Link

https://math.stackexchange.com/q/2778278/290189

You’ve got the right ingredients. (Pair up first term and last term and calculate number of terms.) However, with Gauss’s idea of pairing up the first and the last terms, you may consider condensing your work by one single formula.

\text{required sum} = \frac{(\text{first term} + \text{last term}) \times \text{number of terms}}{2} \\ = \frac{(\text{first term} + \text{last term}) \times (\frac{\text{last term} - \text{first term}}{\text{common difference}}+1)}{2}

Vincent Tam

April 30, 2018

Post id: 2726702
Version in Google Cache

J. M. Bergot asked this early April.

Is 61 the largest prime that is not the sum of prime and a triangular number? (The red box is complaining that this is too short. Has anyone ever looked into this question about 61?)

My answer:

No, the prime number p = 211 also satisfies the property in the question.

\begin{matrix} n & T_n & p - T_n & \text{Is it a prime?} \\ \hline 1 & 1 & 210 & \text{false} \\ 2 & 3 & 208 & \text{false} \\ 3 & 6 & 205 & \text{false} \\ 4 & 10 & 201 & \text{false} \\ 5 & 15 & 196 & \text{false} \\ 6 & 21 & 190 & \text{false} \\ 7 & 28 & 183 & \text{false} \\ 8 & 36 & 175 & \text{false} \\ 9 & 45 & 166 & \text{false} \\ 10 & 55 & 156 & \text{false} \\ 11 & 66 & 145 & \text{false} \\ 12 & 78 & 133 & \text{false} \\ 13 & 91 & 120 & \text{false} \\ 14 & 105 & 106 & \text{false} \\ 15 & 120 & 91 & \text{false} \\ 16 & 136 & 75 & \text{false} \\ 17 & 153 & 58 & \text{false} \\ 18 & 171 & 40 & \text{false} \\ 19 & 190 & 21 & \text{false} \\ 20 & 210 & 1 & \text{false} \end{matrix}

I’ve used the Julia code while any(isprime, x) && p <= Tn[end] p = nextprime(p, 2); x = p .- Tn; x = x[x .> 0]; println("p = $p"); end to search for primes until p = 500501, where Tn = [Int64(k*(k+1)//2) for k in n]; You can set n as large as you like.

Vincent Tam

April 30, 2018

Dead link: math.stackexchange.com/q/2759381/290189 (visible to 10k users)

Does composite zero transformation imply zero tranformation?

Show that if T(T(V)) is a zero transformation then T(V) is a zero transformation as well where V is a finite vector space and T is a linear operator.

Sharath Zotis self-deleted it after it’s closed.

My answer

This is false. Consider matrices V = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}, T = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} representing linear transformations from \Bbb{R}^2 to \Bbb{R}^2.

T(T(V)) = T\begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix},

but T(V) = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} \ne \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}.

OP changed his question to clarify the meaning of T and V in response to a comment.  This renders my answer meaningless, but its original version did answer the question.

 This is false. Consider V = \Bbb{R}^2, T(x,y) = (y,0). Then T(T(x,y)) = T(y,0) = (0,0) for all (x,y) \in V, but T \not\equiv 0.

Link

https://math.stackexchange.com/q/2757509/290189

Hint: observe that $\cos\frac{2\pi}{7} = -\cos\frac{5\pi}{7}$

\begin{align}\cos\frac{\pi}{7}-\cos\frac{2\pi}{7}+\cos\frac{3\pi}{7} &= \cos\frac{\pi}{7}+\cos\frac{3\pi}{7}+\cos\frac{5\pi}{7} \ &= \frac{z+z^{-1}}{2} + \frac{z^3+z^{-3}}{2} + \frac{z^5+z^{-5}}{2} \ &= \frac{1}{2z^5} \frac{z^{12}-1}{z^2-1} \end{align} where $z^7 = -1$.