The Binomial Theorem

RajeshRathod42@gmail.com

The Problem

• Evaluate (A+B)N

– as a polynomial in powers of A and B

– Where N is a positive integer

– A and B are numbers

• Example:

• (A+B)5 = A5+5A4B+10A3B2+10A2B3+5AB4+B5

The Solution

• Let us approach this as a problem of building ‘words’ out of an ‘alphabet’ – Let our alphabet consist of only 2 letters: A, B

– Thus, BABA , AAB, AB, ABBAB are all ‘words’ using our alphabet

• List all the 2 letters word can be constructed using this alphabet: – AA, AB, BA, BB

• And all 3 letter words? – AAA, ABA, BAA, BBA, AAB, ABB, BAB, BBB

• If we have all the possible words with 2 letters • We can derive all 3 letter words by the following process:

– Append each 2 letter word with an A – Append each 2 letter word with a B – The combined output of the 2 steps above is the list of all 3

letter words

• Lets work this out: – The set of all possible words with 2 letters: AA, AB, BA, BB – Append with A to get: AAA, ABA, BAA, BBA – Append with B to get: AAB, ABB, BAB, BBB – Combine the outputs to get:

• AAA, ABA, BAA, BBA, AAB, ABB, BAB, BBB

• Let us express this 3 step process, [2 steps to Append and 1 step to Combine] as an operation represented as a © between 2 SETS:

– [P, Q, R, …, S]©[X, Y]=[PX, QX, RX, …, SX, PY, QY, RY, …, SY]

– Here, P, Q, R, …, S, X, Y are words of our alphabet [A, B] • PX means Appending the word P with the word X

• So, if P represents ABB and X represents BA, PX represents ABBBA

– Thus: • [AA, ABB, BAAB, AB]©[AA,B]=[AAAA, ABBAA, BAABAA, ABAA, AAB,

ABBB, BAABB, ABB]

• The operation © is Associative: • ([P]©[Q])©[R]=[PQ]©[R]=[PQR]=P©[QR]=P©([Q]©[R])

• Where P, Q, R are words in our definition

• The operation © is not Commutative :

– That is P©Q is not same as Q©P.

– For example, if P=AAB and Q=BA:

• [AAB]©[BA]=[AABBA] is not [BA]©[AAB]=[BAAAB]

• Using this notation we can derive the set of all 3 letter words, given the set of all 2 letter words as: – [List of all 2 letter words]©[A, B] – [AA, AB, BA, BB]©[A,B]=[AAA, ABA, BAA, BBA, AAB,

ABB, BAB, BBB]

• As also, the list of all 4-letter-words is: – [List of all 3 letter words]©[A, B] – [AAA, ABA, BAA, BBA, AAB, ABB, BAB,

BBB]©[A,B]=[AAAA, ABAA, BAAA, BBAA, AABA, ABBA, BABA, BBBA, AAAB, ABAB, BAAB, BBAB, AABB, ABBB, BABB, BBBB]

• Clearly, this is true not only for 2, 3 and 4 letter words, but it true for K and (K+1) letter words, K>1

• Thus list of all words with (K+1) letters will be:

– [List of ALL K letter words]©[A, B]

• For example:

– Thus list of all words with 101 letters will be:

– [List of ALL 100 letter words]©[A, B]

• Finally, list of all words with N letters will be: – [List of ALL (N-1) letter words]©[A, B]

• Thus we can build a list of all N letter words as – [All 2 letter words]©[A, B]=[All 3 letter words]

– [All 3 letter words]©[A, B]=[All 4 letter words]

– [All 4 letter words]©[A, B]=[All 5 letter words]

– …

– [All (N-1) letter words]©[A, B]=[All N letter words]

• NOTE: List of all 2 letter words = [A, B]©[A, B]

• Substituting [A, B]©[A, B] for the list of all 2 letter words we obtain: – [[A, B]©[A, B]]©[A, B]=[All 3 letter words]

– [[[A, B]©[A, B]]©[A, B]]©[A, B]=[All 4 letter words]

– [[[[A, B]©[A, B]]©[A, B]]©[A, B]]©[A, B]=[All 5 letter words]

– … =[All N letter words]

• Note that there are 2N N letter words using our alphabet since each letter in the N-long word can independently be A or B

