# 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_PLOT)’ 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

$2 = 1.$