Friday, April 12, 2013

Writing tips

These are some writing tips, I learnt from the ENGR 202S course at Stanford.

Friday, $12^{th}$ April, $2013$:
  1. Hyphen: Hyphens are mostly used to break single words into parts, or to join ordinarily separate words (usually two adjectives) into single words. Spaces should not be placed between a hyphen and either of the words it connects except when using a suspended or "hanging" hyphen (e.g. nineteenth- and twentieth-century writers). Hyphen is also used to line break when a word is left halfway hanging in a line.
  2. En dash: The length of this is the typical length of $n$, which approximately $2$ hyphens. Used to indicate range. For e.g., $10 \text{--}20$.
  3. Em dash: The length of this is the length of a typical $m$, which is approximately $3$ hyphens. Used to demarcates a break of thought or some similar interpolation stronger than the interpolation demarcated by parentheses, which is in turn stronger than a comma.
  4. etc.: $\underbrace{\text{Et}}_{\text{and}} \overbrace{\text{cetera}}^{\text{other things}}$ Always ends with a period and is separated from the previous object with a comma. "For e.g., a, b, c, etc., have been studied by ..."
  5. et al.: $\underbrace{\text{Et}}_{\text{and}} \overbrace{\text{aliaa}}^{\text{other people}}$ Always ends with a period and is separated from the previous object with a comma. "For e.g., a, b, c, et al., have worked ..."
  6. Difference between which and that: Consider the following two sentences. $$\text{We looked at the instructions that were at the end of the manual.}$$ $$\text{We looked at the instructions, which were at the end of the manual.}$$ Both the sentences are correct, but they convey different meanings. $$\text{The first one: The manual had many instructions and we followed the ones given at the end.}$$ $$\text{The second one: We followed the instructions given in the manual. The instructions were given}$$ $$\text{at the end of the manual.}$$ Now some conventions. The above is in US English. In US English, "that" and "which" are never interchangeable and "which" is always preceded by a comma. In UK English, "that" can be replaced by "which" without comma. Hence, in UK English, the comma before which is all the more important not to change the meaning of the sentence.

Friday, $19^{th}$ April, $2013$:

Questions:

"Our fast, new, stable algorithm for large-scale linear inversion problem relies on $\mathcal{H}$-matrix algebra and is applicable when the number of unknowns $m$ is large --- around $100,000$ --- and the number of measurements is of the order of $100 \text{--} 500$."
  1. The above sentence has two "and"s. Is that right?
  2. Should there be a comma before "and"?
  3. When do we need to use a colon and semicolon?

Sunday, April 7, 2013

Eulerian tour theorem

This post looks at Eulerian cycle theorem in graph theory.

We first define a few terms.

A walk is an alternating sequence of vertices and edges, beginning and ending at a vertex with the constraint that each edge between two vertices must be incident on these two vertices. Essentially, it means moving in a graph along the edges in a continuous fashion. For example, a typical walk of length $n$ is of the form $$v_1 e_1 v_2 e_2 v_3 e_3 \cdots v_{n-1} e_{n-1} v_n e_n v_{n+1} \tag{$\star$}$$ where $e_j$ is incident to $v_j$ and $v_{j+1}$.

A cycle is a walk with the beginning and ending vertices being the same, i.e., $(\star)$ with $v_{n+1} = v_1$.

A walk where no edge is repeated is called a trail, i.e., $e_j \neq e_k$ for $j \neq k$ in $(\star)$.

A closed trail, i.e., the beginning and ending vertices being the same and no edge is repeated, i.e., $e_j \neq e_k$ for $j \neq k$ and $v_{n+1} = v_1$ in $(\star)$, is termed as a tour or a circuit.

An Eulerian trail is a trail passing through all the edges in the graph.

An Eulerian tour is a tour passing through all the edges in the graph.

Eulerian tour theorem: Let $G$ be a connected undirected graph, then the following two statements are equivalent:
  1. $G$ has an Eulerian tour.
  2. All the vertices in $G$ are even.
If the graph has loops, i.e., an edge connecting the same vertex the above still holds true, provided we interpret that every loop counts to the degree of the vertex twice.

Proof: We will prove it for graphs without loops since the loops do not make any difference.

$1 \implies 2$. Since $G$ has an Eulerian tour, every time we enter a vertex we need to exit it. Further, the tour doesn't repeat an edge. Hence, every vertex has even degree.

