olvin dsouza

1) We know in case of beta decay if there is excess of protons or neutrons in the nucleus, beta decay take place.
Force involved - weak interaction 
Reason involved - to stabilize, to obtain proton neuton 1:1 ratio in the nucleus.
E. g., Beta minus - Carbon 14 transforms into Nitrogen 14

2) Case of Lambda & Sigma - 

Force involved - weak  
Reason involved - ?

Does anyone know the reason why Lambda Λ0 (uds - quark composition) decays into,
p+ +  π−    
 or
n0  +  π0

Also why Sigma Σ+(uss - quark composition) decays into,     
p+  +  π0
 or
n0  +  π+

What mechanism make them go decay under either the set of proton or neutron? What is the reason involved ?

Introducing The Most Fastest Way to Find Perfect Square Roots.

Creating Successive Series Beforehand. 
0 -- 1  has zero difference.
1 --- 3    has only as difference of one.
3 ---- 6     has a difference of two.
6 ---- 10     has a difference of three.
10 ---- 15    has a difference of four.
.
.
.

This series is a form of infinite successive series where integers at the right hand side of the below series, that is 1, 3, 6, 10 15…. (highlighted below in yellow color) is the result of addition of positive integers in following sequence such as  2, 3, 4, 5, 6, 7, 8….(highlighted below in green color) 

0 -- 1  
         + 2
1 --- 3 
         + 3   
3 ---- 6  
           + 4  
6 ---- 10    
             + 5 
10 ---- 15    

This above successive series is essential on finding key integer c, those key integer c is then used for further calculation of finding square roots. 
While finding square root solutions, the use of this series is to simply avoid calculation timing by creating it beforehand.
Therefore, while solving the problems of finding any square, non square or any higher roots we require this series beforehand.
eg., when we are about to find the square root of 40000. 
40000/ 72 = 555
So, we need check where the 555 appears at the successive series Therefore, we need a ready successive series.
So, as shown above, one can continue go on creating same series by adding 15 + 6 = 21, 21 + 7 = 28 ….. keep record of it and use it whenever required to calculate key integer ‘c' of any other problems of finding square roots. 

Example Solution - ( this method works for all square and non square roots. For finding higher roots check the PDF paper attached).
To find perfect square root of √3249

1) Divide 3249 by 72 to get m.

3249 / 72 = 45.125 …..
Ignore the decimals, we get  45 as answer.

2) Using Successive Series For Finding the Key Integer c .
 0 -- 1  has zero difference. ( ignore this step).
1st Step    1 --- 3    as difference of one.
2nd Step    3 ---- 6     has a difference of two.
3rd Step      6 ---- 10     has a difference of three.
4th Step   10 ---- 15    has a difference of four.
5th Step   15 ---- 21   has a difference of five.
6th Step    21 ---- 28    has a difference of six.
7th Step    28 ---- 36    has a difference of seven.
8th Step    36 ---- 45     has a difference of eight.
9th Step  45  ----  55     has a difference of nine.

To Check, at Which Step 45 Appears.
Now, as per the above successive series,45 appears at ninth step i.e. at 45 to 55 and the series has a difference of nine  therfore, we can take c = 9 
(Note - though 45 appears at both 8th and 9th steps but we must always consider only the second step. So in this case we will consider 9th step).

3) Multiplying c by 6.
From above, we get c = 9
Therefore, 9 × 6 = 54
Now c = 54
Checking whether c is final answer by dividing 3249 by 54 . We found it is not divisible.
Therefore, we will proceed to below 4th step of checking rules.

4) Checking the Rules of Finding ‘c’ .
√X = 3249 is the integer divisible by 3. Therefore it follows the fourth rule of finding c.
Fourth Rule - If the √ X is any odd integer and is divisible by 3, then the final answer c will be adding c by 3 i.e. ‘c + 3'.
54 + 3 = 57
Since 3249 is divisible by 57
Therefore, √3249 = 57
 Note - Rules are available at the paper, check below attachment. Paper also contains explanation on finding higher roots, non square roots.
 

squareroot (1).pdf

On 4/15/2021 at 11:11 PM, John Cuthber said:

Given a product of two primes n, and n+2 the product is going to be very close to (n+1)^2

So you can take the nearest integer to the square root of the product, and the factors will be  that +/- 1

 

Eg.
 imagine I give you the product 19043

the square root of that is 137.996...

Very close to 138

And the integers 1 away from it are 137 and  139.

And those are the factors of 19043.

 

 

I agree, 

What if, 

n = 1661521

And condition is- use only simple calculator.

Try to find sqrt of n.

By the time you calculate or by guessing.

Using my explained method and formula. I can get the two factor of n in less then 1 minute.

No matter how big is  n in digits, calculation time is less than 1 minute.

On 4/15/2021 at 10:34 PM, Ghideon said:

What makes 72 special? Can you show how the method works in the general case n(n+2)=m  where n∈N,m∈N ? 

