In the series: Note 7.

Version: 0.3

By: Albert van der Sel

Doc. Number: Note 7.

For who: for beginners.

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Status: Ready.

Maybe you need to pick up "some" basic "mathematics" rather

So really..., my emphasis is on "rather

So, I am really not sure of it, but I hope that this note can be of use.

Ofcourse, I hope you like my "style" and try the note anyway.

This note: Note 7: The b

Each note in this series, is build "on top" of the preceding ones.

Please be sure that you are on a "level" at least equivalent to the contents up to, and including, note 6.

the sin(x) and cos(x) functions etc..

Here we will meet a new class of functions, namely:

the "base" number. It's also often called the "growthfactor".

You might generalize the equation a bit, if you like. A description like:

f(x)= a b

So, a more general format is:

Note:

In a section below, we will see a notation like N(t) = N

Such notation is quite common in all sorts of sciences. You might plot such function in an "N" and "t"

axis system (instead of the usual XY coordinate system).

Ofcourse, nothing fundamental changes if using (N,t), instead of the usual (x,y) system.

These functions are relatively easy to plot. In figure 1 below, I have plotted 6 simple examples, namely:

f(x)=2

f(x)=5

f(x)=0.3

f(x)=2

f(x)=-2

f(x)=2

Figure 1. f(x)=2

Le's concentrate for a moment on f(x)=2

just a few points, already gives a reasonable picture on how the function "looks like".

Let's first take a look at f(x)=2

value of x: | value of y=f(x) |

x= -2 | y= 1/4, since 2^{-2}=1/4 |

x= -1 | y= 1/2, since 2^{-1}=1/2 |

x= 0 | y= 1, since 2^{0}=1 |

x= 1 | y= 2, since 2^{1}=2 |

x= 2 | y= 4, since 2^{2}=4 |

x= 3 | y= 8, since 2^{3}=8 |

x= 4 | y= 16, since 2^{4}=16 |

In this case, since we have 2

However, would we have created a table for 3

Also, would we have created a table for 7

You can conclude that the larger the "b" (in "b

Question:

Can you make a similar table for f(x)=5

Here are a few observations:

If b > 1, then a fuction as f(x)=b

the faster the function "climbs". Suppose b=10, then when x=3, then f(x)=10

and when x=6, then f(x)=10

In figure 1, you can clearly see that f(x)=5

All exponential functions b

If 0 < b < 1, then a function as f(x)=b

These functions decline fast. For example, for f(x)=0.3

and if x=3, then f(x)=0.3

All exponential functions b

-For example, if we look at f(x)=2

the graphs go nearer and nearer to the line "y=0", but it will never exactly reach "y=0".

-However, if we "lift" such a function upwards, by adding a constant like with f(x)=2

(see subfigure 4 in figure 1 above) then the asymptote has shifted too. In this case, the asymptote now

is the line "y=3".

-Thus note that a function f(x)=b

can be regarded as a simple shift, upwards, or downwards, along the y-axis. It's a translation

Up or Down, of the entire graphic. Thus the Asymptote shifts too (compared to b

As with any exponential function b

In this case, the function must be written as:

So, you might say that we have not much "news" here.

But "e" truly is a special number, as is discussed in the next section.

Figure 2. f(x)=e

Ofcourse, when x=1, we have f(x) = e

Note in figure 2, that e is about 2.7.

Around the year 1683, Jacob Bernoulli found that as "n" goes to infinity in the following expression,

the outcome "converges" to a very specific number. Today we call that specific number 'e'.

Let's do some calculations. If you would calculate (1 + 1/10)

if you would calculate (1 + 1/100)

You see, if we only take n=100, then we are already pretty close to the "true" 'e'.

You get more and more closer, the larger 'n' is.

People with a background in economics or accounting would see a rather familiar formula.

It resembles the formula when you set away an amount of money, to a certain rate of interest, for period of "t" (n).

The formula "(1 + 1/n)

Yes, but what about physics? Just too much examples are available. Here is one that that also resembles the above.

Suppose you have N

which is a very high number of such nuclei, each have a (general) expectation value to decay.

Say that this expectation value is 6 hours, and you single out one of such nuclei, then there still is a chance

that you stare to it for several years. But on average, each nuclei has a good chance to decay in about 6h.

