Talk:Addition
From Academic Kids
When b is replaced with the infinity (??) symbol, the sum is an infinite series. This has a countably infinite number of terms, and represents the limit of the sum of the first n terms, as n grows without bound.
This needs to be explained better. Vera Cruz
The lowercase theta function used on this page needs to be replaced by uppercase theta, as described at Big O notation. --Zero 06:30, 23 Dec 2003 (UTC)
Uncommon Bounds
See also Multiplication.
What if the bounds are fractions? For example the series:
<math>\sum_{i=1}^n 2i-1 = n^2<math>
<math>\left(\frac{a}{b}\right)^2 = \sum_{i=1}^{a/b} 2i-1<math>
<math>\left(\frac{a}{b}\right)^2 = \frac{a^2}{b^2} = \frac{\sum_{i=1}^a 2i-1}{\sum_{i=1}^b 2i-1}<math>
Thus, it can be generalized that <math>\sum_{i=a/b}^{c/d} f(i) = \frac{\sum_{i=a}^c f(i)}{\sum_{i=b}^d f(i)}<math>
Due to the commutative property of addition, <math>\sum_{i=a}^b f(i) = \sum_{i=b}^a f(i)<math>. Thus, with <math>b > a<math>, we iterate in reverse order (that is from the greater bound to the lower bound, or in reverse order) - for example:
<math>\sum_{i=1}^{3} i = 1 + 2 + 3 = 6<math>
<math>\sum_{i=3}^{1} i = 3 + 2 + 1 = 6<math> (note the order)
What if the bounds are negative?
Also, <math>\sum_{i=-1}^{-3} i = -1 + -2 + -3 = -1 - 2 - 3 = -6<math> and
<math>\sum_{i=1}^3 -i = -1 + -2 + -3 = -1 - 2 - 3 = -6<math> (note the sign at the bounds)
If <math>f(-i) = -f(i)\,\!<math>, then the generalization becomes
<math>\sum_{i=-a}^{-b} f(i) = \sum_{i=a}^b f(-i) = \sum_{i=a}^b -f(i) = -\sum_{i=a}^b f(i)<math>
What if the bounds are equal? In this case, the summation will yield the identity element for addition (that is zero or empty sum).
Thus, the generalizations are:
- <math>\sum_{i=a/b}^{c/d} f(i) = \frac{\sum_{i=a}^c f(i)}{\sum_{i=b}^d f(i)}<math>
- <math>\sum_{i=a}^b f(i) = \sum_{i=b}^a f(i)<math>
- <math>\sum_{i=-a}^{-b} f(i) = \sum_{i=a}^b f(-i)<math>
- <math>\sum_{i=a}^{a} f(i) = 0<math>
- <math>\sum_{i=1}^n m = mn<math> (see Multiplication)
- <math>\sum_{i=a}^b m = m(b-a+1)<math> (from the equation before this)
- <math>\sum_{i=a/b}^{c/d} m<math> is disputed because there are two possible definitions
- <math>\frac{m(c-a+1)}{m(d-b+1)} = \frac{c-a+1}{d-b+1}<math> according to #1
- <math>m\left(\frac{c}{d} - \frac{b}{a}+1\right) = m\left(\frac{ac-bd+ad}{ad}\right)<math> according to #6
But we prefer both definitions, i.e.
<math>\sum_{i=a/b}^{c/d} m = m\left(\frac{ac - bd}{ad}+1\right) \mbox{ iff } ad \ne 0<math>
<math>\sum_{i=a/b}^{c/d} m = \frac{c - a + 1}{d - a + 1} \mbox{ iff } d - a + 1 \ne 0<math>
<math>\sum_{i=a/b}^{c/d} m = m\left(\frac{ac-bd}{ad}+1\right) \or \frac{c-a+1}{d-a+1} \mbox{ both iff } a, d \ne 0 \and d-a \ne -1<math>
provided that <math>a, b \in \mathbb{Z} \and a, b \ge 0 \and a \ne b<math> and that the ring is commutative over addition and that no quotient (divisor) is zero.
Critics and corrections are welcome. -- ErikDT