 

 

If you give me n by simply multiplying p×q 

Then the only way i can find p & q  is by sqrt method. It will take some time or lot of time for me to factor using sqrt method by simple calculator.

Addition 72 series is the only method to get the 'last sum of series'.

If i get the 'last sum of series'. - It is the magic number and input it in 0.83 constant. I can factor any small or even larger  1000 + digits n in less then 1 min. (As explained in my method formula)

Take this theorm as alternate to sqrt method.

Which can provide fast result using simple calculator, but with conditions such as following addition 73 series.

On 4/11/2021 at 2:46 PM, John Cuthber said:

I'm not sure i have understood this.

Does this only work with primes where the two factors differ by two- which is interesting but not much use (I think there's an easier way)?

 

If not, please show us how it works with 33,033,660,080,507

(which isn't a product of very big primes, just fairly big.)

Prime numbers differ by two is called twin prime numbers.

This method works only if you get 'n' by following repeated 72  additon series.

If you give me any n this method doesn't work.

Any how, i have found a new method where,
If you multiply any of the below following p & q and give me n.
 
Twin prime × Twin prime = n

Prime number × Composite number = n ( not divisible by 2 or 3, having a gap of 2).

Composite number × Composite number = n ( not divisible by 2 or 3, having a gap of 2).

Then i can factorise it in few minutes. Even if the n is 1000+ digit large.
Very soon i will publish paper on this another method.

2103.0181v2.pdf

There is a paper at vixra (dot) org 

Number theory, citation  2103.0181

Instantly Factorize Any Product Of
Two Small Or Large Twin Prime Numbers.

Simple Method -
72 is the constant integer used in the process to find repeated addition in the
series.
First Step –
Repeated Addition Series.
Following the steps ask your colleague to add 72 and 36 as show below.
 72 * 1 = 72 + 36 = 108
 72 * 2 = 144 + 36 = 180
 72 * 3 = 216 + 36 = 252
 72 * 4 = 288 + 36 = 324 ......... 'Last Sum Of Series'

Counting can be done as many times like 72 *5 , 72 * 6, 72 * 7 ........ and
one time adding 36 for each series.
Series can go up to infinity.
Last sum of series is 324.

Second Step -
Finding ‘r ‘ Total Sum Of Series -
Ask your colleague to add all the sums together with number 35 to get total
sum of series ' r ' as shown below.
108+ 180 + 252 + 324 + 35 = 899 ...... 35 is the constant to be
added at last in total sum of series each time you calculate this series.
Here we get r = 899
Now get this two information from your colleague .
1) Last sum of series ie 324.
2) Total sum of series ie 899 .
You should know this to calculate the formula.
Therefore,
Ask your colleague to show the last sum of series i.e 324 and the total sum of
series i.e 899.
Note - Total sum of series is also a product of some two twin prime numbers
or prime number and composite number or may be of two composite
numbers.
So 899 is the product of p*q = 899 which we don’t know yet and we are going
to factorize it to know p & q using formula explained below.

Third step -
Now, ask your colleague that, 'Can they immediately guess what is the
multiple factors of given total sum of series is, without factorizing ?'
Answer for your colleague must be 'no', since no one can easily guess or
reverse the p * q = n if the ''n ' is any large integer.
But wait, using my new researched method you can factor in few minutes, no
matter what large integer 'n' is.
So without showing your colleague, calculate the process explained below.
Calculation Process – Finding ‘s’.
There are two method to find ' s ’.
1) First method -
Notice the above 'repeated 72 series', those bold highlighted integers 72 * 1,
72 * 2, 72 * 3, 72 * 4 ........
Series of Integers in line i.e 1,2, 3, 4.....
Find the last integer i.e 4
Substitute 4 with 0 of 0.83 (constant).
We get 4.83
Therefore, s = 4.83.
Each time you calculate to find ' s’ always find the last integer in the line as
explained above.
2) Second Method -
You know that last sum of series is 324. ( you got the information from
your colleague)
Taking 324 / 72 = 4.5
Get the left hand side integer before the decimal point i.e 4.
Substitute 4 with 0 of 0.83 (constant).
As per second method, we get s = 4.83
Next,
Apply the ‘r’ and ‘s’ in the below formula.
r / s = m
m/ 6 = n
Where,
‘r’ is the total sum of series.
‘s’ in this case is the substitution of 4 with 0 of 0.83 constant to get as 4.83.
6 is the constant in the formula.
We got r = 899, s = 4.83
Finding ' m ' -
r / s = m
899 / 4.83 = 186.12836....
Notice integer on the left hand side before the decimal point i.e 186
So, consider only those integers as ' m ' and ignore integers on the right
hand side of decimal point.
Therefore, m = 186
Finding ' n ' -
m / 6 = n
186 / 6 = 31
n = 31 ....... is the answer.
Check it dividing 899 by 31.
899/ 31 = 29
So the factors of 899 is 31, 29.  

Immediately show the answer to your colleague.

One can surprise their colleague with this method.

This method works even for factorizing any larger product of multiplied twin primes. One can try checking.

twin prime.pdf