Pumpkin Pi

Fig. 1

Among many images of carved pumpkin, I like the one above (see Fig. 1) the most. It shows Leibniz’s formula for calculating the value of $\pi$. Namely,

$\pi=4\sum\limits_{k=1}^{\infty}\frac{(-1)^{k+1}}{2k-1}$.

To derive this formula, we begin with finding the derivative of $\arctan{x}$:

Let $y = \arctan{x}$, we have $x=tan(y)$, and

$\frac{d}{dx}x=\frac{d}{dx}\tan{y}=\frac{d}{dy}\tan{y}\frac{dy}{dx}=\sec^{2}{y}\frac{dy}{dx} = (1+tan^{2}{y})\frac{dy}{dx}$

$= (1+x^{2})\frac{dy}{dx}$.

Since $\frac{d}{dx}x=1$,

$\frac{d}{dx}\arctan{x}=\frac{1}{1+x^2}\quad\quad\quad(1)$

It follows that by (1) and the Fundamental Theorem of Calculus,

$\int\limits_{0}^{1}\frac{1}{1+x^2}dx = \arctan{x}\bigg|_{0}^{1}=\frac{\pi}{4}$

i.e.,

$\frac{\pi}{4} = \int\limits_{0}^{1}\frac{1}{1+x^2} dx\quad\quad\quad\quad\quad\quad(2)$

From carrying out polynomial long division, we observe

$\frac{1}{1+x^2} = 1 + \frac{-x^2}{1+x^2}$,

$\frac{1}{1+x^2} = 1 - x^2 + \frac{x^4}{1+x^2}$,

$\frac{1}{1+x^2} = 1 - x^2 + x^4 + \frac{-x^6}{1+x^2}$,

$\frac{1}{1+x^2} = 1 - x^2 + x^4 - x^6 + \frac{x^8}{1+x^2}$.

It seems that

$\frac{1}{1+x^2} = \sum\limits_{k=1}^{n}{(-1)^{k+1}x^{2k-2}} + \frac{(-1)^{n} x^{2n}}{1+x^2}\quad\quad\quad\quad\quad\quad(3)$

Assuming (3) is true, we integrate it with respect to $x$ from 0 to 1,

$\int\limits_{0}^{1}\frac{1}{1+x^2}dx=\int\limits_{0}^{1}\sum\limits_{k=0}^{n}(-1)^{k+1}x^{2k-2} dx + \int\limits_{0}^{1}\frac{(-1)^{n} x^{2n}}{1+x^2}dx$

$= \sum\limits_{k=1}^{n}(-1)^{k+1}\int\limits_{0}^{1}x^{2k-2}dx +(-1)^{n}\int\limits_{0}^{1}\frac{x^{2n}}{1+x^2}dx$

$= \sum\limits_{k=1}^{n}(-1)^{k+1}\frac{x^{2k-1}}{2k-1}\bigg|_{0}^{1}+(-1)^{n}\int\limits_{0}^{1}\frac{x^{2n}}{1+x^2}dx$.

As a result of integration,  (2) becomes

$\frac{\pi}{4} = \sum\limits_{k=1}^{n}\frac{(-1)^{k+1}}{2k-1} + (-1)^{n}\int\limits_{0}^{1}\frac{x^{2n}}{1+x^2}dx$,

or,

$\frac{\pi}{4} - \sum\limits_{k=1}^{n}\frac{(-1)^{k+1}}{2k-1} =(-1)^{n}\int\limits_{0}^{1}\frac{x^{2n}}{1+x^2}dx$.

Therefore,

$|\frac{\pi}{4} - \sum\limits_{k=1}^{n}\frac{(-1)^{k+1}}{2k-1}|=|(-1)^{n}\int\limits_{0}^{1}\frac{x^{2n}}{1+x^2}dx | = \int\limits_{0}^{1}\frac{x^{2n}}{1+x^2}dx < \int\limits_{0}^{1}x^{2n}dx$

$=\frac{x^{2n+1}}{2n+1}\bigg|_{0}^{1}= \frac{1}{2n+1}$.

