This could go both in Physics and Math dhaga .. An exciting discovery in finding elegant proof in Math,, inspired by results from physicists. Wow!

Three physicists wanted to calculate how neutrinos change. They ended up discovering an unexpected relationship between some of the most ubiquitous objects in math. Basic Math.

See details in physics dhaga.

## BR Maths Corner-1

### Re: BR Maths Corner-1

xpost from physics dhaga

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Since we were talking about Collatz Conjecture a couple of years (maybe more time has passed, I don't remember), here is some nice result by Terence Tao.

Summary of the result: Prof. Tao proves that if the starting numbers are greater than one quadrillion, 99% of those numbers become smaller than 200.

I think this approach can be used to prove probabilistic results in Number Theory.

Here is the link

https://www.quantamagazine.org/mathemat ... -20191211/

Summary of the result: Prof. Tao proves that if the starting numbers are greater than one quadrillion, 99% of those numbers become smaller than 200.

I think this approach can be used to prove probabilistic results in Number Theory.

Here is the link

https://www.quantamagazine.org/mathemat ... -20191211/

### Re: BR Maths Corner-1

^^Thanks for posting, yes we have talked about this here.. (also known as 3k+1 problem).

Basically it is easy to state (and has been a basis for a few math problems)

- start with any number.

- If it is even divide by half

- If it is odd multiply be 3 and add 1

Repeat above with the new number you got.. what will happen? ( For example if you start with 7 it goes ->22 -> 11->34 ->17 ->52 ->26 -> 13 -> 40 -> 20 -> 10 -> 5 -> 16 -> 8 -> 4 -> 2 ->1 and then repeats 4,2,1 ...). Does this happen all the time?

Basically it is easy to state (and has been a basis for a few math problems)

- start with any number.

- If it is even divide by half

- If it is odd multiply be 3 and add 1

Repeat above with the new number you got.. what will happen? ( For example if you start with 7 it goes ->22 -> 11->34 ->17 ->52 ->26 -> 13 -> 40 -> 20 -> 10 -> 5 -> 16 -> 8 -> 4 -> 2 ->1 and then repeats 4,2,1 ...). Does this happen all the time?

### Re: BR Maths Corner-1

Happy 2020!

Here is something some may find amazing.

By using all the integers from 1 to 9 you get a very good approximation for e which is accurate up to about 10^26 (18457734525360901453873570 ) decimal DIGITS!

The result was found by mathematician R. Sabey in 2004.

It is not too hard to verify this. Try it you will enjoy it.

Note "6.7" above is 6 multiplied by 7 (or42)

Here is something some may find amazing.

By using all the integers from 1 to 9 you get a very good approximation for e which is accurate up to about 10^26 (18457734525360901453873570 ) decimal DIGITS!

The result was found by mathematician R. Sabey in 2004.

It is not too hard to verify this. Try it you will enjoy it.

Note "6.7" above is 6 multiplied by 7 (or42)

### Re: BR Maths Corner-1

Amber G. wrote:Happy 2020!

Here is something some may find amazing.

By using all the integers from 1 to 9 you get a very good approximation for e which is accurate up to about 10^26 (18457734525360901453873570 ) decimal DIGITS!

The result was found by mathematician R. Sabey in 2004.

It is not too hard to verify this. Try it you will enjoy it.

Note "6.7" above is 6 multiplied by 7 (or42)

That's clever, I can see how it works (series expansion, and a creative use of the 9, 4, 6, 7, 3, 2, and 85 - that part was neat). Basically the second term within the brackets, is the inverse of the exponential term - this is the key. I can kind of verify how many digits of precision it would give you - if I went with logarithms, I can figure out what the error would add up to - 3^(2^85) ~= 3^(10^28) ~= 10^(5*10^27) - so I guess we kind of get the 10^26 decimal places of precision right there.

If we designate 3^(2^85) as X, then it's the error between (X-1)/((2!)*X), versus 1/(2!), basically - the remaining terms in the Taylor series should rapidly decline in terms of contribution to the error.

More like creative use of the digits, than a true discovery. I'm impressed at the art behind it.

But you know what, I just figured out that one can do this with all ten digits (0 to 9) by adding a 0 at any of several places in the above equation . This is my big discovery, right now (2020).

### Re: BR Maths Corner-1

sudarshan wrote:Amber G. wrote:Happy 2020!

