Restate Feynman's "Great Identity"

In a letter dated July 7, 1948, Richard Feynman, a Nobel laureate in physics (1965) made a claim:

I am the possessor of a swanky new scheme to do each problem in terms of one with one less energy denominator. It is based on the great identity $\frac{1}{ab} = \int \limits_{0}^{1} \frac{1}{(ax+b(1-x))^2}\;dx$ .

(The Beat of a Different Drum: The Life and Science of Richard Feynman, by Jagdish Mehra, page 262)

Assuming non-zero constants $a, b$ are both real but otherwise arbitrary, let’s check the validity of Feynman’s “great identity”.

If $a \ne b$ ,

$\int \frac{1}{(ax+b(1-x))^2}\;dx$

$= \int \frac{1}{((a-b)x+b)^2}\;dx$

$= \int \frac{1}{a-b}\cdot \frac{a-b}{((a-b)x+b)^2}\;dx$

$= \frac{1}{a-b}\int \frac{((a-b)x+b)'}{((a-b)x+b)^2}\;dx$

$= \frac{1}{a-b}\cdot \frac{-1}{(a-b)x+b}$ .

Using Leibniz’s rule,

$\int\limits_{0}^{1}\frac{1}{(ax+b(1-x))^2}\;dx = \frac{1}{a-b}\cdot\frac{-1}{(a-b)x+b}\bigg| _{0}^{1} =\frac{1}{a-b}(\frac{-1}{a}-\frac{-1}{b})=\frac{1}{ab}$ .

When $a=b$ ,

$\int \frac{1}{(ax+b(1-x))^2}\;dx = \int \frac{1}{b^2}\;dx =\frac{1}{b^2}x$

and,

$\int \limits_{0}^{1}\frac{1}{(ax+b(1-x))^2}\;dx = \frac{1}{b^2}x\bigg|_{0}^{1} = \frac{1}{b^2}=\frac{1}{ab}$ .

All is as Feynman claimed:

$\int\limits_{0}^{1}\frac{1}{(ax+b(1-x))^2}\;dx = \frac{1}{ab}\quad\quad\quad(\star)$

There is something amiss:

If $a$ and $b$ have opposite signs i.e., $ab<0$ then the right hand side of ( $\star$ ) is negative. But the integrand is squared so the integral on the left hand side of ( $\star$ ) is never negative, no matter what $a$ and $b$ may be.

Let’s figure it out !

In its full glory, Leibniz’s rule we used to obtain $(\star)$ is

If the real-valued function $F$ on an open interval $I$ in $R$ has the continuous derivative $f$ and $a, b \in I$ then $\int\limits_{a}^{b} f(x)\; dx = F(b)-F(a)$ .

Essentially, the rule requires the integrand $f$ to be a continuous function on an open interval that contains $a$ and $b$ .

Solving $ax+b(1-x)=0$ for $x$ yields

$x = \frac{b}{b-a}$ ,

the singularity of integrand $\frac{1}{(ax+(1-x))^2}$ in $(\star)$ .

For $ab<0$ , we consider the following two cases:

Case (1-1) $(a>0, b<0) \implies (a>0, b<0, b-a<0)\implies (\frac{b}{b-a}>0$ ,

$\frac{b}{b-a} = \frac{b-a+a}{b-a} = 1+\frac{a}{b-a}<1)\implies 0<\frac{b}{b-a}<1$

Case (1-2) $(a<0, b>0) \implies (a<0, b>0, b-a>0) \implies (\frac{b}{b-a}>0$ ,

$\frac{b}{b-a} = \frac{b-a+a}{b-a} = 1+\frac{a}{b-a}<1) \implies 0<\frac{b}{b-a}<1$

From both cases, we see that

when $ab<0, \frac{1}{(ax+b(1-x))^2}$ has a singularity in $(0, 1) \implies \frac{1}{(ax+b(1-x))^2}$ is not continuous in $(0, 1)$ .

Applying Leibniz’s rule to $\int\limits_{0}^{1} \frac{1}{(ax+b(1-x))^2}\;dx$ regardless of integrand’s singularity thus ensured an outcome of absurdity.

However, $ab>0$ paints a different picture.

Since $ab \implies (a>0, b>0)$ or $(a<0, b<0)$ , we have

Case (2-0) $a=b \implies \frac{1}{(ax+b(1-x))^2}=\frac{1}{b^2} \implies$ no singularity

Case (2-1) $(a>0, b>0, a>b) \implies (a>0, b>0, b-a<0) \implies \frac{b}{b-a}<0$

Case (2-2) $(a>0, b>0, a<b) \implies (a>0, b>0, b-a>0)$

$\implies \frac{b}{b-a} = \frac{b-a+a}{b-a}=1+\frac{a}{b-a}>1$

Case (2-3) $(a<0, b<0, a>b) \implies (a<0, b<0, b-a<0) \implies \frac{b}{b-a}=1+\frac{a}{b-a}>1$

Case (2-4) $(a<0, b<0, a<b) \implies (a<0, b<0, b-a>0) \implies \frac{b}{b-a} <0$

All cases show that when $ab>0$ , the integrand has no singularity in $(0,1)$ .

It means that $\frac{1}{(ax+b(1-x))^2}$ is continuous in $(0, 1)$ and therefore, Leibniz’s rule applies.