$2 \implies 1$. This proceed by induction on the number of edges. The base case is when the number of edges is $2$. This corresponds to a graph with $2$ vertices and $2$ edges between these two vertices. Clearly, this has an Eulerian tour. Assume that all graphs with number of edges less than $m$ and with each vertex having even degree has an Eulerian tour. Now we will show that a graph with $m$ number of edges and with each vertex having even degree has an Eulerian tour. Pick any vertex. There exists a tour starting and ending at this vertex. To see this, if this were not to be possible, then we would end at a vertex from which we cannot escape. However, this can never happen since each vertex has even degree. Hence, we can pick a tour starting and ending at the same vertex. Let this tour be $T$ and be of length $n$ and say $v_1 e_1 v_2 e_2 \ldots v_{n-1} e_{n-1} v_n e_n v_1$. Once we have this we are done. Now remove these edges from the graph. We will then have at most $n$ connected components, say $C_1, C_2, \ldots, C_k$, where $k \leq n$. Now by induction hypothesis, there is a Eulerian tour on each of these connected components. Now we will obtain the Eulerian tour on the entire graph $G$. Now lets get back to the tour $T$ and start at $v_1$. Find the first time this tour intersects one of the connected components. Once it hits the connected component, complete the tour within this connected component and return to the place where the tour hit this connected component. Repeat this till we have visited all the connected components and finally complete the tour. Note that adding loops play no role since if we were to have loop(s) at a vertex, whenever we hit that vertex finish of the loop(s) and continue.

Friday, April 5, 2013

Handshaking lemma

The Handshaking lemma states that any finite undirected graph has an even number of vertices with odd-degree. This follows immediately from a more fundamental result, namely, the degree sum formula.

Degree sum formula: Given an undirected graph $G$, with the set of vertices $V$ and edges $E$, we have$$\sum_{v \in V} \text{deg}(v) = 2\vert E \vert$$ If the graph has loops, i.e., an edge connecting the same vertex the above still holds true, provided we interpret that every loop counts to the degree of the vertex twice. (Sometimes, degree sum formula is also called the handshaking lemma)

Proof: We will prove the result when the graph has no loops. Once we prove it for graphs with no loops, the result is evident for graphs with loops since every time we add a loop, the degree of a vertex increases by $2$ and so does the right hand side.

Let $\mathbb{I}(v_i,v_j)$ be the indicator function that denotes if there is an edge between the vertices $v_i$ and $v_j$. We then have
$$\sum_{v_i \in V;v_j \in V} \mathbb{I}(v_i,v_j) = 2 \vert E \vert \tag{$\star$}$$This is because each edge is incident to two vertices and hence each edge is counted twice in the above summation. However, $(\star)$ can also be rewritten as $$\sum_{v_i \in V} \left(\sum_{v_j \in V} \mathbb{I}(v_i,v_j)\right) = 2 \vert E \vert$$Now note that $\displaystyle \left(\sum_{v_j \in V} \mathbb{I}(v_i,v_j) \right)$ is nothing but the degree of the vertex $v_i$, since each time a vertex is adjacent to the vertex $v_i$, the count increases by $1$. Hence, we have what we want, i.e.,$$\sum_{v \in V} \text{deg}(v) = 2\vert E \vert$$This immediately implies that there has to be an even number of vertices with odd-degree. Let $V_e$ be the set of vertices with even degree and $V_o$ be the set of vertices with odd degree. We then have that$$\sum_{v \in V_e} \text{deg}(v) + \sum_{v \in V_o} \text{deg}(v) = 2\vert E \vert \tag{$\dagger$}$$Since every vertex in $V_e$ has an even degree, we have $\sum_{v \in V_e} \text{deg}(v)$ to be an even number. This and $(\dagger)$ implies $\sum_{v \in V_o} \text{deg}(v) \,\,\, (\perp)$ has to be even. Now each of $\deg(v)$ in $(\perp)$ is odd. If $\vert V_o \vert$ were to be odd, then $\sum_{v \in V_o} \text{deg}(v)$ will be odd (since an odd number of odd integers when added up will give an odd number), contradicting the fact that it has to be even.

Wednesday, April 3, 2013

$\displaystyle \int_0^{\pi} \ln (1-2a \cos(x) + a^2) dx$


This is a nice integral from the $13^{th}$ annual Stanford Mathematics tournament.

