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Saturday, July 25, 2009

Vedic Mathematics Lesson 13: Squaring, Cubing, etc.

Squaring is a special case of multiplication, and as such any technique that has been covered for multiplication can be used for squaring also with no problems. Also, in a previous lesson, we have dealt with the technique for squaring numbers close to a power of 10 using the Yavadunam sutra (Yavadhunam Thavadhunikritya Varga Cha Yojayetu: Whatever The Extent Of The Deficiency, Lessen It Further To That Extent; And Also Set Up The Square Of That Deficiency). Examples of using that method were also dealt with in that lesson.


You can find all the previous posts about Vedic Mathematics below:


Introduction to Vedic Mathematics
A Spectacular Illustration of Vedic Mathematics
10's Complements
Multiplication Part 1
Multiplication Part 2
Multiplication Part 3
Multiplication Part 4
Multiplication Part 5
Multiplication Special Case 1
Multiplication Special Case 2

Multiplication Special Case 3

Vertically And Crosswise I

Vertically And Crosswise II


Squaring can also be performed using the Urdhva-Tiryak (vertically and cross-wise) method. It can be set up as a regular multiplication with the same number on two lines, or the vertical and cross-products can be performed mentally with the number to be squared on just one line. Also, some simplifications are possible because of the performance of squaring rather than general multiplication. For instance, since the cross-products involve the same numbers, the sum of the cross products is just double of the single cross-product. This property is referred to as Duplex, or in the Vedas as Dvandva Yoga. Some examples of this approach are dealt with below:


47

-------

••49 (vertical multiplication, 7 x 7)

•56• (Duplex of 47, 2 x 4 x 7. In normal multiplication, we would use 4 x7 + 4 x 7 rather than the duplex)

16•• (vertical multiplication, 4 x 4)

-------

2209


78

--------

••64 (vertical multiplication, 8 x 8)

112• (duplex of 78, 2 x 7 x 8. In normal multiplication, we would use 7 x 8 + 7 x 8 rather than the duplex)

49•• (vertical multiplication, 7 x 7)

--------

6084


321

--------

•••••1 (vertical multiplication, 1 x 1)

••••4• (Duplex of 21, 2 x 2 x 1)

••10•• (Duplex of 321, 2 x 3 x 1 + 2 x 2. Note that the product of the middle numbers is not doubled)

•12••• (Duplex of 32, 2 x 3 x 2)

•9•••• (vertical multiplication, 3 x 3)

--------

103041


That should make the use of the duplex clear.


The Yavadunam sutra can be used for cubing numbers also. Obviously, this works only when the numbers are close to a power of 10. And there are some minor differences between the application of the sutra to squaring and the application of the sutra to cubing. The procedure is explained as below for numbers that have a small excess with respect to the base:


  • Find the excess of the number to be cubed with respect to the base. Let this be E
  • Add 2 x E to the number to be cubed and set as the left hand part of the answer. Calculate the excess of this number with respect to the base. Let this be F (note that F will be 3 x E)
  • Calculate F x E and set as the middle part of the answer
  • Now calculate the cube of E, and set this as the right hand part of the answer.

Note that the number of digits of each part of the answer has to be the same as the number of zeroes in the base, as before. Any excess digits have to be carried over to the left. If the parts of the answers don’t have the requisite number of digits, they have to be padded with zeroes as necessary.


A few examples of this approach are worked out below for illustration:


1053 = 115|75|125


115 is 105 + 2 x 5 (5 is the excess of 105 over our base, 100. Thus, 5 is E. F becomes 15)

75 is F x E

125 is 53.


After carrying over the 1 from 125 to the left, we get the final answer as 1157625.


10043 = 1012|048|064

1012 is 1004 + 2 x 4 (4 is the excess of 1004 over out base, 1000. Thus, 4 is E. F becomes 12)

48 is F x E

64 is 43.


Note that the numbers have been padded with zeroes to make them 3 digits long, giving us the final answer of 1012048064.


Let us examine the algebraic basis for this method. This will clarify the method further and also help us tackle the case of deficits from the base, rather than excesses. Let the number to be cubed be expressed as (B + E), where B is our base and E is the excess with respect to the base.


