The popular riddle is this: a lily pad appeared on the lake one morning. By the second morning it was two, the third four, etc., doubling each day. On the 10th day, there were already 512, on what day was half? The impetuous man briskly answering that half the water lilies were in the quill on the fifth day is mistaken. As the number of plants doubles each day, half the final count was on the penultimate, ninth day.
For the most part, this gimmick serves nothing but amusement, but from a mathematical point of view it describes relatively accurately the phenomenon of the spread and emergence of life. Imagine, for a moment, that we are in a kind of primordial quill in which, according to scientific theories, life originated, and let us concentrate on the first organism formed-how is this life likely to spread around the world?
This takes a look at another field that differential calculus can use: biology. While it is true that not every biologist will make use of it, it is nevertheless a colourful demonstration. In addition to biology and exotic organisms, we also discover two exotic functions.
We're talking about the first organism to appear on our planet, so it has no predators, it can only spread. And that's what it does, after all, by some definitions, the ability to multiply is a criterion for life. Let's say every once in a while $T$ our organism splits into two exact same organisms. These will divide again in another time $T$ etc. Of course this division does not continue in the same way ad absurdum, probably sometimes our organism will die or mutate, but let's pretend it doesn't happen and try to write the first few numbers of individuals at some time.
Čas | 1T | 2T | 3T | 4T | 5T | 6T | 7T | 8T | 9T | 10T | 11T ----------------------------------------------------------------- Počet | 1 | 2 | 4 | 8 | 16 | 32 | 64 | 128 | 256 | 512 | 1024
In the table above, we got the same development as for the water lilies case. So let's continue the table. Just for clarity, we'll change units and count our organisms not by units, but by thousands. Also, I'm sure some poor people for $11 T have died, so we're going to assume there's about a thousand at the end of Cycle 11. Wait, our table repeats itself in this case! We get exactly the same evolution, only instead of units of organisms we have thousands. So we see that for every $11 T$, the number of animals is going to increase by a thousand times, so pretty soon there's going to be a lot of them.
Let's examine this plot more closely and plot the number of animals over time. This number, or population size, we mark $n$ and time $t$. We get the graph below.
In the graph, the crosses show the actual values of the population count at certain times $T$. We've already talked about population development going the same way, so let's imagine $n$ on the axis in some larger units, like tens of thousands. After that, what is certainly not happening is that the population of our animals suddenly doubles, but each individual splits in a different time. Since there are so many of them, they double up practically all the time. So the evolution of their population over time is not in leaps and bounds, but almost in unison. In this case, to facilitate our understanding of the process, we interlace the points with a line that is dashed in the image. This is a function that roughly describes the number of individuals over time.
How did we come up with this function that we've put points through? If we look at the population table, we see that the number can be represented as $n = 2^{x}$, where $x$ is a natural number. The number $x$ indicates the number of past reproductive cycles. We know that one cycle passes in time $T$, so if time passes $t$, there have been $t/T$ cycles. So we can write $n=2^{t/T}$. Then we plotted the graph using a computer program. For clarity, we'll call the $n=n(t)$ function the population curve.
It may seem that there is nothing more to be done about the task of spreading life, as we have described how their numbers will evolve over time. However, our description ends up plotting the points on the graph. The dashed population curve is already something we call a model, or some simplified description of reality. And this model has its flaws, because, for example, according to it, the number of organisms will grow to infinity \u2014 this is not what happens in the real world. To understand what the model can be improved upon, let's do a differential analysis of the population curve: let's examine its derivative. To do this, we will need the following formula: $$ 2^{\frac{t+c}{T} } = 2^{c/T} \cdot 2^{t/T} \,.$$
The formula above is based on simple rules for power, yet it reveals an interesting truth. Just rewrite $2^{t/T} = n(t)$. While we're at it, let's say $T=1 to simplify the math (this is equivalent to $t$ indicating the number of reproductive cycles passed, even with part started). We get: $$n(t+c) = 2^c \cdot n(t) \,.$$
When we draw a graph of the general function $f(x)$, we get a graph of the function $f(x+c)$ by moving the graph of the function $f(x)$ to the left by $c$. Correspondingly, the $f(x+c)$ function takes its value $c$ earlier. On the other hand, multiplying a function by a value is easy to imagine, all the points move up. The equation above says that these two operations are equivalent to the population curve.