• Using the Associative Property of ©:

– [A, B]©[A, B]©[A, B]=[All 3 letter words]

– [A, B]©[A, B]©[A, B]©[A, B]=[All 4 letter words]

– [A, B]©[A, B]©[A, B]©[A, B]©[A, B]=[All 5 letter words]

– [A, B]©[A, B]© … ©[A, B] {N times}=[All N letter words]

• Clearly, the operation © is indeed analogous to the basic arithmetic operation of “multiply & add”:

• In fact, if in the definition of © as: – [P, Q, R, …, S]©[X, Y]=[PX, QX, RX, …, SX, PY, QY, RY, …, SY]

– We replace “,” by “+” and © by “*” we obtain:

– [P+ Q+ R+ …+ S]*[X+ Y]=[PX+ QX+ RX+ …+ SX+ PY+ QY+ RY+ …+ SY]

– The expression so obtained is mathematically correct if: • P, Q, … X, Y are mathematical objects [real numbers, complex

numbers etc]

• The symbol * denotes multiplication and + denotes addition

• PX is interpreted as P*X, QX is Q*X and so on

• Since [A+B]N = [A+B]*[A+B]*…*[A+B] {N times}

• And © is analogous to multiply and add

• IF we compute the end result of: • [A, B]©[A, B]© … ©[A, B] {N times}

– We should get the same result as if [A+B]N is expanded repeatedly multiplying [A+B] by itself N times

• The only difference will be that instead of “+” we will have “,” separating the individual terms

• The individual terms will be N letter long strings of A and B

• Since [A, B]©[A, B]© … ©[A, B] {N times} – Gives us the list of ALL N-letter words made of the 2 letter alphabet, A and B

• [A+B]N=[A+B]*[A+B]*…*[A+B] {N times}, due to the analogous nature of © and “multiply and add”, should lead to a summation of terms such that: – the set of terms in the summation will be same as the set of all N-long-words

of the alphabet A, B – Each N-long-word should be interpreted as a * of the A’s and B’s in it – Since there are 2N of these N-long-words, there will be 2N terms in the

summation

• So, if we can somehow enumerate ALL the N-long-words using A and B [and we know there will be 2N of them], we can then add them all up to obtain an expression for [A+B]N – Wherein each of the N-long-word should be treated as a product of A’s and B’s

• We now find a way to form all these N-long-words using A and B

• Consider the N letter word as being formed by filling up a row of N positions, with either A or B:

• Any N letter word will have k A’s and [N-k] B’s, where 0<=k<=N

A A B B A A B

1 2 3 4 5 (N-2) (N-1) N

B

N Positions

• Of all the N letter words, how many will have k A’s and [N-k] B’s? [0<=k<=N]

• This is the same as asking, in how many different ways can we choose k positions of the total N positions

• Assume for now, we have a formula that gives this number

– We write this formula as: NCk

• Then the set of ALL N letter words can be partitioned as: – Set of N Letter words with N A’s = NCN

– Set of N Letter words with [N-1] A’s = NC(N-1)

– Set of N Letter words with [N-2] A’s = NC(N-2)

– … – Set of N Letter words with k A’s = NCk

– … – Set of N Letter words with 1 A’s =NC1

– Set of N Letter words with 0 A’s =NC0

• Note that all the above sets are mutually disjoint and their union gives the set of ALL N letter words

• We thus conclude that if we were to expand [A+ B]N using the distribution of * over + then it will lead to the addition of terms [words] such that: – NCN of them have N A’s and 0 B’s – NC(N-1) of them will have (N-1) A’s and 1 B – NC(N-2) of them will have (N-2) A’s and 2 B’s – … – NCk of them will have k A’s and (N-k) B’s – … – NC1 of them will have 1 A and (N-1) B’s – NC0 of them will have 0 A’s and N B’s

• Note that the word AABAB [for example] = A*A*B*A*B and so on (as mentioned earlier) where * is multiplication

• Hence, a word with k A’s and [N-k] B’s is the same as a mathematica term which is the product of k A’s and [N-k] B’s

• Such a word can be represented as a mathematical term AkB(N-k)

• Here we are invoking the fact that: – PQ is to be interpreted as P*Q

– And the * operation IS Commutative

– So that: AAB=ABA=BAA=A2B

• We thus can say that [A+ B]N when expanded consists of the summation of – NCN terms of value ANB0