Moreover, $\forall \epsilon > 0$, we obtain $n > \frac{1}{2}(\frac{1}{\epsilon}-1)$ through solving $\frac{1}{2n+1} < \epsilon$. It means that $\forall \epsilon > 0, \exists n^*=\frac{1}{2}(\frac{1}{\epsilon}-1)$ such that  for all $n > n^*, |\frac{\pi}{4} - \sum\limits_{k=1}^{n}\frac{(-1)^{k+1}}{2k-1}|<\epsilon$, i.e.,

$\lim\limits_{n\to\infty}{ \sum\limits_{k=1}^{n}\frac{(-1)^{k+1}}{2k-1}}=\frac{\pi}{4}$.

Thus

$\pi = 4 \sum\limits_{k=1}^{\infty}\frac{(-1)^{k+1}}{2k-1}\quad\quad\quad(4)$

The numerical value of $\pi$ is therefore approximated according to (4) by the partial sum

$4 \sum\limits_{k=1}^{n}\frac{(-1)^{k+1}}{2k-1}=4(1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\dots+(-1)^{n+1}\frac{1}{2n-1})\quad\quad\quad\quad(5)$

Its value converges to $\pi$ as $n$ increases.

However, (5) is by no means a practical way of finding the value of $\pi$, since its convergence is so slow that many terms must be summed up before a reasonably accurate result emerges (see Fig. 2)

Fig. 2

I doubt Leibniz has ever used his own formula to obtain the value of $\pi$ !

Let me leave you with an exercise: Prove (3)

“Chaplin or Leibniz ?” Revisit

Once the closed form of power summation is derived (see “Little Bird and a Recursive Generator“) namely

$s_{p} = 1^p+2^p+3^p+\dots+n^p=\frac{n(n+1)^p - \sum\limits_{j=2}^{p}\binom{p}{j} s_{p-j+1}}{p+1}, \quad\quad p, n \in N^{+},$

the validity of the inequality in “Chaplin or Leibniz ?” is readily shown:

$1^p+2^p+3^p+\dots+n^p = \frac{n(n+1)^p - \sum\limits_{j=2}^{p}\binom{p}{j} s_{p-j+1}}{p+1}$

$< \frac{(n+1)(n+1)^p - \sum\limits_{j=2}^{p}\binom{p}{j} s_{p-j+1}}{p+1}< \frac{(n+1)^{p+1}}{p+1}$.

i.e.,

$(p+1)(1^p+2^p+3^p+\dots+n^p) < (n+1)^{p+1}, \quad\quad p, n \in N^+$.

Little Bird and a Recursive Generator

When I was asked to prove the following power summation formulas by mathematical induction:

$\sum\limits_{i=1}^{n}i = 1+2+3+...+n=\frac{n (n+1)}{2}$,

$\sum\limits_{i=1}^{n}i^2=1^2+2^2+3^2+...+n^2= \frac{n (n+1) (2n +1)}{6}$,

$\sum\limits_{i=1}^{n}i^3=1^3+2^3+...+n^3 = \frac{n^2 (n+1)^2}{4}$,

I wondered how these closed forms are obtained in the first place!  Did a little bird whisper the formulas into our ears?

Even though there are elementary derivations of the power summation formulas, for example,  the visual derivations in Proof Without Words by MAA, none goes beyond the 3rd power.

In this blog, I will construct a recursive generator capable of generating the closed form of power summation

$\sum\limits_{i=1}^{n}i^p$,

for all $p \in N^{+}$.

Fig. 1

Let $A$ denotes the area of a rectangle, Fig.1 shows

$\sum A_{blue} = A_{n^p \times n} - \sum A_{yellow}$.

Since

$\sum A_{blue} = \sum\limits_{i=1}^{n}i^p$,

$A_{n^p\times n} = n^p n$,

$\sum A_{yellow}= \sum\limits_{i=1}^{n-1}i((i+1)^p-i^p)$,

we have

$\sum\limits_{i=1}^{n}i^p = n^p n - \sum\limits_{i=1}^{n-1}i((i+1)^p-i^p)\quad\quad\quad(1)$.

when $p = 1$, (1) becomes

$\sum \limits_{i=1}^{n}i=n^2-\sum\limits_{i=1}^{n-1}i((i+1)-i)$

$= n^2-(\sum\limits_{i=1}^{n}(i(i+1)-i)-(n(n+1)-n))$

$= n^2-\sum\limits_{i=1}^{n}i+n$,

hence,

$2\sum\limits_{i=1}^{n}i=n^2+n$,

or,

$\sum\limits_{i=1}^{n}i=\frac{n(n+1)}{2}\quad\quad\quad(2)$.

