Amount (math)

Sum ( lat. Summa - total, total) in mathematics is the result of the operation of addition of numerical quantities ( numbers , functions , vectors , matrices , etc. ), or the result of the sequential execution of several operations of addition (summation). Common to all cases are the properties of commutativity , associativity , and also distributivity with respect to multiplication (if multiplication is defined for the quantities under consideration), i.e.

a+b=b+a

{\ displaystyle a + b = b + a}

a + b = b + a

a+(b+c)=(a+b)+c

{\ displaystyle a + (b + c) = (a + b) + c}

a + (b + c) = (a + b) + c

(a+b)\cdot c=a\cdot c+b\cdot c

{\ displaystyle (a + b) \ cdot c = a \ cdot c + b \ cdot c}

(a + b) \ cdot c = a \ cdot c + b \ cdot c

c\cdot (a+b)=c\cdot a+c\cdot b

{\ displaystyle c \ cdot (a + b) = c \ cdot a + c \ cdot b}

c \ cdot (a + b) = c \ cdot a + c \ cdot b

In set theory, the sum (or union) of sets is the set whose elements are all elements of the terms of the sets taken without repetition.

The addition (finding the sum) operation can be defined for more complex algebraic structures . The sum of groups , the sum of linear spaces , the sum of ideals , and other examples. In category theory , the concept of the sum of objects is defined.

Content

Arithmetic sum

Let in the set $\mathbb {N}$ ${\ displaystyle \ mathbb {N}}$ located $a$ ${\ displaystyle a}$ elements forming a subset $A$ ${\ displaystyle A}$ , and $b$ ${\ displaystyle b}$ elements forming a subset $B$ ${\ displaystyle B}$ ( $A\subset \mathbb {N} ,B\subset \mathbb {N}$ ${\ displaystyle A \ subset \ mathbb {N}, B \ subset \ mathbb {N}}$ , a and b are natural numbers). Then the arithmetic sum $a+b$ ${\ displaystyle a + b}$ there will be an amount of elements $c$ ${\ displaystyle c}$ forming a subset $C\subset \mathbb {N}$ ${\ displaystyle C \ subset \ mathbb {N}}$ obtained by the disjoint union of two original subsets $C=A\sqcup B;$ ${\ displaystyle C = A \ sqcup B;}$ .

Algebraic Amount

The sum is mathematically denoted by the capital Greek letter Σ (sigma) .

\sum _{i\mathop {=} m}^{n}a_{i}=a_{m}+a_{m+1}+a_{m+2}+\cdots +a_{n-1}+a_{n}

{\ displaystyle \ sum _ {i \ mathop {=} m} ^ {n} a_ {i} = a_ {m} + a_ {m + 1} + a_ {m + 2} + \ cdots + a_ {n- 1} + a_ {n}}

where: i is the summation index; a _i is a variable denoting each member in the series; m is the lower limit of summation, n is the upper limit of summation. The designation “i = m” under the summation symbol means that the initial (starting) value of the index i is equivalent to m . From this record it follows that the index i is incremented by 1 in each term of the expression and will stop when i = n . ^[one]

In programming, this procedure corresponds to a for loop.

Record Examples

\sum _{i\mathop {=} 1}^{100}i=1+2+3+4+...+99+100

{\ displaystyle \ sum _ {i \ mathop {=} 1} ^ {100} i = 1 + 2 + 3 + 4 + ... + 99 + 100}

\sum _{i\mathop {=} 3}^{6}i^{2}=3^{2}+4^{2}+5^{2}+6^{2}=86

{\ displaystyle \ sum _ {i \ mathop {=} 3} ^ {6} i ^ {2} = 3 ^ {2} + 4 ^ {2} + 5 ^ {2} + 6 ^ {2} = 86 }

Indication of boundaries may be omitted from the record if they are clear from the context:

\sum a_{i}^{2}=\sum _{i\mathop {=} 1}^{n}a_{i}^{2}

{\ displaystyle \ sum a_ {i} ^ {2} = \ sum _ {i \ mathop {=} 1} ^ {n} a_ {i} ^ {2}}

An iterator can be an expression, then the variable is framed with brackets as a function of " $f()$ ${\ displaystyle f ()}$ ". For example, the sum of all natural numbers $k$ ${\ displaystyle k}$ in a certain range:

\sum _{0\leq k<100}f(k)

{\ displaystyle \ sum _ {0 \ leq k <100} f (k)}

Amount $f(x)$ ${\ displaystyle f (x)}$ elements $x$ ${\ displaystyle x}$ many $S$ ${\ displaystyle S}$ :

\sum _{x\mathop {\in } S}f(x)

{\ displaystyle \ sum _ {x \ mathop {\ in} S} f (x)}

Amount $\mu (d)$ ${\ displaystyle \ mu (d)}$ all positive numbers $d$ ${\ displaystyle d}$ being divisors of a number $n$ ${\ displaystyle n}$ :

\sum _{d|n}\;\mu (d)

{\ displaystyle \ sum _ {d | n} \; \ mu (d)}

Several sigma characters can generalize, for example:

\sum _{\ell ,\ell '}=\sum _{\ell }\sum _{\ell '}

{\ displaystyle \ sum _ {\ ell, \ ell '} = \ sum _ {\ ell} \ sum _ {\ ell'}}

Infinite Amount

In mathematical analysis, the concept of a series is defined - the sum of an infinite number of terms.

Examples

1. The sum of the arithmetic progression :

\sum _{i=0}^{n}(a_{0}+b\cdot i)=(n+1){\frac {a_{0}+a_{n}}{2}}

{\ displaystyle \ sum _ {i = 0} ^ {n} (a_ {0} + b \ cdot i) = (n + 1) {\ frac {a_ {0} + a_ {n}} {2}} }

2. The sum of the geometric progression :

\sum _{i=0}^{n}a_{0}\cdot b^{i}=a_{0}\cdot {\frac {1-b^{n+1}}{1-b}}

{\ displaystyle \ sum _ {i = 0} ^ {n} a_ {0} \ cdot b ^ {i} = a_ {0} \ cdot {\ frac {1-b ^ {n + 1}} {1- b}}}

3. $\sum \limits _{k=1}^{n}k^{3}=\left[{\frac {n(n+1)}{2}}\right]^{2}=\left(\sum \limits _{k=1}^{n}k\right)^{2}$ ${\ displaystyle \ sum \ limits _ {k = 1} ^ {n} k ^ {3} = \ left [{\ frac {n (n + 1)} {2}} \ right] ^ {2} = \ left (\ sum \ limits _ {k = 1} ^ {n} k \ right) ^ {2}}$

four. $\sum _{i=0}^{n}{\left({\frac {1}{p}}\right)}^{i}={\frac {p}{p-1}}\left(1-{\frac {1}{p^{n+1}}}\right),\quad p\neq 1,n\geq 0$ ${\ displaystyle \ sum _ {i = 0} ^ {n} {\ left ({\ frac {1} {p}} \ right)} ^ {i} = {\ frac {p} {p-1}} \ left (1 - {\ frac {1} {p ^ {n + 1}}} \ right), \ quad p \ neq 1, n \ geq 0}$

Evidence

\sum _{i=0}^{n}{\left({\frac {1}{p}}\right)}^{i}=\sum _{i=0}^{n}{1\cdot {\frac {1}{p^{i}}}}=1\cdot {\frac {1-{\left({\frac {1}{p}}\right)}^{n+1}}{1-{\frac {1}{p}}}}={\frac {\frac {p^{n+1}-1}{p^{n+1}}}{\frac {p-1}{p}}}={\frac {p^{n+1}-1}{p^{n}(p-1)}}={\frac {p}{p-1}}\left(1-{\frac {1}{p^{n+1}}}\right)