(B + E)3 = B3 + 3B2E + 3BE2 + E3.


Adding twice the excess to the original number and setting it as the left hand side of the answer is the same as deriving B3 + 3B2E. Now, we calculate F as the new excess of this number with respect to the base, and note that F = 3 x E. So, by calculating F x E, we calculate 3 x E2. By setting this as the middle part of the answer, we compute 3BE2. Finally, we cube the excess and set it as the right hand part of the answer, giving us E3, the final part of the algebraic expression.


Now, let us examine the case of (B – D)3, where B is the base and D is a deficit with respect to the base. Expanding (B – D)3 gives us B3 – 3B2D + 3BD2 – D3. This algebraic expression gives us the necessary details to derive the exact steps needed to cube numbers that are just below a power of 10 mechanically and easily. In short, those steps are explained below:


  • Find the deficit of the number to be cubed with respect to the base. Let this be D
  • Subtract 2 x D from the number to be cubed and set as the left hand part of the answer. Calculate the deficit of this number with respect to the base. Let this be F (note that F will be 3 x D)
  • Calculate F x D and set as the middle part of the answer
  • Now calculate the cube of E. Rather than setting this as the right-hand part of the answer, note that this has to be subtracted from the answer. So, use the three-step procedure outlined in this lesson to calculate the final middle and right-hand parts of the answer.


This procedure will become clearer as we work out the following examples.


973 = 91|27|-27

91 is 97 – 2 x 3 (3 is the deficit of 97 with respect to our base, 100. Thus, D is 3 and F becomes 9)

27 is F x D

-27 is -33


Following the 3-step procedure of this lesson, we get the final answer as 912673.


9943 = 982|108|-216

982 is 994 – 2 x 6 (6 is the deficit of 994 with respect to the base, 1000. Thus D is 6 and F becomes 18)

108 is F x D

-216 is -63


The final answer, after application of the 3-step procedure from here is 982107784.


9903 = 970|300|-1000

970 is 990 – 2 x 10 (10 is the deficit of 990 with respect to the base, 1000. Thus D is 10 and F becomes 30)

300 is F x D

-1000 is -103


The final answer, after application of the 3-step procedure from here is 970299000.


Cubing of 2-digit numbers can also be performed using another sutra called the Anurupya Sutra. To use this, follow the procedure below:


  • Put down the cube of the left digit of the number to be cubed as the left most number in a row of 4 numbers
  • Put down the square of the left digit multiplied by the right digit as the second number in the same row of numbers
  • Put down the square of the right digit multiplied by the left digit as the third number in the same row of numbers
  • Put down the cube of the right digit as the right most number in this row of numbers
  • Under the second number in the row above, put down twice the second number
  • Similarly, under the third number in the first row, put down twice the third number
  • Add them up, making sure to carry over excess digits from right to left. That is the final answer.


Note that the first row can also be expressed as writing the numbers from the cube of the first digit to the cube of the second digit such that the numbers in between form the same ratio with respect to each other: in other words, the numbers in the first row are in geometric progression from the cube of the first digit to the cube of the second digit. In fact the constant ratio of the geometric progression is the same as the ratio between the first and second digits of the number to be cubed.


Also note that the procedure above is a direct result of the algebraic identity that (a + b)3 = a3 + 3a2b + 3ab2 + b3. The first line contains the terms a3, a2b, ab2 and b3. The second row contains the remaining 2a2b and 2ab2 (double of the middle two terms of the first row).


To illustrate, let us work through some simple examples:


113 = 1|1|1|1

•••••••••2|2

------------------------

•••••••1|3|3|1


Note that the top row consists of all 1’s because 1 is 1 cubed as well as the square of 1 multiplied by 1. The 2’s in the second row are twice of the numbers in the first row. Adding gives us the final answer of 1331.


253 = 8|20| 50|125

••••••••40|100

----------------------------

15625


Note the carryovers carefully in the example above. From the right most column, 12 is carried over to the left. This gets added to 150, giving 162, of which 2 remains and 16 is carried over further to the left. This gets added to 60, giving 76. 6 remains and 7 is carried over to the left-most column, giving the final sum of 15. Each column should consist of one digit, with all the excess digits carried over to the left until the left-most column, which obviously will not have any carryover out of it.