So let’s restate Feynman’s “great identity”:

$a\cdot b > 0 \iff \frac{1}{ab} = \int\limits_{0}^{1}\frac{1}{(ax+b(1-x))^2}\;dx$

Exercise-1 Evaluate $\int\limits_{0}^{1}\frac{1}{(ax+b(1-x))^2}\;dx$ using Omega CAS Explorer. For example,

(hint : for $ab>0$ , specify $a > b$ or $b>a$ )

Oh! Matryoshka!

Leave a reply

Given polynomial $f(x) = a_0 + a_1 x+a_2 x^2 + ... + a_{n-1}x^{n-1}+a_n x^n$ , we wish to evaluate integral

$\int \frac{f(x)}{(x-a)^p}\;dx, \quad p \in N^+\quad\quad\quad(1)$

When $p = 1$ ,

$\int \frac{f(x)}{x-a} \;dx= \int \frac{f(x)-f(a)+f(a)}{x-a}\;dx$

$= \int \frac{f(x)-f(a)}{x-a}\;dx + \int \frac{f(a)}{x-a}\;dx$

$=\int \frac{f(x)-f(a)}{x-a}\;dx + f(a)\cdot \log(x-a)$ .

Since

$f(x) = a_0 + a_1x + a_2x^2 + ... + a_{n-1}x^{n-1} + a_n x^n$

and

$f(a) = a_0 + a_1 a + a_2 a^2 + ... + a_{n-1}a^{n-1} + a_n a^n$

It follows that

$f(x)-f(a) = a_1(x-a) + a_2(x^2-a^2) + ... + a_{n-1}(x^{n-1}-a^{n-1}) + a_n (x^n-a^n)$ .

That is

$f(x)-f(a) = \sum\limits_{k=1}^{n}a_k(x^k-a^k)$

By the fact (see “Every dog has its day“) that

$x^k-a^k =(x-a)\sum\limits_{i=1}^{k}x^{k-i}a^{i-1}$ ,

we have

$f(x)-f(a) = \sum\limits_{k=1}^{n}a_k(x-a)\sum\limits_{i=1}^{k}x^{k-i}a^{i-1}=(x-a)\sum\limits_{k=1}^{n}(a_k\sum\limits_{i=1}^{k}x^{k-i}a^{i-1})$

or,

$\frac{f(x)-f(a)}{x-a}= \sum\limits_{k=1}^{n}(a_k\sum\limits_{i=1}^{k}x^{k-i}a^{i-1})\quad\quad\quad(2)$

Hence,

$\int\frac{f(x)}{x-a}\;dx = \int \sum\limits_{k=1}^{n}(a_k \sum\limits_{i=1}^{k}x^{k-i}a^{i-1})\;dx + f(a)\log(x-a)$

$=\sum\limits_{k=1}^{n}(a_k \sum\limits_{i=1}^{k}\int x^{k-i}a^{i-1}\; dx)+ f(a)\log(x-a)$

i.e.,

$\int \frac{f(x)}{x-a} = \sum\limits_{k=1}^{n}(a_k\sum\limits_{i=1}^{k}\frac{x^{k-i+1}}{k-i+1}a^{i-1})+ f(a)\log(x-a)$

Let us now consider the case when $p>1$ :

$\int \frac{f(x)}{(x-a)^p}\; dx$

$=\int \frac{f(x)-f(a)+f(a)}{(x-a)^p}\;dx$

$=\int \frac{f(x)-f(a)}{(x-a)^p} + \frac{f(a)}{(x-a)^p}\;dx$

$=\int \frac{f(x)-f(a)}{(x-a)}\cdot\frac{1}{(x-a)^{p-1}} + \frac{f(a)}{(x-a)^p}\;dx$

$= \int \frac{f(x)-f(a)}{x-a}\cdot\frac{1}{(x-a)^{p-1}}\;dx + \int\frac{f(a)}{(x-a)^p}\; dx$

$\overset{(2)}{=}\int \frac{g(x)}{(x-a)^{p-1}}\;dx + \frac{f(a)(x-a)^{1-p}}{1-p}$

where

$g(x) = \frac{f(x)-f(a)}{x-a}=\sum\limits_{k=1}^{n}(a_k\sum\limits_{i=1}^{k}x^{k-i}a^{i-1})$ , a polynomial of order $n-1$ .

What emerges from the two cases of $p$ is a recursive algorithm for evaluating (1):

Given polynomial $f(x) = \sum\limits_{k=0}^{n} a_k x^k$ ,

$\int \frac{f(x)}{(x-a)^p} \;dx, \; p \in N^+= \begin{cases}p=1: \sum\limits_{k=1}^{n}(a_k\sum\limits_{i=1}^{k}\frac{x^{k-i+1}}{k-i+1}a^{i-1})+ f(a)\log(x-a) \\p>1: \int \frac{g(x)}{(x-a)^{p-1}}\;dx + \frac{f(a)(x-a)^{1-p}}{1-p}, \\ \quad\quad\quad g(x) = \frac{f(x)-f(a)}{x-a}=\sum\limits_{k=1}^{n}(a_k\sum\limits_{i=1}^{k}x^{k-i}a^{i-1}). \end{cases}$

Exercise-1 Optimize the above recursive algorithm (hint: examine how it handles the case when $f(x)=0$ )

Integration of Trigonometric Expressions

Leave a reply

We will introduce an algorithm for obtaining indefinite integrals such as

$\int \frac{(1+\sin(x))}{\sin(x)(1+\cos(x))}\;dx$

or, in general, integral of the form

$\int R(\sin(x), \cos(x))\;dx\quad\quad\quad(1)$

where $R$ is any rational function $R(p, q)$ , with $p=\sin(x), q=\cos(x)$ .

