Arithmetic continuation

In mathematical, a arithmetic continuation is a continuation definite on $\ {N \ in \ mathbb NR, N \ geq n_0 \}$ with values in a additive group E such as it exists an element $\ r$ of $\ E$ called reason for lequel :

\ forall N \ geq n_0 \ \ \ u_ {n+1} = u_n + R \,

In practice $E = \ R$ or $\ mathbb C$ . But one can just as easily meet arithmetic continuations with values in a vector Space.

It is said whereas the terms $\ u_n$ are in “ arithmetic progression ”.

Example If the reason $\ r=2$ and $\ u_0=10$ :

$\ u_0=10$
$\ u_1=12$
$\ u_2=14$
$\vdots$

General term

If $E$ is a group and if $(u_n) _ {N \ in \ mathbb NR}$ is an arithmetic continuation of $E$ of reason $r \ in E$ then, for all $n \ in \ mathbb N$ :

u_n = u_0 + n.r \,

More generally, if the continuation is defined on $\ {N \ in \ mathbb NR, N \ geq n_0 \}$ and if N and p belongs to has then:

u_n = u_p + (N - p) .r \,

An arithmetic continuation thus is entirely determined by the data of its first term $u_ {n_0}$ and by its reason R .

Reciprocally, a continuation defined on $\ {N \ in \ mathbb NR, N \ geq n_0 \}$ by

u_n = u_ {n_0} + (N - n_0) .r \,

is an arithmetic continuation of reason R .

In real or complex analysis, the arithmetic continuation is the discrete aspect of the function closely connected.

Direction of variation and convergence

This paragraph relates to the arithmetic continuations with values in

\ R

If R > 0 the continuation is increasing, if R < 0 the continuation is decreasing and if R = 0 the continuation is constant.

In general (if R is nonnull), the arithmetic continuation is divergent. However she admits a limit:

if R > 0 its limit is $+ \ infty$
if R < 0 its limit is $- \ infty$ .
If the reason is null, the continuation is constant and converges towards the constant.

Summon terms

If $E = \ R$ or $\ mathbb C$ and if $(u_n) _ {N \ in \ mathbb NR}$ is an arithmetic continuation of $E$ then, for all $n \ in \ mathbb N$ :

$\ sum_ {0 \ the p \ N} u_p= {(n+1) \ over 2} (u_0+u_n)$

The legend wants that the method of calculating was invented by Carl Friedrich Gauss, raises dissipated that it was a question of occupying and with which one would have entrusted the task to calculate the sum of all the entireties from 1 to 100. By writing the sum twice, in a different order, it obtained:

1 + 2 + 3 +…. + 98 + 99 + 100

100 + 99 + 98 +… + 3 + 2 + 1

Then, noticing that 100 + 1 = 99 + 2 = 98 + 3 =… = 101, it obtained easily

2S = 100 × 101 thus S = 50 × 101.

Caption or reality, this easy way is the method of demonstration to calculate the nap of the terms:

$S = u_0 + u_1 +… + u_n$

S = u_n + u_ {n-1} +… + u_0

Noticing that $u_ {p} + u_ {Np} = u_0 + u_n$ , it comes

$2S = (n+1) \ times (u_0+u_n)$

This property applies to calculate the sum of N first entireties

$1 + 2 + 3… + N = \ frac {N (n+1)}{2}$

and spreads with any sum of consecutive terms of an arithmetic continuation

$u_p + u_ {p+1} +… +u_n = \ frac {(n-p+1) (u_n + u_p)}{2}$

It also spreads with all following values in a vector Space on a characteristic body of different from $2$

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