Eli5, How was number e discovered?
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Jacob Bernoulli was thinking how much money ultimately could be made from compound interest. He figured that if you put $1 in a deposit with 100% interest per year then you would get $2 in a year. Now if you put $1 in a deposit with 50% interest per 6 months and then reinvest it in 6 months in the same way, then at the end of the year you would get not $2 but $2.25 back, despite the fact that the interest rate is “the same” (50% times two equals 100%). Now if you keep dividing the interest periods in smaller and smaller units and reinvesting every time, you would be getting higher and higher returns. It turns out that making the interest payment continuous (that is, if the money gets reinvested constantly), $1 would become approximately $2.72 in a year, that is, the number e.
edit: below is an explanation of how *e* naturally comes up in math and physics, assuming solid end of high school math level, ignore if you are looking for an actual 5 yo explanation, ty.
It’s quite natural to wonder what are the functions where the values (=position, intensity, number of smth) are proportional to the derivative (=speed, slope, growth). Many important phenomenons like bank interest, inflation, virus propagation, cell proliferation, population growth when unchecked, nuclear chain reaction and nuclear decay behave according to that.
So mathematically, that is f’=af. Where a is a constant, the growth rate. Easiest is to take a=1 for starters, so f’=f. You see that if a function f is a solution to this equation, b*f is also a solution, for any constant b, so we can just solve for the simplest case f(0)=1 and just find all other solutions for f(0)=b by multiplying the solutions by b. Finally, if we look for a solution with a Taylor series, i.e. of the form f(x)=f(0)+f'(0)*x +f”(0)/2!*x^2 + … + f^(n) (0)/n!*x^n + …, it all simplifies because the derivatives f^(n) (0) are all 1, so we get a nice solution for f, useful to compute valued to any precision, namely f(x)=sum_n(x^n /n!). In particular we can compute to any accuracy f(1) and we call this number e. The function f we call it exponential or exp.
We can further see that exp(x+y) = exp(x)*exp(y), so we can start from f(1)=e and get f(2)=e^2 and more generally f(n)=f(1+1+…+1)=e^n , using the classical definition of integer powers (multiply n times by). Since we have a way to compute f also for non-integer numbers, with the polynomial development above, we can use this to continuously and naturally extend the definition of powers to all real numbers, so we can just write exp(x)=e^x . And if we come back to the equation above with f’=af and f(0)=b, simple to see f(x)=b*e^(ax) are the solutions we were looking for.
With all that we see that the number e has a really central and natural position in math and physics, and that it was unavoidable that it is found by any population developing calculus sooner or later. We also see there are simple ways to compute numerical approximations of it, for ex with the polynomial development above.
e is defined as the limit n –> infinity of (1+1/n)^n , which is a pretty useful number to know when you’re doing calculus and higher maths. The simplest answer is that the definition integrating things frequently involves taking limits to infinity, so knowing that the expression above converges to a constant makes doing that math much simpler and more precise.
The derivative of y = e^x is e^x, meaning the slope of the function is the same as the answer to the function. This is a very useful property when solving first and second order differential equations because it allows us to build answers off of e^x.
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I don’t know how exactly it was discovered, but in my opinion- this is the most practical derivation of e:
A lot of people think that if something has a 1-in-x chance of happening, then you are guaranteed a hit if you do the thing x times. That’s obviously not the case, because if you did it 2x times, you chances would not be 200%.
Ok, so let’s begin simple. You have a 1/2 chance for heads when you flip a coin. If you flip it twice, there’s a 75% chance that you get at least one heads. (HH, HT, TH, TT are possible outcomes. 3 of 4 include heads).
Now let’s do 1/3 3 times.
AAA, AAB, AAC. ABA, ACA. BAA, CAA.
BBB, BBA, BBC. BAB, BCB. ABB, CBB.
CCC, CCA, CCB. CAC, CBC. ACC, BCC.
ABC, ACB. BAC, BCA. CAB, CBA.
27 combinations. 3^3. You can see how this analysis gets very big very fast. Let’s count a success and something with at least one A. that’s 19/27 or 70.4%.
If you keep going, you end up realizing that as x gets bigger and bigger, your odds become 63.2%. So like- if the odds of winning the lottery jackpot are 1 in 300 million and you buy 300 million tickets, your odds of winning the jackpot are a bit less than 2/3. (Oversimplification warning)
0.632 is 1-1/e.
Don’t tell Grant I’m seeing mathologer too – gives you the intuition for your question and takes it to the next level – https://youtu.be/-dhHrg-KbJ0
Btw – if you’ll indulge me – why is it legal to convert (1 + (pi * i) / (n * pi * i))^(n * pi * i) to (1 + (pi * i) / M)^M where M is a positive integer heading to infinity? I would have thought the i in there couldn’t be so easily substituted out.
There’s many ways it was discovered. One way that comes to mind is how if you have n balls labelled 1,2,3,….n which correspond to boxes 1,2,3,…..n, then
The ratio of no of ways of arranging 1 in each box to no of ways of arranging one in each box such that none of them are in their corresponding box. Extend this to infinity and you get e.
Although fairly complicated it’s slightly intuitive.
There’s also the find a function whose derivative is the function itself and the Bernoulli compound interest story
Others have answered how it was discovered but I’d like to point out that it may be easier to think of it as “defined” instead of discovered. There are several ways e is defined and I think the easiest way to understand is that f(x)= e^x is defined as a function where the slope is always equal to f(x). That is, the derivative of e^x is e^x. With that definition you can calculate the value of e.
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This comment is so long not because the subject is necessarily super complex, but because I wanted to explain every single detail as much as possible. Hope it helps!
Say you put 1$ into a bank account with 100% interest a year. So after a year, the bank pays you an additional 100% of your money, so they give you an additional 1$, leaving you with your initial 1$ and the 1$ you got from interest, which makes 2$.
Now imagine a second bank offering the same 100% interest. However, the bank offers you a special deal: instead of paying you the 100% at the end of the year, they already pay you 50% after 6 months and another 50% after 6 more months. So after 6 months, you have your initial 1$ and get 50% of that from the bank as interest already, meaning you are now at 1,50$. Now comes the actual point of the deal: after waiting another 6 months, you will receive the other 50% of your interest. However, it will be applied to your *current* bank balance, meaning the bank will give you 50% of 1,50$, not 50% of 1$. This amounts to 0.75$, leaving you at 2,25$. So this deal is actually better than the first banks offer!
Where did the additional 0,25$ come from? Well, it comes from the fact that on its second payment, the bank paid you interest not just on your initial investment of 1$, but also on the 0,50$ you got from the first interest payment. You can compare it to a snowball: the reason that a tiny rolling snowball can become so big is the fact that as it picks up new snow by rolling, it increases in size, allowing it to pick up more snow and increase in size even more and even faster. To come back to interest:.
Your snowball is your initial 1$. Say you let it roll 1 meter and then another meter. Then, it will grow faster on the second meter because it has gained additional surface area from the first meter. Analogously, our 1$ initial investment will grow faster if we let it increase in size before applying interest to it, which is precisely how the two interest payments at 50% grow faster than the one time interest payment at 100%.
Now, coming back to e. Seeing as the process of making the interest payments more frequent lets you earn more money from them, mathematicians asked the following question: if we let the bank pay interest more and more often (say, weekly or daily), how much more money can we get out of it exactly?
The answer to this is “e” as many. That is, no matter how much more frequent you make the bank pay its interest, you can not go beyond getting more than about 2.7172… times your initial dollar. This number is a constant and is called eulers number, or “e” for short. Hope this helped and feel free to ask questions!
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