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Please note that I believe this post isn't a duplicate as I am trying to find the simplest explanation, not a rigorous proof or theorem, and many others may want the same.





The original is as follows:

This is just a quick question about the mathematical constant $e$. Basically, I learned that $$e = \lim_{n \to \infty} \left( 1 + \frac{1}{n} \right)^n \approx 2.71828...$$

(Originally miswritten as $e = \lim_{n \to \infty} \left( 1 + \frac{1}{n} \right)^\infty \approx 2.71828...$)

but I can't get a grasp of how this works. I only started with basic calculus, so I don't get the expansions and stuff, but I kind of want a simple explanation of how $e \neq 1$ because it seems like it should tend to $1^\infty$=1.

Thanks in advance!

Bulby2024
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  • It is $e = 1 + 1/2! + 1/3! + 1/4! ... to \infty$. You forgot the factorial and summation sign. Your limit does go to 1. It is not the correct definition of e. – devssh Nov 19 '24 at 12:26
  • Hey, you might like this video: https://youtu.be/drsd7Mc-gvI?si=qX8QN0X77Vb5PGWd – Alma Arjuna Nov 19 '24 at 12:27
  • Thanks! can you summarize the idea in the video though? – Bulby2024 Nov 19 '24 at 12:28
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    I think you are missing the exponent $n$, and you make $n$ tends, not $x$. Also, note that $1^\infty$ is an indeterminate form, but given that $1+\frac{1}{n} > 1$, it will certainly not tend towards $0$ – caduk Nov 19 '24 at 12:30
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    @devssh Yes I know that $e$ also equals that, but I don't get why this definition does not expand to 1, as it is another way of representing $e$ in compound interest. – Bulby2024 Nov 19 '24 at 12:30
  • @caduk Oh... I meant 1. But why not? – Bulby2024 Nov 19 '24 at 12:31
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    The limit you wrote does not make any sense. There is no $x$ and even if you change it to $\lim_{n\to \infty}$ the limit is not $e$. Check the formula you are thinking of and you will see it is more subtle. – lulu Nov 19 '24 at 12:32
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    $1^\infty$ is an indeterminate form similar to $\frac{0}{0}$. There are too many transcendental numbers near 1. Example: sqrt(2), cube root(2), fourth root(2) ... 100th root (2) ... – devssh Nov 19 '24 at 12:34
  • @lulu Got it, fixed. – Bulby2024 Nov 19 '24 at 12:35
  • Ah that makes sense @devssh – Bulby2024 Nov 19 '24 at 12:35
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    Nope, still wrong. The limit is $e=\lim_{n\to \infty} \left(1+\frac 1n\right)^n$. You can't just replace some of the $n's$ with $\infty$. And you left the $x$ in place. Details are important. – lulu Nov 19 '24 at 12:36
  • @lulu Thanks! I know I messed up, because I was editing here and there a bit too fast haha – Bulby2024 Nov 19 '24 at 12:42
  • I tried to explain the problem here, and I'm sure this is duplicate thread of several earlier ones. Anyway, the comparison to compound interest is a good one. – Jyrki Lahtonen Nov 19 '24 at 13:01
  • I'm not sure about the wisdom of fixing the last typo. It did not quite render the answers useless, but they look a bit out of synch now :-( – Jyrki Lahtonen Nov 19 '24 at 13:03
  • @JyrkiLahtonen Haha you might think it is, but I'm trying to get an easy-to-follow answer with nothing complex.

    After all, why explain simple things/phenomenons using complex tools?

     – Bulby2024 Nov 19 '24 at 13:03
  • @JyrkiLahtonen Yup I might include the original ver. – Bulby2024 Nov 19 '24 at 13:03

3 Answers3

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Not quite the right formula. The, constant $e$ is defined as the limit

$$ e = \lim_{n \to \infty} \left( 1 + \frac{1}{n} \right)^n. $$ There are many ways to think about $e$ geometrically, but one way for this is to think about interpreting the expression

$$ \left( 1 + \frac{1}{n} \right)^n $$

by imagining a square of side length $1$. If each side grows by a factor of $1 + \frac{1}{n}$ over $n$ steps, the total growth after $n$ steps is given by the product of the growth factors, or $\left( 1 + \frac{1}{n} \right)^n$. So, as $n$ increases, the number of steps increases, but each individual step becomes smaller and smaller. The value, $e$, can be thought of as the result of continuous growth in this process (and it is hence so intimately related to growth and change).

Edit: Another way to see why it is related to growth is by looking at this formula, and that of compound interest. Hope this answers your question somewhat. There are many ways to think about $e$, and there are some good videos on YouTube. Read about a good intuitive explanation here

If you want a bit more of an advanced explanation of why $e$ is related to growth and change so intimately, it is because $e^x$ is the eigenfunction of the derivative operator, which in some sense is like the operator that captures how something changes. You could read more into this, which may lead you down a bit of a functional analysis rabbithole, but definately an interesting one.