Compute the integral
$$\int_0^{\pi} \ln \left(1-2a \cos(x) + a^2\right) dx$$

Solution:

Moor xu has presented $5$ different solutions here. We look at another way to compute the integral.
Let $$I(a) = \displaystyle \int_0^{\pi} \ln \left(1-2a \cos(x) + a^2\right) dx$$ Some preliminary results on $I(a)$. Note that we have $$I(a) = \underbrace{\displaystyle \int_0^{\pi} \ln \left(1+2a \cos(x) + a^2\right) dx}_{\spadesuit} = \overbrace{\dfrac12 \displaystyle \int_0^{2\pi} \ln \left(1-2a \cos(x) + a^2\right) dx}^{\clubsuit}$$ $(\spadesuit)$ can be seen by replacing $x \mapsto \pi-x$ and $(\clubsuit)$ can be obtained by splitting the integral from $0$ to $\pi$ and $\pi$ to $2 \pi$ and replacing $x$ by $\pi+x$ in the second integral.
Now let us move on to our computation of $I(a)$.
\begin{align}
I(a^2) & = \int_0^{\pi} \ln \left(1-2a^2 \cos(x) + a^4\right) dx = \dfrac12 \int_0^{2\pi} \ln \left(1-2a^2 \cos(x) + a^4\right) dx\\
& = \dfrac12 \int_0^{2\pi} \ln \left((1+a^2)^2-2a^2(1+ \cos(x))\right) dx = \dfrac12 \int_0^{2\pi} \ln \left((1+a^2)^2-4a^2 \cos^2(x/2)\right) dx\\
& = \dfrac12 \int_0^{2\pi} \ln \left(1+a^2-2a \cos(x/2)\right) dx + \dfrac12 \int_0^{2\pi} \ln \left(1+a^2+2a \cos(x/2)\right) dx
\end{align}
Now replace $x/2=t$ in both integrals above to get
\begin{align}
I(a^2) & = \int_0^{\pi} \ln \left(1+a^2-2a \cos(t)\right) dt + \int_0^{\pi} \ln \left(1+a^2+2a \cos(t)\right) dt = 2I(a)
\end{align}
Now for $a \in [0,1)$, this gives us that $I(a) = 0$. This is because we have $I(0) = 0$ and $$I(a) = \dfrac{I(a^{2^n})}{2^n}$$ Now let $n \to \infty$ and use continuity to conclude that $I(a) = 0$ for $a \in [0,1)$. Now lets get back to our original problem. Consider $a>1$. We have
\begin{align*}
I(1/a) & = \int_0^{\pi} \ln \left(1-\dfrac2{a} \cos(x) + \dfrac1{a^2}\right)dx\\
& = \int_0^{\pi} \ln(1-2a \cos(x) + a^2) dx - 2\int_0^{\pi} \ln(a)dx\\
& = I(a) - 2 \pi \ln(a)\\
& = 0  \tag{Since $1/a < 1$, we have $I(1/a) = 0$}
\end{align*}
Hence, we get that
$$I(a) = \begin{cases} 2 \pi \ln(a) & a \geq 1 \\ 0 & a \in [0,1] \end{cases}$$

Saturday, March 30, 2013

Ostrowski's theorem

In this blog post, we look at yet another fascinating way in which prime numbers appear in mathematics. The result below is called the Ostrowski's theorem, named after the Russian mathematician Alexander Ostrowski.

Before we go into the theorem, let us attempt to define absolute value on the set of rationals. We call a function $f: \mathbb{Q} \mapsto \mathbb{R}^+ \cup \{0\}$ an absolute value, if the following properties $(\star)$ are satisfied:
  1. Non-negativity: $f(0) = 0$ and $f(x) > 0$ for all $x > \mathbb{Q} \backslash \{0\}$.
  2. Multiplicativity: $f(x \cdot y) = f(x) \cdot f(y)$.
  3. Sub-additivity (or) Triangle inequality: $f(x+y) \leq f(x) + f(y)$.
Note: From $(1)$ and $(2)$, we have that $f(-1) = f(1) = 1$. This also gives us $f(-x) = f(x)$ and $f \left(x^k \right) = f \left(x \right)^k$. In addition, from $(3)$, we also get that $f(n) \leq \vert n \vert$, for $n \in \mathbb{Z}$.