Let

$t = \tan(\frac{x}{2})\quad\quad(2)$

Solving (2) for $x$ , we have

$x = 2\cdot\arctan(t)\quad\quad\quad(3)$

which provides

$\frac{dx}{dt} = \frac{2}{1+t^2}\quad\quad\quad(4)$

and,

$\sin(x) =2\sin(\frac{x}{2})\cos(\frac{x}{2})\overset{\cos^(\frac{x}{2})+\sin^2(\frac{x}{2})=1}{=}\frac{2\sin(\frac{x}{2})\cos(\frac{x}{2})}{\cos^2(\frac{x}{2})+\sin^2(\frac{x}{2})}=\frac{2\frac{\sin(\frac{x}{2})}{\cos(\frac{x}{2})}}{1+\frac{\sin^2(\frac{x}{2})}{\cos^2(\frac{x}{2})}}=\frac{2\tan(\frac{x}{2})}{1+\tan^2(\frac{x}{2})}$

yields

$\sin(x) = \frac{2 t}{1+t^2}\quad\quad\quad(5)$

Similarly,

$\cos(x) = \cos^2(\frac{x}{2})-\sin^2(\frac{x}{2})=\frac{\cos^2(\frac{x}{2})-\sin^2(\frac{x}{2})}{\cos^2(\frac{x}{2})+\sin^2(\frac{x}{2})}=\frac{1+\frac{\sin^2(\frac{x}{2})}{\cos^2(\frac{x}{2})}}{1+\frac{\sin^2(\frac{x}{2})}{\cos^2(\frac{x}{2})}}=\frac{1-\tan^2(\frac{x}{2})}{1+\tan^2(\frac{x}{2})}$

gives

$\cos(x)=\frac{1-t^2}{1+t^2}\quad\quad\quad(6)$

We also have (see “Finding Indefinite Integrals” )

$\int f(x)\;dx \overset{x=\phi(t)}{=} \int f(\phi(t))\cdot\frac{d\phi(t)}{dt}\;dt$ .

Hence

$\int R(\cos(x), \sin(x))\;dx \overset{(2), (4), (5), (6)}{=} \int R(\frac{1-t^2}{1+t^2}, \frac{2t}{1+t^2})\cdot\frac{2}{1+t^2}\;dt$ ,

and (1) is reduced to an integral of rational functions in $t$ .

Example-1 Evaluate $\int \csc(x)\;dx$ .

Solution: $\csc(x) = \frac{1}{\sin(x)}\implies \int \csc(x)\;dx = \int \frac{1}{\sin(x)}\;dx$

$= \int \frac{1}{\frac{2t}{1+t^2}}\cdot\frac{2}{1+t^2}\;dt=\int\frac{1}{t}\;dt = \log(t) = \log(\tan(\frac{x}{2}))$ .

Example-2 Evaluate $\int \sec(x)\;dx$ .

Solution: $\sec(x) = \frac{1}{\cos(x)}\implies \int \sec(x)\; dx =\int \frac{1}{\cos(x)}\;dx$

$= \int \frac{1}{\frac{1-t^2}{1+t^2}}\cdot \frac{2}{1+t^2}\; dt=\int \frac{2}{1-t^2}\;dt=\int \frac{2}{(1+t)(1-t)}\;dt=\int \frac{1}{1+t} + \frac{1}{1-t}\;dt$

$=\int \frac{1}{1+t}\;dt - \int \frac{-1}{1-t}\;dt$

$=\log(1+t) -\log(1-t) =\log\frac{1+t}{1-t}=\log(\frac{1+\tan(\frac{x}{2})}{1-\tan(\frac{x}{2})})$ .

According to CAS (see Fig. 1),

Fig. 1

However, the two results are equivalent as a CAS-aided verification (see Fig. 2) confirms their difference is a constant (see Corollary 2 in “Sprint to FTC“).

Fig. 2

Exercise-1 According to CAS,

Show that it is equivalent to the result obtained in Example-1

Exercise-2 Try

$\int \frac{1}{\sin(x)+1}\;dx$

$\int \frac{1}{\sin(x)+\cos(x)}\;dx$

$\int \frac{1}{(2+\cos(x))\sin(x)}\;dx$

$\int \frac{1}{5+4\sin(x)}\;dx$

$\int \frac{1}{2\sin(x)-\cos(x)+5}\;dx$

and of course,

$\int \frac{1+\sin(x)}{\sin(x)(1+\cos(x))}\;dx$

Double Feature on Christmas Day

Leave a reply

Feature One (Pascal’s Identity):

$\binom{n-1}{r-1} + \binom{n-1}{r} = \binom{n}{r}\quad\quad\quad(1)$

Proof :