Here is something some may find amazing.

By using all the integers from 1 to 9 you get a very good approximation for e which is accurate up to about 10^26 (18457734525360901453873570 ) decimal DIGITS!

The result was found by mathematician R. Sabey in 2004.

<equation e ≈ (1+9^4^(7*6))^3^2^85)>It is not too hard to verify this. Try it you will enjoy it.

Note "6.7" above is 6 multiplied by 7 (or42)

That's clever, I can see how it works (series expansion, and a creative use of the 9, 4, 6, 7, 3, 2, and 85 - that part was neat). Basically the second term within the brackets, is the inverse of the exponential term - this is the key. I can kind of verify how many digits of precision it would give you - if I went with logarithms, I can figure out what the error would add up to - 3^(2^85) ~= 3^(10^28) ~= 10^(5*10^27) - so I guess we kind of get the 10^26 decimal places of precision right there.

If we designate 3^(2^85) as X, then it's the error between (X-1)/((2!)*X), versus 1/(2!), basically - the remaining terms in the Taylor series should rapidly decline in terms of contribution to the error.

More like creative use of the digits, than a true discovery. I'm impressed at the art behind it.

But you know what, I just figured out that one can do this with all ten digits (0 to 9) by adding a 0 at any of several places in the above equation . This is my big discovery, right now (2020).

"Adding 0" is nice.. ..... Thanks.

---

To me, the very definition of e (base of natural logarithm) is that it is a limiting value of (1+1/x)^x when x is very large. This is intuitive and this is how I was introduced to e and this is how I introduce e for the first time to elementary students. (compound interest problem when interest is calculated "continuously"). As you pointed out, x = 9^(4^42) = 9^(2^84) = 3^(2^85) is a VERY large number so by very definition, the whole expression is very close to e.

### Re: BR Maths Corner-1

That "compound interest calculated continuously" seems like a very intuitive definition for e. Of course, the interest rate would have to be just right (as in - 1/N for a period of N years/ days/ seconds/ whatever - of course, with N being a very long interval - tending to infinity).

### Re: BR Maths Corner-1

OK, so to define e for a lay person (just solidifying my own understanding of the concept):

(Pi is easy, ratio of circumference of a circle to its diameter, pretty intuitive)

e~=2.718282

So you go to a bank, and deposit Re. 1. They give you 100% interest per annum, for a year.

If this is simple interest, at the end of 1 year, you get Rs. 2. If this is compound interest, even then you get Rs. 2 after a year.

Now the bank changes its policy, they will give you 50% interest, every six months, for the duration of a year.

If this is simple interest, at the end of 1 year, you still get Rs. 2. If this is compound interest, however, you will get Rs. 2.25. So compound interest over simple, will yield you an additional 25 paise.

The bank now changes its policy to - 33.33% interest (100%/3), every four months, for the duration of a year.

Simple interest - still Rs. 2 after a year. Compound interest - roughly Rs. 2.37.

25% interest every 3 months for a year? Rs. 2, vs. Rs. 2.44 (roughly).

10% interest, 10 times a year, for a year? Rs. 2, vs. Rs. 2.59 (roughly).

1% interest, 100 times a year, for a year? Rs. 2, vs. Rs. 2.70 (roughly).

If they split up and paid the interest every second, for a year? Rs. 2, vs. Rs. 2.718282... (very close to e).

So as the interest is split and compounded over more and more intervals within the year (i.e., as the calculation becomes more and more continuous), we approach e (2.718282...).

So the "compound interest advantage over simple" is Rs. 0.718282..., or e-2.

Never thought of it that way before, but it is a great way to explain e to a lay person. Thanks for that pointer.

(Pi is easy, ratio of circumference of a circle to its diameter, pretty intuitive)

e~=2.718282

So you go to a bank, and deposit Re. 1. They give you 100% interest per annum, for a year.

If this is simple interest, at the end of 1 year, you get Rs. 2. If this is compound interest, even then you get Rs. 2 after a year.

Now the bank changes its policy, they will give you 50% interest, every six months, for the duration of a year.

If this is simple interest, at the end of 1 year, you still get Rs. 2. If this is compound interest, however, you will get Rs. 2.25. So compound interest over simple, will yield you an additional 25 paise.

The bank now changes its policy to - 33.33% interest (100%/3), every four months, for the duration of a year.