The natural question is can be categorize/classify functions that satisfy the properties described above. To do so, let us look at some examples of function that satisfy the properties $(\star)$.

Examples:

$\mathbf{1}$: The usual absolute value function, $f(x) = \vert x \vert$, which measures the distance of a point on the real line from the origin, satisfies the properties $(\star)$.

$\mathbf{2}$: The function given by $f(x) = \vert x \vert^{\alpha}$, where $\alpha \in [0,1]$, also satisfies the properties $(\star)$.  Note that $\alpha = 0$ is the trivial function $$f(x) = \begin{cases} 0 & \text{if } x = 0\\ 1 & \text{if } x \neq 0\end{cases}$$ The case $\alpha = 1$, gives us the usual absolute value.

$\mathbf{3}$: $p$-adic absolute value: We shall now define a function on rational numbers as follows. Consider $a \in \mathbb{Q}$ and any prime number $p$. We have $a = \dfrac{m}n$, where $m,n \in \mathbb{Z}$. Note that we can always write any rational number as
$$a = \dfrac{m}n = p^{\beta} \dfrac{r}s$$where $(r,p) = 1 = (s,p)$. Essentially, we are pulling out all the powers of $p$ from both the numerator and denominator. Now define the function $f_p(a)$ as $f_p(a) = p^{-\beta}$ for $a\in \mathbb{Q} \backslash \{0\}$ and $f_p(0) = 0$.

Claim: The function described above satisfies the properties $(\star)$.
Proof:
Even before we check the properties, we need to check if the function is well-defined. It is not hard to check this, i.e., if $a = \dfrac{m_1}{n_1} = \dfrac{m_2}{n_2}$, where $\dfrac{m_1}{n_1} = p^{\beta_1} \dfrac{r_1}{s_1}$, $(r_1,p) = 1 = (s_1,p)$ and $\dfrac{m_2}{n_2} = p^{\beta_2} \dfrac{r_2}{s_2}$, $(r_2,p) = 1 = (s_2,p)$, then $\beta_1 = \beta_2$.

Trivially non-negativity is satisfied. Multiplicativity follows immediately from the fact that if $(k,l) = 1$ and $(t,l) = 1$, then $(tk,l) = 1$. Sub-additivity requires to be checked. Let $a = p^{\beta_a} \dfrac{m_a}{n_a}$ and $b = p^{\beta_b} \dfrac{m_b}{n_b}$. Since the function is well-defined, we can always choose $m_a,n_a$ and $m_b, n_b$ such that $(m_a,n_a) = 1$ and $(m_b,n_b) = 1$. We have
$$f_p(a) = \dfrac1{p^{\beta_a}} \text{ and } f_p(b) = \dfrac1{p^{\beta_b}}$$
Without loss of generality, let us assume $\beta_a \leq \beta_b$. We then have $$a+b = p^{\beta_a} \dfrac{m_a}{n_a} + p^{\beta_b} \dfrac{m_b}{n_b} = p^{\beta_a} \dfrac{m_a \cdot n_b + p^{\beta_b - \beta_a} m_b \cdot n_a}{n_a \cdot n_b}$$ Now consider $\dfrac{m_a \cdot n_b + p^{\beta_b - \beta_a} m_b \cdot n_a}{n_a \cdot n_b}$. No power of $p$ divides the denominator always. If $\beta_a < \beta_b$, then no power of $p$ divides the numerator as well. Hence, $f_p(a + b) = \dfrac1{p^{\beta_a}} = f_p(a)$. If $\beta_a = \beta_b$, then it is possible that some power of $p$ could divide the numerator. Then we have $f_p(a+b) = \dfrac1{p^{\beta_a + r}}$ where $r > 0$. Hence, either way, we have $$f_p(a+b) \leq \dfrac1{p_a} = \max\{f_p(a),f_p(b)\}.$$ Hence, we in-fact have a strong sub-additivity condition that
$$f_p(a+b) \leq \max \{f_p(a), f_p(b)\}$$Absolute values that satisfy this stronger property are called ultra-metric or non-Archimedean absolute values. We shall denote the $p$-adic absolute by the $\vert \cdot \vert_p$.

$\mathbf{4}$: The $p$-adic absolute value raised to the power $\alpha$, where $\alpha \geq 0$ also satisfies the properties $(\star)$, i.e., $\vert x \vert_p^{\alpha}$. The only thing that needs some work to be checked is the sub-additivity. The argument for sub-additivity is the same as before and this absolute value is also a non-Archimedean absolute value.

We now have $4$ examples of functions satisfying the properties $(\star)$. Strictly speaking, we only have $2$ examples. Examples $1$ and $3$ are special cases of examples $2$ and $4$ respectively, with $\alpha = 1$.

Ostrowski's theorem states that examples $2$ and $4$ gives us all the functions that satisfy $(\star)$, i.e., all possible absolute values on $\mathbb{Q}$ are given by examples $2$ and $4$.

Ostrowski's theorem is a powerful statement. In general, all classification theorems are powerful statements. If we have a property $(\dagger)$, we can construct objects satisfying property $(\dagger)$. Classification theorems enable us to classify all possible objects satisfying $(\dagger)$. This is usually done by finding example objects satisfying $(\dagger)$ and once we feel that there can be no more objects, we set above proving that any object satisfying property $(\dagger)$ falls in one of the categories/examples.

Now let us go above proving Ostrowski's theorem. We will prove that examples $2$ and $4$ are the only possible absolute values for the set of natural numbers. Hence, these could only be the possible set of absolute values for the rationals as well. We will split the theorem into two cases.

Case $1$: $f(n) \geq 1$, $\forall n \in \mathbb{Z}^+$. For this case, we shall show that $f(x) = \vert x \vert^{\alpha}$, where $\alpha \in [0,1]$ is the only such function.

We know that $f(1)$ has to be $1$. Consider $n \in \mathbb{Z}^+$ and not equal to $1$. Consider another $m \in \mathbb{Z}^+$. We can write $m$ in base $n$ as $$m = m_0 + m_1 n + m_2 n^2 + \cdots + m_k n^k$$where $m_l \in \{0,1,2,\ldots,n-1\}$. From sub-additivity of $f$, we have $$f(m) \leq f(m_0) + f(m_1) f(n) + f(m_2) f(n)^2 + \cdots + f(m_k) f(n)^k$$ Since $m_l \in \{0,1,2,\ldots,n-1\}$, we have $f(m_l) \leq n-1$. Hence, we get that \begin{align*} f(m) & \leq (n-1) \left(1 + f(n) + f(n)^2 + \cdots + (f(n))^k\right)\\ & < (n-1)(k+1) f(n)^k \leq (n-1)(\log_n(m)+1) f(n)^{\log_n(m)} \end{align*} Now replacing $m$ by $m^l$, we get that $$f(m^l) = f(m)^l \leq (n-1)(l \cdot \log_n(m)+1) f(n)^{l \cdot \log_n(m)}$$ Taking the $l^{th}$ root on both sides, we get that $$f(m) \leq (n-1)^{1/l} (1+l \cdot \log_n(m))^{1/l}f(n)^{\log_n(m)}$$ Now let $l \to \infty$, to get that $$f(m) \leq f(n)^{\log_n(m)}, \text{i.e.,} f(m)^{1/\log(m)} \leq f(n)^{1/\log(n)}$$ Since this inequality is symmetric in $n$ and $m$, we obtain $$f(n)^{1/\log(n)} = \text{constant}$$ for all $n \in \mathbb{Z}^+$. This gives us $$f(n) = \text{constant}^{\log(n)} = e^{\alpha \cdot \log(n)} = n^{\alpha}$$ Triangle inequality and the fact that $f(n) \geq 1$, $\forall n \in \mathbb{Z}^+$ gives us $\alpha \in [0,1]$.

Case $2$: $f(n) < 1$, for some $n \in \mathbb{Z}^+$. For this case, we shall show that $f(x) = \vert x \vert_p^{\alpha}$, where $\alpha \in \mathbb{R}^+$ is the only such function.