By definition,

$\binom{n-1}{r-1} + \binom{n-1}{r}$

$= \frac{(n-1)!}{(n-1-(r-1))!(r-1)!} + \frac{(n-1)!}{(n-1-r)!r!}$

$= \frac{(n-1)!}{(n-r)!(r-1)!} + \frac{(n-1)!}{(n-1-r)!r!}$

$=\frac{r(n-1)!}{(n-r)!r(r-1)!} + \frac{(n-1)!(n-r)}{(n-r)(n-r-1)!r!}$

$= \frac{r(n-1)!}{(n-r)!r!} + \frac{(n-1)!(n-r)}{(n-r)!r!}$

$=\frac{(r + n-r)(n-1)!}{(n-r)!r!}$

$=\frac{n(n-1)!}{(n-r)!r!}$

$= \frac{n!}{(n-r)!r!}$

$= \binom{n}{r}$ .

Feature Two (Binomial Theorem):

$(a+b)^n = \sum\limits_{r=0}^{n}\binom{n}{r}a^{n-r}b^r\quad\quad\quad(2)$

Proof :

$(a+b)^1 = a+b, \sum\limits_{r=0}^{1}a^{1-r}b^r = a+b \implies (a+b)^n=\sum\limits_{r=0}^{n}a^{n-r}b^r, n=1$ .

If $(a+b)^{k-1} = \sum\limits_{r=0}^{n}\binom{k-1}{r}a^{k-1-r}b^r$ then

$(a+b)^k = (a+b)(a+b)^{k-1} = (a+b)\sum\limits^{k-1}_{r=0}\binom{k-1}{r}a^{k-1-r}b^r$

$= \sum\limits^{k-1}_{r=0}\binom{k-1}{r}a^{k-r}b^r + \sum\limits^{k-1}_{r=0}\binom{k-1}{r}a^{k-1-r}b^{r+1}$

$=a^k + \sum\limits^{k-1}_{r=1}\binom{k-1}{r}a^{k-r}b^r +\sum\limits^{k-2}_{r=0}\binom{k-1}{r}a^{k-1-r}b^{r+1} +b^k$

$=a^k + \sum\limits^{k-1}_{j=1}\binom{k-1}{j}a^{k-j}b^j +\boxed{ \sum\limits^{k-2}_{r=0}\binom{k-1}{r}a^{k-1-r}b^{r+1}} +b^k$

$\boxed {j=r+1\implies r = j-1, r:0\rightarrow k-1\implies j:1 \rightarrow k-1}$

$=a^k + \sum\limits^{k-1}_{j=1}\binom{k-1}{j}a^{k-j}b^j+\sum\limits^{k-1}_{j=1}\binom{k-1}{j-1}a^{k-1-(j-1)}b^{(j-1)+1} +b^k$

$=a^k + \sum\limits^{k-1}_{j=1}(\boxed{\binom{k-1}{j}+\binom{k-1}{j-1}}) a^{k-j}b^j +b^k$

$\overset{(1)}{=}a^k + \sum\limits^{k-1}_{j=1}{\binom{k}{j}a^{k-j}b^j}+b^k$

$\overset{\binom{k}{0}=1, \binom{k}{k}=1}{=}\binom{k}{0}a^{k-0}b^0+\sum\limits^{k-1}_{j=1}\binom{k}{j} a^{k-j}b^j +\binom{k}{k} a^{k-k}b^k$

$= \sum\limits^k_{j=0}\binom{k}{j}a^{k-j}b^j$ .

Exercise-1 Prove $\binom{n}{1} + \binom{n}{2} + \dots + \binom{n}{n-1} + \binom{n}{n} = 2^n-1$ . hint: $2 = 1+1$

Exercise-2 Prove $1-\binom{n}{1} + \binom{n}{2} - \dots +(-1)^{n-1}\binom{n}{n-1} +(-1)^n = 0$ .

Exercise-3 Generate

I think I shall never see
A poem as lovely as a tree – Joyce Kilmer

hint: (1)

Deriving Lagrange's multiplier

Leave a reply

We wish to consider a special type of optimization problem:

Find the maximum (or minimum) of a function $f(x,y)$ subject to the condition $g(x,y)=0\quad\quad(1)$

If it is possible to solve $g(x)=0$ for $y$ so that it is expressed explicitly as $y=\psi(x)$ , by substituting $y$ in (1), it becomes

Find the maximum (or minimum) of a single variable function $f(x, \psi(x))$ .

In the case that $y$ can not be obtained from solving $g(x,y)=0$ , we re-state the problem as:

Find the maximum (or minimum) of a single variable function $z=f(x,y)$ where $y$ is a function of $x$ , implicitly defined by $g(x, y)=0\quad\quad\quad(2)$

Following the traditional procedure of finding the maximum (or minimum) of a single variable function, we differentiate $z$ with respect to $x$ :

$\frac{dz}{dx} = f_x(x,y) + f_y(x,y)\cdot \frac{dy}{dx}\quad\quad\quad(3)$

Similarly,

$g_x(x,y) + g_y(x,y)\cdot \frac{dy}{dx}=0\quad\quad\quad(4)$

By grouping $g(x, y)=0$ and (3), we have

$\begin{cases} \frac{dz}{dx}= f_x(x, y)+g_x(x, y)\cdot \frac{dy}{dx}\\ g(x,y) = 0\end{cases}\quad\quad\quad(5)$

