Series

Some remarks concerning sequences and series are in the following paragraphs....

Exponential (Euler) series

Series defined by the following relation makes an interesting group:

E({a_k}) = ∑ a_k * x^k/k!

Let us denote m the period of the sequence {a_k}.

Basic sequences

m = 1

For the zero sequence {a_k}, E{0} = 0. And for E{1} or E{-1} we use the Euler relation: E(x) = e^x = x^0/0! + x^1/1! + x^2/2! + x^3/3! ...

    E(-1) = -e       E{1}  = e 

m = 2

From the relations:

    E{0,1} = (e - 1/e)/2 = 1.175201194      E{1, 0} =  (e + 1/e)/2 = 1.543080635

we get:

  E(p,q) = p*(e + 1/e)/2 + q*(e - 1/e)/2 = 1/2*[(p+q)*e + (p-q)/e]  

m = 3

    E{1,0,0} = 1.168058313
    E{0,1,0} = 1.041865365
    E{0,0,1} = 0.508358157(=1.016716314 / 2)

m = 4

    E{1,0,0,0} = 1.04169147 (=a)
    E{0,1,0,0} = 1.00833609 (=b)
    E{0,0,1,0} = 0.50138916 (=1.00277832 / 2 =c)
    E{0,0,0,1} = 0.16686511 (=1.00119066 / 4 =d)

And the known series for sinus and cosine:

    E{0,1,0,-1} = sin 1 [rad] = (ei-1/ei)/2i = 0.841470985 (=b-d)
    E{1,0,-1,0} = cos 1 [rad] = (ei+1/ei)/2 = 0.540302306 (=a-c)

Nevertheless (inheriting from m = 2), we have:

    E{0,1,0,1} = E{0, 1} = (e - 1/e)/2 = 1.175201194 (=b+d) = sinh 1 (hyperbolic sine)         
    E{1,0,1,0} = E{1, 0} = (e + 1/e)/2 = 1.543080635 (=a+c) = cosh 1 (hyperbolic cosine)

From the sum of the expressions E{1,0,-1,0} + E{1,0,1,0} = 2*E{1,0,0,0} (=2*a) we get (eⁱ+1/eⁱ)/2 + (e + 1/e)/2.
And similarly from E{0,1,0,-1} + E{0,1,0,1} = 2*E{0,1,0,0} (=2*b) we have (eⁱ/i-1/eⁱ/i)/2 + (e - 1/e)/2 i.e.:

  E(1,0,0,0) = (e¹+ eⁱ+ e^-1+ e^-i)/4       E(0,1,0,0) = (e¹+ eⁱ/i- e^-1- e^-i/i)/4  

Next E{1,0,1,0} - E{1,0,-1,0} = 2*E{0,0,1,0} (=2*c) and E{0,1,0,1} - E{0,1,0,-1} = 2*E{0,0,0,1} (=2*d) i.e.:

  E(0,0,1,0) = (e¹- eⁱ+ e^-1- e^-i)/4       E(0,0,0,1) = (e¹- eⁱ- e^-1+ e^-i)/4  

• (a=) E(1,0,0,0) = cos 1 + cosh 1, (b=) E(0,1,0,0) = sin 1 + sinh 1.
• Euler was able to decompose number (e^x-1/e^x)/2 in product of number x and factors of the form (1+(x/nπ)²) and similarly also number (e^ix-1/e^ix)/2i in product of number x and factors of the form (1-(x/nπ)²).

Sums through power sequences

From the sums with simple power sequences we get the expressions whose coefficients are Stirling's second-order numbers:

            E(n) = x*ex 
            E(n2) = (x2+x)*ex 
            E(n3) = (x3+3*x2+x)*ex 
            E(n4) = (x4+6*x3+7*x2+x)*ex 
            E(n5) = (x5+10*x4+25*x3+15*x2+x)*ex 
        

Sums through special sequences

The result of the sum for Bernoulli's sequence B_n:

E(B_n) = x / e^x-1

For numbers E_n/2ⁿ of Euler's polynomials E_n(x):

E(E_n/2ⁿ) = 2e^x/2/(e^x+1)