Fastest n'th root technique / Solving any equation in single line Mentally

Mental Calculation, Mathematics, and Puzzles
1

Hi all ,
Might you wondered how Shakuntala devi calculate 13’th root of any number in head. Is it possible for anyone to master these art ? I was in search of such technique - and finally found some vedic math book. But with vedic math
one can not compute root beyond cube root. Then I found book called modern approach to speed math secret by indian mathematician V.B. Jadhav (refer amazon.in/Modern-Approach-Speed-Math-Secret-ebook/dp/B00D2UE1AA).

It explore lot of new technique and shortcut. Beside it explain whole speed math just by using zero. Reveal secret
behind Trachtenberg system and I also learned whole vedic math through this book. I wondered are there other book like this ?

Can you list some good book ?

Modern Approach to Speed Math Secret & Trachtenberg
2

Hi Rock, welcome to the forum.

In 5 posts you have plugged the same book.
In order not to be recognised as a spammer I suggest you stop doing that.

One of the best books I know is this one: Local SEO for Lawyers in 2024

3

nice book… Like to read soon

4

I have a question for you.
After reading the book Modern Approach to Speed Math Secret, what kinds of calculations have sped up the most for you?
Can you give examples?

5

Thanks kinma for suggesting good book.

The 'modern approach to speed secret' speed up almost all calculation.But It specially gives best method to compute power and n'th root . I had learned one new theorem / shortcut called as golden lemma. Using this principle one can easily compute inverse , power , multiplication , division  of any polynomial or number. The method is systematic and innovative. I also learned one line method  to extract n'th root of any number , VJ's cross divisibility test , VJ's matrix method etc.

Author has developed number system called global number system . This system is similiar to vedic mathematics but it also unifies other speed math system like vedic math , Trachtenberg system.
I like cross divisibility lemma which can be used to derive any divisibility test for any number. Hence it eliminate necessity to remember divisibility test. Also there is no need to remember Trachtenberg formulae . It can be easily obtained by using systematic method (for more detail refer http://www.youtube.com/watch?v=b3PFjsUgULM)

6

Are we talking taking the nth root to get an exact match or can we get approximate answers?
The latter has wider uses.
For example today I was working on a mental problem that needs the 10th root of 10.
Answer is close to 1.26.

Can you give examples of calculations that you do?

7 
I am talking about exact match. Suppose we need answer upto 3 significant digit. Then

expand (100a +10b +c )^10 according to decreasing power of 10 (This expansion can be easily obtained by VJ’s golden lemma) and get first 3 term by leaving 10’s power . Here we get
a^10
10*(a^9 * b)
(This term is obtained by differentiating previous term i.e. a^10 with ‘a’ and integrating it with respect b)

Again differentiating previous term with a integrating with b , differentiating with b and integrating with c , we get

45 a^8 b^2 + 10a^9 c

Now from these three term we can get exact root of equation as follow :

  1. Find highest integer for which a^10 <10
    a^10 < 10
    => a = 1
    now remainder comes out -10- a^10 = 10-1 =9
    prepend it to next digit of 10.000 i.e. 0 then next gross dividend becomes 90

  2. Similarly proceeding ahead
    10 * a^9 * b < 90
    i.e. 10b <90
    => b = 8 , next gross dividend = 100

  3. 45 a^8 b^2 + 10 *a^9 c < 100
    => 45     64 + 10c < 100
    i.e. 10c < -2780
    => c = -279 next gross dividend = 100

Since c is decimal digit , c can not be negative so decrease b until we get c positive
finally for b=2 we will get c as positive integer.

Therefore , root up to 2 significant digit is 1.2

(Note :- Here I am not able to explain full method and why it works due to some limitation. But whole above
procedure can be done in single line. Same method can be used to compute root of any polynomial. For more detail refer VedicMaths.Org - 11. Novel Algorithm for 'Nth Root of Number' using Multinomial Expansion or Ken’s recently published ‘Crowning Gem’ book at VedicMaths.Org - Usage Problems )

8

Hi Rock,

I am reading the book now.
I have to admit, some of the things he writes about are fascinating.

The method of calculating the n-th root seems difficult for now, but maybe if I read the chapter it will make more sense.

For now I like the method of calculating let’s say 38 X 42 by using:

4|-2
4|2 X

(Using Ronald Doerflers notation)
His notation would be:
_
42
42 X

(there should be a dash over the 2, meaning -2, one’s complement, so 40-2 instead of 38)

9

Hi kinma,
With same technique one can easily compute any power of any number. Also nice to know about complement. Complement has wide use. In speed math , they helps to convert larger digit to smaller , thus they are responsible for
boosting speed of operation. Because larger digit means larger carry and carry slows down speed of operation.

Are you reading 'crowning Gem ’ or ‘Modern approach to speed math Secret’ ? If yu are reading ‘Crowning Gem’ then i would like to ask some question after your reading

10

In the book Modern Approach he uses it a lot and I must say that I like it!

I am reading Modern Approach…

Can you ask the question before I read Crowning Gem?
I’ll try to do my best to understand your question.