The fact that $\frac{dz}{dx}= 0$ at any stationary point $x^*$ means for all $(x^*, y^*)$ where $g(x^*, y^*)=0$ ,

$\begin{cases} f_x(x^*, y^*)+g_x(x^*, y^*)\cdot \frac{dy}{dx}\vert_{x=x^*}=0 \\ g(x^*,y^*) = 0\end{cases}\quad\quad\quad(6)$

If $g_y(x^*,y^*) \ne 0$ then from (4),

$\frac{dy}{dx}\vert_{x=x^*} = \frac{-g_x(x^*, y^*)}{g_y(x^*, y^*)}$

Substitute it into (6),

$\begin{cases} f_x(x^*, y^*)+f_y(x^*, y^*)\cdot (\frac{-g_x(x^*, y^*)}{g_y(x^*, y^*)})=f_x(x^*, y^*)+g_x(x^*, y^*)\cdot (\frac{-f_y(x^*, y^*)}{g_y(x^*, y^*)})\\ g(x^*,y^*) = 0\end{cases}\quad\quad\quad(7)$

Let $\lambda = \frac{-f_y(x^*, y^*)}{g_y(x^*, y^*)}$ , we have

$f_y(x^*, y^*) + \lambda g_y(x^*, y^*) =0\quad\quad\quad(8)$

Combining (7) and (8) gives

$\begin{cases} f_x(x^*, y^*)+\lambda g_x(x^*, y^*) = 0 \\ f_y(x^*, y^*)+\lambda g_y(x^*, y^*)=0 \\ g(x^*, y^*) = 0\end{cases}$

It follows that to find the stionary points of $z$ , we solve

$\begin{cases} f_x(x, y)+\lambda g_x(x, y) = 0 \\ f_y(x, y)+\lambda g_y(x, y)=0 \\ g(x, y) = 0\end{cases}\quad\quad\quad(9)$

for $x, y$ and $\lambda$ .

This is known as the method of Lagrange’s multiplier.

Let $F(x,y,\lambda) = f(x,y) + \lambda g(x,y)$ .

Since

$F_x(x,y,\lambda) = f_x(x,y) + \lambda g_x(x,y)$ ,

$F_y(x,y,\lambda)=f_y(x,y) + \lambda g_y(x,y)$ ,

$F_{\lambda}(x,y,\lambda) = g(x, y)$ ,

(9) is equivalent to

$\begin{cases} F_x(x, y, \lambda)=0 \\ F_y(x,y,\lambda)=0 \\ F_{\lambda}(x, y) = 0\end{cases}\quad\quad\quad(10)$

Let’s look at some examples.

Example-1 Find the minimum of $f(x, y) = x^2+y^2$ subject to the condition that $x+y=4$

Let $F(x, y, \lambda) = x^2+y^2+\lambda(x+y-4)$ .

Solving

$\begin{cases}F_x=2x-\lambda=0 \\ F_y = 2y-\lambda = 0 \\ F_{\lambda} = x+y-4=0\end{cases}$

for $x, y, \lambda$ gives $x=y=2, \lambda=4$ .

When $x=2, y=2, x^2+y^2=2^2+2^2=8$ .

$\forall (x, y) \ne (2, 2), x+y=4$ , we have

$(x-2)^2 + (y-2)^2 > 0$ .

That is,

$x^2-4x+4 + y^2-4y+4 = x^2+y^2-4(x+y)+8 \overset{x+y=4}{=} x^2+y^2-16+8>0$ .

Hence,

$x^2+y^2>8, (x,y) \ne (2,2)$ .

The target function $x^2+y^2$ with constraint $x+y=4$ indeed attains its global minimum at $(x, y) = (2, 2)$ .

I first encountered this problem during junior high school and solved it:

$(x-y)^2 \ge 0 \implies x^2+y^2 \ge 2xy$

$x+y=4\implies (x+y)^2=16\implies x^2+2xy +y^2=16\implies 2xy=16-(x^2+y^2)$

$x^2+y^2 \ge 16-(x^2+y^2) \implies x^2+y^2 \ge 8\implies z_{min} = 8$ .

I solved it again in high school when quadratic equation is discussed:

$x+y=4 \implies y =4-x$

$z=x^2+y^2 \implies z = x^2+(4-x)^2 \implies 2x^2-8x+16-z=0$

$\Delta = 64-4 \cdot 2\cdot (16-z) \ge 0 \implies z \ge 8\implies z_{min} = 8$

In my freshman calculus class, I solved it yet again:

$x+y=4 \implies y=4-x$

$z = x^2+(4-x)^2$

$\frac{dz}{dx} = 2x+2(4-x)(-1)=2x-8+2x=4x-8$

$\frac{dz}{dx} =0 \implies x=2$

$\frac{d^2 z}{dx^2} = 4 > 0 \implies x=2, z_{min}=2^2+(4-2)^2=8$

Example-2 Find the shortest distance from the point $(1,0)$ to the parabola $y^2=4x$ .

We minimize $f = (x-1)^2+y^2$ where $y^2=4x$ .