The "half-life" is defined to be that time, that halve of the material has decayed (with a very high probability).

The amount of original (not decayed) nuclei, can be described by:

N(t)= N

In this case the e

Many

all sorts of events where an "ensemble" growths or declines.

What is that with "e"? I don't think it's only math or physics. I bet that it's a matter for philosophy too.

For example, a certain initial amount of bacteria, under favourable circumstances for those living creatures,

start to grow according to an exponential relation. So, the number increases and that number is often denoted by "N".

However, instead of the usual x-axis and y-axis, the researchers plot the relation as N related to time (t).

Ofcourse, there is no fundamental difference in taking an N-axis and t-axis. It's exactly the same as

in using a x-axis and y-axis.

In many comparable studies as was sketched above, the relation used, is expressed as:

-"b" is initial (starting) amount at t=0. Similar for N

-"N" is the new amount after "t" has passed, in which "b" has growthed (or declined) to "N".

-"g" is the base number, or also called the "growthfactor".

There is really no fundamental difference when you compare equation 4, to for example f(x)=2

Ocourse, instead of "x", we now consider the time "t" as the variable, which is plotted horizontally,

just fully similar as to the x-axis.

Furher, we have the exact same behaviour with "g", as we already have seen in chapter 1.

That is:

For g > 1: we have a rising function (comparable to e.g. 2

For 0 < g < 1: we have a declining function (comparable to e.g. 0.3

Most often, "g" is also called the "growth factor".

However, in many cases there will indeed be growth (g > 1), but in many other cases, an initial amount

will decline over t, just like for example "0.3

-Thus, if you see the relation N = b g

that the

-Thus, if you see the relation N = b g

that the

Also note that the starting value, "b", is the amount at "t=0" (or some moment

You can easily check that with:

N = b g

The time "t" could have been measured in seconds, or months, or years...

In Physics, many physical processes are measured in seconds, or fractions thereof.

In Biology, for example the study of the growth of bacteria, might be sampled in hours, or days.

Let's take a look at a few examples.

Suppose that you have collected the number of internet users, of a certain city, from the year 2000, up to 2004.

Suppose You have found the following data:

N |
Year |

30 |
2000 |

240 |
2001 |

1920 |
2002 |

15360 |
2003 |

122880 |
2004 |

So, what to make of this data? If the Number at each year, divided by the former number of the

year before that, produces a constant factor, then we are likely to have found

and exponential relation.

Let's try:

122880/15360=8, 15360/1920=8, 1920/240=8 etc..

It seems that we have found, per year, we have a constant factor of "8".

You may say that we have found a "growthfactor" of "8". Each year, the former N

has been multiplied with "8".

Please notice, that this factor works

so it's not a linear relation.

If you would look at two years, like this example, 122880/1920= 64 = 8

You can also try something like that over 3 years, which gives us 8

The relation thus, smells like something which has a factor like 8

Also note that we started in the year 2000, up to 2004, meaning that we have 4 timesteps,

namely 2000->2001, 2001->2002, 2002->2003, 2003->2004.

Let's assume that our relation is:

we have started the measurements.

Would it work?. Let's find out. Using the formula, then over 4 years, the initial value "30"

must have grown to:

N = 30 * 8

Indeed, this matches the data of the table above.

Suppose we have 1 kg (1000 g) of a certain radioactive substance.

This stuff decays into other producs, and suppose we have some sort of technique

to measure the remaining amount of that radioactive substance, over "t".

Suppose further that we have found the following measurements:

Amount |
At time: |

1000g |
t=0h |

700g |
t=1h |

490g |
t=2h |

343g |
t=3h |

240g |
t=4h |

Let's see how the amount have changed per hour:

240/343=0.7, 343/490=0.7 etc.. Here, I used the data over 3 hours.

It seems that our relation is:

N = 1000 * 0.7

In example 1, we found a growthfactor of "8".

For convience, the data of example 1 is repeated here:

N |
Year |

30 |
2000 |

240 |
2001 |

1920 |
2002 |

15360 |
2003 |

122880 |
2004 |

Our methodology was to take the value of N

value of N

And we repeated that procedure for other consecutive years.

But what if, you want to evaluate the growthfactor over e.g. 2 years, instead of one?