11

Hi kinma ,
I had not read ‘Crowning Gem’ but I would like to read it in future. I am not able to buy because of its high price.
If I will read I will look for

  1. What is master formula to compute any power in single line ?

  2. Is it Vedic or author are just trying to put it under the name of Vedic math ?
    Is it uses recurrence relation ?
    ( Because all shortcut uses recurrence relation and repetitive pattern to simplify computation.
    It is highly sophisticated technique used by modern math. )
    I am sure he must be using recurrence relation.

  3. Is it possible to extend one line power expansion method to simplify inter-base conversion ?
    (Converting number from one base to other is called inter-base conversion.)

Complement Application

  1. Time addition
    Suppose I have to add time 3 h 40 m and 4 h 59 m .
    Then just add 340 and 459 in decimal addition
    it give 799 i.e. 7 h 99 m
    but minute should be less than 60
    so add time constant 40 to number
    that gives 799 + 40 = 813 ( i.e. 8h 13m)

    What is time constant and why it works ?
    Time constant is ten’s complement of Minute base .
    Minute Base = Base for minute addition = 60
    (Because minute use 0, 1, 2, …59 numerals)

    Time constant = (Minute Base)" = (60)" = (10-6)0 = 40

    (Here we are addition minute by rules of decimal number system, So we need to take complement in Base-10
    i.e. decimal number system)

  2. Binary addition :
    Suppose we need to add binary number 01 and 11 . Then just add it as if they are decimal numbers
    i.e. 01 + 11 = 12
    Since in binary digit can not be greater than base i.e. 2 , so add base constant = 8 to all invalid digits
    12
    + 8

      20

Again 2 is invalid, So continue process of adding 8 until all digit are valid

       12 
      + 8

      20
   + 8

     100

Here all digit are valid. So process end here.
Therefore 01 + 11 = 100

What is base constant ?
Base constant = (Binary base ) " = (2) " = 8

Similarly , we can add two octal number by using decimal addition concept . For it base constant is given by
Base constant = (Octal base) " = (8)" = 2

But why above method works ?
Number wrapping concept are involved.

0 h 61 m = (0 +1) h (61-60) m
Omitting h and m

              0 61     (Last two digit refers to minute and remaining for hour)
              1(-60)  

              0 61
             + (60)"
       -------------------
             0 61
            + 40
        ---------------- 
            1 01

      Similarly , inter-base conversion can be explained.
12

I like the approach from the book ‘Modern approach…’.
However, I disagree with the statement that this would be the fastest technique for calculating n’th root.
Have you tried logarithms? IMO, they are a lot faster.
Mentally, it is difficult to go beyond 5 decimal places though, so the answers may not always be exact.
Then again, the 10th root of 10 is an irrational number and by definition cannot be written out in an exact way.

I like the approach of writing out (100a +10b + c)^10 to calculate the 10th power of something.

But how would you calculate 2.2^8.3?
Without logarithms this seems difficult to do.
Can you do this using Vedic techniques?

Why is this important?

Again, why is this important?
Outside of mathematics and information science hardly anybody works with different bases.

13

I like your approach to study particular field and curiosity to know “why something is important ?” It seems you like to work on area that has deep impact on science and would like to know such area.

Some question I just put due to my curiosity .

"Is it Vedic or author are just trying to put it under the name of Vedic math ? "
=> I am Vedic math researcher. Now a day many people are including modern math under the name of Vedic math ,
so just I like to know due to my curiosity. It is not important.

Is it possible to extend one line power expansion method to simplify inter-base conversion ?
(Converting number from one base to other is called inter-base conversion.)

=> Every math equation has number of application. Today many scientific application require inter-base conversion.
So if we can develop better number system or develop an faster method for inter-base conversion then it will help
in many scientific scenario. Ken’s 'Crowning Gem ’ specifically tries to give better method for operation like
power computation and I had only listed some of possible research that can be done based on ken’s book.
Once again as you said "Inter base conversion is not important besides information or computer science ".
Kinma , I never think about applicability when I do research on particular area. "Whether something is
important or not " is relative. Almost everything is relative. If there are resource / method , there will be
application. Example- John Nash’s "Nash equilibrium " had very few application in beginning but afterward
it was applied in divorce area and finally John Nash got Nobel prize in 1994 . No one can tell what will happen
in future or what method will be widely used in all area. So I don’t judge specific method is widely applicable or
or not. If you like to work on area that is widely applicable then try yourself you will find. Are you looking for
such area / work ?

How to calculate 2.2^8.3?
=> First I would like to thanks for raising this question.
It is possible to calculate 2.2^8.3 by using Golden Lemma , but it is not as easy as computing a^n (where a is any real number and n is natural number).
Why it is difficult ?
=> its difficult due to difficulty in computing fractional power of ’ decimal base’ or Decimal digits. I will show systematic solution to this problem after 2-3 days.

14

I agree. Sometimes just the thought of being able to find things out is amazing.