Simple interest - still Rs. 2 after a year. Compound interest - roughly Rs. 2.37.

25% interest every 3 months for a year? Rs. 2, vs. Rs. 2.44 (roughly).

10% interest, 10 times a year, for a year? Rs. 2, vs. Rs. 2.59 (roughly).

1% interest, 100 times a year, for a year? Rs. 2, vs. Rs. 2.70 (roughly).

If they split up and paid the interest every second, for a year? Rs. 2, vs. Rs. 2.718282... (very close to e).

So as the interest is split and compounded over more and more intervals within the year (i.e., as the calculation becomes more and more continuous), we approach e (2.718282...).

So the "compound interest advantage over simple" is Rs. 0.718282..., or e-2.

Never thought of it that way before, but it is a great way to explain e to a lay person. Thanks for that pointer.

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^ Actually, it seems "e" was obtained in exactly the above way in 1683 by Bernoulli. Some websites even have the exact example I wrote above, just with "$" instead of "Rs." I had no clue all these days.

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^^^ Very nicely explained. Thanks.

As I said, this is how I, always, introduce e to young people (who never heard of it before) and this is how I learned about it (fairly young age). It is quite intuitive. Of course it does not have to be compound interest.. anything growing exponentially (like population, germs, or lotus flowers in a lake - example I fondly remember my father giving it to me) will do. Similarly anything decaying (like radioactive decay).. here one gets (1/e).

Recently I asked my granddaughter (not going to school yet) who was asking "Alexa" to "Increase the volume by 100%" for her favorite song -- How about if you asked it "Increase the volume by 50% twice" .. or "Increase the volume by 10% 10 times" ..(or "1% hundred times ..or even 1 zillionth of a percent 100 zillion times).... Yes eventually the volume would be 271.28...% .

Another use of this method, young people get the understanding of concepts like convergent series and limiting value without doing any higher math. For example one can intuitively see that e is > 2 (or even 2.5 etc.. by taking sufficient smaller periods). But one can also see that e will always be less than 3. . Generally one sees this in ordinary math books by showing the series e=1+1+ 1/2+1/6+1/24+ 1/5! ... is less than 3.

(One normally does by comparing with geometric series of similar way)

However, thinking of "decay" and having starting with 1/e = 1 -1+ 1/2 - 1/6 + (small) - (smaller) + (even smaller)..

One can see that 1/e is between 1/2 and (1/2-1/6= 1/3).. IOW e is between 2 and 3 !!

****

As I said, this is how I, always, introduce e to young people (who never heard of it before) and this is how I learned about it (fairly young age). It is quite intuitive. Of course it does not have to be compound interest.. anything growing exponentially (like population, germs, or lotus flowers in a lake - example I fondly remember my father giving it to me) will do. Similarly anything decaying (like radioactive decay).. here one gets (1/e).

Recently I asked my granddaughter (not going to school yet) who was asking "Alexa" to "Increase the volume by 100%" for her favorite song -- How about if you asked it "Increase the volume by 50% twice" .. or "Increase the volume by 10% 10 times" ..(or "1% hundred times ..or even 1 zillionth of a percent 100 zillion times).... Yes eventually the volume would be 271.28...% .

Another use of this method, young people get the understanding of concepts like convergent series and limiting value without doing any higher math. For example one can intuitively see that e is > 2 (or even 2.5 etc.. by taking sufficient smaller periods). But one can also see that e will always be less than 3. . Generally one sees this in ordinary math books by showing the series e=1+1+ 1/2+1/6+1/24+ 1/5! ... is less than 3.

(One normally does by comparing with geometric series of similar way)

However, thinking of "decay" and having starting with 1/e = 1 -1+ 1/2 - 1/6 + (small) - (smaller) + (even smaller)..

One can see that 1/e is between 1/2 and (1/2-1/6= 1/3).. IOW e is between 2 and 3 !!

****

### Re: BR Maths Corner-1

^^^ To establish the value of e with accuracy .. (both upper and lower bound) .. one can take Sudarshanji's example above.

If we use "negative 10% interest" (loose money - like some of my investments).. 10 times a year.. and start with Rs 1.. I will have (.9)^10 about 0.349 (=1/2.87)..

So for "continuous interest".. e is strictly between 2.59 and 2.87 !!

(And this is all without using any higher math other than use of a calculator or multiplying 1.1 (or .9) 10 times)

10% interest, 10 times a year, for a year? [Rs 1 will become] Rs. 2.593 (roughly).