This particular fact can be used to calculate the derivative. The derivative is just the slope of the curve at some point, we can think of it as a tangent. The same rules therefore apply to this derivative, namely: $$\frac{\mathrm{d}}{\mathrm{d}t} n(t+c) = 2^c \frac{\mathrm{d}}{\mathrm{d}t} n(t) $$
$$ n'(c) = 2^c \frac{\mathrm{d}}{\mathrm{d}t} n(t=0) \,.$$
There's nothing stopping us from changing the markings and writing $c=t$. Here's what we get: $$\begin{align} n'(t) &= 2^t k\,,\\ n'(t) &= n(t) k\,. \end{align}$$
The logarithm function will not become officially known until the next chapter.
So the $n(t)$ derivative, except for some constant, is equal to itself! To know the derivative completely, we have to determine the constant $k$, which is the derivative of the $n(t)$ function at the point $t=0$. This value later turns out to be equal to $\log (2)\doteq 0{,}7$ (the natural logarithm of two). We can't explain this with our current methods yet (we haven't even gotten to know the logarithm function yet), so we'll save that for another time. But for qualitative analysis, writing $k$ will be enough.
To better understand the population curve of organisms, let's imagine that a particular individual multiplies into a total of three more in one division. Then, by thinking, its population curve can be described by $n_3(t)= 3^t$. We will then express its derivative according to the same procedure as above as follows: $$n_3'(t) = n_3(t) \cdot k_3 \,.$$
Similarly, for a different number of divisions and a number other than $3, we could create corresponding functions. The constant $k_x$ then indicates the slope of the $n_x(t)$ function at zero. It's clear that $k_x < k_y$ for $x < y$, by different choices $x$ we can change the $k$ tendency in different continuously. So surely for some $x$, we have $k_x = $1. For such a $n_e$ feature: $$ n_e'(t)= n_e(t)\,.$$
Or its derivative is equal to itself, which is definitely a prominent feature for the function. Let's call the number for which this applies $e$ (we don't know anything about it yet). How is this relevant to our example? Remember, originally, in the population curve, we had $t/T$, where $T$ was the time that the organisms doubled. However, we could have chosen a different time $T$, $$n(t) = e^t\,,$$
which describes the evolution of the number of individuals over time as well as previous functions, only has the convenient property that $$n'(t) = n(t)\,.$$
A class of functions that can be described by a prescription in the form $f(x) = a^x$, where $a$ is any number, we call an exponential function. What stands out among them is the $e^x$ function, which itself is sometimes called the exponential. This exponential is sometimes referred to as $\exp (x) \equiv e^x$. We have examined the properties of exponential functions above in the example of the $2^x$ function, which can be summarized as follows:
Notice that above we could talk about the function $e^x$ and describe its properties without knowing exactly what the value of $e$ was, it was enough for us that some $e$ existed. This approach is often used in mathematics, made possible by mathematical marking and algebra.
$$f(x+\Delta x) \doteq f(x) + \Delta x f'(x)\,.$$
When we use the notion of derivative and we factor out, we get the following form: $$f(x+\Delta x) \doteq (1+\Delta x) f(x) \,.$$
We assume here that $\Delta x$ is a very small number. We see that when we want to move a little bit to the right in terms of our functional value, we just have to multiply the function by some coefficient. What if we used the formula above twice in a row? We then get the following: $$\begin{align*} f(x+\Delta x + \Delta x) &\doteq (1+\Delta x)(1+\Delta x) f(x)\\ f(x+2\Delta x ) &\doteq (1+\Delta x)^2 f(x)\,. \end{align*}$$
From similar considerations, we can see that in general for some whole $n$ there is: $$f(x+n\cdot\Delta x) \doteq (1+\Delta x)^n f(x) \,.$$
We see that this way we can choose $n$ very large and express that value far from $f(x)$. Let's use this approximation method to express $f(1)$, as $f(1)=e$. We do that by substituting $x=0$ and $\Delta x= 1/n$. What should we choose $n$? As large as possible, for the larger the smaller the pieces we move, so the more accurate the result we get. We get a completely accurate result if we do a $n$ limit going to infinity. So let's look at the result (using that $f(0)=1$): $$f(1) = e = \lim_{n\to \infty} \left( 1+\frac{1}{n}\right)^n \,.$$
Unfortunately, we cannot say exactly what this limit is equal to with our capabilities. But we can use a calculator and put in, say, $n=10\,000$, and we get that $e\doteq $2,718. The $e$ number is called Euler's number and is as irrational as $\pi$ or $\sqrt{2}$. So it has infinite decimal development, so this approximation using a calculator can be enough for us: we couldn't figure out the whole decimal development anyway.
