Monthly Archives: March 2021

We all bleed the same color

In “Mathematical Models in Biology”, Leah Edelstein-Keshet presents a model describing the number of circulating red blood cells (RBC’s). It assumes that the spleen filters out and destroys a fraction of the cells daily while the bone marrow produces a amount proportional to the number lost on the previous day:

\begin{cases} R_{n+1} = (1-f)R_n+M_n\\ M_{n+1} = \gamma f R_n\end{cases}(1)

where

R_n - number of RBC’s in circulation on day n,

M_n - number of RBC’s produced by marrow on day n,

f - fraction of RBC’s removed by the spleen,

\gamma - numer of RBC’s produced per number lost.

What would be the cell count on the n^{th} day?


Observe first that (1) is equivalent to

R_{n+2} = (1-f)R_{n+1}+M_{n+1}\quad\quad\quad(2)

where

M_{n+1} = \gamma f  R_n.\quad\quad\quad(3)

Let n = -1,

M_0=\gamma f R_{-1} \implies R_{-1} = \frac{M_0}{\gamma f}.\quad\quad\quad(4)

Substituting (3) into (2) yields

R_{n+2} = (1-f)R_{n+1}+\gamma f R_{n}.

We proceed to solve the following initial-value problem using ‘solve_rec‘ (see “Solving Difference Equations using Omega CAS Explorer“):

\begin{cases} R_{n+2}=(1-f)R_{n+1}+\gamma f R_{n}\\ R_{0}=1, R_{-1} = \frac{1}{\gamma f}\end{cases}

Evaluate the solution with f=\frac{1}{2}, g=1, we have

R_n = \frac{4}{3} + \frac{(-1)^{n+1}2^{-n}}{3}.\quad\quad\quad(5)

Plotting (5) by ‘plot2d(4/3 + (-1)^(n+1)*2^(-n)/3, [n, 0, 10], WEB_IMAGE)’ fails (see Fig. 1) since plot2d treats (5) as a continuous function whose domain includes number such as \frac{1}{2}.

Fig. 1

Instead, a discrete plot is needed:

Fig. 2

From Fig. 2 we see that R_{n} converges to a value between 1.3 and 1.35. In fact,

\lim\limits_{n \rightarrow \infty}  \frac{4}{3} + \frac{(-1)^{n+1}2^{-n}}{3} = \frac{4}{3}\approx 1.3333....

ONE

X had a Q&A session with Buddha.

X: What is 0.9\;?

Buddha: 0.9 is \frac{9}{10}\;.

X: What is 0.99\;?

Buddha: 0.99 is \frac{99}{100}\;.

X: What is 0.999\;?

Buddha: 0.999 is \frac{999}{1000}\;.

\ddots

X: What is 0.99999....\;?

Buddha: What are the dots ?

X: The dots are all 9‘s. It means \forall n \ge 1, the n^{th} digit after the decimal point is 9.

Buddha: 0.99999... is

\sum\limits_{i=1}^{\infty}\frac{9}{10^i}

X: What is \sum\limits_{i=1}^{\infty}\frac{9}{10^i}\;?

Buddha: It is

1

A Sophism in Calculus

Evaluating indefinite integral \int \frac{1}{x}\;dx using integration by parts gives

\int \frac{1}{x}\;dx =\int x'\cdot \frac{1}{x}\;dx=x\cdot\frac{1}{x}-\int x\cdot(\frac{1}{x})'\;dx=1-\int x\cdot\frac{-1}{x^2}\;dx=1+\int \frac{1}{x}\;dx.

That is,

\int \frac{1}{x}\;dx = 1 + \int \frac{1}{x}\;dx.

Substracting \int \frac{1}{x}\;dx from both sides we conclude

0=1.\quad\quad\quad(1)

The expression

\int f(x)\;dx = g(x)

means

\left(\int f(x)\; dx - g(x)\right)' = 0;

i.e.,

\int f(x)\;dx- g(x) = C

where C is a constant.

Therefore,

\int \frac{1}{x}\;dx = 1+ \int \frac{1}{x}\;dx \implies \int \frac{1}{x}\;dx - (1 + \int \frac{1}{x}\;dx) =C.

Or,

C = -1.

Moreover, define \leftrightarrow as

f(x) \leftrightarrow g(x), if f(x)-g(x)=C, \quad\quad\quad(2)

it can be shown that

f(x) \leftrightarrow h(x), h(x) \leftrightarrow g(x) \implies f(x) \leftrightarrow g(x) \quad\quad\quad(2-1)

and

f(x) \leftrightarrow g(x) \Longleftrightarrow f(x)+h(x) \leftrightarrow g(x) + h(x).\quad\quad\quad(2-2)

If we proceed to evaluate \int \frac{1}{x}\;dx as follows:

\int \frac{1}{x}\;dx \leftrightarrow \int x'\cdot \frac{1}{x}\;dx \leftrightarrow x\cdot\frac{1}{x}-\int x\cdot(\frac{1}{x})'\;dx \leftrightarrow 1-\int x\cdot\frac{-1}{x^2}\;dx \leftrightarrow 1+\int \frac{1}{x}\;dx,

we have (by (2-1))

\int \frac{1}{x}\;dx \leftrightarrow 1 + \int \frac{1}{x}\; dx.

Substracting \int \frac{1}{x}\;dx from both sides (by (2-2)) we conclude

0 \leftrightarrow 1.\quad\quad\quad(3)

Unlike (1), (3) is true:

0-1 = -1, a constant.

By the way,

\int \frac{1}{x}\;dx \leftrightarrow \log(x).

See “Introducing Lady L” for details.

Exercise-1 Can you spot the fallacies?

For x>0,

x^2 = x\cdot x = \underbrace{x+x+x...+x}_{x\;\; x's}.

Differentiating both sides, we have

2x=\underbrace{1+1+1+... + 1}_{x\;\;1's}=x.

Dividing both sides by x (since x>0) yields

2 = 1.