Mathematics Euler's Number-e | and Natural Logarithm

# Euler’s Number-e | and Natural Logarithm

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In this Article we will try to understand the idea of Euler’s Number-e and Natural Logarithm. Lets begin with the concept of ‘e’.[mathjax]

# Euler’s Number-e

The Definition of Euler’s Number-e states – $e$ is the maximum possible result after continuous compounding 100% growth over the unit time period. Now Lets understand the Concept.

Things grow in nature in a fashion which is related to their own numbers or size, and when decay, they decay with the relationship with their own numbers, see here. Let consider a Number of any natural event (such as birth or growth of any living thing) in a particular time period gets double. The time period is considered as a unit time period. Say, In the beginning, the Number was $1$. $1^{st}$ unit time period later the number became $2$ with $100\text{%}$ growth rate. After the $2^{nd}$ unit time period, the Number $2$ gets doubled and became $4$. And at the end of the $3^{rd}$ unit time period, the Number is $8$. After every unit time period, the Number grows $100\text{%}$ of the previous Number. So the growth after one unit time: $1+1*100\text{%} =2$, after $2^{nd}$ unit time: $2+2*100\text{%} = 4$, after $3^{rd}$ unit time: $4+4*100\text{%} = 8$. [we know 100% = 1]

Suppose, the unit time period is One year. After every $1^{year}$, the Number gets double or the growth rate is $100\text{%}$ after one year.

If we calculate $50\text{%}$ growth after $\frac{1}{2}$ time period of $1^{year}$, then after 6 months the Number is $1+(1*\frac{1}{2}) = 1.5$.  after 2nd-half-or-after $1^{year}$, the number is $1.5+(1.5*\frac{1}{2}) = 2.25$. Which better than the previous growth after One year or $1$ unit time period.

Now, if we calculate the growth every quarter ($\frac{1}{4}$) of the year then the growth is:
$$1+(1*\frac{1}{4}) = 1.25 \quad\text{(after 1st quarter.)}$$

$$1.25+(1.25*\frac{1}{4}) = 1.5625 \quad\text{(after 2nd quarter.)}$$

$$1.5625+(1.5625*\frac{1}{4}) = 1.9531 \quad\text{(after 3rd quarter.)}$$

$$1.9531+1.9531*\frac{1}{4} = 2.4414 \quad\text{(after 1 year)}.$$

This time the result is improved then which was found after $1^{year}$ calculating semesterly.

Here we find, the formula of growth after unit time period, which is, $(1+\frac{1}{4})^{4} = 2.4414$,
Or, $(1+\frac{1}{n})^{n}$. Here $n$ is the division of unit time periodor simply times, and $1$ is the initial number of the entity before the time period.

Now if we count the growth for every month of a year, the growth after one year:
$(1+\frac{1}{12})^{12} = 2.6130$  (1 year has twelve months). The growth is improved than which was of calculating quarterly.

As we can see here that the smaller we divide the unite time period the better the growth rate we find. But it is not possible to reach into infinite growth rate. This is the fact which Euler’s Number-e tells us. Let See..

if we count for every day the growth after One year: $(1+\frac{1}{365})^{365} = 2.7146$.
Similarly for every hour, the growth after One year: $(1+\frac{1}{8760})^{8760} = 2.71813…$.
For every minute, the growth after One year: $(1+\frac{1}{525600})^{525600} = 2.71827…$.
Counting for every second, the growth after 1 year: $(1+\frac{1}{31536000})^{31536000} = 2.71828…$.

For $1,000,000,000$ division, the growth after 1 unit time period:

$$(1+\frac{1}{1,000,000,000})^{1,000,000,000} = 2.71828182…$$

So, it’s clear that, how infinitesimal the division is, the growth is always around $2.718…$ and doesn’t grow any further. This is a constant number which called Euler’s Number-e. Swiss Mathematician Leonhard Euler first discovered this natural phenomenon. Hence,

$$e=(1+\frac{1}{n})^{n}.$$

The Approximate value of $e$ is ∼$2.718$. This is an irrational number means the numbers after decimal point do not repeat and go on forever. Therefore the number $e$ can also be calculated as the sum of the infinite series.

$$e= \sum\limits_{n=0}^{\infty}{\frac{1}{n!}} = 1+ \frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+…+\frac{1}{n!}+…\\\quad\text{goes on forever…}$$

In mathematics, $e$ is so important because it is the natural language of growth, a graph below shows the function of $y=e^{x}$. Where at the point $x =1$, value $y = e^{x}$, the gradient at that point $e^{x}$ and area under the curve also $e^{x}$.

The only number $e$ has this unique property. And for having this property it becomes the natural language of Calculus. Calculus is the math of rate of change, growth, and areas. Therefore in Calculus if we write things in terms of $e$, math becomes much simpler. Otherwise, we have to deal with lots of messy constants and maths become really difficult to solve.

## Natural Logarithm

Natural logarithm is logarithm of base Euler’s Number-e. From the Definition we know that Logarithm is different way of writing exponentials or powers. For example, if we have a term $2^{3}=8$, which means $2$ is our growth rate, and $3$ is the times, which gives $8$ the final result.

In case of Logarithm, if we want to express the term, we will write – $log_{2}8=3$. Which is read “$log$ base $2$ of $8$ equals to $3$”.

Here $2$ is the growth rate means how fast it is growing. $8$ is the end up result after growth. And $3$ is the time it takes to be grown to $8$.

You might consider this lecture, to understand the concept of logarithm:

The natural logarithm or $log$ base $e$ is nothing but $log_{2.718…}$. Because, $e=2.718…$.

But the way we write the natural logarithm is, $lnX$, which is similar to $log_{2.718…}X$ or $log_{e}X$.

To read more recent Math articles go here »

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