{\ displaystyle \ sum _ {i = 0} ^ {n} {\ left ({\ frac {1} {p}} \ right)} ^ {i} = \ sum _ {i = 0} ^ {n} {1 \ cdot {\ frac {1} {p ^ {i}}}} = 1 \ cdot {\ frac {1 - {\ left ({\ frac {1} {p}} \ right)} ^ {n +1}} {1 - {\ frac {1} {p}}}} = {\ frac {\ frac {p ^ {n + 1} -1} {p ^ {n + 1}}} {\ frac {p-1} {p}}} = {\ frac {p ^ {n + 1} -1} {p ^ {n} (p-1)}} = {\ frac {p} {p-1} } \ left (1 - {\ frac {1} {p ^ {n + 1}}} \ right)}

five. $\sum _{i=0}^{n}ip^{i}={\frac {np^{n+2}-(n+1)p^{n+1}+p}{(p-1)^{2}}},\quad p\neq 1$ ${\ displaystyle \ sum _ {i = 0} ^ {n} ip ^ {i} = {\ frac {np ^ {n + 2} - (n + 1) p ^ {n + 1} + p} {( p-1) ^ {2}}}, \ quad p \ neq 1}$

Evidence

\sum _{i=0}^{n}ip^{i}=\sum _{i=1}^{n}ip^{i}=p\cdot \sum _{i=1}^{n}ip^{i-1}=p\cdot \sum _{i=0}^{n-1}(i+1)p^{i}=p\cdot \left(\sum _{i=0}^{n-1}{ip^{i}}+\sum _{i=0}^{n-1}p^{i}\right)=p\cdot \sum _{i=0}^{n}ip^{i}-p\cdot np^{n}+p\cdot {\frac {1-p^{n}}{1-p}}\Rightarrow

{\ displaystyle \ sum _ {i = 0} ^ {n} ip ^ {i} = \ sum _ {i = 1} ^ {n} ip ^ {i} = p \ cdot \ sum _ {i = 1} ^ {n} ip ^ {i-1} = p \ cdot \ sum _ {i = 0} ^ {n-1} (i + 1) p ^ {i} = p \ cdot \ left (\ sum _ { i = 0} ^ {n-1} {ip ^ {i}} + \ sum _ {i = 0} ^ {n-1} p ^ {i} \ right) = p \ cdot \ sum _ {i = 0} ^ {n} ip ^ {i} -p \ cdot np ^ {n} + p \ cdot {\ frac {1-p ^ {n}} {1-p}} \ Rightarrow}

\Rightarrow (1-p)\sum _{i=0}^{n}ip^{i}={\frac {-np^{n+1}(1-p)+p-p^{n+1}}{1-p}}\Rightarrow \sum _{i=0}^{n}ip^{i}={\frac {np^{n+2}-(n+1)p^{n+1}+p}{(1-p)^{2}}}

{\ displaystyle \ Rightarrow (1-p) \ sum _ {i = 0} ^ {n} ip ^ {i} = {\ frac {-np ^ {n + 1} (1-p) + pp ^ {n +1}} {1-p}} \ Rightarrow \ sum _ {i = 0} ^ {n} ip ^ {i} = {\ frac {np ^ {n + 2} - (n + 1) p ^ { n + 1} + p} {(1-p) ^ {2}}}}

6. $\sum _{i=0}^{n}p^{i}=(p-1)\sum _{i=0}^{n-1}((n-i)p^{i})+n+1,\quad p\neq 1$ ${\ displaystyle \ sum _ {i = 0} ^ {n} p ^ {i} = (p-1) \ sum _ {i = 0} ^ {n-1} ((ni) p ^ {i}) + n + 1, \ quad p \ neq 1}$

Evidence

(p-1)\sum _{i=0}^{n-1}((n-i)p^{i})+n+1=(p-1)\sum _{i=0}^{n}((n-i)p^{i})+n+1=(p-1)\left(n\cdot \sum _{i=0}^{n}p^{i}-\sum _{i=0}^{n}ip^{i}\right)+n+1=

{\ displaystyle (p-1) \ sum _ {i = 0} ^ {n-1} ((ni) p ^ {i}) + n + 1 = (p-1) \ sum _ {i = 0} ^ {n} ((ni) p ^ {i}) + n + 1 = (p-1) \ left (n \ cdot \ sum _ {i = 0} ^ {n} p ^ {i} - \ sum _ {i = 0} ^ {n} ip ^ {i} \ right) + n + 1 =}

=(p-1)\left(n\cdot {\frac {1-p^{n+1}}{1-p}}-{\frac {np^{n+2}-(n+1)p^{n+1}+p}{(1-p)^{2}}}\right)+n+1=