If we eliminate $y^2$ in $f$ , then $f = (x-1)^2+4x$ . Solving $\frac{df}{dx} = 2x+2=0$ gives $x=-1$ , Clearly, this is not valid for it would suggest that $y^2=-4$ from $y^2=4x$ , an absurdity.

By Lagrange’s method, we solve

$\begin{cases} 2(x-1)-4\lambda=0 \\2y\lambda+2y = 0 \\y^2-4x=0\end{cases}$

Fig. 1

The only valid solution is $x=0, y=0, k=-\frac{1}{2}$ . At $(x, y) = (0, 0), f=(0-1)^2+0^2=1$ . It is the global minimum:

$\forall (x, y) \ne (0, 0), y^2=4x \implies x>0$ .

$(x-1)^2+y^2 \overset{y^2=4x}{=}(x-1)^2+4x=x^2-2x+1+4x=x^2+2x+1\overset{x>0}{>}1=f(0,0)$

Example-3 Find the shortest distance from the point $(a, b)$ to the line $Ax+By+C=0$ .

We want find a point $(x_0, y_0)$ on the line $Ax+By+C=0$ so that the distance between $(a, b)$ and $(x_0, y_0)$ is minimal.

To this end, we minimize $(x_0-a)^2+(y_0-b)^2$ where $Ax_0+By_0+C=0$ (see Fig. 2)

Fig. 2

We found that

$x_0=\frac{aB^2-bAB-AC}{A^2+B^2}, y_0=\frac{bA^2-aAB-BC}{A^2+B^2}$

and the distance between $(a, b)$ and $(x_0, y_0)$ is

$\frac{|Aa+Bb+C|}{\sqrt{A^2+B^2}}\quad\quad\quad(11)$

To show that (11) is the minimal distance, $\forall (x, y) \ne (x_0, y_0), Ax+By+C=0$ .

Let $d_1 = x-x_0, d_2=y-y_0$ , we have

$x = x_0 + d_1, y=y_0 + d_2, d_1 \ne 0, d_2 \ne 0$ .

Since $Ax+By+C=0$ ,

$A(x_0+d_1)+B(y_0+d_2)+C=Ax_0+Ad_1+By_0+Bd_2+C=0$

That is

$Ax_0+By_0+C+Ad_1+Bd_2=0$ .

By the fact that $Ax_0+By_0+C=0$ , we have

$Ad_1 + Bd_2 =0\quad\quad\quad(12)$

Compute $(x-a)^2+(y-b)^2 - ((x_0-a)^2+(y_0-b)^2)$ (see Fig. 3)

Fig. 3

yields

$\boxed{-\frac{2d_2BC}{B^2+A^2}-\frac{2d_1AC}{B^2+A^2}}+[\frac{d_2^2B^2}{B^2+A^2}]-\underline{\frac{2bd_2B^2}{B^2+A^2}}+(\frac{d_1^2B^2}{B^2+A^2})-\frac{2ad_2AB}{B^2+A^2}-\underline{\frac{2bd_1AB}{B^2+A^2}}+[\frac{d_2A^2}{B^2+A^2}]+(\frac{d_1^2A^2}{B^2+A^2})-\frac{2ad_1A^2}{B^2+A^2}$

After some rearrangement and factoring, it becomes

$\frac{-2C}{A^2+B^2}(Ad_1+Bd_2)+\frac{-2B}{A^2+B^2}(Ad_1+Bd_2)+\frac{-2A}{A^2+b^2}(Ad_1+Bd_2) + d_1^2+d_2^2$

By (12), it reduces to

$d_1^2 + d_2^2$ .

This is clearly a positive quantity. Therefore,

$\forall (x, y) \ne (x_0, y_0), Ax+By+C=0 \implies (x-a)^2+(y-b)^2> (x_0-1)^2+(y_0-b)^2$

A Feynmanistic Derivation

Leave a reply

The operator $\Delta$ introduced in my previous post has many properties. The most notable ones are:

(1) $\Delta c = 0, c$ is a constant

(2) $\Delta(c\cdot f(x)) = c\cdot \Delta f(x)$

(3) $\Delta (f_1(x) + f_2(x) + ... + f_n(x) ) = \Delta f_1(x) + \Delta f_2(x) + ... + \Delta f_n(x)$

(4) $\Delta(f(x)\cdot g(x)) = (f(x)+1)\Delta g(x) + g(x) \Delta f(x)$

(5) $\Delta\frac{f(x)}{g(x)} = \frac{g(x)\Delta f(x) - f(x)\Delta g(x)}{g(x)(g(x)+1)}$

We will add to this list an additonal property:

(6) $\Delta^{p} x^m = 0, p > m$

To derive (6), let us consider

$\Delta^{k-1} x^m, k=2,3,...m$ .

When $k=2$ ,

$\Delta^{k-1} x^m = \Delta^{2-1}x^m=\Delta x^m$

$= (x+1)^m-x^m$

$= mx^{m-1} + \binom{m}{2}x^{m-1} +...$

$= mx^{m-1} + \frac{m(m-1)}{2!}x^{m-2} + ...$

$= mx^{m-1} + 1\cdot \frac{m(m-1)}{2!}x^{m-2} + ...$

$=m x^{m-(2-1)} + (2-1)\cdot \frac{m(m-1)}{2!}x^{m-2}+ ...$

$= (m-k+2) x^{m-(k-1)} + (k-1)\cdot \frac{m(m-k+1)}{2!}x^{m-k} + ...$ .