– NC(N-1) terms of value A(N-1)B1

– NC(N-2) terms of value A(N-2)B2

– …

– NCk terms of value AkB(N-k)

– …

– NC1 terms of value A1B(N-1)

– NC0 terms of value A0BN

• Thus: • [A+ B]N = NCN*AN*B0 + NC(N-1)*A(N-1) *B1 + NC(N-2) *A(N-2) *B2 + … +

NCk*Ak*B(N-k) + … + NC1*A1*B(N-1) + NC0*A0*BN

• More concisely

– [𝐴 + 𝐵]𝑁= 𝐶𝑁 𝑘𝐴(𝑁−𝑘)𝐵𝑘𝑁

𝑘=0

• Also, we know from basic Combinatorics that

– 𝑁𝐶𝑘 =𝑁!

𝑘! 𝑁−𝑘 !

• Thus the famous result:

• [𝐴 + 𝐵]𝑁= 𝑁!

𝑘! 𝑁−𝑘 !𝐴(𝑁−𝑘)𝐵𝑘𝑁

𝑘=0

• For N being a positive integer, and A, B numbers [real or complex]

Fractional Exponents

• Applying the binomial theorem of 𝐴 = 1, 𝐵 = 𝑥:

– [1 + 𝑥]𝑁= 𝑁!

𝑘! 𝑁−𝑘 !𝑥𝑘𝑁

𝑘=0

• Note that 𝑁!

𝑘! 𝑁−𝑘 !=

𝑁 𝑁−1 …(𝑁− 𝑘−2 )(𝑁− 𝑘−1 )

𝑘! and so:

• (1 + 𝑥)𝑛= 1 +𝑛𝑥1

1!+

𝑛(𝑛−1)

2!𝑥2 +

𝑛(𝑛−1)(𝑛−2)

3!𝑥3 +⋯+

𝑛 𝑛−1 … 𝑛−[𝑘−2] 𝑛−[𝑘−1]

𝑘!𝑥𝑘 +⋯

• Note that for 𝑘 ≥ (𝑛 + 1) the coefficients of 𝑥𝑘 become 0 since they contain a 0 in the numerator. Hence:

• (1 + 𝑥)𝑛= 𝑛 𝑛−1 …(𝑛−[𝑘−1])

𝑘!𝑥𝑘∞

𝑘=0 where the upper limit of summation is changed to ∞ instead of 𝑁.

• Let us define 𝑓 𝑛 for any positive rational number 𝑛:

• 𝑓 𝑛 = 𝑛 𝑛−1 …(𝑛−[𝑘−1])

𝑘!𝑥𝑘∞

𝑘=0

• Note that for a non-integral value of 𝑛 the coefficients never become 0. In fact they can attain negative values as well.

• For 𝑛 =3

2 for example:

• 𝑓3

2= 1 +

3

2𝑥 +

3

2(3

2−1)

2!𝑥2 +

3

2(3

2−1)(

3

2−2)

3!𝑥3 +⋯ , so that:

• 𝑓3

2= 1 +

3

2𝑥 +

3

8𝑥2 −

1

16𝑥3 +⋯

• Then we have 𝑓 𝑚 = 𝑚 𝑚−1 …(𝑚−[𝑘−1])

𝑘!𝑥𝑘∞

𝑘=0 , 𝑚 being any positive rational number

• And of course for positive integral values of 𝑛 𝑎𝑛𝑑 𝑚, 𝑓 𝑛 = (1 + 𝑥)𝑛 and 𝑓 𝑚 = (1 + 𝑥)𝑚

• Now if we consider the product 𝑓 𝑛 . 𝑓 𝑚 for 𝑛,𝑚 positive rational numbers:

• 1 +𝑛𝑥1

1!+

𝑛 𝑛−1

2!𝑥2 +

𝑛 𝑛−1 𝑛−2

3!𝑥3 +⋯ (1 +

𝑚𝑥1

1!+

𝑚 𝑚−1

2!𝑥2 +

𝑚 𝑚−1 𝑚−2

3!𝑥3 +⋯)

• For fractional 𝑛,𝑚 the coefficients do not got to 0 and hence the product is an infinite series multiplied as:

• 𝑓 𝑛 . 𝑓 𝑚 = 𝑓 𝑛 . 1 + 𝑓 𝑛 .𝑚𝑥1

1!+ 𝑓 𝑛 .

