View answer 4 votes It's probably easier to use$$\cos^2\alpha = \frac{1}{\sec^2\alpha} = \frac{1}{1+\tan^2\alpha} = \frac{1}{1+\frac{1}{\cot^2\alpha}}$$to find \cos\alpha, this gives \cos\alpha=\pm\frac7{25}. The ... View answer Accepted answer 3 votes If a=b, then write k_1 e^{ax}-k_2e^{bx} as ke^{ax}. If k and a are opposite sign, or if either is 0, then x=ke^{ax} has exactly one solution. Otherwise,$$x=ke^{ax}$$if and only if$$ax=... View answer Accepted answer 3 votes Obviously, you could "cheat" and use the rhs to inform your factorisation. But that's not what you want. If you write$a=e^{ix}$and$b=e^{iy}$, then you have $$\frac{1}{2i}\left(a-\frac 1a +... View answer Accepted answer 3 votes The rational root theorem says that if you have a polynomial a_n x^n + ... + a_0 with integer coefficients, then all rational roots must be of the form \pm\frac{p}{q}, where p divides a_0 and ... View answer 3 votes If a power of 2 starts with 123, then it must be between 1.23\times 10^n and 1.24\times 10^n for some n. So you want k and n for which$$1.23\times 10^n\leq 2^k < 1.24\times 10^n$$. This ... View answer Accepted answer 3 votes Not a coincidence. The same holds for circles - the circumference is the derivative of the area. Start with a sphere of radius r. If the skin has thickness dr, the volume of the skin will be ... View answer 3 votes That is a correct way to solve it. A better correct way to solve it, since you're dealing with uniform random variables on the unit square, is to ask "what is the area of the shape defined by X^2+Y^... View answer Accepted answer 3 votes Well, a basis for a vector space is a linearly independent spanning set. Do you know that \{\cos^2(x), \sin^2(x)\} span V? Hint: elements of V are of the form a_1\cos^2(x) + a_2\sin^2(x) + ... View answer 3 votes Well, one way is to use Burnside's Counting Lemma. There may be easier ways, but I like Burnside's counting lemma. There are, as you noted, 924 ways to colour the grid if you ignore symmetry. The ... View answer 3 votes You can't conclude$$\lim_{n\rightarrow\infty} \int_a^b g(x,n)dx$$doesn't exist just because$$\lim_{n\rightarrow\infty} g(x,n)$$doesn't. For example,$$\lim_{n\rightarrow\infty} \sin(nx)$$doesn't ... View answer Accepted answer 2 votes Your way is fine. Here's another way: proceed using induction. The proposition is true if m=0:$${\binom{n-0}{k}}=\binom{n+1}{k+1}-\binom{n-0}{k+1}$$because$${\binom{n}{k}}+\binom{n}{k+1}=\binom{n+... View answer Accepted answer 2 votes if$\sin\alpha\cos\beta>\frac12$then$\frac12\sin(\alpha+\beta)+\frac12\sin(\alpha-\beta)>\frac12$However,$\frac12\sin(\alpha+\beta)\leq\frac12$, so this is only possible if$\sin(\alpha-\...

If T is undecideable within formal system F, it means (1): F does not contain a proof of T, AND F does not contain a proof of ~T. Suppose the question "T is undecideable" is undecideable. ...

Yes. Proof: Assume $\prod x_h$ is maximised, and there exists $i,j$ such $x_i+1\leq x_j-1$ (that is, $x_i$ and $x_j$ differ by at least 2). Without loss of generality, let $i=1$ and $j=2$. Let $y_h$ ...

a positive semidefinite matrix has all eigenvalues non-negative. However, $A$ has a negative eigenvalue $\lambda_{min}$.

I would agree with your instincts. To show 1, you just need an example. The function $f(x) = x^2$ on $[-1,2]$ will do. Hint: what are the local maxima? And don't forget to prove $f$ is convex.

"$(1,0,0),(0,1,0),(0,0,1)$ are all eigenvectors corresponding to eigenvalue 1" What you have found is not all eigenvectors, but a basis for the vector space of the eigenvectors. If you want ...

If polynomial $p(x) = ax^n + bx^{n-1} + ...$ has roots $x_i$, then $x_1+\dots+x_n=-\frac{b}{a}$, as you noted. Now, though, you don't want $p(x)=0$, you want $p(x)=mx+b$. So, to use Vieta's theorem, ...

Well, $1-a^N=(1-a)(1+a+a^2+a^3+...+a^{N-1})$, and each of the terms in the second bracket is less than or equal to 1, so $1-a^N\leq(1-a)N$ Therefore, $\frac{a(1-a^N)}{1-a}\leq\frac{a(1-a)N}{1-a}\leq ... View answer 2 votes That formula is not the right one to use. That would apply if each dice had two faces, 0 and 1, and then$p$is the probability it shows 1. Examples of things like this: tossing a coin$n$times ($...

In this notation, numbers not mentioned are fixed. So, if p = (1,3,2), then p sends 1 to 3, 3 to 2, 2 to 1, 4 to 4, 5 to 5, 6 to 6, 7 to 7, etc.

Let $f(a)=0$ be monic, degree $n$. If the polynomial is not already a polynomial over the integers, it has a term whose coefficient is rational, with denominator $q$ when fully reduced. Suppose that's ...

If $x=t+s$ and $y=t-s$, you can solve these as a system of equations for $t$ and $s$. This gives $t=\frac12(x+y)$ and $s=\frac12(x-y)$. Then you can substitute these into $z=t^2+s^2$ to get $z$ as a ...