MxJ
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    Ohhh! Thanks so much! Can this be shown visually, just curious? – Bulby2024 Nov 19 '24 at 12:33
  • I found this which I think will be worth a read – MxJ Nov 19 '24 at 12:38
  • Nice what a difference that makes. $1^\infty = $undefined. But $\left(1 + \frac{1}{\infty} \right)^\infty$ is defined as e correctly. That extra term really makes a difference because a zero counters an infinity. – devssh Nov 19 '24 at 12:40
  • Great! Yeah, I was responding to the comments a bit too fast, accidentally wrote the wrong limit... – Bulby2024 Nov 19 '24 at 12:40
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    @devssh Just be careful with convention. How you write it with infinity in the denominator and exponent is wrong convention, as you should write $\lim_{n \rightarrow \infty} \ldots$ as you think of it as you approach infinity – MxJ Nov 19 '24 at 12:43
  • Yes, the convention is more important. I wrote it like this without limit to highlight the confusion people have. – devssh Nov 19 '24 at 12:48
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    @devssh Actually, $\left(1 + \frac{1}{\infty} \right)^\infty$ as a form is still indeterminate, because $\lim_{n \to \infty} \left(1 + \frac{1}{\ln n}\right)^n = \infty$ – caduk Nov 19 '24 at 12:56
  • @caduk try this in a calculator 1.0001^10000. Thats why the limit convention is important because it highlights that this is something that can only be tried in a calculator by plugging in a sufficient approximate of n like 10,000. But theoretically it is garbage to prove as more rigorous convention is used and $\infty$ can never truly substitute "n" without giving contradiction. – devssh Nov 19 '24 at 13:01
  • It is similar to how $\frac{0}{0}$ is undefined but you use L'Hopital's rule and suddenly you MAY/MIGHT get a correct answer. The same reason applies for $1^\infty$ – devssh Nov 19 '24 at 13:04
  • $(1 + 0)^\infty$ the 0 represents how sure you are that the 1 is actually a 1. If there is a delta it will shoot $1^\infty$ to 2.718.. , if there is more delta it will make $1^\infty$ to equal 5.436... or 50 or 1.35 . It is similar to how 0/0 can prove 1=2 and 2=50. To understand this better you need to understand delta $\delta$ and impose other conditions and more rigorous convention. $1^\infty$ is gibberish similar to 0/0 or dividing by zero. As for e=2.71 that is just a number we get in base 10. If you change to base 64 or base 16 or base 2 you will get different answers. e has other define – devssh Nov 19 '24 at 13:35
  • Basically it is true that 1.000... 100 zeros ...0002 to the power infinity equals infinity and 0.999... 100 nines ... 998 to power infinity will equal zero. Thus $1^\infty$ is considered indeterminate form similar to $0^\infty$ - being sure that the zero is indeed a zero and x/0 - dividing any number by zero. – devssh Nov 20 '24 at 01:08
  • The problem does not just stem from the precision of 1 in decimal places. There is also another factor which is related to $\infty$ which can also be written as $\frac{1}{0}$ or $\frac{50}{0}$. So $1^\infty$ can be written as $1^\frac{1}{0}$ or $1^\frac{50}{0}$. In limit $n \to \infty$ is fine to substitute $\infty$ with an astronomically large number like the molar mass of 1kg of atoms which is on the order of 10^23 because this big number is still infinitely smaller than infinity and a finite value that can be calculated by experiments in physics. – devssh Nov 21 '24 at 08:34
  • The finite value would prevent us from swapping $\frac{1}{0}$ with $\frac{50}{0}$ or any arbitrary number as it would be $\frac{1}{x}$ where x is a finite number close to zero but not exactly equal to zero avoiding the divide by zero contradiction. – devssh Nov 21 '24 at 09:34
  • The number 0 itself is not a problem if it is written as $0 = \frac{0}{1}$ or $0 = \frac{0}{50}$ but number 0 itself gives a divide by zero error on $0 = \frac{1}{\infty}$ because it can be rewritten as $0 = \frac{1}{\frac{1}{0}}$ or $0 = \frac{1}{\frac{50}{0}}$. Which is what we work around in calculus and limits and math in general. 0 does not exist. But 0 has to exist. As the perfect subtraction of 1-1 – devssh Nov 21 '24 at 09:47
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If you're familiar with binomial expansion, then it will give you a very simple explanation of why $\ \lim_\limits{n\rightarrow\infty}\left(1+\frac{1}{n}\right)^n\ $ cannot be $1$. The binomial theorem tells us that \begin{align} (1+x)^n&=\sum_{k=0}^n{n\choose k}x^k\\ &\ge\sum_{k=0}^1{n\choose k}x^k\ \ \ \text{ if }\ \ x\ge0\\ &= 1+nx\ , \end{align} and putting $\ x=\frac{1}{n}\ $ in this inequality gives $\ \left(1+\frac{1}{n}\right)^n\ge2\ .$ Thus, since every term of the sequence $\ \left\{\left(1+\frac{1}{n}\right)^n\right\}\ $ is at least $2$, its limit cannot be less than that.

