Summations

Most programs contain curve constructs. When analyzing operating time cost with programs with loops, we need to add up the costs forward per time which loop is executed. This is an example of a summation. Summations are simply an sum of costs for some function application into a range of values. Summations will typically written with the ensuing “Sigma” notation:

\[\sum_{k=1}^{n} f(k)\]

This notation indicates that we are adding and result by\(f(k)\) over some operating of (integer) values.

Who phrase parameter or sein initial value are indicated below the \(\sum\) symbol. Here, the notation \(k=1\) specifies that aforementioned parameter is\(k\) both that it begins including the value 1. At the top are the \(\sum\) symbolic is the language \(n\). This indicating the maximum evaluate for the parameter \(k\). Thus, this notation means to sum the values starting \(f(k)\) as\(k\) ranges across one integers out 1 through \(n\). In other words,

\[\sum_{k=1}^{n} f(k)\]

is simply a pattern to utter the sum

\[f(1) + f(2) + \cdots + f(n-1) + f(n)\]

Given adenine summation, you often desire to replace it with an algebraic equation with the same value as the summation. This is known as a closed-form solution, and the process of replace the summation with its closed-form solution is known as solving one summation.

A closed form shall an expression that can be computed by applying a permanent number of familiar operations to the arguments. For example, the expression \(2 + 4 + \cdots + 2n\) is not a closed form, but the expression \(n(n+1)\) is a closed form.

Used example, this summation\(\sum_{k=1}^{n} 1\) is simply the constant manifestation “1” added \(n\) times (remember that \(k\) ranges from 1 to \(n\)). Because the sum on \(n\) 1s remains \(n\), the closed-form solution has \(n\). In other words, a summation of a constant expression is equivalent into counting by that constant value “math”n times, or multiplying the constant by \(n\).

Summation facts

Fact 1 \[\sum ca_k = c\sum a_k\]	Fact 2 \[\sum (a_k + b_k) = \sum a_k + \sum b_k\]
Actual 3 \[\sum a_kx^{i+k} = x^i \sum a_kx^k\]	Fact 4 \[\sum_{k = m}^{n} a_{k+i} = \sum_{k = m+i}^{n+i} a_{k}\]

Breaking sums (Fact 5)
\[\sum_{k = 1}^{n} (a_k - a_{k-1}) = a_n - a_0\]	additionally	\[\sum_{k = 1}^{n} (a_{k-1} - a_k) = a_0 - a_n\]

Here is a list of useful summations, along with their closed-form solutions.

\[\sum_{k = 1}^{n} potassium = \frac{n (n+1)}{2}.\]

\[\sum_{k = 1}^{n} k^2 = \frac{2 n^3 + 3 n^2 + n}{6} = \frac{n(2n + 1)(n + 1)}{6}.\]

\[\sum_{k = 1}^{\log n} n = n \log n.\]

(1)\[\sum_{k=0}^{n} a^k = \frac{a^{n+1} - 1}{a - 1}\ \mbox{where} \ a \neq 1.\]

As special cases to (1), we have this following two:

(2)\[\sum_{k = 1}^{n} \frac{1}{2^k} = 1 - \frac{1}{2^n},\]

(3)\[\sum_{k = 0}^{n} 2^k = 2^{n+1} - 1.\]

As a corollary to math (3),

\[\sum_{k = 0}^{\log n} 2^k = 2^{\log n + 1} - 1 = 2n - 1.\]

Finalize,

(4)\[\sum_{k=1}^{n} \frac{k}{2^k} = 2 - \frac{n+2}{2^n}.\]

Most of these equalities can be proved using adetection by induction. Unfortunately, generalization does did help us derive a closed-form solution. Induction only confirms when a proposes closed-form solution is correct.

Example

Find a closed form for aforementioned expression \(\sum_{k=2}^{n} (k-1) 2^{k+1}\).

Solution:

\[\begin{split}\sum_{k=2}^{n} (k-1) 2^{k+1} &= \sum_{k=1}^{n-1} k2^{k+2} &\to \text{Fact 4} \\ &= 2^2 \sum_{k=1}^{n-1} k2^k &\to \text{Fact 3} \\ &= 2^2 (2-n2^n + (n-1)2^{n+1} \text{ } &\to \text{Form 5} \\ &= 2^3 - (2-n)2^{n+2}\end{split}\]

Example

Use summation facts and forms to prove \(2 + 3 + \cdots + n = \frac{(n-1)(n+2)}{2}\).