When p = 2,

$\sum\limits_{i=1}^{n}i^2=n^2 n-\sum\limits_{i=1}^{n-1}i((i+1)^2-i^2)$

$= n^3-2\sum \limits_{i=1}^{n}i^2-\sum\limits_{i=1}^{n}i+2n^2+n$.

Substituting (2) for $\sum\limits_{i=1}^{n}i$,  we have

$3\sum\limits_{i=1}^{n}i^2=n^3-\frac{n(n+1)}{2}+2n^2+n$

or,

$\sum\limits_{i=1}^{n}i^2= \frac{n(n+1)(2n+1)}{6}$.

In general, $\forall p \in N^+$,

$\sum\limits_{i=1}^{n}i^p = n^p n-\sum\limits_{i=1}^{n-1}i((i+1)^p-i^p)$

$= n^{p+1}-(\sum\limits_{i=1}^{n}i((i+1)^p-i^p)-n((n+1)^p-n^p))$

$= n^{p+1}- \sum\limits_{i=1}^{n}(i(\sum\limits_{j=0}^{p}\binom{p}{j}i^{p-j}-i^p)) +n((n+1)^p-n^p)$

$=n(n+1)^p-\sum\limits_{i=1}^{n}(i\sum\limits_{j=1}^{p}\binom{p}{j}i^{p-j})$

$= n(n+1)^p - \sum\limits_{i=1}^{n}\sum\limits_{j=1}^{p}\binom{p}{j}i^{p-j+1}$

$= n(n+1)^p - \sum\limits_{i=1}^{n}(p i^p + \sum\limits_{j=2}^{p}\binom{p}{j}i^{p-j+1})$

$= n(n+1)^p - p\sum\limits_{i=1}^{n}i^p-\sum\limits_{i=1}^{n}\sum\limits_{j=2}^{p}\binom{p}{j}i^{p-j+1}$

$= n(n+1)^p-p\sum\limits_{i=1}^{n}i^p-\sum\limits_{j=2}^{p}\sum\limits_{i=1}^{n}\binom{p}{j}i^{p-j+1}$

$= n(n+1)^p-p\sum\limits_{i=1}^{n}i^p-\sum\limits_{j=2}^{p}(\binom{p}{j}\sum\limits_{i=1}^{n}i^{p-j+1})$.

Solving for $\sum\limits_{i=1}^{n}i^p$, we obtain

$\sum\limits_{i=1}^{n}i^p = \frac{n(n+1)^p -\sum\limits_{j=2}^{p}(\binom{p}{j}\sum\limits_{i=1}^{n}i^{p-j+1}) }{p+1}$.

Let

$s_p \triangleq \sum\limits_{i=1}^{n}i^p$,

then

$s_p = \frac{n(n+1)^p - \sum\limits_{j=2}^{p}\binom{p}{j} s_{p-j+1}}{p+1}\quad\quad\quad(3)$

What we have here is a recursive generator capable of  generating  power summation formulas for virtually all integrer powers. Implemented in Omega CAS Explorer,  the figures below illustrate the awesomeness of this generator:

Fig. 2

Fig. 3

Chaplin or Leibniz ?

Before I do proofs via the 3-step mathematical induction, the classic Charlie Chaplin movie clip often comes to my mind. It nudges me to seek better ways. For example, to prove

$(k+1)(1^k+2^k+3^k+\dots+n^k)<(n+1)^{k+1}, \quad k, n \in N^+$,

Instead of applying the 3-step mathematical induction, let us look at Fig. 1.

Fig. 1

If A denotes the total area of the blue rectangles, Fig.1 shows

$A=1\cdot 1^k+1\cdot 2^k+1\cdot 3^k+\dots+1\cdot n^k = 1^k+2^k+3^k+\dots+n^k$.

It also shows that

A < the  area under the curve $x^k = \int\limits_{0}^{n+1}x^k\; dx$.

By the fundamental theorem of calculus,

$\int\limits_{0}^{n+1}x^k\;dx = \frac{x^{k+1}}{k+1}\bigg|_{0}^{n+1}=\frac{(n+1)^{k+1}}{k+1}$.

