MATH-500

Convert the point $(0,3)$ in rectangular coordinates to polar coordinates. Enter your answer in the form $(r,\theta),$ where $r > 0$ and $0 \le \theta < 2 \pi.$

We have that $r = \sqrt{0^2 + 3^2} = 3.$ Also, if we draw the line connecting the origin and $(0,3),$ this line makes an angle of $\frac{\pi}{2}$ with the positive $x$-axis. [asy] unitsize(0.8 cm); draw((-0.5,0)--(3.5,0)); draw((0,-0.5)--(0,3.5)); draw(arc((0,0),3,0,90),red,Arrow(6)); dot((0,3), red); label("$(0,3)$", (0,3), W); dot((3,0), red); [/asy] Therefore, the polar coordinates are $\boxed{\left( 3, \frac{\pi}{2} \right)}.$

\left( 3, \frac{\pi}{2} \right)

Precalculus

2

test/precalculus/807.json

Define \[p = \sum_{k = 1}^\infty \frac{1}{k^2} \quad \text{and} \quad q = \sum_{k = 1}^\infty \frac{1}{k^3}.\]Find a way to write \[\sum_{j = 1}^\infty \sum_{k = 1}^\infty \frac{1}{(j + k)^3}\]in terms of $p$ and $q.$

We count the number of times $\frac{1}{n^3}$ appears in the sum \[\sum_{j = 1}^\infty \sum_{k = 1}^\infty \frac{1}{(j + k)^3},\]where $n$ is a fixed positive integer. (In other words, we are conditioning the sum on $j + k$.) We get a term of $\frac{1}{n^3}$ each time $j + k = n.$ The pairs $(j,k)$ that work are $(1,n - 1),$ $(2,n - 2),$ $\dots,$ $(n - 1,1),$ for a total of $n - 1$ pairs. Therefore, \begin{align*} \sum_{j = 1}^\infty \sum_{k = 1}^\infty \frac{1}{(j + k)^3} &= \sum_{n = 1}^\infty \frac{n - 1}{n^3} \\ &= \sum_{n = 1}^\infty \left( \frac{n}{n^3} - \frac{1}{n^3} \right) \\ &= \ …

p - q

Intermediate Algebra

5

test/intermediate_algebra/1994.json