Daily math problems for October 23, 2026 — Euclidean Geometry, Infinite Series, Geometry & more

$6+4+2+2+4+6=24$ cm.

At $n=4$ the coefficient is ${\displaystyle \frac{1}{4!}}={\displaystyle \frac{1}{24}}$.

(Quick check: $4!=4\cdot 3\cdot 2\cdot 1=24$.)

$A={\displaystyle \frac{1}{2}}\cdot 10\cdot 6=30$.

Setting $y^{\mathrm{\prime }}=0$: $0=2y+6⟹y=-3$.

Equivalently, $y_p=-{\displaystyle \frac{b}{a}}=-{\displaystyle \frac{6}{2}}=-3$.

With $(x_0,y_0)=(0,0)$ and the line $3x+4y-10=0$:

$d={\displaystyle \frac{\mathrm{\mid }3(0)+4(0)-10\mathrm{\mid }}{\sqrt{3^2+4^2}}}={\displaystyle \frac{\mathrm{\mid }-10\mathrm{\mid }}{\sqrt{25}}}={\displaystyle \frac{10}{5}}=2$.

${\displaystyle \frac{1-\cos x}{x^2}}={\displaystyle \frac{2{\sin }^2(x\mathrm{/}2)}{x^2}}={\displaystyle \frac{1}{2}}\cdot \text{\hspace{0.17em}⁣}{\left({\displaystyle \frac{\sin (x\mathrm{/}2)}{x\mathrm{/}2}}\right)}^2\to {\displaystyle \frac{1}{2}}$.

A $3$-cycle has sign $(-1{)}^{3-1}=+1$. (Equivalently, it can be written as the product of two transpositions: $(123)=(13)(12)$.)

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

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

$R=\sqrt{1+3}=2$, so $f(x)=2\sin (x+\pi \mathrm{/}3)$. Max value: $2$.