lonza leggiera
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  • An elegant demonstration – MxJ Nov 20 '24 at 14:09
  • Thanks! That is the best explanation yet! – Bulby2024 Nov 21 '24 at 05:51
  • You address the main problem correctly that the value has to be at least two by saying the value is at least 1 + n/n which is at least 2. The binomial formula looks complicated but still it is just the simple summation of product expansion of an exponential power. Of course we cannot substitute $1 + \frac{\infty}{\infty}$ and expect it to cancel out to 1. – devssh Nov 21 '24 at 08:41
  • Actually the binomial expansion is how to convert between the compound interest definition of e to and from the sum of factorial definition of e. The full binomial expansion of $(1 + \frac{1}{n})^n$ is $e = \frac{1}{0!} + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} \to \infty$. In this answer we shortened the calculation to the first 2 terms giving that the value must be at least/greater than equal to 2. If you calculate 3 terms it will give the value must be at least/greater than equal to 2.5 and so on. The 4th term calculation gives the value must be at least 2.66667 – devssh Nov 21 '24 at 08:56
  • Because e is a sum of products it is not an algebraic period. Pi is an algebraic period number which can be expressed as the infinite product of numbers. But e cannot be represented as the infinite product of natural numbers and must have summation or product of rational/irrational numbers. – devssh Nov 21 '24 at 09:09
  • Binomial expansion - the binomial refers to the two terms being raised Binomial(1, $\frac{1}{n}$). If there were more terms it would go to multinomial expansion. If we remove the $1^n$ term, then we could do it even simpler with a "unary nomial" expansion, but the cross product will probably create two terms at least anyway(2b and $b^2$). – devssh Nov 21 '24 at 09:32
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Where do you get the idea that it should go to $1$?
Just try with the following values for $n$:

n $(1+\frac{1}{n})^n$
$1$ $2$
$2$ $1.5^2=2.25$
$3$ $(\frac{4}{3})^3 = \frac{64}{27} \approx 2.370370...$
$4$ $(\frac{5}{4})^4 = \frac{625}{256} = 2.44140625$

You see the calculated value getting higher and higher, so why would you believe it to go to $1$? (I understand very well that you see that $\frac{1}{n}$ going to zero, but the $n$ as exponent really gives that "small" deviation real power! :-) )

Dominique
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  • Do the calculation. $(1 + 0.5)^2$ and $(1 + 0.333)^3$ and (1.1)^(10) and (1.01)^(100) – devssh Nov 19 '24 at 12:44
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    Yeah... I know experimentally it does converge, but I don't get why... – Bulby2024 Nov 19 '24 at 12:45
  • @devssh: please correct the MathJax in your comment. – Dominique Nov 19 '24 at 12:47
  • @Bulby: it's not the intention of my answer to show that it converges: it's to show that it gets away from value $1$. There is no way an experiment like this can "show" that the limit converges. – Dominique Nov 19 '24 at 12:48
  • By the way, why are some people so obsessed to downvote my answer? – Dominique Nov 19 '24 at 12:49
  • 1 + some infinitely small delta when raised to power of infinity had to be greater than 1 but less than infinity. If the delta was bigger the whole value would have gone to infinity. So all those small delta added up to a number 2.718... which we named e. An imperceptibly small change to semantics would change this number. Which is why the more mainsteam definition of e is a summation and factorial, not this form which is $1^\infty$ which is hardly a proper numbers calculation without the limit convention that was mentioned earlier like 1.001^1000 in a calculator. – devssh Nov 19 '24 at 12:55
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    The downvotes (not one of mine) are because this does not answer the question. It only provides numerical evidence for the first few iterations. That's not even close to the OPs wondering why the limit isn't 1. – Ethan Bolker Nov 19 '24 at 14:57
  • @EthanBolker: let me explain my actions here: I followed mathematics in secondary school, and although I was following the 8-hours curriculum (we were known as a math geniuses of the school), I was shocked seeing the lack of understanding in mathematics in my class: lots of my classmates looked at definitions purely without even trying to understand the meaning of the things they were learning: continuity as an $\epsilon, \delta$ thing, so were limits, ..., but no one bothered understanding that a limit had a real-life: – Dominique Nov 19 '24 at 15:03
  • A limit is the answer on the question "What should the value of a function be in order to make this function continuous?", "continuity" means that you don't need to lift your pen in order to draw the function, ..., and the $\epsilon, \delta$ definitions were nothing but well-elaborated ways to write those real-life properties into math definitions. – Dominique Nov 19 '24 at 15:05
  • So my answer here goes back to that "real-life meaning" of a limit: here you have a limit in infinity (meaning "What would be the value of that function at infinity?", to which I answer "Well, you start at $1$, and you start moving ($2$, $3$, $4$, ...) and what do you see? Well, you see that you are moving away from $1$, not towards $1$ as the question author seemed to think at first. – Dominique Nov 19 '24 at 15:08
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    @Dominique Ethan was saying that comments are for partial answers and questions. Since you gave an answer to the question which does not categorically answer the question, any reviewer/spectator will downvote it as irrelevant/unanswered incorrectly. – devssh Nov 20 '24 at 01:01