When $k=3$ ,

$\Delta^{k-1} x^m = \Delta^{3-1} x^m = \Delta^2 x^m = \Delta (\Delta x^m)$

$= m((m-1)x^{m-2} + \binom{m-1}{2}x^{m-3} + ... ) + \binom{m}{2}((m-2) x^{m-3} + ... )$

$= m(m-1)x^{m-2} + m(\binom{m-1}{2} + \binom{m}{2}(m-2))x^{m-3} + ...$

$= m(m-1)x^{m-2} +(\frac{m(m-1)(m-2)}{2!} + \frac{m(m-1)(m-2)}{2!})x^{m-3} + ...$

$= m(m-1)x^{m-2} + 2\cdot \frac{m(m-1)(m-2)}{2!}x^{m-3} + ...$

$= m(m-1)x^{m-(3-1)} + (3-1) \cdot \frac{m(m-1)(m-2)}{2!}x^{m-3} + ...$

$= m(m-k+2)x^{m-(k-1)} + (k-1) \cdot \frac{m(m-1)(m-k+1)}{2!} x^{m-k}+ ...$ .

When $k = 4$ ,

$\Delta^{k-1} x^m = \Delta^{4-1}x^m = \Delta^3 x^m = \Delta(\Delta^2 x^m)$

$=m(m-1)((m-2)x^{m-3} + \binom{m-2}{2}x^{m-4} + ...) + 2\cdot\frac{m(m-1)(m-2)}{2!}((m-3)x^{m-4}+...)$

$=m(m-1)(m-2)x^{m-3} + \frac{m(m-1)(m-2)(m-3)}{2!}x^{m-4}+2\cdot \frac{m(m-1)(m-2)(m-3)}{2!}x^{m-4} + ...$

$= m(m-1)(m-2)x^{m-3} + 3\cdot \frac{m(m-1)(m-2)(m-3)}{2!}x^{m-4} + ...$

$= m(m-1)(m-2)x^{m-(4-1)} + (4-1) \cdot \frac{m(m-1)(m-2)(m-3)}{2!}x^{m-4} + ...$

$= m(m-1)(m-k+2)x^{m-(k-1)} + (k-1) \cdot \frac{m(m-1)(m-2)(m-k+1)}{2!} x^{m-k} + ...$ .

When $k = 5$ ,

$\Delta^{k-1}x^m = \Delta^{5-1} x^m = \Delta^4 x^m = \Delta(\Delta^3 x^m)$

$= m(m-1)(m-2)((m-3)x^{m-4} + \binom{m-3}{2} x^{m-5}+ ...)$

$+ 3\cdot \frac{m(m-1)(m-2)(m-3)}{2!}((m-4)x^{m-5} +...)$

$= m(m-1)(m-2)(m-3)x^{m-4} +\frac{m(m-1)(m-2)(m-3)(m-4)}{2!}x^{m-5}$

$+ 3\cdot\frac{m(m-1)(m-2)(m-3)(m-4)}{2!}x^{m-5} + ...$

$= m(m-1)(m-2)(m-3)x^{m-4} + 4\cdot\frac{m(m-1)(m-2)(m-3)(m-4)}{2!} x^{m-5} + ...$

$= m(m-1)(m-2)(m-3)x^{m-(5-1)} + (5-1)\cdot\frac{m(m-1)(m-2)(m-3)(m-4)}{2!}x^{m-5} + ...$

$= m(m-1)(m-2)(m-k+2)x^{m-(k-1)} + (k-1)\cdot\frac{m(m-1)(m-2)(m-3)(m-k+1)}{2!}x^{m-k} + ...$

$\vdots$

When $k=m$ ,

$\Delta^{k-1} x^m = \Delta^{m-1}x^m$

$= m(m-1)(m-2)(m-3) \cdots 2 x + (m-1)\cdot \frac{m(m-1)(m-2)(m-3) \cdots 1}{2!} + 0$

$= m! x + (m-1)\cdot \frac{m!}{2!}$

$= m! x + (m-1)\cdot \frac{m!}{2}$

$= m!(x+\frac{m-1}{2})$ .

Therefore,

$\Delta^m x^m = \Delta(\Delta^{m-1}x^m)=\Delta(m!(x+\frac{m-1}{2})= m! \Delta(x+\frac{m+1}{2}))= m! (\Delta x + \Delta(\frac{m-1}{2}))= m! (1+0)= m!.$

It follows that

$p>m \implies p-m>0 \implies p-m \ge 1 \implies \Delta^p x^m = \Delta^{p-m}(\Delta^{m}x^m) = \Delta^{p-m}m!= 0.$

Introducing Operator Delta

1 Reply

The $r^{th}$ order finite difference of function $f(x)$ is defined by

$\Delta^r f(x) = \begin{cases} f(x+1)-f(x), r=1\\ \Delta(\Delta^{r-1}f(x)), r > 1\end{cases}$