𝑚 𝑚−1

2!𝑥2 +

𝑓 𝑛 .𝑚 𝑚−1 𝑚−2

3!𝑥3 +⋯

• This leads to another infinite series in powers of 𝑥𝑘, 𝑘 ≥ 0, where the coefficients are functions of 𝑛 and 𝑚

• 𝑓 𝑛 . 𝑓 𝑚 = 𝐶0 + 𝐶1𝑥1 + 𝐶2𝑥

2 + 𝐶3𝑥3…

• For example, for 𝑥2 the coefficient 𝐶2 is: 𝑛 𝑛−1

2!+𝑛𝑚

1!+𝑚 𝑚−1

2!=

(𝑛+𝑚) 𝑛+𝑚−1

2!

• And for 𝑥3 the coefficient is 𝐶3 is:

•𝑛 𝑛−1 𝑛−2

3! +

𝑛 𝑛−1

2!

𝑚

1!+

𝑛

1!

𝑚 𝑚−1

2! +

𝑚 𝑚−1 𝑚−2

3!

– Which simplifies to: (𝑛+𝑚) 𝑛+𝑚−1 𝑛+𝑚−2

3!

• Is there an expression we can derive for 𝐶𝑘? • Expressing as 𝐶𝑘𝑛 and 𝐶𝑘𝑚 the coefficients of 𝑥𝑘 in 𝑓 𝑛 and

𝑓 𝑚 respectively. • Examining the product 𝑓 𝑛 . 𝑓 𝑚 =

• 1 + 𝐶1𝑛𝑥1 + 𝐶2𝑛𝑥

2 + 𝐶3𝑛𝑥3…+⋯𝐶𝑘𝑛𝑥

𝑘 +⋯ . (1 + 𝐶1𝑚𝑥1 +

𝐶2𝑚𝑥2 + 𝐶3𝑚𝑥

3 +⋯+ 𝐶𝑘𝑚𝑥𝑘 +⋯)

– 𝐶𝑘= 𝐶𝑘𝑛 + 𝐶(𝑘−1)𝑛𝐶1𝑚 + 𝐶(𝑘−𝑟)𝑛𝐶𝑟𝑚+…+𝐶1𝑛𝐶(𝑘−1)𝑚+𝐶𝑘𝑚

– Note: (𝑘 + 1) terms add up to form 𝐶𝑘

• So while the product 𝑓 𝑛 . 𝑓(𝑚) has infinite terms [since 𝑓 𝑛 and 𝑓(𝑚) have infinite terms], each of the coefficients 𝐶𝑘 therein is the addition of (𝑘 + 1) terms as above

• Remember: 𝐶𝑛𝑘 =𝑛 𝑛−1 …(𝑛−[𝑘−1])

𝑘! and 𝐶𝑚𝑘 =

𝑚 𝑚−1 …(𝑚−[𝑘−1])

𝑘!

• Now in the case of 𝑛 and 𝑚 being positive integers, 𝑓(𝑛) and 𝑓(𝑚) actually represent (1 + 𝑥)𝑛 and (1 + 𝑥)𝑚 respectively [as per the binomial theorem for positive integral exponents]

• Hence: 𝑓 𝑛 . 𝑓 𝑚 = (1 + 𝑥)𝑛. (1 + 𝑥)𝑚= (1 + 𝑥)(𝑛+𝑚) and now since (𝑛 + 𝑚) is also a positive integer we know that [again using the binomial theorem]:

• (1 + 𝑥)𝑛+𝑚= 1 +(𝑛+𝑚)𝑥1

1!+

(𝑛+𝑚)(𝑛+𝑚−1)

2!𝑥2 +

(𝑛+𝑚)(𝑛+𝑚−1)(𝑛+𝑚−2)

3!𝑥3 +⋯+

(𝑛+𝑚) 𝑛+𝑚−1 … 𝑛+𝑚−[𝑘−2] 𝑛+𝑚−[𝑘−1]

