Month: November 2020

Pronunciation and origin of my name.

November 19, 2020 Incoherent ramblings , , , , , ,

The question of how to pronounce my names is frequently asked.

Joot isn’t Dutch, but is Estonian.  I’m not sure what sort of linguistic crossover there is between the two languages, if any(**).

Dad was Estonian, and he wanted us all to have Estonian spellings of our names (Peeter, Krista, Erik, Karin.)  Dad pronounced my name with the standard North American pronunciation for Peter.  However, my Vanaema (grandmother) pronounced Peeter with enunciation of all the e’s in a way that I can’t actually vocalize myself.  Estonian words have lots of doubled vowels (google finds me Kuulilennuteetunneliluuk as an example (a nice long palindrome (*))).  Unlike doubled vowels in English, if they are there, it’s because they should all be pronounced.

If somebody named Peter says that I spell my name wrong, I rebut by calling them pet-er, since the long e requires vowel doubling per English spelling conventions (i.e. my name is spelled correctly, but their spelling is wrong.)

My last name Joot is pronounced as like “Yoat”, like oat. I can’t recall the subtleties of how Vanaema pronounced Joot, but I’m sure she also somehow enunciated both o’s.  When I was a kid, I was very inflexible about the pronunciation of my name, and insisted on “Yoat”, not “Jewt”.  That inflexibility was too much work, and I mellowed out considerably over time.  I now flexible and respond to anything that approximates any possible pronunciation that I can recognize, and no longer correct anybody.

People correct the spelling of names for me all the time, as they couldn’t possibly be spelled right as is.

 

(*)

Originally I thought I saw an article that said that kuulilennuteetunneliluuk also meant palindrome, but cannot find that anymore.  Instead, googling this word, I find it translated as “the hatch a bullet flies out of when exiting a tunnel“.  If kuulilennuteetunneliluuk actually meant palindrome, that would be the most amazing word for palindrome in any language!  I’m very sad that I appear to have gotten the meaning wrong.  My hope for the future of linguistics, is that Estonians will start using kuulilennuteetunneliluuk as a word for palindrome, giving it a second meaning through popular use.  If that trend can be started, eventually the Estonian language has the best word for palindrome in any language.

 

(**)

On the other hand, Dad said he could understand most of Finnish when spoken (but said that Finns couldn’t understand him.)  I’m guessing that this means Finnish was probably a root of Estonian, but dialect could also be a factor, as I’ve since met Finns that said they could understand some Estonian.  Dad talked about the dialect variation from Estonian town to town at the beginning of the 1900’s, which was apparently so bad that understanding somebody from a few towns away could be difficult.  By the time he was born, radio was starting to obliterate that dialect variation.  He also wouldn’t have heard that dialect variation first hand, since he escaped the Soviet invasion of Estonia with my grandmother when he was only 3.  His refugee journey started in Finland (who had a pact with the Soviets to kick out refugees after some fixed time (i.e.: the Soviet’s said “kick out refugees, or else we’ll invade you too!”)   After a few years in Sweden, Dad and Vanaema eventually landed in Canada.

EBCDIC, thou art evil.

November 18, 2020 Mainframe , , ,

Here’s a bit of innocuous code. It was being compiled with gcc -fexec-charset=1047, so all the characters and strings were being treated as EBCDIC. For example ‘0’ = ‘\xF0’.

    if (c >= '0' && c <= '9')                                                                                                         
         c -= '0';                                                                                                                    
    else if (c >= 'A' && c <= 'Z')                                                                                                    
         c -= 'A' - 10;                                                                                                               
    else if (c >= 'a' && c <= 'z')                                                                                                    
         c -= 'a' - 10;                                                                                                               
    else                                                                                                                              
         break;

Specifying the charset is not enough to port this code to the mainframe.  The problem is that EDCDIC is completely braindead, and DOESN’T PUT THE FRIGGEN LETTERS TOGETHER!

The letters are clustered in groups:

  • a-i
  • j-r
  • s-z
  • A-I
  • J-R
  • S-Z

with whole piles of crap between each range of characters, so comparisons like c >= ‘A’ && c <= ‘Z’ are useless, as are constructions like (c-‘A’-10) since c in J-R or S-Z will break that.

Now I have a big hunt and destroy task ahead of me.  I can fix this code, but where else are problems like this lurking!

Guessing the nth Fibonacci number formula

November 17, 2020 math and physics play , , , ,

[If mathjax doesn’t display properly for you, click here for a PDF of this post]

My last two posts:

  1. The nth term of a Fibonacci series.
  2. More on that cool Fibonacci formula.

were both about a cool formula for the n-th term of the Fibonacci series.  Looks like I’m not done playing with this beastie.  A recap:

Definition 1.1: Fibonacci series.

With \( F_0 = 0 \), and \( F_1 = 1 \), the nth term \( F_n \) in the Fibonacci series is the sum of the previous two terms
\begin{equation*}
F_n = F_{n-2} + F_{n-1}.
\end{equation*}

Theorem 1.1: Nth term of the Fibonacci series.

\begin{equation*}
F_n = \frac{ \lr{ 1 + \sqrt{5} }^n – \lr{ 1 – \sqrt{5} }^n }{ 2^n \sqrt{5} }.
\end{equation*}

 

The guess.

We can rearrange the formula for the nth Fibonacci number as a difference equation
\begin{equation}\label{eqn:fibonacci:260}
F_n – F_{n-1} = F_{n-2}.
\end{equation}
This is a second order difference equation, so my naive expectation is that there are two particular solutions involved. We know the answer, so it’s not too hard to guess that the particular form of the solution has the following form
\begin{equation}\label{eqn:fibonacci:280}
F_n = \alpha a^n + \beta b^n.
\end{equation}
Given this guess, can we take some of the magic out of the formula, by just solving for \( \alpha, \beta, a, b \)? Let’s try that
\begin{equation}\label{eqn:fibonacci:300}
F_0 = \alpha + \beta = 0,
\end{equation}
\begin{equation}\label{eqn:fibonacci:320}
\begin{aligned}
F_1 &= \alpha a + \beta b \\
&= \alpha \lr{ a – b } \\
&= 1,
\end{aligned}
\end{equation}
and
\begin{equation}\label{eqn:fibonacci:340}
\begin{aligned}
F_n
&= F_{n-1} + F_{n-2} \\
&=
\alpha \lr{ a^{n-1} + a^{n-2} }
-\alpha \lr{ b^{n-1} + b^{n-2} } \\
&=
\alpha a^{n-2} \lr{ 1 + a }
-\alpha b^{n-2} \lr{ 1 + b },
\end{aligned}
\end{equation}
so
\begin{equation}\label{eqn:fibonacci:360}
\begin{aligned}
a^2 &= a + 1 \\
b^2 &= b + 1.
\end{aligned}
\end{equation}
If we complete the square we find
\begin{equation}\label{eqn:fibonacci:380}
\lr{ a – \inv{2} }^2 = 1 + \inv{4} = \frac{5}{4},
\end{equation}
or
\begin{equation}\label{eqn:fibonacci:400}
a, b = \inv{2} \pm \frac{\sqrt{5}}{2}.
\end{equation}
Out pop the golden ratio and it’s complement. Clearly we need to pick alternate roots for \( a \) and \( b \) or else we’d have zero for every value of \( n > 0 \). Suppose we pick the positive root for \( a \), then to find the scaling constant \( \alpha \), we just compute
\begin{equation}\label{eqn:fibonacci:420}
\begin{aligned}
1
&=
\alpha \lr{ \frac{ 1 + \sqrt{5}}{2} – \frac{ 1 – \sqrt{5} }{2} } \\
&= \alpha \sqrt{5},
\end{aligned}
\end{equation}
so our system \ref{eqn:fibonacci:280} has the solution:
\begin{equation}\label{eqn:fibonacci:520}
\begin{aligned}
a &= \frac{1 + \sqrt{5}}{2} \\
b &= \frac{1 – \sqrt{5}}{2} \\
\alpha &= \inv{\sqrt{5}} \\
\beta &= -\inv{\sqrt{5}}.
\end{aligned}
\end{equation}

We now see a path that will systematically lead us from the Fibonacci difference equation to the final result, and have only to fill in a few missing steps to understand how this could be discovered from scratch.

Motivating the root-fives.

I showed this to Sofia, and she came up with a neat very direct way to motivate the \( \sqrt{5} \). It follows naturally (again knowing the answer), by assuming the Fibonacci formula has the following form:
\begin{equation}\label{eqn:fibonacci:440}
F_n = \inv{x} \lr{
\lr{ \frac{1 + x}{2}}^n
–
\lr{ \frac{1 – x}{2}}^n
}.
\end{equation}
We have only to plug in \( n = 3 \) to find
\begin{equation}\label{eqn:fibonacci:460}
\begin{aligned}
2 x
&= \inv{4} \lr{ 1 + 3 x + 3 x^2 + x^3 – \lr{ 1 – 3 x + 3 x^2 – x^3 } } \\
&= \inv{2} \lr{ 3 x + x^3 },
\end{aligned}
\end{equation}
or
\begin{equation}\label{eqn:fibonacci:480}
8 = 3 + x^2,
\end{equation}
so
\begin{equation}\label{eqn:fibonacci:500}
x = \pm \sqrt{5}.
\end{equation}
Again the \( \sqrt{5} \)’s pop out naturally, taking away some of the mystery of the cool formula.

More on that cool Fibonacci formula

November 15, 2020 math and physics play

[If mathjax doesn’t display properly for you, click here for a PDF of this post]

In my previous post, I explored the following cool formula for the nth term of the Fibonacci series. In this post, I’ll show why there are no square root fives after evaluation. A recap:

Definition 1.1: Fibonacci series.

With \( F_0 = 0 \), and \( F_1 = 1 \), the nth term \( F_n \) in the Fibonacci series is the sum of the previous two terms
\begin{equation*}
F_n = F_{n-2} + F_{n-1}.
\end{equation*}

Theorem 1.1: Nth term of the Fibonacci series.

\begin{equation*}
F_n = \frac{ \lr{ 1 + \sqrt{5} }^n – \lr{ 1 – \sqrt{5} }^n }{ 2^n \sqrt{5} }.
\end{equation*}

How the square root fives cancel out.

One of the interesting things in this Fibonacci formula, is the \( \sqrt{5} \)’s that are all over the place, while the formula represents only integer values. Expanding the formula in binomial series shows us exactly why those terms all vanish. Consider the first few values of \( n \) explicitly.
\begin{equation}\label{eqn:fibonacci:160}
\begin{aligned}
F_1
&= \frac{ 1 + \sqrt{5} – \lr{ 1 – \sqrt{5} } }{ 2^1 \sqrt{5} } \\
&= \frac{ 2 \sqrt{5} }{ 2^1 \sqrt{5} } \\
&= 1,
\end{aligned}
\end{equation}
\begin{equation}\label{eqn:fibonacci:180}
\begin{aligned}
F_2
&= \frac{ 1 + 2 \sqrt{5} + 5 – \lr{ 1 – 2 \sqrt{5} + 5 } }{ 2^2 \sqrt{5} } \\
&= \frac{ 4 \sqrt{5} }{ 2^2 \sqrt{5} } \\
&= 1,
\end{aligned}
\end{equation}
\begin{equation}\label{eqn:fibonacci:200}
\begin{aligned}
F_3
&= \frac{ 1 + 3 \sqrt{5} + 3 (5) + \sqrt{5} 5 – \lr{ 1 – 3 \sqrt{5} + 3(5) – \sqrt{5} 5 } }{ 2^3 \sqrt{5} } \\
&= \frac{ 2 \lr{ 3 \sqrt{5} + \sqrt{5} 5 } }{ 2^3 \sqrt{5} } \\
&= \frac{ 3 + 5 }{ 2^2 } \\
&= 2.
\end{aligned}
\end{equation}
In the general case, we have
\begin{equation}\label{eqn:fibonacci:220}
\begin{aligned}
2^n \sqrt{5} F_n
&=
\sum_{k = 0}^n
\binom{n}{k}
{\sqrt{5}}^k
–
\sum_{k = 0}^n \binom{n}{k} (-\sqrt{5})^k \\
&=
2 \sum_{1 \le k \le n, \mbox{$k$ is odd}} \binom{n}{k} (\sqrt{5})^k \\
&=
2 \sqrt{5} \sum_{m = 0}^{\lfloor (n-1)/2 \rfloor} \binom{n}{2 m + 1} 5^m,
\end{aligned}
\end{equation}

so (for any \( n > 0 \)),
\begin{equation}\label{eqn:fibonacci:240}
F_n =
\inv{2^{n-1}} \sum_{m = 0}^{\lfloor (n-1)/2 \rfloor } \binom{n}{2 m + 1} 5^m.
\end{equation}
Since only the odd powers of \( \sqrt{5} \) in the binomial expansions survive, the root in the basement is obliterated every time, leaving only integers upstairs, and a power of two factor downstairs. It is still somewhat remarkable seeming that there is always a perfect cancellation of all the factors of two in the basement.

The nth term of a Fibonacci series.

November 13, 2020 math and physics play , , , ,

[If mathjax doesn’t display properly for you, click here for a PDF of this post.]

I’ve just started reading [1], but already got distracted from the plot by a fun math fact. Namely, a cute formula for the nth term of a Fibonacci series. Recall

Definition 1.1: Fibonacci series.

With \( F_0 = 0 \), and \( F_1 = 1 \), the nth term \( F_n \) in the Fibonacci series is the sum of the previous two terms
\begin{equation*}
F_n = F_{n-2} + F_{n-1}.
\end{equation*}

We can quickly find that the series has values \( 0, 1, 1, 2, 3, 5, 8, 13, \cdots \). What’s really cool, is that there’s a closed form expression for the nth term in the series that doesn’t require calculation of all the previous terms.

Theorem 1.1: Nth term of the Fibonacci series.

\begin{equation*}
F_n = \frac{ \lr{ 1 + \sqrt{5} }^n – \lr{ 1 – \sqrt{5} }^n }{ 2^n \sqrt{5} }.
\end{equation*}

This is a rather miraculous and interesting looking equation. Other than the \(\sqrt{5}\) scale factor, this is exactly the difference of the nth powers of the golden ratio \( \phi = (1+\sqrt{5})/2 \), and \( 1 – \phi = (1-\sqrt{5})/2 \). That is:
\begin{equation}\label{eqn:fibonacci:60}
F_n = \frac{\phi^n – (1 -\phi)^n}{\sqrt{5}}.
\end{equation}

How on Earth would somebody figure this out? According to Tattersal [2], this relationship was discovered by Kepler.

Understanding this from the ground up looks like it’s a pretty deep rabbit hole to dive into. Let’s save that game for another day, but try the more pedestrian task of proving that this formula works.

Start proof:

\begin{equation}\label{eqn:fibonacci:80}
\begin{aligned}
\sqrt{5} F_n
&=
\sqrt{5} \lr{ F_{n-2} + F_{n-1} } \\
&=
\phi^{n-2} – \lr{ 1 – \phi}^{n-2}
+ \phi^{n-1} – \lr{ 1 – \phi}^{n-1} \\
&=
\phi^{n-2} \lr{ 1 + \phi }
-\lr{1 – \phi}^{n-2} \lr{ 1 + 1 – \phi } \\
&=
\phi^{n-2}
\frac{ 3 + \sqrt{5} }{2}
-\lr{1 – \phi}^{n-2}
\frac{ 3 – \sqrt{5} }{2}.
\end{aligned}
\end{equation}
However,
\begin{equation}\label{eqn:fibonacci:100}
\begin{aligned}
\phi^2
&= \lr{ \frac{ 1 + \sqrt{5} }{2} }^2 \\
&= \frac{ 1 + 2 \sqrt{5} + 5 }{4} \\
&= \frac{ 3 + \sqrt{5} }{2},
\end{aligned}
\end{equation}
and
\begin{equation}\label{eqn:fibonacci:120}
\begin{aligned}
(1-\phi)^2
&= \lr{ \frac{ 1 – \sqrt{5} }{2} }^2 \\
&= \frac{ 1 – 2 \sqrt{5} + 5 }{4} \\
&= \frac{ 3 – \sqrt{5} }{2},
\end{aligned}
\end{equation}
so
\begin{equation}\label{eqn:fibonacci:140}
\sqrt{5} F_n = \phi^n – (1-\phi)^n.
\end{equation}

End proof.

References

[1] Steven Strogatz and Don Joffray. The calculus of friendship: What a teacher and a student learned about life while corresponding about math. Princeton University Press, 2009.

[2] James J Tattersall. Elementary number theory in nine chapters. Cambridge University Press, 2005.