By well-ordering principle, we consider the smallest such $n \in \mathbb{Z}^+$ such that $f(n) < 1$. If $n$ is not a prime, then $n= ab$ for some $a,b \in \{2,3,\ldots,n-1,n\}$. This means $f(a) \cdot f(b) = f(n) < 1$. This means either $f(a) \in (0,1)$ or $f(b) \in (0,1)$, which contradicts the minimality of $n$. Hence, $n$ has to be a prime. Now consider any $m \in \mathbb{Z}^+$. We write $m$ in base $p$ as $$m = m_0 + m_1 p + m_2 p^2 + \cdots m_k p^k$$ We then have \begin{align*}f(m) & \leq f(m_0) + f(m_1) f(p) + f(m_2) f(p^2) + \cdots f(m_k) f(p^k)\\ & < (p-1) + (p-1) f(p) + (p-1) f(p) + \cdots + (p-1) f(p) = (k+1)(p-1) f(p)\\ & \leq (1+\log_p(m))(p-1)f(p) \end{align*} As before replace $m$ by $m^l$ to get $$f(m^l) = f(m)^l < (1+ l \log_p(m))(p-1) f(p)$$ This gives us $$f(m)^l < (1+l \log_p(m)) (p-1) f(p)$$ and taking the $l^{th}$ root on both sides and letting $l \to \infty$, gives us $$f(m) \leq 1$$ for all $m \in \mathbb{Z}^+$ and since $f(-m) = f(m)$, we also have $f(m) \leq 1, \forall m \in \mathbb{Z}$. The only thing left for us to show is that if $(m,p) = 1$, then $f(m) = 1$. (This is so since if $(m,p) \neq 1$, we can always write $m = p^{\gamma} \times u$, where $(u,p) = 1$ and make use of multiplicativity.) If $(m,p) = 1$, we have $(m^n,p^n) = 1$ for all integers $n$. Hence, there exists sequence of integer pairs $(x_n,y_n) \in \mathbb{Z}^2$ such that $$m^n x_n + p^n y_n = 1$$ We then have $$f(m^n x_n + p^n y_n) = f(1) = 1$$ This gives us $$1 \leq f(m)^n f(x_n) + f(p)^n f(y_n) \leq f(m)^n + f(p)^n$$ Letting $n \to \infty$, we have $f(p)^n \to 0$, since $f(p) < 1$. Hence, we have $f(m) = 1$, whenever $(m,p) = 1$. Hence, if $n = p_1^{\beta_1} p_2^{\beta_2} \cdots p_k^{\beta_k}$, and if $p_1$ is the smallest prime such that $f(p_1) < 1$, we then have $$f(n) = f(p_1^{\beta_1}) \cdot f(p_2^{\beta_2} p_3^{\beta_3} \cdots p_k^{\beta_k}) = f(p_1)^{\beta_1} \times 1 = f(p_1)^{\beta_1},$$ where $f(p_1)$ is some quantity less than $1$ given by $f(p_1) = p_1^{-\alpha}$, where $\alpha > 0$. This gives the $p$-adic absolute value.

Monday, March 25, 2013

$\displaystyle \int_0^{2\pi} \frac{\sin^2\theta}{a+b\cos\theta}\ d\theta$ using Catalan numbers

A recent post on math.stackexchange asked for a proof of the following statement:
$$\int_0^{2\pi} \frac{\sin^2\theta}{a+b\cos\theta}\ d\theta = \frac{2\pi}{b^2} \left(a-\sqrt{a^2-b^2} \right) $$
A typical way to evaluate this integral is to rewrite the integral as
$$\int_0^{2\pi} \frac{\sin^2\theta}{a+b\cos\theta}\ d\theta = \oint_{\vert z \vert = 1} \dfrac{\left( \dfrac{z-1/z}{2i}\right)^2}{a+b \left(\dfrac{z+1/z}2\right)} \dfrac{dz}{iz}$$ and then do some algebraic massaging and make use of residue theorem to evaluate this integral. In this post, we look at an elementary method to evaluate this integral. (I wrote this as an answer to that question on math.stackexchange.)

Proof:

