Math Puzzle Solutions!

PUZZLE ONE via Badal Joshi.

What is \sum_{n=2}^{\infty}\sum_{k=2}^{\infty}\frac{1}{n^k}? Hence, what is \sum_{s\in{}S}\frac{1}{s-1} where S = \{ a^n | a, n \in \mathbb{N}; a,n \ge 2 \}?

 \sum_{n=2}^{\infty}\sum_{k=2}^{\infty}\frac{1}{n^k} = \sum_{n=2}^{\infty} \left[ \left( \sum_{k=0}^{\infty} \frac{1}{n^k} \right) -1 -\frac{1}{n} \right] \\
 = \sum_{n=2}^{\infty} \left[ \frac{n}{n-1} - \frac{n+1}{n} \right] \\
 = \lim_{n \rightarrow \infty} \left[ \frac{2}{2-1} - \frac{n+1}{n} \right] \\
 = 2-1 = 1

The first line just adds and subtracts two terms from the inner sum. The second line uses the formula for geometric series. The third gives the value of the telescoping series.

Let T := \mathbb{N} \setminus (S\cup\{1\}), and notice that \mathbb{N}\setminus \{1\} = \bigcup_{m=1}^{\infty}T^{m} and S = \bigcup_{n=2}^{\infty} T^{n}, where both these unions are disjoint.

\sum_{s \in S}\frac{1}{s-1} = \sum_{n=2}^{\infty}\sum_{t \in T}\frac{1}{t^{n}-1} \\
 = \sum_{n=2}^{\infty} \sum_{t \in T} \sum_{m=1}^{\infty} \left( \frac{1}{t^{n}} \right)^{m} \\
 = \sum_{n=2}^{\infty} \sum_{m=1}^{\infty} \sum_{t \in T} \left( \frac{1}{t^{m}} \right)^{n} \\
 = \sum_{n=2}^{\infty} \sum_{k=2}^{\infty} \frac{1}{k^{n}} = 1

The first line comes from S = \bigcup_{n=2}^{\infty} T^{n}, the second uses the formula for geometric series, the third is a rearrangement, and the fourth uses \mathbb{N}\setminus \{1\} = \bigcup_{m=1}^{\infty}T^{m}.

PUZZLE TWO

Consider the sequence of functions (f_n:[0,1]\rightarrow\mathbb{R})_{n \in \mathbb{N}} given by f_{n}(x) = e^{-n^{\alpha}x^{\beta}}n^{\gamma}x^{\delta} where \alpha , \beta , \gamma , \delta are all positive, and \delta - \frac{\gamma\beta}{\alpha} > -1.

Compute \lim_{n\rightarrow\infty}\int_{0}^{1}f_{n}(x) d\!x.

Okey doke, this one's fairly tough unless you took my hint, which short-circuits all the technicalities. First, you wanna notice that the integrand is non-negative over [0,1], so for all n \in \mathbb{N}, \int_{0}^{1} f_{n}(x) d\!x \geq 0.

We now provide an upper bound for each integral, and show that the upper bounds tend to zero. This is enough to show that \lim_{n \rightarrow \infty} \int_{0}^{1} f_{n} d\!x = 0.

Let a := \gamma - \frac{\alpha}{\beta}(\delta +1) < \gamma - \frac{\alpha}{\beta}\frac{\gamma\beta}{\alpha}=\gamma - \gamma = 0.

Make the substitution u := n^{\alpha}x^{\beta}. For each n \in \mathbb{N}, this gives:

\int_{0}^{1} f_{n}(x) d\!x \\
 = \frac {n^{\gamma - \frac{\alpha}{\beta}(\delta +1)} }{\beta} \int_{0}^{n^{\alpha}} e^{-u}u^{\frac{\delta +1}{\beta} -1} d\!u \\
 \leq \frac{n^{-a}}{\beta} \int_{0}^{\infty} e^{-u}u^{\frac{\delta +1}{\beta} -1} d\!u \\
 = \frac{n^{-a}}{\beta} \Gamma \left( \frac{\delta +1}{\beta} \right) \rightarrow 0

as n \rightarrow 0.

PUZZLE THREE

\int^{1}_{0}\frac{x^{4}(1-x)^{4}}{1+x^{2}} d\!x

It's long division time!

\frac{x^{4}(1-x)^{4}}{1+x^{2}} = x^{6}-4x^{5}+5x^{4}-4x^{2}+4-\frac{4}{1+x^{2}}\mathtext{,}

so the integral is

\int_{0}^{1}x^{6} -4x^{5} +5x^{4} - 4x^{2} +4 -\frac{4}{1+x^{2}} d\!x \\
 = \left[ \frac{x^{7}}{7}-\frac{4x^{6}}{6}+x^{5}-\frac{4x^{3}}{3}+4x-4\arctan(x) \right]_{0}^{1} \\
 = \frac{1}{7} -\frac{4}{6} +1 -\frac{4}{3} +4 - \frac{4\pi}{4} \\
 = \frac{22}{7} - \pi

Which, in my opinion, is kinda cool.

PUZZLE FOUR

You're given a line, and two points on the line, A & C. Using only a compass, and using your compass only four times, find the point B on the line segment AC such that length(AB) = 2length(BC).

Okay, so here's what you're given:

The first circle you wanna draw is centred on C and passes through A; we label the other point it intersects the line as D:

The second circle is centred on A and passes through D; we label its other intersection with the line as E:

Circle number three centres on E and passes through C, intersecting circle two at G and H, and the line at F.

The last circle centres at G and passes through C, intersecting the line again at a point we'll call B. I claim length(AB) = 2length(BC).

To see this, we're gonna draw two line segments, CG, FG, and GH, the last of which intersects our original line two create point J.

Triangles FGC and GJC are similar, and right-angled. (FGC is right since, inscribed in a circle, one of its sides is a diameter; GJC is right because GH is a perpendicular.)

length(CG) = length(AC).

length(CF) = 6length(AC).

Thus, length(CJ) = (1/6)length(CG) = (1/6)length(AC)

Now, length(BC) = 2length(JC) = (1/3)length(AC).

B separates A and C on a line, so this means length(AB) = 2length(BC).