GREETING CARD MAKING COMPETITION

 GREETING CARD MAKING COMPETITION



A blend of exceptional creativity and imagination was displayed by the students  during the “Greeting Card Making Competition” held on September 02, 2022 in the secondary Wing. The theme is teacher's day wishes.





Comments

Post a Comment

Popular posts from this blog

INVENTION OF ZERO

Image
 INVENTION OF ZERO The concept of zero has not always been around, however, the introduction of zero bought a lot of changes not only in math but also in the general life of people. Zero has so many different names, for example, ‘null’, ‘nil’, ‘0’ as a digit, ‘sunya’ in Sanskrit, and so on. It is fascinating how the origin of zero bought changed and now it is used as a prime digit in mathematics. Before learning about the modern zero, let’s learn about the origin of zero in India,   Origin of Zero in India The origin of zero in India came from a well-known astronomer and mathematician of his time, Aryabhatta. The well-known scientist used zero as a placeholder number. In the 5th century, Aryabhatta introduced zero in the decimal  number system  and hence, introduced it in mathematics. After Aryabhatta, Brahmagupta described rules for zero in the 7th century. The most evident proof of the origin of zero in mathematics is mentioned in the oldest manuscript of India known as the ‘Bakshali
Read more

RAMANUJAN'S INFINITE SEQUENCE OF SQUARE ROOT (NESTED RADICAL).

Image
         INFINITE SEQUENCE OF SQUARE ROOT        INFINITE SEQUENCE :                   An infinite sequence is a list or string of discrete objects, usually numbers, that can be paired off one-to-one with the set of positive integer s {1, 2, 3, ...}. Examples of infinite sequences are  N  = (0, 1, 2, 3, ...) and  S  = (1, 1/2, 1/4, 1/8, ..., 1/2  n  , ...).  INFINITE SEQUENCE OF SQUARE ROOT: Consider,                  y = x + n                (x +n)² = n² + x(x+2n)               (x + n) = √[ n² +x(x+2n)]               (x + 2n) = √[ n² +x(x+2n)(x+3n)]               ( x + 3n) = √[ n² +x(x+2n)(x+3n)(x+4n)]               ...               ...               ...               ...               ...                x+n = √[ n² +x √ n²+(n+x) √ n² +(n+2x).... ]   Remarkably the answer is exactly 3.    We have verified the pattern holds up to 4, with the next term under the radical being 6 = √6 2 . Let’s justify why the pattern will continue. Assume the pattern holds where  n  – 2 is the coef
Read more