# 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}}.

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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! ...

#### 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] = (e^{i}-1/e^{i})/2i = 0.841470985 (=b-d)
E{1,0,-1,0} = cos 1 [rad] = (e^{i}+1/e^{i})/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 (e^{i}+1/e^{i})/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 (e^{i}/i-1/e^{i}/i)/2 + (e - 1/e)/2 i.e.:

#### E(1,0,0,0) = (e^{1}+ e^{i}+ e^{-1}+ e^{-i})/4
E(0,1,0,0) = (e^{1}+ e^{i}/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) = (e^{1}- e^{i}+ e^{-1}- e^{-i})/4
E(0,0,0,1) = (e^{1}- e^{i}- 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 (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}).

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Stirling triangle

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

E(n) = z*e^{z}
E(n^{2}) = (z^{2}+z)*e^{z}
E(n^{3}) = (z^{3}+3*z^{2}+z)*e^{z}
E(n^{4}) = (z^{4}+6*z^{3}+7*z^{2}+z)*e^{z}
E(n^{5}) = (z^{5}+10*z^{4}+25*z^{3}+15*z^{2}+z)*e^{z}

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Bernoulli numbers

Result of the sum, where the specified sequence is the Bernouli number serie, is also known:

#### E(B_{n}) = z / e^{z}-1