{\ displaystyle = (p-1) \ left (n \ cdot {\ frac {1-p ^ {n + 1}} {1-p}} - {\ frac {np ^ {n + 2} - (n +1) p ^ {n + 1} + p} {(1-p) ^ {2}}} \ right) + n + 1 =}

={\frac {np^{n+2}-np-np^{n+1}+n-np^{n+2}+np^{n+1}+p^{n+1}-p+pn-n+p-1}{p-1}}=

{\ displaystyle = {\ frac {np ^ {n + 2} -np-np ^ {n + 1} + n-np ^ {n + 2} + np ^ {n + 1} + p ^ {n + 1 } -p + pn-n + p-1} {p-1}} =}

={\frac {p^{n+1}-1}{p-1}}=\sum _{i=0}^{n}p^{i}

{\ displaystyle = {\ frac {p ^ {n + 1} -1} {p-1}} = \ sum _ {i = 0} ^ {n} p ^ {i}}

It is worth noting that with

p=10\

{\ displaystyle p = 10 \}

we get

\sum _{i=0}^{n}10^{i}=9\cdot \sum _{i=0}^{n-1}((n-i)10^{i})+n+1

{\ displaystyle \ sum _ {i = 0} ^ {n} 10 ^ {i} = 9 \ cdot \ sum _ {i = 0} ^ {n-1} ((ni) 10 ^ {i}) + n +1}

, and this is a sequence of equalities of the following form:

1=9\cdot 0+1,\quad 11=9\cdot 1+2,\quad 111=9\cdot 12+3,\quad 1111=9\cdot 123+4,\quad 11111=9\cdot 1234+5

{\ displaystyle 1 = 9 \ cdot 0 + 1, \ quad 11 = 9 \ cdot 1 + 2, \ quad 111 = 9 \ cdot 12 + 3, \ quad 1111 = 9 \ cdot 123 + 4, \ quad 11111 = 9 \ cdot 1234 + 5}

Indefinite amount

Undefined amount $a_{i}$ ${\ displaystyle a_ {i}}$ by $i$ ${\ displaystyle i}$ this function is called $f(i)$ ${\ displaystyle f (i)}$ denoted by $\sum _{i}^{}a_{i}$ ${\ displaystyle \ sum _ {i} ^ {} a_ {i}}$ , what $\forall i:f(i+1)-f(i)=a_{i}$ ${\ displaystyle \ forall i: f (i + 1) -f (i) = a_ {i}}$ .

Newton-Leibniz formula

If an undefined amount is found $\sum _{i}^{}a_{i}=f(i)$ ${\ displaystyle \ sum _ {i} ^ {} a_ {i} = f (i)}$ then $\sum _{i=a}^{b}a_{i}=f(b+1)-f(a)$ ${\ displaystyle \ sum _ {i = a} ^ {b} a_ {i} = f (b + 1) -f (a)}$ .

Etymology

The Latin word summa is translated as “main point”, “essence”, “total”. From the 15th century the word begins to be used in the modern sense, the verb “summarize” appears ( 1489 ).

This word has penetrated into many modern languages: sum in Russian, sum in English, somme in French.

Euler was the first to introduce a special symbol for the sum (S) in 1755 . Alternatively, the Greek letter Sigma Σ was used. Later, due to the connection between the concepts of summation and integration, S was also used to denote the integration operation.

Encoding

Unicode has a sum symbol U + 2211 ∑ n-ary summation (HTML ∑ · ∑ ).

Notes

↑ Chapter 2: Sums // Concrete Mathematics: A Foundation for Computer Science (2nd Edition) : [ eng. ] . - Addison-Wesley Professional, 1994. - ISBN 978-0201558029 . (inaccessible link)

Literature

Fichtenholtz G. M. The course of differential and integral calculus. - 7th. - M .: Nauka, 1969. - T. 1. - 608 p. - 100,000 copies.

[1] Chapter 2: Sums // Concrete Mathematics: A Foundation for Computer Science (2nd Edition) : [ eng. ] . - Addison-Wesley Professional, 1994. - ISBN 978-0201558029 . (inaccessible link)

Amount (math)

Content

Arithmetic sum

Algebraic Amount

Infinite Amount

Examples

Indefinite amount

Newton-Leibniz formula

Etymology

Encoding

See also

Notes

Literature

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