## Exponential Series:

An exponential series is a mathematical expression that represents the sum of an infinite sequence of terms, where each term is a power of a fixed number raised to a non-negative integer power.

You know that the infinite series, Σt_{n}, is said to be convergent when the sum of its first n terms is less than or equal to a finite quantity, however, large n may be. Thus, series, 1 + 1/3 + 1/3^{2} + … to ∞, is convergent which converges to 1/[1 – (1/3)] i.e., to 3/2.

The infinite series, 1 + 1/1! + 1/2! + 1/3! + … to ∞, is convergent and its sum is denoted by the symbol ‘e’.

Thus, one may write e = 1 + 1/1! + 1/2! + 1/3! + … to ∞ Now, e = 2 + 1/2! + 1/3! + 1/4! + … to ∞ ∴ e > 2 ……….(i) Again, 3! = 3 x 2 x 1 = 6 > 2 ^{2} ∴ 1/3! < 1/2 ^{2} 4! = 4 x 3 x 2 x 1 = 24 > 2 ^{3} ∴ 1/4! < 1/2 ^{3} Similarly, 1/5! < 1/2 ^{4} , 1/6! < 1/2^{5} and so on.∴ e = 2 + 1/2! + 1/3! + 1/4! + … to ∞ ……….(ii) < 2 + 1/2 + 1/2 ^{2} + 1/2^{3} + … to ∞< 2 + [(1/2) / {1 – (1/2)}] < 3 ……….(iii) From (ii) and (iii), 2 < e < 3 The exact value of e cannot be found out. It is incommensurable (irrational), which can easily be proved by the method of contradiction. If possible, let us suppose that e is rational. Then, one may write e = p/q, where p and q ∈ N (since 2 < e < 3) and q ≠ 1. Then, p/q = 1 + 1/1! + 1/2! + 1/3! + … + 1/q! + 1/(q + 1)! + 1/(q + 2)! + … to ∞ Multiplying both sides by q! we get p (q – 1)! = 2 (q!) + q!/2! + q!/3! + … + 1 + 1/(q + 1) + 1/[(q + 1) (q + 2)] + … = k + f where k = 2 (q!) + q!/2! + q!/3! + … + 1 = a positive integer and f = 1/(q + 1) + 1/[(q + 1) (q + 2)] + … to ∞ < 1/(q + 1) + 1/(q + 1) ^{2} + … to ∞ < [1/(q + 1)] / [1 – 1/(q + 1)] < 1/q But f = 1/(q + 1) + 1/[(q + 1) (q + 2)] + … ⇒ f > 1/(q + 1) ∴ 1/(q + 1) < f < 1/q Thus, f is a positive proper fraction. But p (q – 1)! is a positive integer. Thus, from (iii), it is found that a positive integer = a positive integer + a positive proper fraction, which is absurd. So, e is an incommensurable (irrational) number. The approximate value of e may be taken as 2.7182818 (correct to seven places of decimals). Let n > 1, then 1/n < 1 Using the binomial theorem, we have (1 + 1/n) ^{n} = 1 + n (1/n) + [{n (n – 1)}/2!] (1/n)^{2} + [{n (n – 1) (n – 2)}/3!] (1/n)^{3} + …= 1 + 1/1! + (1 – 1/n)/2! + [(1 – 1/n) (1 – 2/n)]/3! + …, which is an identity. Now, when n approaches infinity, each of 1/n, 2/n, 3/n, etc., will approach zero. Thus, in the limiting case, when n is very large, (1 + 1/n) ^{n} = 1 + 1/1! + 1/2! + 1/3! + … = e |

