संकलन

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संख्याओं के किसी क्रम को जोड़ने की संक्रिया संकलन (Summation) कहलाती है। इसका परिणाम योग (sum) या कुलयोग (total) कहलाती है।

प्रतीक (notation)

कैपितल सिग्मा (Capital-sigma)

यह निम्नलिखित तरीके से परिभाषित है-

<math>\sum_{i=m}^n x_i = x_m + x_{m+1} + x_{m+2} +\cdots+ x_{n-1} + x_n. </math>
एक उदाहरण-
<math>\sum_{k=2}^6 k^2 = 2^2+3^2+4^2+5^2+6^2 = 90.</math>

संकलन से संबंधित सर्वसमिकाएँ (Identities)

सामान्य

<math>\sum_{n=s}^t C\sdot f(n) = C\sdot \sum_{n=s}^t f(n)</math>, where C is a constant
<math>\sum_{n=s}^t f(n) + \sum_{n=s}^{t} g(n) = \sum_{n=s}^t \left[f(n) + g(n)\right]</math>
<math>\sum_{n=s}^t f(n) - \sum_{n=s}^{t} g(n) = \sum_{n=s}^t \left[f(n) - g(n)\right]</math>
<math>\sum_{n=s}^t f(n) = \sum_{n=s+p}^{t+p} f(n-p)</math>
<math>\sum_{n=s}^j f(n) + \sum_{n=j+1}^t f(n) = \sum_{n=s}^t f(n)</math>
<math>\left(\sum_{i=k_0}^{k_1} a_i\right)\left(\sum_{j=l_0}^{l_1} b_j\right) = \sum_{i=k_0}^{k_1}\sum_{j=l_0}^{l_1} a_ib_j</math>
<math>\sum_{i=k_0}^{k_1}\sum_{j=l_0}^{l_1} a_{i,j} = \sum_{j=l_0}^{l_1}\sum_{i=k_0}^{k_1} a_{i,j}</math>
<math>\sum_{n=0}^t f(2n) + \sum_{n=0}^t f(2n+1) = \sum_{n=0}^{2t+1} f(n)</math>
<math>\sum_{n=0}^t \sum_{i=0}^{z-1} f(z\sdot n+i) = \sum_{n=0}^{z\sdot t+z-1} f(n)</math>
<math>\sum_{n=s}^t \ln f(n) = \ln \prod_{n=s}^t f(n)</math>
<math>c^{\left[\sum_{n=s}^t f(n) \right]} = \prod_{n=s}^t c^{f(n)}</math>

बहुपद ब्यंजकों का संकलन

<math>\sum_{i=m}^n 1 = n-m+1</math>
<math>\sum_{i=1}^n \frac{1}{i} = H_n</math> (See Harmonic number)
<math>\sum_{i=m}^n i = \frac{(n-m+1)(n+m)}{2}</math> (see arithmetic series)
<math>\sum_{i=0}^n i = \sum_{i=1}^n i = \frac{n(n+1)}{2}</math> (Special case of the arithmetic series)
<math>\sum_{i=1}^n i^2 = \frac{n(n+1)(2n+1)}{6} = \frac{n^3}{3} + \frac{n^2}{2} + \frac{n}{6}</math>
<math>\sum_{i=1}^n i^3 = \left(\frac{n(n+1)}{2}\right)^2 = \frac{n^4}{4} + \frac{n^3}{2} + \frac{n^2}{4} = \left[\sum_{i=1}^n i\right]^2</math>
<math>\sum_{i=1}^n i^4 = \frac{n(n+1)(2n+1)(3n^2+3n-1)}{30} = \frac{n^5}{5} + \frac{n^4}{2} + \frac{n^3}{3} - \frac{n}{30}</math>
<math>\sum_{i=0}^n i^p = \frac{(n+1)^{p+1}}{p+1} + \sum_{k=1}^p\frac{B_k}{p-k+1}{p\choose k}(n+1)^{p-k+1}</math> where <math>B_k</math> denotes a Bernoulli number


The following formulas are manipulations of <math>\sum_{i=1}^n i^3 = \left(\sum_{i=1}^n i\right)^2</math> generalized to begin a series at any natural number value (i.e., <math>m \in \mathbb{N}</math>):

<math>\left(\sum_{i=m}^n i\right)^2 = \sum_{i=m}^n (i^3 - im(m-1))</math>
<math>\sum_{i=m}^n i^3 = \left(\sum_{i=m}^n i\right)^2 + m(m-1)\sum_{i=m}^n i</math>

चरघातांकी पदों के योग

In the summations below x is a constant not equal to 1

<math>\sum_{i=m}^{n-1} x^i = \frac{x^m-x^n}{1-x}</math> (साँचा:Nowrap; see geometric series)
<math>\sum_{i=0}^{n-1} x^i = \frac{1-x^n}{1-x}</math> (geometric series starting at 1)
<math>\sum_{i=0}^{n-1} i x^i = \frac{x-nx^n+(n-1)x^{n+1}}{(1-x)^2}</math>
<math>\sum_{i=0}^{n-1} i 2^i = 2+(n-2)2^{n}</math> (special case when x = 2)
<math>\sum_{i=0}^{n-1} \frac{i}{2^i} = 2-\frac{n+1}{2^{n-1}}</math> (special case when x = 1/2)

द्विपद गुणांकों वाले संकलन (summations involving binomial coefficients)

There exist enormously many summation identities involving binomial coefficients (a whole chapter of Concrete Mathematics is devoted to just the basic techniques)। Some of the most basic ones are the following.

<math>\sum_{i=0}^n {n \choose i} = 2^n</math>
<math>\sum_{i=1}^{n} i{n \choose i} = n2^{n-1}</math>
<math>\sum_{i=0}^{n} i!\cdot{n \choose i} = \lfloor n!\cdot e \rfloor</math>
<math>\sum_{i=0}^{n-1} {i \choose k} = {n \choose k+1}</math>
<math>\sum_{i=0}^n {n \choose i}a^{(n-i)} b^i=(a + b)^n</math>, the binomial theorem

वृद्धि दर

The following are useful approximations (using theta notation):

<math>\sum_{i=1}^n i^c = \Theta(n^{c+1})</math> for real c greater than −1
<math>\sum_{i=1}^n \frac{1}{i} = \Theta(\log n)</math> (See Harmonic number)
<math>\sum_{i=1}^n c^i = \Theta(c^n)</math> for real c greater than 1
<math>\sum_{i=1}^n \log(i)^c = \Theta(n \cdot \log(n)^{c})</math> for non-negative real c
<math>\sum_{i=1}^n \log(i)^c \cdot i^d = \Theta(n^{d+1} \cdot \log(n)^{c})</math> for non-negative real c, d
<math>\sum_{i=1}^n \log(i)^c \cdot i^d \cdot b^i = \Theta (n^d \cdot \log(n)^c \cdot b^n)</math> for non-negative real b > 1, c, d

सन्दर्भ

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