Maybe in another thread you want to explain a bit about true (ancient) Vedic Mathematics?
I would love to hear more.

Not many people do base conversion or even non decimal arithmetics.
As a child - nurturing my nerd side - when my father would take me and my brother to fancy restaurants, we would do long division in binary or hexadecimal. On drink coasters!
Then we would explain it to the waiters, at least we tried to.

I agree btw about applicability. A lot of mathematics gets invented without any application.
Then later a whole new part of sience gets invented using the new maths.

Would love to see it. And if you do, I will show how I do it so we can compare notes.

Stay curious.

15

Some other approaches to compute 2.2 ^ 8.3

  1. Cube 2.2,
    find the 10th root of the result,
    and then multiply it by 2.2^8.

  2. Find 83 power of 2.2 up to desired significant digit.
    Then find 10’th root of result

16

How would you calculate this mentally?
Because this is one of these things that are easier said than done.

2.2^83 is bigger than 10^28. Most people have difficulties working with such large numbers.
The answer to 2.2^8.3 is almost 700. If you have to go to 10^28 intermediately to ultimately go to 700 you are bound to make mistakes.

While your answer is mathematically true, the calculation seem difficult to do mentally.
This subforum is set up to solve issues like this. Therefore I am very interested to know how you would do a calculation like this mentally.

On the Yahoo Mental Calculation forum a couple of weeks ago, somebody asked how to calculate 2.2^8.3.
I wrote down how I would do this: Yahoo | Mail, Weather, Search, Politics, News, Finance, Sports & Videos

Let’s copy/paste my answer.
Let me know if this is unclear, please.

In this case I would do 2^8 first = 256 (memorized)

Then I would move to 2.2^8 = 2^8 * 1.1^8 by doing the evolution for 11^8:
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
1 7 21 35 21 7 1
1 8 28 56 56 28 8 1

Doing all the carrying leaves us with: 1.1 ^ 8 ~= 2,144…
256 * 2.144 ~= 550

Alternatively, one could square 11 = 121, then square it again = 14641…, then square 14641. To be honest, I would not mentally square 14641. Too many numbers for my brain.
Rounding down I would square 146. = 21316, then realise I need about 3 per mill more to get from 146 to 14641.

In the square I need 6 per mill more. I would round 21316 to 213, and add the 6 per mill = 2143. since I rounded down, I will round the answer up to 2144.

Putting back the decimal point gives me 2.144.
256 * 2,144 ~= 550. We have calculated 2.2^8 ~= 550.

Now I need to guess 2.2 ^.3. I would first calculate 2.2^.1 then calculate that answer to the power of three. Or (2.2^.1)^3.

My father used to ask me how fast an interest of x% would double one’s money.
I know by heart that 7% doubles in about 10 years. It’s a little over 10 years.
So 8% in 10 years results in about 2.2 times your money.

We therefore need a little over 8%.
Getting 8% in 3 years leads to about 26% more. We need a little more, so let’s take 26.5%
550 plus 26.5% = 695,75

17

Nice solution.

Skip my second approach . Mental computation step for 2.2 ^ 8.3 is given as below

Mental step

  1. First compute 2 ^ 8.3

2^8 = 256 ,
2^0.1 = 1.07 (10’th root of 2 . Compute it as I said earlier)

Therefore , 2 ^ 0.3 = 1.07 ^ 3 = 1.23

Therefore , 2 ^8.3 = 256 * 1.23 = 315 (Vertically crosswise or 256 *125 -512)

  1. Then compute 2.2 ^8.3 by Golden lemma or proportionately formula ( Vedic math),

                                            315  ---  (i)

    •  315 * 8.3 /1  * 0.1   =    261  ---   (ii)

    •  261 * 7.3 /2  * 0.1   =      95  ---   (iii)

    •    95 * 6.3 /3  * 0.1   =      20  ----  (iv)

    •    20 * 5.3 /4  *  0.1  =       2   ----- (v)     (Here we can stop and declare answer)

                                             693

Also for 22 ^ 8.3 , we need to compute 20 ^ 8.3 first and then proceed as above .

18

Excellent solution.
I like the way you do this!

I have a question about the Golden lemma.
How is his different from a Taylor series?

19

Can you tell what is difference between Taylor’s series and Multinomial thereorem . Golden lemma is advanced and
efficient way to expand multinomial expansion. We can also expand any algebraic function by Taylor’s series. Actually
Taylor’s series / McCluarin’s series is special case of Golden lemma. For more detail read all comment by author at
http://www.linkedin.com/groups/Binomial-Theorem-Reformulated-7432677.S.5830362692673630210?trk=groups_most_popular-0-b-cmr&goback=%2Egmp_7432677

20

Can you be a bit more specific about your question? The Taylor series is a binomial expansion of a calculation.
Does this answer your question?

Here is the Taylor expansion, which is basically the same calculation as you did (the 315 X 1.1^8.3 part):
Taylor (1+x)^8.3

So, again I am curious to know what the difference is between the Golden lemma and binomial or multinomial expansion.