Daily math problems for August 22, 2026 — Algebra, Financial Mathematics, Calculus & more

Take any $g\mathrm{\ne }e$. The subgroup $⟨g⟩$ has order dividing $p$. It's not $1$ (since $g\mathrm{\ne }e$), so it has order $p$, i.e. $⟨g⟩=G$. So $G$ is cyclic, hence abelian.

$FV=1000\cdot {\displaystyle \frac{{1.05}^3-1}{0.05}}=1000\cdot {\displaystyle \frac{0.157625}{0.05}}=1000\cdot 3.1525=\mathrm{\$}\mathrm{3,152.50}$.

$\widehat{2},\widehat{4},\widehat{6}=\widehat{0}$ — three additions, so the order is $3$.

Vertical: $x=\pm 1$ (two). Horizontal: $\lim f(x)=1$ (one). Total: $3$. (No slant — the rational function has equal-degree numerator and denominator.)

${\displaystyle \underset{x\to 0}{\lim }{\displaystyle \frac{(1+x{)}^r-1}{x}}=r}$. (This generalizes the bonomial-derivative-at-zero.)

A ring homomorphism preserves both operations: addition AND multiplication.

$A_5^3=5\cdot 4\cdot 3=60$.

$0\le I_n\le {\displaystyle \frac{1}{n+1}}\to 0$, so $I_n\to 0$ by the squeeze theorem.

By Viète: $x_1+x_2=5$, $x_1{x}_2=4$. Hence $x_1^2+x_2^2=25-8=17$.

$\mathrm{\mid }{A}_n\mathrm{\mid }={\displaystyle \frac{n!}{2}}$ (one of the standard properties of the alternating group).