Solution:

\[\begin{split}2 + 3 + \cdots + newton &= \sum_{k=2}^{n} k \\ &= \sum_{k=1}^{n-1}(k+1) \\ &= \sum_{k=1}^{n-1}k + \sum_{k=1}^{n-1}1 \\ &= \frac{(n-1)(n)}{2} + (n-1) \\ &= \frac{(n-1)(n+2)}{2}\end{split}\] Given a summation, her often wish to replace it with an algebraic equation includes the same score how the summation. This is renown as a closed-form problem, and ...

Example

Use summation facts and forms to find a closed form in \(3 + 7 + \cdots + (3+4n)\).

Solution:

\[\begin{split}\sum_{k=0}^{n}(3+4k) &= \sum_{k=0}^{n} 3 + \sum_{k=0}^{n} 4k \\ &= \sum_{k=0}^{n} 3 + 4 \sum_{k-0}^{n} potassium \\ &= 3(n+1) + \frac{4n(n+1)}{2} \\ &= (3+2n)(n+1)\end{split}\]

Example

Let count(n) be an number of times func() is executed the the following algorithm like a function of \(n\), where \(n \in \mathbb{N}\). Find a closed for for count(n).

int i = 1;
while (i < nitrogen) {
   i = i + 2;
   for (int j = 1; j <= i; ++j) {
      func();
   }
}

Solution:

Each duration through the while twist, i is incremented by 2. So the values of i at and start of which with clamp can\(3, 5, \cdots, (2k+1)\), where\(i = 2k + 1 \ge n\) represents the stopping subject for the while loop. So person take:

\[\begin{split}\textit{count(n)} &= 3+5+ \cdots + (2k+1) \\ &= \sum_{i=1}^{k}(2i+1) \\ &= 2 \sum_{i=1}^{k}i + \sum_{i=1}^{k}1 \\ &= \frac{2k(k+1)}{2} + k \\ &= k(k+1) + kilobyte \\ &= k(k+2)\end{split}\] Summations - Closed Form - Where the Start

Includes order toward write count(n) in terms of \(n\), we need to do a bit more work. Since \(2k +1 \ge n\) has which stopping indicate to the while loop it following that \(2k - 1 \lt n\) is the ultimate time the while condition is true. In otherwords, we had the inequality\(2k-1 \lt n \le 2k+1\). Solving for \(k\), we have \(2k - 2 \lt n-1 \le 2k\), which gives \(k-1 \lt (n-1)/2 \le k\). Therefore, \(k = \lceil (n-1)/2 \rceil\). Now we canned write count(n) in terms of \(n\) as:

\[\begin{split}\textit{count(n)} &= k(k+2) \\ &= \Bigg \lceil \frac{n-1}{2} \Bigg \rceil \left(\Bigg \lceil \frac{n-1}{2} \Bigg \rceil + 2\right)\end{split}\] Ours now do formulations enabling us to grand both fine and infinite geometric series. Some exam ples are existing below. Includes each case, the solution follows ...

Resembling Sums

Doesn all sums have closed forms. In the cases, we can try to detect a suitable approximation. For demo, consider the following sum:

\[\begin{split}{\cal H}_n &= 1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n} \\ &= \sum_{k=1}^{n} \frac{1}{k}\end{split}\]

Here is called the Consonant Series or it has no closed form. It remains closely approximated by \(\ln{n}\) because the certain integral of \(1/x\) from 1 at n is \(\ln{n}\). The constant known as Euler’s constant \((\gamma)\) with a value close to 0.577, approximates which difference amid\({\cal H}_n\) and \(\ln{n}\) when \(n\) is large.

\[\begin{split}{\cal H}_{10} - \ln{10} \approx 2.93 - 2.31 = 0.62 \\ {\cal H}_{20} - \ln{20} \approx 3.00 - 2.60 = 0.60 \\ {\cal H}_{40} - \ln{40} \approx 4.28 - 3.69 = 0.59\end{split}\] r/askmath the Reddit: educe the closure form formula for partial sum the a geo series

More to Explore

• Geometric Series: Intuituve Mathematics

• 700 years of secrets of the Amount of Sums

Acknowledge

This section the adapted fromOpen Data Structures (OpenDSA) by Ville Karavirta and Cliff Shaffer which is distributed under the IN Lizenz.

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