In this example, the growthfactor will be 8x8 = 8

by using the N values of those "n" years, like for example "122880/1920 = 64 = 8

So, for the growthfactor, over "n" timeunits, that factor will be g

-If you have a data table, and you go for example from a unit of one year to a unit of "n" years,

then your growthfactor changes to:

g → g

This is true not only for years, but works the same for going from months to years,

or from units of 10 sec to units of 100 sec etc..

-If you have a data table, and you go for example from a unit of time, to "n" smaller units,

then your growthfactor changes to:

g → g

This is true not only for years going to months, but works the same for going from months to weeks,

or from units of 100 sec to units of 10 sec etc..

If something growths or declines with a certain percentage per unit of time, then you can

express that as a growthfactor as well.

If some initial value increases with, say, with 17% per unit of time, then the associated growthfactor is:

In this example, your exponential formula would be: N=N

In case the initial value at t=0 would be, for example "1000", we would have N=1000 * 1.17

If some initial value decreases with, say, with -17% per unit of time, then we can apply

the same formula, but this time we take into account the negative sign of that percentage.

So in the example of 17% decrease, we would have g = (1 - 17/100) = 0.83.

In the last example, your exponential formula would be: N=N

In case the initial value at t=0 would be, for example "1000", we would have N=1000 * 0.83

Without any proof, we state that:

Well, it's not uncommon to have such functions. Suppose we have these two:

u(x)=√ x

v(x)=x

Then u(v(x))= √ (x

So, if we let "v" operate on x first, en then let "u" operate in "v(x)", we have x again. So, if both functions

are applied, then nothing happens to "x". The funcion "u(v(x)" maps "x" onto itself.

If we look at these two again:

u(x)=√ x

v(x)=x

Then people also often say, that "v" is the

When two functions are related

but the use the

Thus: f(f

The functions "f" and " f

Note: (optional reading)

In other mathematical disciplines, like that of studying vectors, matrices etc.., when a "mapping" or "operator"

has an inverse operator,

where I is the Identity operator, or

Let's plot both √ x and it's inverse x

See figure 3. Note that if you would also plot the line "y=x", then it becomes visible that both functions

are "mirrored" through "y=x".

Ofcourse, when we say "x is mapped onto itself", it means the line y=x, since that function is f(x)=x,

thus for example f(f

Figure 3. x

the logarithmic function using the number '10' as it's "base", is very helpful.

A couple of examples may illustrate that.

The "claim" here, is that if g(x)=e

But, if that is true, then h(x)=ln(x), must be the inverse function of g(x)=e

We know how the the curve of e

Indeed, ofcourse it's possible to draw such an inverse curve, since you only need to mirror e

Sure, it would be great to

Figure 4. g(x)=e

Thus, if they are each others inverse, then it should be true that:

ln(e

lim _{h -> 0} |
f(x + h) - f(x) ---------------- h |

It's important to know, that the upper equation really is the "heart" of finding derivatives.

If needed, please check note 5 again, where the relation is fully explainend.

Now, using "e

lim _{h -> 0} |
e^{(x + h)} - e^{x}---------------- h |

Remember from note 1, that a

Thus:

lim _{h -> 0} |
e^{x} e^{h} - e^{x}---------------- h |

Now, we can "factor out" e

lim _{h -> 0 } |
e^{x} (e^{h} - 1)---------------- h |

So, we may write that as:

e^{x} lim _{h -> 0 } |
(e^{h} - 1)------------ h |

Note: Do you see that we have managed to pull out e

While the Limit itself converges to "1" (I should have proved that too), we have found the very remarkable fact, that:

f '(x)= e

d e^{x}------ dx |
= e^{x} |

This fact, that the derivative of a function

Indeed. It's a very remarkable fact. But we also know that e

So, it might be argued, that it's very remarkable status, should not surprise us at all.

for "quick" introductions in math. But I really like you to remember the following:

f '(x)= 1/x

(for x>0)

or:

d ln(x) ------ dx |
= 1/x |

Note that the function 1/x, shows asymptotic behaviour if "x" approaches "0".

We have seen 1/x before, so we were already aware of that fact.

The next note is a super quick intro on the "primitive integral" and "primitive functions".