From this definition, we have

$\Delta f(x) = \Delta^1 f(x) = f(x+1)-f(x)$

and,

$\Delta^2 f(x) = \Delta (\Delta^{2-1} f(x))$

$= \Delta (\Delta f(x))$

$= \Delta( f(x+1)-f(x))$

$= (f(x+2)-f(x+1)) - (f(x+1)-f(x))$

$= f(x+2)-2f(x)+f(x+1)$

as well as

$\Delta^3 f(x) = \Delta (\Delta^2 f(x))$

$= \Delta (f(x+2)-2f(x)+f(x+1))$

$= (f(x+3)-2f(x+1)+f(x+2)) - (f(x+2)-2f(x)+f(x+1))$

$= f(x+3)-3f(x+2)+3f(x+1)-f(x)$

The function shown below generates $\Delta^r f(x), r:1\rightarrow 5$ (see Fig. 1).

delta_(g, n) := block(
    local(f),
    
    define(f[1](x), 
           g(x+1)-g(x)),

    for i : 2 thru n do (
        define(f[i](x), 
               f[i-1](x+1)-f[i-1](x))
    ),

    return(f[n])
);

Fig. 1

Compare to the result of expanding $(f(x)-1)^r=\sum\limits_{i=0}^r(-1)^i \binom{r}{i} f(x)^{r-i}, r:1\rightarrow 5$ (see Fig. 2)

Fig. 2

It seems that

$\Delta^r f(x) = \sum\limits_{i=0}^r(-1)^i \binom{r}{i} f(x+r-i)\quad\quad\quad(1)$

Lets prove it!

We have already shown that (1) is true for $r= 1, 2, 3$ .

Assuming (1) is true when $r=k-1 \ge 4$ :

$\Delta^{k-1} f(x) = \sum\limits_{i=0}^{k-1}(-1)^i \binom{r}{i} f(x+k-1-i)\quad\quad\quad(2)$

When $r=k$ ,

$\Delta^k f(x) = \Delta(\Delta^{k-1} f(x))$

$\overset{(2)}{=}\Delta (\sum\limits_{i=0}^{k-1}(-1)^i \binom{k-1}{i}f(x+k-1-i))$

$=\sum\limits_{i=0}^{k-1}(-1)^i \binom{k-1}{i}f(x+1+k-1-i)-\sum\limits_{i=0}^{k-1}(-1)^i \binom{k-1}{i}f(x+k-1-i)$

$=(-1)^0 \binom{k-1}{0}f(x+k-0)$

$+\sum\limits_{i=1}^{k-1}(-1)^i \binom{k-1}{i}f(x+k-i)-\sum\limits_{i=0}^{k-2}(-1)^i \binom{k-1}{i}f(x+k-1-i)$

$-(-1)^{k-1}\binom{k-1}{k-1}f(x+k-1-(k-1))$

$\overset{\binom{k-1}{0} = \binom{k-1}{k-1}=1}{=}$

$f(x+k)+ \sum\limits_{i=1}^{k-1}(-1)^i \binom{k-1}{i}f(x+k-i)-\sum\limits_{i=0}^{k-2}(-1)^i \binom{k-1}{i}f(x+k-1-i) -(-1)^{k-1}f(x)$

$=f(x+k)+ \sum\limits_{i=1}^{k-1}(-1)^i \binom{k-1}{i}f(x+k-i)+\sum\limits_{i=0}^{k-2}(-1)^{i+1}\binom{k-1}{i}f(x+k-1-i) -(-1)^{k-1}f(x)$

$\overset{j=i+1, i:0 \rightarrow k-2\implies j:1 \rightarrow k-1}{=}$

$f(x+k)+ \sum\limits_{i=1}^{k-1}(-1)^i\binom{k-1}{i}f(x+k-i) + \sum\limits_{j=1}^{k-1}(-1)^j \binom{k-1}{j-1}f(x+k-j)-(-1)^{k-1}f(x)$

$= f(x+k)+ \sum\limits_{i=1}^{k-1}(-1)^i\binom{k-1}{i}f(x+k-i) + \sum\limits_{i=1}^{k-1}(-1)^i\binom{k-1}{i-1}f(x+k-i)+(-1)^k f(x)$

$= f(x+k) + \sum\limits_{i=1}^{k-1}(-1)^i f(x+k-i) (\binom{k-1}{i} + \binom{k-1}{i-1})+(-1)^k f(x)$

$\overset{\binom{k-1}{i} + \binom{k-1}{i-1}=\binom{k}{i}}{=}$

$f(x+k)+ \sum\limits_{i=1}^{k-1}(-1)^i \binom{k}{i} f(x+k-i)+(-1)^k f(x)$

$= (-1)^0 \binom{k}{0}f(x+k-0)+\sum\limits_{i=1}^{k-1}(-1)^i \binom{k}{i} f(x+k-i)+(-1)^k \binom{k}{k} f(x+k-k)$

$= \sum\limits_{i=0}^{k}(-1)^i \binom{k}{i} f(x+k-i)$

M	T	W	T	F	S	S
« Jan
					1	2
3	4	5	6	7	8	9
10	11	12	13	14	15	16
17	18	19	20	21	22	23
24	25	26	27	28	29

Vroom

A Laboratory for Advanced Computing

Restate Feynman's "Great Identity"

Oh! Matryoshka!

Integration of Trigonometric Expressions

Double Feature on Christmas Day

Deriving Lagrange's multiplier

A Feynmanistic Derivation

Introducing Operator Delta