Therefore,

$1^k+2^k+3^k+\dots+n^k < \frac{(n+1)^{k+1}}{k+1}$,

i.e.,

$(k+1)(1^k+2^k+3^k+\dots+n^k)<(n+1)^{k+1}, \quad\quad k, n \in N^+$.

This proof suggests another inequality:

$(k+1)(1^k+2^k+3^k+\dots+n^k)>n^{k+1}, \quad\quad k, n \in N^+$.

a x + b y + c = 0 : Why It Applies to All Straight Lines

In the traditional teaching of Analytical Geometry, the governing equation for a straight line has the following five forms, along with limitations for the first four:

[1]  Point-Slope form: $y - y_1 = k (x-x_1)$ where $(x_1, y_1)$ is a point on the line, and $k$ is the slope. The limitation for this form is that it can not represent line perpendicular to the x-axis since it has no slope.

[2]  Slope-Intercept form: $y = k x + b$ where $k$ is the slope, $b$ is the intersect the line made on y-axis. Its limitation is that it can not represent line perpendicular to the x-axis.

[3] Two-Point form: $\frac{y-y1}{y_2 - y_1} = \frac{x - x_1}{x_2 - x_1}$ where $(x_1, y_1), (x_2, y_2)$ are two points on the line. However, this form can represent neither line perpendicular nor parallel to x-axis due to the fact when $x_1 = x_2$ or $y_1 = y_2$, the form breaks down from dividing by zero.

[4] Point-Intercept form: $\frac{x}{a} + \frac{y}{b} = 1$ where $a, b$ are the intersects the line made on x-axis and y-axis respectively, and $a\neq 0, b\neq0$. Again, this form can represent neither line perpenticular nor parallel to the x-axis. It does not work for any line that passes the point of origin either.

[5] General form: $a x +b y +c = 0 (a^2+b^2 \neq 0)$, this form can represent all lines.

Here I am presenting a proof to show [5] is indeed capable of representing all straight lines.

In a rectangular coordinate system, given two distinct points $(x_1, y_1), (x_2, y_2)$, and any point $(x, y)$ on the line connecting $(x_1, y_1)$ and $(x_2, y_2)$, the area of triangle with vertices $(x_1, y_1), (x_2, y_2)$ and $(x, y)$ must be zero!

Recall a theorem proved in my blog “Had Heron Known Analytic Geometry“, it means for such $(x_1, y_1), (x_2, y_2)$ and $(x, y)$,

$\left|\begin{array}{ccc} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x & y & 1 \end{array}\right|= 0$.

Therefore, we can define the line connecting two distinct points as a set of $(x, y)$ such that the area of the triangle with vertices $(x_1, y_1), (x_2, y_2)$ and $(x, y)$ is zero, mathematically written as

$A \triangleq \{ (x, y) | \left|\begin{array}{ccc} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x & y & 1 \end{array}\right|= 0, (x_1-x_2)^2+(y_1-y_2)^2 \neq 0\}$.

Since $\forall (x, y) \in A$,

$\left|\begin{array}{ccc} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x & y & 1 \end{array}\right|= x_1 y_2-x y_2-x_2 y_1+x y_1+x_2 y-x_1y =$

$(y-y_1)(x_1-x_2)-(x-x_1)(y_1-y_2)=0\quad\quad\quad\quad(1)$

is an algebraic representation of the line connecting two distinct points $(x_1, y_1)$ and $(x_2, y_2)$.

When $x_1=x_2$, (1) becomes

$(x-x_1)(y_1-y_2)=0$,

and when $x_1 = x_2, y_1-y_2 \neq 0$, we have

$x = x_1$,

a line perpendicular to the horizontal axis.

When $y_1=y_2$, (1) becomes

$y = y_1$,

a line parallel to the horizontal axis.

Evaluate (1) with $x_2=0, y_2=0$ yields:

$(y-y_1) x_1 -(x-x_1)y_1=0$.

Collecting terms in (1), and letting

$a=y_1-y_2$,

$b=x_2-x_1$,

$c=x_1y_2-x_2y_1$,

(1) can be expressed as

$ax + by + c = 0$.

In fact, we can prove the following theorem:

$B \triangleq \{ (x, y) | \exists a, b, a^2+b^2 \neq 0, a x +b y+c=0\} \implies A=B$.