For all real values of x, eTheorem 1: ^{x} = 1 + x/1! + x^{2}/2! + x^{3}/3! + … to ∞ (exponential series) Let n > 1, then 1/n < 1. Now, using the binomial theorem, we have, for all real values of x,Proof-(1 + 1/n) ^{nx} = 1 + nx (1/n) + [nx (nx – 1)/2!] [1/n^{2}] + [nx (nx – 1) (nx – 2)/3!] [1/n^{3}] + …⇒ [(1 + 1/n) ^{n}]^{x} = 1 + x + x (x – 1/n)/2! + x (x – 1/n) (x – 2/n)/3! + …when n → ∞, 1/n → 0, 2/n → 0, 3/n → 0 etc. Thus, in the limiting case, when n is very large, [(1 + 1/n) ^{n}]^{x} = 1 + x/1! + x^{2}/2! + x^{3}/3! + … to ∞⇒ e ^{x} = 1 + x/1! + x^{2}/2! + x^{3}/3! + … to ∞ |

If a > 0, then aTheorem 2: ^{x} = 1 + (x/1!) log_{e} a + (x^{2}/2!) (log_{e} a)^{2} + (x^{3}/3!) (log_{e} a)^{3} + … to ∞ From the exponential theorem, we have for all values of zSolution-e ^{z} = 1 + z/1! + z^{2}/2! + z^{3}/3! + … to ∞Replacing z by cx, we have e ^{cx} = 1 + cx/1! + c^{2}x^{2}/2! + c^{3}x^{3}/3! + … to ∞ ……….(i)Let e ^{c} = a, the c log_{e} e = log_{e} a ⇒ c = log_{e} a∴ e ^{cx} = (e^{c})^{x} = a^{x}Thus, from (i), a ^{x} = 1 + (x/1!) log_{e} a + (x^{2}/2!) (log_{e} a)^{2} + (x^{3}/3!) (log_{e} a)^{3} + …We have, e ^{x} = 1 + x/1! + x^{2}/2! + x^{3}/3! + x^{4}/4! + … Replacing x by -x, e ^{-x} = 1 – x/1! + x^{2}/2! – x^{3}/3! + x^{4}/4! – … ∴ (e ^{x} + e^{-x})/2 = 1 + x^{2}/2! + x^{4}/4! + … and (e^{x} – e^{-x})/2 = x + x^{3}/3! + x^{5}/5! + …Also, e = 1 + 1/1! + 1/2! + 1/3! + 1/4! + … e ^{-1} = 1 – 1/1! + 1/2! – 1/3! + 1/4! – …∴ (e + e ^{-1})/2 = 1 + 1/2! + 1/4! + … and (e – e^{-1})/2 = 1/1! + 1/3! + 1/5! + … |

Find the coefficient of xExample 1:^{10} in the expansion of e^{2x}. eSolution-^{2x} = 1 + 2x/1! + (2x)^{2}/2! + (2x)^{3}/3! + …⇒ e ^{2x} = 1 + 2x/1! + 2^{2}x^{2}/2! + 2^{3}x^{3}/3! + …∴ coefficient of x ^{10} is 2^{10}/10! Find the coefficient of bExample 2:^{n} in the series (1 + b)/1! + (1 + b)^{2}/2! + (1 + b)^{3}/3! + … to ∞ The given series is,Solution-(1 + b)/1! + (1 + b) ^{2}/2! + (1 + b)^{3}/3! + …= [1 + (1 + b)/1! + (1 + b) ^{2}/2! + (1 + b)^{3}/3! + …] – 1= e ^{1+b} – 1= e ^{1} . e^{b} – 1∴ coefficient of b ^{n} is e (1/n!) |

Prove that logExample 3:_{e} 3 + (log_{e} 3)^{2}/2! + (log_{e} 3)^{3}/3! + … to ∞ = 2 The given series is-Solution-log _{e} 3 + (log_{e} 3)^{2}/2! + (log_{e} 3)^{3}/3! + … to ∞= (1 + log _{e} 3 + (log_{e} 3)^{2}/2! + …) – 1= e ^{loge 3} – 1= 3 -1 = 2 |

Calculate the sum of the series 2/1! + 4/3! + 6/5! + … to ∞Example 4: The given series is,Solution-S _{∞} = (1/0! + 1/2! + 1/4! + …) + (1/1! + 1/3! + 1/5! + …)S _{∞} = (e^{1} + e^{-1})/2 + (e^{1} – e^{-1})/2S _{∞} = (e + e^{-1} + e – e^{-1})/2S _{∞} = 2e/2 S _{∞} = e |

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