FTC saves the day!

Problem-1 Given

f(x) = \int\limits_{0}^{2x}f(\frac{t}{2})\;dt +\log(2)\quad\quad\quad(\star)

where f(x) is a continuous function, find f(x).



p=2x \implies \frac{dp}{dx} = 2.\quad\quad\quad(1-1)

By (\star),

\frac{df(x)}{dx} =\frac{d}{dx} \int\limits_{0}^{p} f(\frac{t}{2})\;dt + \frac{d\log(2)}{dx} = \underline{\frac{d}{dp}\left(\int\limits_{0}^{p} f(\frac{t}{2})\;dt\right)} \cdot \frac{dp}{dx}\overset{\textbf{FTC}}{=}\underline{f(\frac{p}{2})}\cdot \frac{dp}{dx}\overset{(1-1)}{=}2f(x),


\frac{df(x)}{dx} = 2f(x).\quad\quad\quad(1-2)

Moreover, we see from (\star) that

f(0) = \int\limits_{0}^{0}f(\frac{t}{2})\;dt + \log(2) = 0 + \log(2) = \log(2).\quad\quad\quad(1-3)

Solving initial-value problem

\begin{cases} \frac{df(x)}{dx} = 2f(x)\\ f(0)=\log(2)\end{cases}


f(x) = \log(2)\cdot e^{2x}.

We use Omega CAS Explorer to verify:

Fig. 1-1

Problem-2 Given

\int\limits_{0}^{1}f(u\cdot x) \;du = \frac{1}{2} f(x) +1\quad\quad\quad(\star\star)

where f(x) is a continuous function, find f(x).


Let p=u\cdot x,

u=\frac{p}{x} \implies \frac{du}{dp} = \frac{1}{x}\quad\quad\quad(2-1)

u=0\implies p=0; u=1\implies p=x.\quad\quad\quad(2-2)

\int\limits_{0}^{1}f(u\cdot x)\;du\overset{(2)}{=} \int\limits_{0}^{x}f(p)\cdot\frac{du}{dp}\cdot dp\overset{(1)}{=}\int\limits_{0}^{x}f(p)\frac{1}{x}\;dp=\frac{1}{x}\int\limits_{0}^{x}f(p)\;dp.\quad\quad\quad(2-3)

By (2-3), we express (\star\star) as

\frac{1}{x}\int\limits_{0}^{x}f(p)\;dp = \frac{1}{2}f(x)+1,


\int\limits_{0}^{x} f(p)\;dp = \frac{x}{2}f(x)+x.

It follows that

\underline{\frac{d}{dx}\left(\int\limits_{0}^{x}f(p)\;dp\right)}=\frac{d}{dx}\left(\frac{x}{2}f(x)+x\right)\overset{\textbf{FTC}}{\implies}\underline{f(x)}=\frac{1}{2}\left(f(x) + x\frac{d f(x)}{dx}\right)+1.\;(2-4)

Solving differential equation (2-4) (see Fig. 2-1) gives

f(x) = c x + 2.

Fig. 2-1

The solution is verified by Omega CAS Explorer:

Fig. 2-2

Exercise-1 Solving \begin{cases} \frac{df(x)}{dx} = 2f(x)\\ f(0)=\log(2)\end{cases} using a CAS.

Exercise-2 Solving (2-4) without using a CAS.


An Epilogue to “Truth vs. Intellect”

This post illustrates an alternative of compute the approximate value of \pi.

We begin with a circle whose radius is r, and let L_{n}, L_{n+1} denotes the side’s length of regular polygon inscribed in the circle with 2^n and 2^{n+1} sides respectively, n=2, 4, ....

Fig. 1

On one hand, we see the area of \Delta ABC as

\frac{1}{2}\cdot AB\cdot BC = \frac{1}{2}\cdot AB\cdot L_{n+1}.

On the other hand, it is also

\frac{1}{2}\cdot AC\cdot BE = \frac{1}{2}\cdot 2r\cdot \frac{L_n}{2}=\frac{1}{2}\cdot r\cdot L_n.


\frac{1}{2}AB\cdot L_{n+1}= \frac{1}{2}r\cdot L_n.


AB^2\cdot L_{n+1}^2 = r^2\cdot L_n^2\quad\quad\quad(1)

where by Pythagorean theorem,

AB^2= (2r)^2 - L_{n+1}^2.\quad\quad\quad(2)

Substituting (2) into (1) gives

(4r^2-L_{n+1}^2)L_{n+1}^2 = L_n^2\implies 4r^2L_{n+1}^2 - L_{n+1}^4 = r^2 L_n^2.