If we use "negative 10% interest" (loose money - like some of my investments).. 10 times a year.. and start with Rs 1.. I will have (.9)^10 about 0.349 (=1/2.87)..

So for "continuous interest".. e is strictly between 2.59 and 2.87 !!

(And this is all without using any higher math other than use of a calculator or multiplying 1.1 (or .9) 10 times)

### Re: BR Maths Corner-1

106 years ago, on this date (Jan 16, 1913) Srinivas Ramanujan wrote his famous letter...

Attached with the letter were some of his results..Hope people enjoy solving them..

Attached with the letter were some of his results..Hope people enjoy solving them..

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Featuring Jayadev Athreya on a new discovery about platonic solids, in particular dodecahedra.

### Re: BR Maths Corner-1

sudarshan wrote:... If we designate 3^(2^85) as X, then it's the error between (X-1)/((2!)*X), versus 1/(2!), basically - the remaining terms in the Taylor series should rapidly decline in terms of contribution to the error.

Nice.

### Re: BR Maths Corner-1

^^^ A few comments..(FWIW)

In most books, e is generally *defined* as lim x->infinity (1+1/x)^x which if expanded using binomial theorem :

1 + 1 + (1/2!) (x(x-1)/x^2) + (1/3!) x(x-1)(x-2)/x^3 +...

when x tends to infinity this gives the familiar series ,,, e=1+1+1/2+1/3!+...

Also (1-1/x)^x is 1/e.

In most books, e is generally *defined* as lim x->infinity (1+1/x)^x which if expanded using binomial theorem :

1 + 1 + (1/2!) (x(x-1)/x^2) + (1/3!) x(x-1)(x-2)/x^3 +...

when x tends to infinity this gives the familiar series ,,, e=1+1+1/2+1/3!+...

Also (1-1/x)^x is 1/e.

### Re: BR Maths Corner-1

On this day in 1905 Indian Mathematician Dattatreya Ramachandra Kaprekar was born. He liked numbers, and was known as “Ganitanand”. And he had lot of "fun" problems ( recreational mathematician) He is known for describing several classes of natural numbers which included Kaprekar, Harshad and Self numbers. ..

Here is sample of "Kaprekar" number (or series)..

Start with any 4 digit number (except all the digits are same)

write all digits in ascending order and subtract from this a number written in descending order..

Repeat this process...

You will reach "Kaprekar" constant of order 4 which is

For example you start with numbers say 3,3,4,7 Now do this

7433–3347 = 4086

8640–0468 = 8172

8721–1278 = 7443

7443–3447 = 3996

9963–3699 = 6264

6642–2466 = 4176

7641–1467 = 6174

No matter where you start, you will reach 6174 (unless you start with 1111 (or 2222 etc) then you will get 0)

***

(Interestingly 6174 is also a "harshad" number because it is divisible by sum of it's digits.

Here is sample of "Kaprekar" number (or series)..

Start with any 4 digit number (except all the digits are same)

write all digits in ascending order and subtract from this a number written in descending order..

Repeat this process...

You will reach "Kaprekar" constant of order 4 which is

For example you start with numbers say 3,3,4,7 Now do this

7433–3347 = 4086

8640–0468 = 8172

8721–1278 = 7443

7443–3447 = 3996

9963–3699 = 6264

6642–2466 = 4176

7641–1467 = 6174

No matter where you start, you will reach 6174 (unless you start with 1111 (or 2222 etc) then you will get 0)

***

(Interestingly 6174 is also a "harshad" number because it is divisible by sum of it's digits.

### Re: BR Maths Corner-1

dup-- deleted.

Last edited by Amber G. on 18 Jan 2020 08:22, edited 1 time in total.

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Shwetank wrote:[>>> Youtube>>> G9_l8QASobI[/youtube]

Featuring Jayadev Athreya on a new discovery about platonic solids, in particular dodecahedra.

Thanks for posting it. I enjoyed playing with these regular polyhedrons..

Here is an interesting fact I recently noticed (rediscovered).. (Not too hard to prove - will not put the solution here.. go ahead and put it if you find a good explanation)

In any convex solid which is made out of hexagons and pentagons only (say a soccer ball or "bucky ball type molecule) or any other shape... number of pentagons will always be 12...

For example here:

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