$$I = \int_0^{2 \pi} \dfrac{\sin^2(x)}{a+b \cos(x)} dx \implies aI = \int_0^{2 \pi} \dfrac{\sin^2(x)}{1+\dfrac{b \cos(x)}a}dx$$
\begin{align}
aI & = \sum_{k=0}^{\infty}\left(\dfrac{(-b)^k}{a^k} \int_0^{2 \pi}\sin^2(x) \cos^k(x) dx \right) = \sum_{k=0}^{\infty}\left(\dfrac{b^{2k}}{a^{2k}} \int_0^{2 \pi}\sin^2(x) \cos^{2k}(x) dx \right)
\end{align}
Note that we have thrown away the odd terms, since for odd $k$, the integral $\displaystyle \int_0^{2 \pi}\sin^2(x) \cos^k(x) dx$ is zero.
\begin{align}
\dfrac{\displaystyle \int_0^{2 \pi}\sin^2(x) \cos^{2k}(x) dx}4 & = \int_0^{\pi/2}\sin^2(x) \cos^{2k}(x) dx\\
&
\int_0^{\pi/2}\cos^{2k}(x) dx - \int_0^{\pi/2}\cos^{2k+2}(x) dx\\
& = \dfrac{2k-1}{2k} \dfrac{2k-3}{2k-2} \cdots \dfrac12 \dfrac{\pi}2 - \dfrac{2k+1}{2k+2} \dfrac{2k-1}{2k} \dfrac{2k-3}{2k-2} \cdots \dfrac12 \dfrac{\pi}2\\
& = \dfrac1{2k+2} \dfrac{2k-1}{2k} \dfrac{2k-3}{2k-2} \cdots \dfrac12 \dfrac{\pi}2\\
& = \dfrac{\pi}{2^{k+2}} \dfrac{(2k-1)(2k-3)\cdots3 \cdot 1}{(k+1)!} = \dfrac{\pi}{2^{2k+2}} \dfrac{(2k)!}{k! (k+1)!}
\end{align}
Hence,
$$\dfrac{aI}{\pi} = \sum_{k=0}^{\infty} \left(\dfrac{b}{2a} \right)^{2k} \underbrace{\dfrac{(2k)!}{k! (k+1)!}}_{\text{Catalan numbers}}$$
Now $$\sum_{k=0}^{\infty} \dfrac{\dbinom{2k}k x^{k}}{k+1} = \dfrac{1-\sqrt{1-4x}}{2x} \,\,\,\,\,\, \forall \vert x \vert < \dfrac14$$ This is the generating function for the Catalan numbers. Mike Spivey has a nice little post on Catalan numbers, where you can find the expression and generating function of the Catalan numbers. Hence, in our case, we get that
$$\sum_{k=0}^{\infty} \left(\dfrac{b}{2a} \right)^{2k} \underbrace{\dfrac{(2k)!}{k! (k+1)!}}_{\text{Catalan numbers}} = \dfrac{1-\sqrt{1-4 \cdot \left(\dfrac{b}{2a} \right)^2}}{2 \cdot \left(\dfrac{b}{2a} \right)^2} \,\,\,\,\,\,\,\,\forall \dfrac{b}{2a} < \dfrac12$$
Hence,
$$\dfrac{aI}{\pi} = \dfrac{1-\sqrt{1-\left(\dfrac{b}a\right)^2}}{\dfrac{b^2}{2a^2}} = \dfrac{a-\sqrt{a^2-b^2}}{\dfrac{b^2}{2a}} \implies I = \dfrac{2\pi}{b^2} (a-\sqrt{a^2-b^2})$$

Friday, March 15, 2013

Generalization of Millin series

The series $$\sum_{k=0}^{\infty} \dfrac1{F_{2^k}} = \dfrac{7-\sqrt5}2$$where $F_m$ denotes the $m^{th}$ Fibonacci number is called the Millin series. In this post, we will look at a generalization of this result for recurrences of the form
$$a_{n+2} = b a_{n+1} + a_n$$
where $a_0=0$, $a_1 = 1$ and $b \in \mathbb{R}$. Note that in the case of the Fibonacci numbers, we have $b = 1$.

Claim: For any $b \in \mathbb{R}$, we have $$\sum_{k=0}^{\infty} \dfrac1{a_{2^k}} = 1 + \dfrac2b + \dfrac{b - \sqrt{b^2+4}}2 \,\,\,\,\, (\heartsuit)$$

Note that if we prove the above claim, then setting $b=1$ gives us
$$\sum_{k=0}^{\infty} \dfrac1{F_{2^k}} = 1 + 2 + \dfrac{1-\sqrt5}2 = 3 + \dfrac{1-\sqrt5}2 = \dfrac{7-\sqrt5}2$$which is the Millin series.