That is,

L_{n+1}^4-4r^2L_{n+1}^2+r^2 L_n^2 = 0.

Let p = L_{n+1}^2, we have

p^2-4r^2 p + r^2 L_n^2=0.\quad\quad\quad(3)

Solving (3) for p yields

p = 2r^2 \pm r \sqrt{4 r^2-L_n^2}.

Since L_n^2 must be greater than L_{n+1}^2 (see Exercise 1), it must be true (see Exercise 2) that

L_{n+1}^2=2r^2 - r \sqrt{4r^2-L_n^2}.\quad\quad\quad(4)

Notice when r=\frac{1}{2}, we obtain (5) in “Truth vs. Intellect“.

With increasing n,

L_n\cdot 2^n \approx \pi\cdot 2r \implies \pi \approx \frac{L_n 2^n}{2r}.\quad\quad\quad

We can now compute the approximate value of \pi from any circle with radius r:

Fig. 2 r=2

Fig. 3 r=\frac{1}{8}

Exercise 1 Explain L_{n}^2 > L_{n+1}^2 geometrically.

Exercise 2 Show it is 2r^2-r\sqrt{4r^2-L_n^2} that represents L_{n+1}^2.

Truth vs. Intellect

It was known long ago that \pi, the ratio of the circumference to the diameter of a circle, is a constant. Nearly all people of the ancient world used number 3 for \pi. As an approximation obtained through physical measurements with limited accuracy, it is sufficient for everyday needs.

An ancient Chinese text (周髀算经,100 BC) stated that for a circle with unit diameter, the ratio is 3.

In the Bible, we find the following description of a large vessel in the courtyard of King Solomon’s temple:

He made the Sea of cast metal, circular in shape, measuring ten cubits from rim to rim and five cubits high, It took a line of thirty cubits to measure around it. (1 Kings 7:23, New International Version)

This infers a value of \pi = \frac{30}{10} = 3.

It is fairly obvious that a regular polygon with many sides is approximately a circle. Its perimeter is approximately the circumference of the circle. The more sides the polygon has, the more accurate the approximation.

To find an accurate approximation for \pi, we inscribe regular polygons in a circle of diameter 1. Let L_{n}, L_{n+1} denotes the side’s length of regular polygon with 2^n and 2^{n+1} sides respectively, n=2, 4, ....

Fig. 1

From Fig. 1, we have

\begin{cases} L_{n+1}^2 = x^2 + (\frac{1}{2} L_n)^2\quad\quad\quad(1) \\ (\frac{1}{2})^2 = (\frac{1}{2}L_n)^2 + y^2\quad\quad\;\quad(2)\\ x+y = \frac{1}{2}\;\quad\quad\quad\quad\quad\quad(3) \end{cases}

It follows that

y\overset{(2)}{=}\sqrt{(\frac{1}{2})^2-(\frac{1}{2} L_n)^2} \overset{(3)}{ \implies} x=\frac{1}{2}-\sqrt{(\frac{1}{2})^2-(\frac{1}{2}L_n)^2}.\quad\quad\quad(4)

Substituting (4) into (1) yields

L_{n+1}^2 = \left(\frac{1}{2}-\sqrt{(\frac{1}{2})^2-(\frac{1}{2}L_n)^2}\right)^2+(\frac{1}{2}L_n)^2

That is,

L_{n+1}^2 = \frac{1}{4}\left(L_n^2 + \left(1-\sqrt{1-L_n^2}\right)^2\right).

Further simplification gives

L_{n+1}^2 = \frac{1}{2}\left(1-\sqrt{1-L_n^2}\right),\quad\quad\quad(5)

Starting with an inscribed square (L_2^2 =\frac{1}{2}), we compute L_{n+1}^2 from L_{n}^2 (see Fig. 2). The perimeter of the polygon with 2^{n+1} sides is 2^{n+1} \cdot L_{n+1}.

Fig. 2


\lim\limits_{n \rightarrow \infty} 2^n \cdot L_{n} = \pi.

Exercise-1 Explain, and then make the appropriate changes:

Hint: (5) is equivalent to L_{n+1}^2 = \frac{L_n^2}{2\left(1+\sqrt{1-L_{n}^2}\right)}.