𝑘!𝑥𝑘 +⋯

• Hence in the case of 𝑛,𝑚 being positive integers we can derive that 𝐶𝑘, the coefficient of 𝑥𝑘 in 𝑓 𝑛 . 𝑓(𝑚) is same as:

•(𝑛+𝑚) 𝑛+𝑚−1 … 𝑛+𝑚−[𝑘−2] 𝑛+𝑚−[𝑘−1]

𝑘! and thus:

• 𝐶𝑘= 𝐶𝑘𝑛 + 𝐶(𝑘−1)𝑛𝐶1𝑚 + 𝐶(𝑘−𝑟)𝑛𝐶𝑟𝑚+…+𝐶1𝑛𝐶(𝑘−1)𝑚+𝐶𝑘𝑚

• = (𝑛+𝑚) 𝑛+𝑚−1 … 𝑛+𝑚−[𝑘−2] 𝑛+𝑚−[𝑘−1]

𝑘!

• We now have a value for 𝐶𝑘 in terms of 𝑛,𝑚 assuming their positive integer values – Of course for 𝑘 > 𝑛 +𝑚, the 𝐶𝑘 values will be 0 since the

numerator will have a 0 term

Now comes a key argument!

• 𝐶𝑘 which is given in its SUM FORM [for integral or fractional 𝑛, 𝑚]: • 𝐶𝑘𝑛 + 𝐶(𝑘−1)𝑛𝐶1𝑚 + 𝐶(𝑘−𝑟)𝑛𝐶𝑟𝑚+…+𝐶1𝑛𝐶(𝑘−1)𝑚+𝐶𝑘𝑚

• is, for integral 𝑛,𝑚, (𝑛+𝑚) 𝑛+𝑚−1 … 𝑛+𝑚−[𝑘−2] 𝑛+𝑚−[𝑘−1]

𝑘! [FACTOR FORM]

• Each term of 𝐶𝑘, in the Sum Form is as 𝐶(𝑘−𝑟)𝑛𝐶𝑟𝑚, which is a finite polynomial in 𝑛 and 𝑚

• So one can imagine that using successive steps of algebraic manipulations the Sum Form of 𝐶𝑘 can be transformed to the Factor Form of 𝐶𝑘 for integral values of 𝑛 and 𝑚 – These successive steps will be using Commutative, Associative and Distributive Laws of

numbers

• NOW: – The Commutative, Associative and Distributive laws are exactly the same for rational numbers

as for integers and so – if the same successive steps are applied on the Sum Form of 𝐶𝑘 for fractional values of 𝑛 and

𝑚, the result will be the SAME Factor Form of 𝐶𝑘

• Thus 𝐶𝑘= (𝑛+𝑚) 𝑛+𝑚−1 … 𝑛+𝑚−[𝑘−2] 𝑛+𝑚−[𝑘−1]

𝑘! for fractional 𝑛,𝑚

• Hence for 𝑛,𝑚 positive rational numbers: 𝑓 𝑛 . 𝑓 𝑚 =

(𝑛+𝑚) 𝑛+𝑚−1 … 𝑛+𝑚−[𝑘−2] 𝑛+𝑚−[𝑘−1]

𝑘!∞𝑘=0 𝑥𝑘

– Note that for k=0, the numerator is taken as 1

• And in fact this can be extended to:

• 𝑓 𝑛 . 𝑓 𝑚 …𝑓(𝑝) =

(𝑛+𝑚…+𝑝) 𝑛+𝑚…+𝑝−1 … 𝑛+𝑚…+𝑝−[𝑘−2] 𝑛+𝑚…+𝑝−[𝑘−1]

𝑘!∞𝑘=0 𝑥𝑘

• Let each of 𝑛,𝑚,…𝑝 be = 𝑙 ℎ for a positive integers 𝑙, ℎ and let us multiply 𝑓(𝑙 ℎ ) by itself ℎ times:

• 𝑓(𝑙 ℎ ). 𝑓(𝑙 ℎ )… . 𝑓(𝑙 ℎ ) [ℎ 𝑡𝑖𝑚𝑒𝑠] =

(𝑙 ℎ +𝑙

ℎ …+𝑙ℎ ) 𝑙

ℎ +𝑙ℎ …+𝑙

ℎ −1 … 𝑙ℎ +𝑙

ℎ …+𝑙ℎ −[𝑘−2] 𝑙

ℎ +𝑙ℎ …+𝑙

ℎ −[𝑘−1]

