I cannot find intuitive explanation for formula $e^x = \lim_{n \to \infty} (1 + \frac{1}{n} \cdot x)^n.$ I will write down my understanding of the exponential function so you can understand the question better.
Multiplication.
For hole numbers, multiplication is repeated addition (subtraction). Let's say we want to find $5 \cdot 3 = x.$ We know (or better, we defined this to be) $x = 3 + 3 + 3 + 3 + 3 = 15.$
For rational numbers it's simillar but it has two steps: $\frac{3}{2}7 = x.$ Problem on the left tells us to find number $y$ such that when I add it to itself gives me $7$ then I have to add that number to zero 3 times. So $y = 3.5.$ Now, I have to find $y + y + y.$ So it's kinda back and fort game, but it still can be desribed using repeated addition (subtraction).
Finally for irrational numbers multiplication can be approximated with error of $\epsilon \rightarrow 0$ using arithmetic in $\mathbb{Q}.$
Exponentiation.
Simillary as before we can do repeated multiplication. This arithmetically maybe isn't clear as before, but we can still do it (and for example doubling of something can easly be understood). Reasoning is simillar as before, just turn addition into multiplication. For example, number $7^{\frac{3}{2}} = x$ we can find as follows: find a number $y$ which multiplied by itself gives $7$ (so $y = 2.645…$). Then find $y \cdot y \cdot y.$ Thus $x = 18.520…$ For irrational exponent (base) we can approximated it very well.
Now, we can investigate doubling of a money in hours, for example (suppose "money" can be any real number). In other words we are investigating function $f(x) = 2^x$ where $x$ is time passed in hours after 0th second.
Let's say we are starting from $2^0 = 1$ dollar. Right after 1 hour we have $2$ dollars, etc. Also 1 hour before starting moment we had $0.5$ dollars and so on. This starting from 1 (*) will be crucial.
We can observe that if more time passes doubling become larger, of course. Take intervals of one hour and see resulting change in amount of money, for example $ \frac{2^{0} – 2^{- 1}}{1 \ \text{hour}} < \frac{2^1 – 2^0}{1 \ \text{hour}} <\frac{2^{10} – 2^9}{1 \ \text{hour}}.$
Of course, change in money over small time interval (dx) is proportional to amount of money we already have, $2^x$ for some fixed $x.$ After hypothetical calculation we get $\frac{d}{dx}2^x \approx 0.6931 \cdot 2^x$ and simillary for any positive base, $\frac{d}{dx}b^x = kb^x.$ We want to find base $a$ such that $k = 1$ (then obviously $\frac{d}{dx}a^x = 1 \cdot a^x = a^x$). After algebraic manipulation we can find that magical number and denote it by "e": $e:= a = \lim_{n \to \infty} (1 + \frac{1}{n})^n.$
- I am interested in intuition why this is true. Simply: why we add to starting moment, 1 (*), this fraction $\frac{1}{n}$ and then multiply it by itself $n$ times for large $n$?
- Simillary, why we add $x$ times this fraction $\frac{1}{n}$ and then again multilpy that $n$ times (I understand $e^x = e^{1 \cdot x}$ but I do not see it clearly.
Of course if you give good argument for $e^x = \lim_{n \to \infty}(1 + \frac{x}{n})^n$ then setting $x = 1$ we get basic formula for $e.$ Please, I am looking for intuition and logic, not just algebra. Thanks!
Also, is this limit well behaved, i. e. is inequality
$|e – \lim_{n \to \infty}(1 + (\frac{1}{n + 1})^{n + 1}| < |e – \lim_{n \to \infty} ((1 + \frac{1}{n})^n|$ true?
Best Answer
$1$ is not important, it's just a matter of scale. You can start with million dollars and call it "1", meaning "1 million". It's just what you have at the beginning.
Consider that you take interest $\alpha$, that is, after $1$ time unit, you would have $1+\alpha$ money, after two time-units, you would have $(1+\alpha)^2$ and so on -- if it is compounded once per time unit.
However, if you split the time unit into $n$ smaller time-intervals and take the interest more often, you will have $(1+\frac{\alpha}{n})$ money after the first $\frac{1}{n}$-th time, then $(1+\frac{\alpha}{n})^2$ after $\frac{2}{n}$ time, and so on. After the full time unit, you would have $(1 + \frac{\alpha}{n})^n$.
Algebraically speaking, if you want to prove that $(1+\frac{x}{n})^n$ is similar to $\big((1+\frac{1}{n})^n\big)^x$ (for large $n$), you can think of it as $$ (1+\frac{x}{n})^n = (1+\frac{1}{\frac{n}{x}})^n = (1+\frac{1}{\frac{n}{x}})^{x\frac{n}{x}} = \big((1+\frac{1}{\frac{n}{x}})^{\frac{n}{x}}\big)^x \simeq \big((1 + \frac{1}{M})^M\big)^x $$ for some large $M$.