Monthly Archives: August 2017

A Case of Pre-FTC Definite Integration

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Fig. 1

By recalling the general formula for power summation derived in “Little Bird and a Recursive Generator“, namely

\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}\quad\quad\quad(1)

We can obtain the result

\int\limits_{a}^{b}x^{p} dx = {{b^{p+1}-a^{p+1} }\over {p+1}}

where b \ge  a \ge 0 and p \in Z_{0}^{+}, without the Fundamental Theorem of Calculus.

To this end, we divide the interval [0, b] into n sub-intervals of equal length as shown in Fig. 1. Let s_{n} denotes the sum of the areas of the rectangles. The value s_{n} tends to as n increases, is the definite integral of x^{p} from x=0 to x=b, i.e.,

\int\limits_{0}^{b} x^{p} dx = \lim\limits_{n \to \infty} s_{n}.

Since

s_{n} = \sum\limits_{i=1}^{n-1}\frac{b}{n}(i\frac{b}{n})^p= \frac{b^{p+1}}{n^{p+1}}\sum\limits_{i=1}^{n-1}i^p

It follows from (1) that

s_{n} = \frac{b^{p+1}}{n^{p+1}}\cdot\frac{(n-1)n^{p}- \sum\limits_{j=2}^{p}\binom{p}{j}\sum\limits_{i=1}^{n-1}i^{p-j+1}  }{p+1}

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

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

For  0 \leq  p < 2, s_{n} reduces to

s_{n} = \frac{b^{p+1}(1-\frac{1}{n})}{p+1},

and the result immediately follows:

\int\limits_{0}^{b}x^{p} dx = \lim\limits_{n \to \infty} s_{n}=\lim \limits_{n \to \infty}\frac{b^{p+1}(1-\frac{1}{n})}{p+1}=\frac{b^{p+1}}{p+1}.

For p \ge 2, we let

t_{n} = \frac {\sum\limits_{i=1}^{n-1}i^{p-j+1}}{n^{p+1}}.

Clearly,

0 < t_{n} < \frac{(n-1) n^{p-j+1}}{n^{p+1}}=\frac{n-1}{n^{j}}.

Since j \ge 2,

\lim\limits_{n \to \infty}\frac{n-1}{n^j}=0.

i.e.,

\lim\limits_{n \to \infty} t_{n} = 0.

Therefore,

\int\limits_{0}^{b}x^{p} dx = \lim\limits_{n \to \infty} s_{n}=\lim\limits_{n \to \infty} \frac{b^{p+1}}{p+1}((1-\frac{1}{n})-\sum\limits_{j=2}^{p}\binom{p}{j}t_{n})

=\frac{b^{p+1}}{p+1}(\lim\limits_{n \to \infty}{(1-\frac{1}{n})}-\sum\limits_{j=2}^{p}\binom{p}{j}\lim\limits_{n \to \infty}{t_{n}})=\frac{b^{p+1}}{p+1}.

Applying this result to the area from 0 to a  we have

\int\limits_{0}^{a}x^{p} dx = \frac {a^{p+1}}{p+1},

and by subtraction of the areas,

\int\limits_{a}^{b}x^{p} dx = \int\limits_{0}^{b}x^{p} dx -\int\limits_{0}^{a}x^{p} dx = \frac {b^{p+1}-a^{p+1}}{p+1}.

 

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Pumpkin Pi

pi.jpg

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)

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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)