5
$\begingroup$

My friends and myself were discussing Pascal's Triangle, specifically the following property of it.

First, consider the Pascal's Triangle - $$1\\ 1\ 1\\ 1\ 2\ 1\\ 1\ 3\ 3\ 1\\ 1\ 4\ 6\ 4\ 1\\ 1\ 5\ 10\ 10\ 5\ 1\\ 1\ 6\ 15\ 20\ 15\ 6\ 1\\ ..................\\ .....................$$

Now, one interesting observation that can be easily spotted is that initial rows of the Pascal's Triangle follow the form $11^n$ and then a few of the following rows follow $101^n$ and so on.

Now, if we deform the above argument as the following:

  • The zeroth row is of form $(1.1)^0$
  • The first row is of the form $(1.1)^1$
  • Similarly the second row as $(1.1)^2$ the following one as $(1.1)^3$ and then the fourth row as $(1.1)^4$
  • Now for the immediate next row if we follow the pattern shown above we get overflow due to digits being carried forward and hence we represent the fifth row as $(1.01)^5$ yielding $1.0510100501$.
  • Now again we keep on proceeding with the introduction of more zeroes after the decimal point for preventing overflow and preserving the form of the Pascal's Triangle.

Consider the following idea, the $n^\text{th}$ row can be represented as following $(1.\overbrace{000.....0}^{n}1)^n$. Now I grossly miscalculated this, previously. As $n \to \infty$ this value becomes equal to $1$.

Due to the fatal error above, as pointed out in the answers, I rephrase the initial question as: Is there any way to make this series converge to e with the adding of zeroes suitably as needed?

Also, later on studying the Pascal's Triangle further I found another interesting relation.

Consider, $f(n)$ to be the product of all digits in the $n^\text{th}$ row of the Pascal's Triangle. Also, consider the beginning to be the zeroth row. Then, with algebraic manipulation we obtain: $$\frac{f(n)}{f(n-1)} = \frac{n^{(n-1)}}{(n-1)!}$$ from which we can further deduce that $\frac{f(n+1) \times f(n-1)}{f(n)^2}$ converges to $e$ as $n \to \infty$ i.e., $$\lim_{n \to \infty} \frac{f(n+1) \times f(n-1)}{f(n)^2} = e$$

Since, I have updated the problem statement, I am not very sure if the following questions hold .

  • can we connect the initial observation with above observation?
  • is there some correlation between the number $e$ and the product of its digits? (Well this seems quite silly now :))

Update: Thanks for pointing out the mistake.

$\endgroup$
10
  • $\begingroup$ It's not entirely clear the exact process you're describing, but it seems like you're rediscovering the formula (or definition) $e = \lim_{n\to\infty} (1+\frac1n)^n$. $\endgroup$ Commented Apr 28, 2020 at 18:11
  • $\begingroup$ Please see the update $\endgroup$
    – Arjo
    Commented Apr 28, 2020 at 18:56
  • $\begingroup$ The formula $\dfrac{f(n)}{f(n-1)} = \dfrac{n^{n-1}}{(n-1)!}$ actually contradicts the claim right before that, that this fraction would approximate $e^{n-1}$. Rather, via Stirling's formula, this formula implies that $\dfrac{f(n)}{f(n-1)}$ approximates $\dfrac{e^{n-1}}{\sqrt{2\pi(n-1)}}$ (in the sense that the ratio of thiese quantities goes to $1$ for $n\to \infty$). $\endgroup$ Commented Apr 28, 2020 at 18:56
  • $\begingroup$ @TorstenSchoeneberg I believe the value does not converge can you please check with the Stirling Approximation? $\endgroup$
    – Arjo
    Commented Apr 28, 2020 at 19:02
  • 1
    $\begingroup$ The limit $ (1.\overbrace{000.....0}^{n}1)^n \to 1$, not $\to e$. $\endgroup$
    – antkam
    Commented Apr 28, 2020 at 19:30

2 Answers 2

5
$\begingroup$

First of all are you familiar with the Binomial Theorem? It is the reason why e.g. the row $14641$ looks like $(1.1)^4 = 1.4641$ -- just plug in $x = 1, y= 0.1$ into $(x+y)^n$.

Having said that, as you correctly pointed out, for later rows you need to "add zeros" to basically space out the numerals. This is equivalent to plugging in $x= 1, y = 0.01$ into $(x+y)^n$ for row $5$.

Now you also know that $\lim_{n \to \infty} (1 + 1/n)^n = e$; however this crucially relies on the $y = 1/n$ part of the sum dropping as $1/n$. If it drops faster or slower, the limit will be smaller than or larger than $e$. So in your "adding zeros" scheme it all depends on how fast you have to add zeros.

If your rule is always add just enough zeros so that the successive terms in a row do not "overlap", then even in the limit the first digit (the only digit before the decimal point) is always $1$, which is the left edge of the triangle. This rules out the limit being $e = 2.718...$ which has a leading digit $2$. In other words, following your rule, you would have to add zeros so fast (as a function of row number $n$) that the $y$ term in the sum drops faster than $1/n$.

In fact in your case $y = 10^{-D}$ where $D =$ no. of digits in the decimal representation of the biggest coefficient in the row, i.e. the center coefficient ${n \choose \lfloor n/2 \rfloor}$. This coefficient is known to grow exponentially fast, i.e. your $y$ would drop exponentially fast. As a further fact, your limit is actually $1$ because for really large $n$, you'd have to add soooo many zeroes that the first term (which is $n$) would make the decimal number look like $1.00000n...$

$\endgroup$
4
  • $\begingroup$ Please check the updated problem statement, I guess you misinterpret it. $\endgroup$
    – Arjo
    Commented Apr 28, 2020 at 19:06
  • 1
    $\begingroup$ If you add at least as many zeros as $D$, your limit will be $1$, not $e$. In particular, if your $y = 10^{-n}$ i.e. $(1.0000...0001)^n$ with $n$ (or $n-1$) zeros, the limit will certainly be $1$, not $e$. $\endgroup$
    – antkam
    Commented Apr 28, 2020 at 19:12
  • 1
    $\begingroup$ It's also quite obvious that "adding zeroes" to prevent overflow and "make it match an $n$-th row of the triangle" cannot make the limit go to $e$, since after all, each row of the triangle starts with $1$ whereas the decimal expansion of $e$ starts with $2.$. So whatever you try to make a sequence of digits match rows will make that sequence of digits not approach $e$. $\endgroup$ Commented Apr 28, 2020 at 19:18
  • $\begingroup$ Yeah I guess that was a blunder on my side. I guess there is indeed no way. $\endgroup$
    – Arjo
    Commented Apr 28, 2020 at 19:54
4
$\begingroup$

One of the definitions of $e$ is the limit

$$\lim_{n\to\infty}\left(1+\frac1n\right)^n.$$

Hence with $n=10^m$ you get better and better approximations $$(1+10^{-m})^{10^m}.$$

For instance

$$(1+0.0001)^{10000}=2.718146\cdots$$ where the first three decimals are exact.

On the other hand, this number is

$$\binom{10000}{0}+\binom{10000}{1}0.0001+\binom{10000}{2}0.00000001+\cdots\binom{10000}{10000}10^{-40000}$$ $$=1+1.0000+0.49995000+0.166616670000+\cdots$$

You indeed have the sum of Pascal's triangle entries with shifts, but the shifts are insufficient to separate the values and there are overlaps.

Compare to

$$(1+0.00000000001)^{10000}=1.00000010000000499950016661667\cdots$$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .