Archive for Agosto, 2021

Palindromic numbers…

lunedì, Agosto 30th, 2021

A palindrome – as you can read here from wikipedia – is a word, number, phrase, or other sequence of characters which reads the same backward as forward, such as madam or racecar.

A curious series of palindromic numbers is obtained by multiplying by themselves numbers composed of strings of 1 (the 1’s must be at least two to have a product that has more than one digit):

The series of palindromes ends here. If you get to ten 1’s per factor, the product is no longer a palindrome:

The reason why the series is interrupted at ten digits has to do with the ordering by columns of the partial products of the multiplications. The addition of partial products is performed by shifting each subsequent row by one column to the left. This determines first the increase and then the decrease of the sums that generate the final product. As long as the 1’s in the column are at most nine, the mechanism works:

But if the 1’s in one column become ten, then carrying 1 to the next column breaks the symmetry and prevents the product of the multiplication from being a palindrome number.

…quando non siamo sicuri di niente e curiosi di tutto

venerdì, Agosto 27th, 2021

L’anima la si ha ogni tanto. Nessuno la ha di continuo e per sempre. Giorno dopo giorno, anno dopo anno possono passare senza di lei. A volte nidifica un po’ più a lungo sole in estasi e paure dell’infanzia. A volte solo nello stupore dell’essere vecchi. Di rado ci da una mano in occupazioni faticose, come spostare mobili, portare valigie o percorrere le strade con scarpe strette. Quando si compilano moduli e si trita la carne di regola ha il suo giorno libero. Su mille nostre conversazioni partecipa a una, e anche questo non necessariamente, poiché preferisce il silenzio. Quando il corpo comincia a dolerci e dolerci, smonta di turno alla chetichella. È schifiltosa: non le piace vederci nella folla, il nostro lottare per un vantaggio qualunque e lo strepito degli affari la disgustano. Gioia e tristezza non sono per lei due sentimenti diversi. È presente accanto a noi solo quando essi sono uniti. Possiamo contare su di lei quando non siamo sicuri di niente e curiosi di tutto. Tra gli oggetti materiali le piacciono gli orologi a pendolo e gli specchi, che lavorano con zelo anche quando nessuno guarda. Non dice da dove viene e quando sparirà di nuovo, ma aspetta chiaramente simili domande. Si direbbe che così come lei a noi, anche noi siamo necessari a lei per qualcosa.

Wisława Szymborska, da Qualche parola sull’anima.

A (apparently) simple problem …

giovedì, Agosto 26th, 2021

So when I say a simple mathematical problem most would think that I am kidding but I am not, there are many unsolved mathematical problems in the world but this is so simple but yet unsolved.

The problem is called “3n + 1 problem” or “Collatz conjecture”. To understand the problem first we need to understand what it is, so basically just pick natural number if the number is odd then we mutiple the number with 3 and add 1, if the number is even we divide it by 2. We apply these conditions to resultant value. in other words:

In modular arithmetic notation, define the function f as follows:

{\displaystyle f(n)={\begin{cases}{\frac {n}{2}}&{\text{if }}n\equiv 0{\pmod {2}}\\[4px]3n+1&{\text{if }}n\equiv 1{\pmod {2}}.\end{cases}}}

Now form a sequence by performing this operation repeatedly, beginning with any positive integer, and taking the result at each step as the input at the next.

The Collatz conjecture is: This process will eventually reach the number 1, regardless of which positive integer is chosen initially.

To demonstrate the problem let’s consider a number 5; since it is odd we apply 3n+1:

\(3 \cdot 5 + 1 = 16\left( {{\text{even}}} \right);16/2 = 8;8/2 = 4;4/2 = 2;2/2 = 1\)

So we get a value of one but we apply conditions fruther we will be stuck in loop which is 4, 2, 1.

This video sums up the problem well

If the conjecture is false, it can only be because there is some starting number which gives rise to a sequence that does not contain 1. Such a sequence would either enter a repeating cycle that excludes 1, or increase without bound. No such sequence has been found.

Since I am from engineering background here is the Python code for Collatz conjecture:

def collatz(n):
    while n > 1:
        print(n, end=' ')
        if (n % 2):
            # n is odd
            n = 3*n + 1
        else:
            # n is even
            n = n//2
    print(1, end='')
 
 
n = int(input('Enter n: '))
print('Sequence: ', end='')
collatz(n)

The above code is the demonstration of Collatz conjecture…

The Ramanujan Summation…

mercoledì, Agosto 25th, 2021
Srinivasa Aiyangar Ramanujan

Ramanujan summation – as you can read from Wikipedia – is a technique invented by the mathematician Srinivasa Ramanujan for assigning a value to divergent infinite series. Although the Ramanujan summation of a divergent series is not a sum in the traditional sense, it has properties that make it mathematically useful in the study of divergent infinite series, for which conventional summation is undefined.

According to Ramanujan, if you add all the natural numbers, that is 1, 2, 3, 4, and so on, all the way to infinity, you will find that it is equal to -1/12. Yup, -0.08333333333.

\( \xi \left( { – 1} \right) = 1 + 2 + 3 + 4 + \ldots = – \frac{1}{{12}}\)

Don’t believe me? Keep reading to find out how I prove this, by proving two equally crazy claims:

\(1 – 1 + 1 – 1 + 1 – 1 + \ldots = 1/2 \) \(1 – 2 + 3 – 4 + 5 – 6 + \ldots = 1/4\)

First off, the bread and butter. This is where the real magic happens, in fact the other two proofs aren’t possible without this.

I start with a series, A, which is equal to 1–1+1–1+1–1 repeated an infinite number of times. I’ll write it as such:

\(A = 1 – 1 + 1 – 1 + 1 – 1 + \ldots \)

We can write, now:

\(1 – A = 1 – \left[ {1 – 1 + 1 – 1 + 1 – 1 + \ldots } \right]\)

so, we have:

\(1-A=A\)

and then:

\(A=\frac{1}{2}.\)

This little beauty is Grandi’s series, called such after the Italian mathematician, philosopher, and priest Guido Grandi.

Now, let to prove that:

\(1 – 2 + 3 – 4 + 5 – 6 + \ldots = 1/4\)

We start the same way as above, letting the series B =1–2+3–4+5–6⋯. Then we can start to play around with it. This time, instead of subtracting B from 1, we are going to subtract it from A. Mathematically, we get this:

\(A – B = \left( {1 – 1 + 1 – 1 + 1 – 1 + \ldots } \right) – \left( {1 – 2 + 3 – 4 + 5 – 6 + \ldots } \right)\)

and then:

\(A – B = \left( {1 – 1 + 1 – 1 + 1 – 1 + \ldots } \right) – 1 + 2 – 3 + 4 – 5 + 6 – \ldots \)

Then we shuffle the terms around a little bit, and we see another interesting pattern emerge.

\(A – B = \left( {1 – 1} \right) + \left( { – 1 + 2} \right) + \left( {1 – 3} \right) + \left( { – 1 + 4} \right) + \left( {1 – 5} \right) + \left( { – 1 + 6} \right) + \ldots \) \(A – B = 0 + 1 – 2 + 3 – 4 + 5 + \ldots \)

Once again, we get the series we started off with, and from before, we know that A = 1/2, so we use some more basic algebra and prove our second mind blowing fact of today.

\(A – B = B \Leftrightarrow A = 2B \Rightarrow 1/2 = 2B \Leftrightarrow B = 1/4\)

And voila! This equation does not have a fancy name, since it has proven by many mathematicians over the years while simultaneously being labeled a paradoxical equation. Nevertheless, it sparked a debate amongst academics at the time, and even helped extend Euler’s research in the Basel Problem and lead towards important mathematical functions like the Riemann Zeta function.

Now for the icing on the cake, the one you’ve been waiting for, the big cheese. Once again we start by letting the series C = 1+2+3+4+5+6⋯, and you may have been able to guess it, we are going to subtract C from B.

\(B – C = \left( {1 – 2 + 3 – 4 + 5 – 6 + \ldots } \right) – 1 – 2 – 3 – 4 – 5 – 6 – \ldots \)

Because math is still awesome, we are going to rearrange the order of some of the numbers in here so we get something that looks familiar, but probably wont be what you are suspecting.

\(B – C = \left( {1 – 1} \right) + \left( { – 2 – 2} \right) + \left( {3 – 3} \right) + \left( { – 4 – 4} \right) + \ldots \)

and then,

\(B – C = 0 – 4 + 0 – 8 + 0 – 12 + \ldots = – 4 – 8 – 12 – \ldots \)

At this point, we can write:

\(B-C=-4(1+2+3+\ldots)\)

that is to say

\(B-C=-4C\)

and finally

\(B=-3C\)

And since we have a value for B=1/4, we simply put that value in and we get our magical result:

\(C=-1/12\)

or, in other words:

\( \xi \left( { – 1} \right) = 1 + 2 + 3 + 4 + \ldots = – \frac{1}{{12}}\)

Now, why this is important… Well for starters, this result is used in string theory. More. The Ramanujan Summation also has had a big impact in the area of general physics – specifically in the solution to the phenomenon know as the Casimir Effect. Hendrik Casimir predicted that given two uncharged conductive plates placed in a vacuum, there exists an attractive force between these plates due to the presence of virtual particles bread by quantum fluctuations. In Casimir’s solution, he uses the very sum we just proved to model the amount of energy between the plates. And there is the reason why this value is so important.

Find all nonnegative integers a and b such that…

martedì, Agosto 24th, 2021

The problem is the following, and is a modified version of the 2009 British Mathematical Olympiad issue: Find all nonnegative integers a and b such that \(\sqrt{a}+\sqrt{b}=\sqrt{2020}.\)

Before looking at one possible way of solving this problem, which requires nothing more than school-level arithmetic, I want to explain why I like this problem, and problems of this nature, so much. Growing up, I loved detective/mystery novels, tv shows, movies, whatever. Being able to construct a solution based on some scattered information seemed almost like a superpower to me.

Let’s begin our solution to the problem. The first step most people would take is to square both sides of the equation. This is something we are always told to do when dealing with square roots. It’s certainly what I tried. Let’s do this with the equation as it is, so we have:

\(\left(\sqrt{a}+\sqrt{b}\right)^2=2020\)

or

\(a+b+2\sqrt{a}{b}=2020\)

Ok, so this is a little nicer, maybe. We have reduced the number of square roots, which is good, but we have also made an unintentional problem for ourselves.

This is the first thing this problem taught me. Sometimes, before jumping in and just applying a method that seems right, ask yourself:
(1) Is this indeed the best method you know for dealing with this problem?
(2) If this is the best method you know, is the problem set up in the best way for you to apply it?

Let me elaborate here. There is nothing wrong with what we have done. However, in squaring the equation, in the form it is was given, we have added in the variable square root of the product \(ab\). This has made things harder for us since we have now unintentionally muddled together the information given by the variables a and b.

What if instead we first rearranged our equation as

\(\sqrt{a}=\sqrt{2020}-\sqrt{b}\)

Then we can again square both sides, only this time we get:

\(a=2020-2\sqrt{2020b}+b\)

Notice that this time we have again reduced the number of square roots we had, but we have additionally kept the variables a and b separate. This may seem like a simple difference but it makes all the difference with how we can proceed. We now make note of the fact that all terms of our new equation are clearly integers, except potentially \(2\sqrt{2020b}\).

Here comes the second thing I learned from this problem, deductive reasoning. Since we know that adding or subtracting two integers always results in an integer, and since we can re-write our equation as

\(b-a+2020=2\sqrt{2020b}\)

we can deduce that \(2\sqrt{2020b}\) must indeed be an integer. This is a huge clue!

We first notice that

\(2020=2^2 \cdot 5 \cdot 101\)

Thus we can say that

\(\sqrt{2020b}=2\sqrt{505b}\)

So, since \(\sqrt{2020b}\) is an integer, we must have that \(\sqrt{505b}\) is an integer. It means that we must have, whit \(c \in \mathbf{N} \):

\(505b=c^2\)

But, the \(\sqrt{505}\) is not an integer, so this only makes sense if \(b=505d^2\) for \(d \in \mathbf{N}\).

So, we have:

\(505b=\left(41d\right)^2\)

or

\(b=505d^2.\)

The third thing I learned from this problem: Don’t waste your time! What I mean by this is the following, there was nothing special about moving the \(b\) across the equals sign in our original equation, we could just have easily started with the equation:

\(\sqrt{b}=\sqrt{2020}-\sqrt{a}\)

and everything would follow in the exact same way. In particular, we would arrive at a similar conclusion that \(a=505 e^2\) for some integer \(e\).

We call this a symmetric argument and really, all we are saying is that since it is clear that everything will work the exact same, if we were to replace the variable \(b\) with the variable \(a\), we are not willing to write the argument down again and instead we will just skip to the conclusion. With \(e,d \in \mathbf{N}\):

\(a= 505 e^2\) \(b=505 d^2\)

So, going back to our original equation, we have:

\(\sqrt{505e^2}+\sqrt{505d^2}=\sqrt{2020}=2\sqrt{505}\)

or

\(e\sqrt{505}+d\sqrt{505}=2\sqrt{505}\)

then

\(e+d=2\)

This is certainly a much easier equation to deal with. There are only three solutions to this equation for d and e. In particular, we can have:

\(d=0, e=2 \Rightarrow a=2020, b=0\)

\(d=1, e=1 \Rightarrow a=505, b=505\)

\(d=2, e=0 \Rightarrow a=0, b=2020\)

…un mostro diabolico

venerdì, Agosto 6th, 2021
una persona che è evaporata

Il portico è quello della filiale della Sumitomo Bank Company. Nel Memoriale del Museo della Pace di Hiroshima, quell’ombra sul gradino viene descritta come “una persona che è evaporata”. Nel 1996, quell’ombra è stata identificata come quella della signora Mitsuno Ochi (nata nel 1903) che all’epoca dell’esplosione aveva 42 anni e i cui discendenti sono tuttora viventi.
Hiroshima è la dimostrazione che l’essere umano può essere un mostro diabolico.

L’atroce dilemma…

giovedì, Agosto 5th, 2021

C’è sul Corriere in edicola una intervista, a firma di Tommaso Labate, assai interessante a Ignazio La Russa. Il nostro racconta i dettagli della giornata trascorsa insieme a Giorgia Meloni a Villa Certosa, residenza estiva di Silvio Berlusconi in Sardegna. Vi ho appreso aneddoti assai gustosi. Pare, per dire, siano andati sulla macchinetta elettrica, che Berlusconi guida “stile Formula 1”; ancora: hanno visto la collezione di farfalle – Berlusconi “ha delle farfalle vive e delle farfalle imbalsamate”, dice La Russa estasiato – e hanno pranzato accanto al vulcano finto. A tavola la Meloni ha preso solo il salmone, ma poi Berlusconi l’ha invitata ad assaggiare anche la parmigiana, che – sostiene La Russa – “era molto buona”. Un racconto, ripeto, assai gustoso, fresco, scoppiattante; un racconto che, una volta letto, lascia nel lettore l’atroce dilemma: “in questi casi, scusate, si dice mecojoni o ‘sticazi?”.