𝑘!∞𝑘=0 𝑥𝑘

• Now the left hand side of the above equation = 𝑓(𝑙 ℎ )ℎ • The right side contains the term 𝑙 ℎ + 𝑙

ℎ …+ 𝑙ℎ wherein 𝑙 ℎ occurs ℎ

times so that 𝑙 ℎ + 𝑙ℎ …+ 𝑙

ℎ = 𝑙 giving the right hand side as:

• (𝑙) 𝑙−1 … 𝑙−[𝑘−2] 𝑙−[𝑘−1]

𝑘!∞𝑘=0 𝑥𝑘

• Note that the term (𝑙) 𝑙−1 … 𝑙−[𝑘−2] 𝑙−[𝑘−1]

𝑘!

becomes 0 for 𝑘 = 𝑙 + 1 so we can write the

RHS as: (𝑙) 𝑙−1 … 𝑙−[𝑘−2] 𝑙−[𝑘−1]

𝑘!𝑙𝑘=0 𝑥𝑘

• In the above, since 𝑙 is a positive integer, we know by the binomial theorem that it represents (1 + 𝑥)𝑙. Thus:

• 𝑓(𝑙 ℎ )ℎ=(1 + 𝑥)𝑙and so: 𝑓 𝑙ℎ = (1 + 𝑥)

𝑙ℎ

• Thus: (1 + 𝑥)𝑟= 𝑓 𝑟 = 𝑟 𝑟−1 …(𝑟−[𝑘−1])

𝑘!𝑥𝑘∞

𝑘=0

for a positive rational number 𝑟. Equivalently:

• (1 + 𝑥)𝑟=

1 + 𝑟𝑥 +𝑟(𝑟−1)

2!𝑥2 +

𝑟(𝑟−1)(𝑟−2)

3!𝑥3+

𝑟(𝑟−1)(𝑟−2)(𝑟−3)

4!𝑥4…

+𝑟 𝑟−1 𝑟−2 …(𝑟− 𝑘−1 )

𝑘!𝑥𝑘 +⋯ where 𝑟 is a

positive rational number (including positive integers, of course!)

Lets try it out!

• 𝑥 and 𝑟 both positive integers: 𝑥 = 2; 𝑟 = 3

• (1 + 2)3= 1 + 3.2 +3.(3−1)

2!22 +

3.(3−1)(3−2)

3!23 +

3.(3−1)(3−2)(3−3)

4!24 +

3.(3−1)(3−2)(3−3)(3−4)

5!25 +⋯

• = 1 + 3.2 + 3. 22 + 23 + 0 + 0 +⋯ = 27

• Works – of course this is the same as binomial theorem for positive integer exponents

• Also, in the infinite series expansion, the coefficients after the 4th terms are all 0.

Lets try it out!

• 𝑥 and 𝑟 both positive fractions: 𝑥 =1

3; 𝑟 =

1

2

• (1 +1

3)1

2= 1 +1

2.1

3+

1

2.1

2−1

2!(1

3)2 +

1

2.1

2−1

1

2−2

3!(1

3)3 +

1

2.1

2−1

1

2−2

1

2−3

4!(1

3)4 +

1

2.1

2−1

1

2−2

1

2−3

1

2−4

5!(1

3)5 +⋯

• = 1 + 0.16666 − 0.01388 + 0.00231 − 0.00048 +0.00011 −⋯ = 1.15472

• Check: (1 +1

3)1

2= (1.333333)1

2 = 1.15470054 ≈ 1.15472

• So, taking 6 terms of the infinite series we get pretty close to the ‘real’ answer

Lets try it out!

• 𝑥 and 𝑟 both positive fractions: 𝑥 =11

10; 𝑟 =

1

2

• (1 +11

10)1

2= 1 +1

2.11

10+

1

2.1

2−1

2!(11

10)2 +

1

2.1

2−1

1

2−2

3!(11

10)3 +

1

2.1

2−1

1

2−2

1

2−3

4!(11

10)4 +

1

2.1

2−1

1

2−2

1

2−3

1

2−4

5!(11

10)5 +⋯

• = 1 + 0.55 − 0.15125 + 0.08318 − 0.05719 +0.04403 −⋯ = 1.46878

• Check: (1 +11

10)1

2= (2.1)1

2 = 1.449138 ≈? 1.46878

• So, taking 6 terms of the infinite series we are close but not as close!