To prove $A=B$, we need to show

$\forall (x, y) \in A \implies (x, y) \in B\quad\quad\quad\quad(2)$

$\forall (x, y) \in B \implies (x, y) \in A\quad\quad\quad\quad(3)$

We have already shown (2) by setting the values of $a, b$ and $c$ earlier.

We will prove (3) now:

$\forall (x_1, y_1), (x_2, y_2)$ and $(x, y) \in B$, we have

$\begin{cases}a x_1 + b y_1 +c =0 \\ a x_2 + b y_2 +c =0 \\ a x + b y+c =0\end{cases}$.

Written in matrix form,

$\left(\begin{array}{ccc} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x & y & 1 \end{array}\right)$ $\left(\begin{array}{rrr} a \\ b \\ c \end{array}\right)= 0$.

If

$\left|\begin{array}{ccc} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x & y & 1 \end{array}\right| \neq 0$,

then by Cramer’s rule,

$\left(\begin{array}{rrr} a \\ b \\ c \end{array}\right)$ is a column vector of zeros,

i.e.,

$a=b=c=0$

$a,b$ are not all zero.

Hence,

$\left|\begin{array}{ccc} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x & y & 1 \end{array}\right| = 0$

which implies:

$(x, y) \in A$.

The consequence of $A=B$ is that every point $(x, y)$ on a line connecting two distinct points satisfies equation $a x + b y + c =0$ for some $a, b (a^2+b^2\neq 0)$.

Stated differently,

$a x + b y +c = 0$ where $a, b$ are not all zero is the governing equation of any straight line.

In my previous two posts, “An Algebraic Proof of Heron’s Formula” and “An Alternative Derivation of Heron’s Formula,” I proved Heron’s formula for the area of a triangle with three given sides.  Based on Heron’s formula, we can now prove a theorem concerning the area of any triangle in a rectangle coordinate system, namely,

The area of a triangle with vertices at $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ in a rectangle coordinate system can be expressed as

$|\frac{1}{2}D|\quad\quad\quad\quad\quad\quad\quad\quad\quad(1)$

where $D$ is the determinant of matrix:

$\left(\begin{array}{ccc} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{array}\right)$

I offer the following proof:

Fig. 1

By Heron’s formula, the area of triangle in Fig. 1

$A=\sqrt{s (s-a) (s-b) (s-c)}\quad\quad\quad\quad\quad(2)$

where $a, b, c$ are three sides of the triangle and, $s=(a+b+c)/2$.

Therefore,

$A^2=s(s-a)(s-b)(s-c).$

where

$a^2=(x_2-x_3)^2+(y_2-y_3)^2,$

$b^2=(x_1-x_3)^2+(y_1-y_3)^2,$

$c^2=(x_1-x_2)^2+(y_1-y_2)^2.$

Let $B=|\frac{1}{2} D|$,  we have

$B^2 ={|\frac{1}{2} D|}^2=(\frac{1}{2}D)^2.$

Compute $A^2-B^2$ using Omega CAS Explorer (see Fig. 2) , the result shows

$A^2-B^2=0.$

Fig. 2

Since $A> 0, B\geq 0$ implies

$A+B >0,$

$A^2-B^2=(A-B)(A+B)=0$ implies

$A-B=0$,

i.e.,

$A = B.$

Hence, (1) and (2) are equivalent.

I would like to learn any other alternative proof.

Higham’s Parametric Curve

Fig. 1

The parametric curve (see Fig. 1) has always intrigued me. It was to my delight to finally find its equations from “MATLAB guide” written by the Highams:

$x = \int\limits_{0}^{t} \sin {\omega^2}\; d\omega$

$y = \int\limits_{0}^{t} \cos {\omega^2}\; d\omega$

Below is the curve plotted by Omega CAS Explorer:

Fig. 2

After several failed trials, I realized that ‘nticks’ must be provided in ‘plot2d’ in order to produced the image correctly.

Next, I tried ‘draw2d’ function (see Fig. 3), but the tics and numbers are too close to the image of the curve.

Fig. 3

To better position the image,  I specified ‘xrange’ and ‘yrange’ to put more space between the image and the tics and numbers. Cropping the resulting image to obtain the Fig. 1 at the top of this post.

Fig. 4

I would like to ask all the maxima Jedis out there,

Without specify ‘xrange’ and ‘yrange’, is there an option that I can set to turn off the tics and numbers ?