A few more examples of cubing using this method are provided for further illustration:


323 = 27|18|12|8

•••••••••36|24

----------------------

32768


473 = 64|112|196|343

••••••••224|392

----------------------------

103823


183 = 1| 8| 64 |512

16|128

------------------------

5832


Expanding out (a + b)4 gives us a4 + 4a3b + 6a2b2 + 4ab3 + b4. Using this algebraic expansion, we can derive a similar method for finding the fourth powers of 2 digit numbers as below:


  • Put down the fourth power of the left digit of the number to be raised to the fourth power as the left most number in a row of 5 numbers
  • Put down the cube of the left digit multiplied by the right digit as the second number in the same row of numbers
  • Put down the square of the right digit multiplied by the square of the left digit as the third number in the same row of numbers
  • Put down the cube of the right digit multiplied by the left digit as the fourth number in this row of numbers
  • Put down the fourth power of the right digit as the last number in this row of numbers
  • Under the second number in the row above, put down thrice the second number
  • Similarly, under the third number in the first row, put down five times the third number
  • Under the fourth number in the first row, put down thrice the fourth number
  • Add them up, making sure to carry over excess digits from right to left. That is the final answer.


Let us illustrate by taking a simple example:


114 = 1|1|1|1|1

3|5|3

---------------------

1|4|6|4|1


Note that just as in the case of calculating cubes, the numbers in the top row will be in geometric progression from the fourth power of the first digit to the fourth power of the second digit. A few more examples will illustrate this progression as well as give us insight into the method so that we can use it quickly and with ease.


214 = 16| 8| 4|2|1

24|20|6

--------------------------

194481


Note the carryovers in this example carefully. The right most and the 4th columns do not generate any carryover. The 3rd column adds up to 24, leading to the retention of the 4 and the carryover of the 2, which makes the second column 34. Once again, the 4 is retained, and the 3 is carried over. Adding the 3 to 16 gives us 19 which is put down on the left since there are no more columns to carry numbers over to the left.


634 = 1296| 648| 324|162|81

1944|1620|486

--------------------------------------

15752961


Note how the knowledge that the numbers in the top row are in geometric progression leads to some facility in the calculation of the numbers. Since 3 (the second digit of the number to be raised to the fourth power) is half of 6 (the first digit of the number), we know that each number in the top row will be half of the previous number in that row, proceeding all the way from 1296 (64) to 81 (34). The carryovers are more complicated in this example, but keeping the column dividers enables us to work them out pretty easily (just in case you are confused, the numbers carried over to the left are 8, 65, 200 and 279).


We will conclude this lesson with a couple more examples to illustrate the method:


154 = 1| 5| 25|125|625

15|125|375|

-------------------------------

50625


324 = 81| 54| 36|24|16

162|180|72|

-------------------------------

1048576


Extending this method further to find the fifth and higher powers of 2-digit numbers is quite trivial. Note that the coefficients of the binomial terms of higher and higher powers follow the Pascal triangle, so they are very easy to calculate (for instance, the coefficients for (a + b)5 are 1, 5, 10, 10, 5, 1, and those of (a + b)6 are 1, 6, 15, 20, 15, 6 and 1). Once the first row is written down in geometric progression using successively lower powers of a and higher powers of b, all we have left to do is to fill out the middle of the second row with one less than the coefficients in the expansion, and then add the columns up correctly, accounting for carryover.


Hope this lesson has been a useful extension of the principles of general multiplication to some special cases that are likely to be encountered somewhat frequently. Computing cubes, fourth powers, etc., is not easy to do mentally, even with the simplifications introduced by these methods, but hopefully it is better than performing long multiplication over and over again, with the attendant errors that is likely to bring along. Practice will make the application of these methods more a habit than a chore. So, happy computing, and good luck!

2 comments:

Anonymous said...

How would you square 299??

Naween Raj said...

As we know that square of 29 is 841. And hence square of 299 is 89401. Also square of 2999 is 8994001.


Moreover we can use square of (300-1) instead of square of299.

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