• Taking 20 terms instead of 6 and then we reach 1.459846

• Taking 50 terms instead of 6 and then we reach 1.494788 – now we are farther from what we had with 20 terms!

• Lets try 100 terms: 3.321921 !!

• Worse still, 101 terms gives: −0.58017 !

• Lets look at how the sum behaves as we take more and more values!

Sum Value vs Number of Terms!!

-0.5

0

0.5

1

1.5

2

2.5

3

3.5

4

1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 39 41 43 45 47 49 51 53 55 57 59 61 63 65 67 69 71 73 75 77 79 81 83 85 87 89 91 93 95 97 99

Series1

Lesson

• Our infinite series formula: • (1 + 𝑥)𝑟=1 + 𝑟𝑥 +

𝑟(𝑟−1)

2!𝑥2 +

𝑟(𝑟−1)(𝑟−2)

3!𝑥3+

𝑟(𝑟−1)(𝑟−2)(𝑟−3)

4!𝑥4…

+𝑟 𝑟−1 𝑟−2 …(𝑟− 𝑘−1 )

𝑘!𝑥𝑘 +⋯

• Works always if 𝑟 is a positive integer • Works for |𝑥| < 1 for if 𝑟 is a positive fraction • And this stems from the fact that the infinite sum

will converge if and only if |𝑥| < 1

Utilising the Series

• So, how do we use this infinite sum in real life to evaluate (𝑎 + 𝑏)𝑟 where 𝑎, 𝑏 are real number and 𝑟 is a positive fraction?

• If 𝑎 < 𝑏:

• (𝑎 + 𝑏)𝑟=(𝑏𝑎

𝑏+ 1 )𝑟= 𝑏𝑟(

𝑎

𝑏+ 1)𝑟 and now

since 𝑎

𝑏<1 the infinite series will converge and can

be used.

• Similarly, if 𝑎 > 𝑏 then (𝑎 + 𝑏)𝑟=𝑎𝑟(𝑏

𝑎+ 1)𝑟

What if 𝑟 is a negative fraction?

• Notice that up until the derivation of:

• 𝑓 𝑛 . 𝑓 𝑚 …𝑓(𝑝) =

(𝑛+𝑚…+𝑝) 𝑛+𝑚…+𝑝−1 … 𝑛+𝑚…+𝑝−[𝑘−2] 𝑛+𝑚…+𝑝−[𝑘−1]

𝑘!∞𝑘=0 𝑥𝑘

• There is nothing in the argument that is specific to 𝑛 and 𝑚 being positive rational numbers

• In fact the argument is valid for 𝑛,𝑚 being negative rational numbers, real numbers or even complex numbers since they all follow the same rules of Commutation, Association, Distribution etc. – In fact as long as 𝑛,𝑚 are members of a Commutative Ring the above

derivation is true – Of course the question is whether the infinite series 𝑓(𝑛) converges or

not, for a class of inputs • E.g. for |𝑥| < 1, the 𝑓(𝑛) converges for any positive rational value of 𝑛

• So for a rational number 𝑟 > 0, we can say:

• 𝑓 𝑟 . 𝑓 −𝑟 =

(𝑟+[−𝑟]) 𝑟+[−𝑟]−1 … 𝑟+[−𝑟]−[𝑘−2] 𝑛+𝑚−[𝑘−1]

𝑘!∞𝑘=0 𝑥𝑘

– And remembering that for 𝑘 = 0, the numerator = 1

• The series sum then reduces to 1 since there is a factor (𝑟 + [−𝑟]) in each numerator but for when 𝑘 = 0 when the numerator is 1

• So we get: 𝑓 𝑟 . 𝑓 −𝑟 = 1 • We proved that 𝑓 𝑟 = (1 + 𝑥)𝑟 for 𝑟 > 0 • Hence: 𝑓 −𝑟 = (1 + 𝑥)−𝑟for 𝑟 > 0 • Thus for all rational values of 𝑟, 𝑓 𝑟 = (1 + 𝑥)𝑟 , 𝑥 < 1