In this chapter, we presented an exponential function on the problem of organism population growth. In the next chapter, we're going to build on that and show why this function is important to solve so-called differential equations. This will give us a better insight into the original problem of organism growth.
Now we know roughly what differential equations are, so let's try to construct a suitable equation to fit the problem of animals in the primordial quill. Let's forget that we know what the population curve looks like, let's describe it differently, using its increment (change).
$$\begin{align} n(t + \mathrm d t) \doteq n(t) + \mathrm d t \cdot n'(t) &= n(t) 2^{\mathrm d t /T}\\ n(t) + \mathrm d t \cdot n'(t) &= n(t) 2^{\mathrm d t}\,. \end{align}$$
Here we made an adjustment where we laid down $T=1$, or $\mathrm d t$ we interpreted as some small number of reproductive cycles. So we've got some differential equation and we can continue to make adjustments. Let's try to express the population curve derivative: $$\begin{align} n(t) + \mathrm d t \cdot n'(t) &= n(t) 2^{\mathrm d t}\\ \mathrm d t \cdot n'(t)&=n(t) (2^{\mathrm d t} -1) \\ n'(t)&=n(t) \frac{ 2^{\mathrm d t} -1}{\mathrm d t} = n(t) \cdot k \,. \end{align}$$
We wrote the $k$ number again using the logarithm. This expression is as yet unknown to us, but we will soon be able to illuminate it.
We found an interesting fact that the population curve derivative is proportional to the population curve itself over some constant $k$. Our calculations are going to be more accurate the smaller $\mathrm d t$ is going to be, so we have to do a limit transition for $\mathrm d t \to 0$, which is what $k$ equals. Again, a calculator or some more advanced mathematical methods can be used, and we find that $k=\log (2) \doteq 0{,}7$ (we have not yet introduced the $\log $) function).
The function we propose as a solution is $N(t) = e^{a\cdot t}$, where $a$ is some unknown constant. Let's try to install it: $$N'(t) = \frac{\mathrm{d} e^{at}}{\mathrm{d} at } \frac{\mathrm{d} (at) }{\mathrm{d} t} = ae^{at} \,.$$
For the derivative, we used the derivative of the compound function that we had seen in second chapter. We can specify the constant $a$ by comparing $N(t)$ with the previous result for $n(t)$. We get $a=k. Overall: $$ N(t) =e^{ \log (2) t} \overset{?}{=} 2^t \,.$$
Do we get the same result as in the previous chapter? To get it, this would have to be true: $$ e^{\log 2 t} = (e^{\log 2})^t = 2^t \Rightarrow e^{\log 2} = 2 \,.$$
You can try performing functions, for example, at geogebra.com.
We're getting into a bit of tricky territory here because we haven't introduced the logarithm function yet, so we can't know that. However, if the $2^t$ and $e^{\log (2) t}$ functions were brought up (even with $\log (2) \doteq 0{,}7$), we would receive identical graphs. So that's why we got the same result as last time, just by a different method.
You may think that we have arrived at the same result, but we have employed more obscure methods and have made little clarification. But differential analysis has given us a whole new, independent perspective on the problem being studied. It may seem more complex, but this is the first problem we have studied from this point of view, it would be strange if we didn't have difficulties! The seasoned mathematician has already solved countless of these problems, and they do not cause him problems.
In any case, a new point of view is always useful because it allows us to get knowledge that we would not otherwise get. Moreover, some problems cannot be solved without differential methods. That's what their value is all about.
Looking for a population curve, we encountered exponential functions. As their escort, there was the occasional feature marked $\log$, which we quietly passed over. Not only do both functions describe our particular problem well, but they occur in fundamental descriptions of various other events in our world. Therefore, we can no longer postpone the formal performance of the $\log$ function. In the next chapters, we will meet them further and thus understand them in even greater depth.
