# Series

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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({ak}) = ∑ ak * xk/k!

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

### Basic sequences

#### m = 1

For the zero sequence {ak}, 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:

#### 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

Notes:
• (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 (ex-1/ex)/2 in product of number x and factors of the form (1+(x/nπ)2) and similarly also number (eix-1/eix)/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 Bn:

#### E(Bn) = x / ex-1

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