Daily math problems for July 27, 2026 — Calculus, Analytic Geometry, Algebra & more

$\bullet$ Differentiate by the chain rule:

$$f'(x)=6\pi\cos(2\pi x+\tfrac{\pi}{4})$$

$\bullet$ Differentiate again:

$$f''(x)=-12\pi^2\sin(2\pi x+\tfrac{\pi}{4})$$

$\bullet$ Evaluate at $x=\pi/3$ and $x=\pi/6$:

$$f'\!\left(\tfrac{\pi}{3}\right)=6\pi\cos\!\left(\tfrac{2\pi^2}{3}+\tfrac{\pi}{4}\right)$$

$$f''\!\left(\tfrac{\pi}{6}\right)=-12\pi^2\sin\!\left(\tfrac{\pi^2}{3}+\tfrac{\pi}{4}\right)$$

$\bullet$ Add the two:

$$6\pi\cos\!\left(\tfrac{2\pi^2}{3}+\tfrac{\pi}{4}\right)-12\pi^2\sin\!\left(\tfrac{\pi^2}{3}+\tfrac{\pi}{4}\right)$$

$m_1 \cdot m_2 = -1 \Rightarrow m_2 = -\dfrac{1}{2}$.

A $3$-cycle has sign $(-1)^{3-1} = +1$. (Equivalently, it can be written as the product of two transpositions: $(1\;2\;3) = (1\;3)(1\;2)$.)

$\mathbb{Z}_n$ is a field iff every non-zero element is invertible, iff every element from $1$ to $n - 1$ is coprime to $n$, iff $n$ is prime.

By parts: $\int x \ln x\,dx = \dfrac{x^2}{2}\ln x - \int \dfrac{x}{2}\,dx = \dfrac{x^2}{2}\ln x - \dfrac{x^2}{4}$. Evaluating on $[0,1]$ (with $x^2 \ln x \to 0$ as $x \to 0^+$) gives $0 - \tfrac{1}{4} - 0 = -\tfrac{1}{4}$.

The positive divisors of $12$ are $\{1, 2, 3, 4, 6, 12\}$, so $|A| = 6$.