Proof:  Before we prove this, let us first obtain a result for the finite sum counterpart in $(\heartsuit)$.
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Lemma: If we have a sequence given by the recurrence,
$$a_{n+2} = ba_{n+1} + a_n,$$ with $a_0 =0 $ and $a_1 = 1$, we then have
$$\boxed{\color{blue}{\displaystyle \sum_{k=0}^{N} \dfrac1{a_{2^k}} = 1 + \dfrac2b - \dfrac{a_{2^N-1}}{a_{2^N}}}}$$

Proof: Let us write out a few terms of this sequence, we get
$$a_0 = 0, a_1 = 1, a_2 = b, a_3 = b^2 + 1, a_4 = b^3 + 2b, \cdots$$
The proof is by induction on $N$. For $N=1$, we have the left hand side to be $$\dfrac1{a_1} + \dfrac1{a_2} = 1 + \dfrac1b$$ while the right hand side is $$1 + \dfrac2b - \dfrac{a_1}{a_2} = 1 + \dfrac2b - \dfrac1{b} = 1 + \dfrac1b$$
For $N=2$, we have the left hand side to be $$\dfrac1{a_1} + \dfrac1{a_2} + \dfrac1{a_4} = 1 + \dfrac1b + \dfrac1{b^3 + 2b}$$ while the right hand side is $$1 + \dfrac2b - \dfrac{a_3}{a_4} = 1 + \dfrac2b - \dfrac{b^2+1}{b^3+2b} = 1 + \dfrac1b + \dfrac1b - \dfrac{b^2+1}{b^3+2b} = 1 + \dfrac1b + \dfrac1{b^3+2b}$$ Hence, it holds for $N=1$ and $N=2$. Now lets go ahead with induction now. Assume the result is true for $N=m$, i.e., we have
$$\sum_{k=0}^{m} \dfrac1{a_{2^k}} = 1 + \dfrac2b - \dfrac{a_{2^m-1}}{a_{2^m}}$$
Now $$\sum_{k=0}^{m+1} \dfrac1{a_{2^k}} = 1 + \dfrac2b - \dfrac{a_{2^m-1}}{a_{2^m}} + \dfrac1{a_{2^{m+1}}}$$ Hence, we want to show that
$$ - \dfrac{a_{2^m-1}}{a_{2^m}} + \dfrac1{a_{2^{m+1}}} = -\dfrac{a_{2^{m+1}-1}}{a_{2^{m+1}}}, \,\, \text{i.e.}, \,\, \dfrac1{a_{2^{m+1}}} + \dfrac{a_{2^{m+1}-1}}{a_{2^{m+1}}} =  \dfrac{a_{2^m-1}}{a_{2^m}}, \,\, \text{i.e.}, \,\, a_{2^m}(1+a_{2^{m+1}-1}) = a_{2^m-1} a_{2^{m+1}} \,\,\,\, (\star)$$
which can be verified using the recurrence. In fact $(\dagger)$, a slightly more general version of $(\star)$, which is easier to check is true.
$$a_{2k}(1+a_{4k-1}) = a_{2k-1} a_{4k} \,\,\,\, (\dagger),\,\, \text{i.e.}, \,\, a_{2k-1} a_{4k} - a_{2k} a_{4k-1} = a_{2k} \,\,\,\, (\dagger)$$
Hence, we get that
$$\boxed{\color{red}{\displaystyle \sum_{k=0}^{N} \dfrac1{a_{2^k}} = 1 + \dfrac2b - \dfrac{a_{2^N-1}}{a_{2^N}}}}$$
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Now letting $N \to \infty$, we get what we want immediately. This is so since from the recurrence we get that
$$\dfrac{a_{n+2}}{a_{n+1}} = b + \dfrac{a_n}{a_{n+1}}$$
If we have $\displaystyle \lim_{n \to \infty} \dfrac{a_n}{a_{n+1}} = L$, then we get that
$$\dfrac1L = b + L$$ and since $L>0$, we have $L = \dfrac{\sqrt{b^2+4}-b}2$. Hence,
$$\boxed{\color{red}{\displaystyle \sum_{k=0}^{\infty} \dfrac1{a_{2^k}} = \lim_{N \to \infty} \displaystyle \sum_{k=0}^{N} \dfrac1{a_{2^k}} = 1 + \dfrac2b - \lim_{N \to \infty} \dfrac{a_{2^N-1}}{a_{2^N}} = 1 + \dfrac2b - L = 1 + \dfrac2b + \dfrac{b-\sqrt{b^2+4}}2}}$$