New blog URL: https://lstu.fr/vt

Reason: better math support

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.

Post id: 2760540

Backup version on Evernote

I wanted to answer, but OP deleted own post before I clicked the submit button.

My solution:

Post id: 2760507

Backup version on Evernote

The comment is too quick for spam to exist on Math.SE.

Post id: 2726702

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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 also satisfies the property in the question.

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 , where `Tn = [Int64(k*(k+1)//2) for k in n];`

You can set as large as you like.

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

Does composite zero transformation imply zero tranformation?

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

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

My answer

This is false. Consider matrices , representing linear transformations from to .

but

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

This is false. Consider , . Then for all , but .

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 $z^7 = -1$.