We write the logarithmic functions generally in the form $f(x)=\log_a x$ and read the logarithm of the base $a$ of $x$. Again, this is a class of functions that can be written in this form, different functions are given by changing the $a$parameter (this is obviously constant for the function. Logarithms are always accompanied by exponential functions because they are inverse to them. This means that: $$\begin{align} &1.\:\: \log_a ( a^x) = x \,, \\ &2.\:\: a^{\log_a(x)} = x \,. \end{align}$$
Inversi can also be expressed as, for every $x$, if we put it in the exponential and then the result in the logarithm, we get $x$ again. How do we define logarithm? If we have an equation $$ \log_a (x) = y \,,$$
so $y$ is the number of times we have to raise $a$ to raise $x$. For obvious reasons, we can't put $0$ or any negative number in the logarithm as $x$.
Euler's $e$ number is also important for logarithms. We define the so-called natural logarithm as the logarithm of the base $e$. We refer to it as the logarithm of naturalis, or $\ln (x)$, or simply the logarithm of $\log (x)$. Also notable is the base- $10 logarithm (for measuring intensity and the so-called logarithmic scale), and the base- $2 logarithm binary, which finds use in theories of information.
The following two identities apply trivially to exponential functions: $$\begin{align} &1.\:\: e^{a+b}=e^ae^b \,, \\ &2.\:\: (e^a)^b = e^{ab} \,. \end{align}$$
Logarithmic functions are associated with exponential ones, so similar rules should apply to them. Indeed, these rules apply $$\begin{align} &1.\:\: \log(x) + \log(y) = \log (xy) \,, \\ &2.\:\: \log(x^y) = y\log (x) \,. \end{align}$$
How do you prove these relationships? Simply by recognizing that every $x$ can be written as $e^a$ and every $y$ as $e^b$. Then all you have to do is use the relationships that apply to exponential functions. For example, for the first logarithmic rule, we have the left side: $\log e^a + \log e^b = a+b$ and the right side $\log( e^a e^b ) = \log (e^{a+b}) = a+b$. Both sides equal, so the rule stands.
We've been writing about logarithmic functions for a long time, but we have no idea what they might look like yet. Below is therefore a sample graph of natural logarithm. Notice, in particular, the slow rise (the rise keeps falling and falling until the logarithm is almost horizontal) and also that it is not defined for negative numbers. As $x$ goes to zero, the logarithm goes to minus infinity (which, of course, cannot be perfectly displayed on the graph).
If you're hearing about logarithm for the first time, that's probably a lot of information at once. You don't have to worry if you don't remember everything right away, you can always look up the information again, and if you need it often, it's bound to stick in your memory. But the three main things to remember are:
The derivative of logarithm remains to be calculated. We start with the derivative definition: $$\begin{align} \frac{\mathrm d }{\mathrm d x} \log x &= \lim_{h \to 0} \frac{ \log(x+h) - \log (x)}{h} = \lim_{h \to 0} \frac{1}{h} \log \left( \frac{x+h}{x} \right)\\ &= \lim_{h \to 0} \frac{1}{h} \log \left( 1 + \frac{h}{x} \right) = \lim_{h \to 0} h^{-1} \log \left( 1 + \frac{x^{-1}}{h^{-1}} \right) = \log \left[ \lim_{h \to 0} \left( 1 + \frac{x^{-1}}{h^{-1}} \right)^{h^{-1}} \right] \,. \end{align}$$
On the first line here, we used both logarithmic rules. On the second line, we set the stage for a limit that resembles the limit we met at the exponential. Indeed, just remember that $h^{-1}=1/h$ for a positive $h$ going to zero can be written as a large $n$ going to infinity. So we write: $$\begin{align} \frac{\mathrm d }{\mathrm d x} \log x &= \log \left[ \lim_{h \to 0} \left( 1 + \frac{x^{-1}}{h^{-1}} \right)^{h^{-1}} \right] = \log \left[ \lim_{n \to \infty} \left( 1 + \frac{x^{-1}}{n} \right)^{n} \right] = \log \left[ e^{x^{-1}}\right]\\ \frac{\mathrm d }{\mathrm d x} \log x &= x^{-1} = \frac{1}{x} \,. \end{align}$$
NB: in general, the following relationship involving the limit above applies: $$f(x) = e^x = \lim_{n\to \infty} \left( 1+\frac{x}{n}\right)^n \,.$$
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