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<main class="main"><div id="content" class="pretext-content"><section xmlns:svg="http://www.w3.org/2000/svg" class="section" id="section-45"><h2 class="heading hide-type">
<span class="type">Section</span> <span class="codenumber">7.2</span> <span class="title">Summation notation</span>
</h2>
<section class="subsection" id="subsection-98"><h3 class="heading hide-type">
<span class="type">Subsection</span> <span class="codenumber">7.2.1</span> <span class="title">Basic definitions</span>
</h3>
<p id="p-1291">Summation notation is a compact form for writing sums. It is written with three pieces. \[ \sum_{\fbox{\(k=1\)}}^\fbox{\(5\)} \fbox{\(2k\)} \] The part below the greek letter sigma (\(\Sigma\)) tells us to start with \(k=1\text{.}\) We increment \(k\) by \(1\) until we get to the value above the sigma. In this case we let \(k\) take on the values \(1, 2, 3, 4, 5\text{.}\) We next substitute those values of \(k\) into the expression to the right of the sigma successively and take the sum all of the resulting terms. Here is a table of the resulting values:</p>
<div class="displaymath">
\begin{align*}
k \amp\amp 2k\\
1 \amp\amp 2\\
2 \amp\amp 4\\
3 \amp\amp 6\\
4 \amp\amp 8\\
5 \amp\amp 10
\end{align*}
</div>
<p class="continuation">which evaluates to</p>
<div class="displaymath">
\begin{equation*}
2+4+6+8+10=30
\end{equation*}
</div>
<p class="continuation">and so we write</p>
<div class="displaymath">
\begin{equation*}
\sum_{k=1}^5 2k=30\text{.}
\end{equation*}
</div>
<p class="continuation">The variable \(k\) is called <dfn class="terminology">the index of summation</dfn>. The number above the sigma is called <dfn class="terminology">the limit of summation</dfn>. The example shows us how to write a sum of even numbers. We can write the sum of odd numbers, too.</p>
<div class="displaymath">
\begin{equation*}
\sum_{k=1}^8 (2k-1)=1+3+5+7+9+11+13+15=64 \text{.}
\end{equation*}
</div>
<p class="continuation">It is not necessary to use \(k\) as a variable or to start the summation at \(k=1\text{.}\) It is easy to verify that</p>
<div class="displaymath">
\begin{equation*}
\sum_{k=1}^8(2k-1)=\sum_{l=0}^7(2l+1)=64 \text{.}
\end{equation*}
</div>
<p class="continuation">The number above the sigma can also be written as a variable:</p>
<div class="displaymath">
\begin{gather*}
\sum_{k=1}^n a_k=a_1+a_2+\cdots+a_{n-1}+a_n\\
\sum_{k=0}^n a_kb_k=a_0b_0+a_1b_1+a_2b_2+\cdots+a_{n-1}b_{n-1}+a_nb_n\\
\sum_{k=1}^n (2k-1) = 1+3+5+7+\cdots+(2n-1)
\end{gather*}
</div>
<p class="continuation">Next we take the last expression and vary \(n\text{,}\) the limit of summation. Here is a table that includes some small values of \(n\text{:}\)</p>
<figure class="table table-like" id="table-11"><figcaption><span class="type">Table</span><span class="space"> </span><span class="codenumber">7.2.1<span class="period">.</span></span><span class="space"> </span></figcaption><div class="tabular-box natural-width"><table class="tabular">
<tr>
<td class="l m b0 r0 l0 t0 lines">\(\sum_{k=1}^1(2k-1)=1\)</td>
<td class="l m b0 r0 l0 t0 lines">\(n=1\)</td>
</tr>
<tr>
<td class="l m b0 r0 l0 t0 lines">\(\sum_{k=1}^2(2k-1)=4\)</td>
<td class="l m b0 r0 l0 t0 lines">\(n=2\)</td>
</tr>
<tr>
<td class="l m b0 r0 l0 t0 lines">\(\sum_{k=1}^3(2k-1)=9\)</td>
<td class="l m b0 r0 l0 t0 lines">\(n=3\)</td>
</tr>
<tr>
<td class="l m b0 r0 l0 t0 lines">\(\sum_{k=1}^4(2k-1)=16\)</td>
<td class="l m b0 r0 l0 t0 lines">\(n=4\)</td>
</tr>
<tr><td class="l m b0 r0 l0 t0 lines">\(\vdots\)</td></tr>
<tr><td class="l m b0 r0 l0 t0 lines">\(\sum_{k=1}^n(2k-1)=1+3+5+\cdots+2n-1=?\)</td></tr>
</table></div></figure><article class="exercise exercise-like" id="exercise-49"><a data-knowl="" class="id-ref exercise-knowl original" data-refid="hk-exercise-49"><h6 class="heading">
<span class="type">Checkpoint</span><span class="space"> </span><span class="codenumber">7.2.2</span><span class="period">.</span>
</h6></a></article><div class="hidden-content tex2jax_ignore" id="hk-exercise-49"><article class="exercise exercise-like"><p id="p-1292">What simple expression involving \(n\) can replace the question mark?</p></article></div>
<p id="p-1293">The summands can be symbolic, too. If \(f(x)\) is any function, we may write</p>
<div class="displaymath">
\begin{gather*}
\sum_{i=1}^r f(i)=f(1)+f(2)+\cdots+f(r-1)+f(r)
\end{gather*}
</div></section><section class="subsection" id="subsection-99"><h3 class="heading hide-type">
<span class="type">Subsection</span> <span class="codenumber">7.2.2</span> <span class="title">Alternating sums</span>
</h3>
<p id="p-1294">A sum is <dfn class="terminology">alternating</dfn> if the signs of consecutive terms are alternate between positive and negative. An example: \(1-2+3-4+5\text{.}\) These sums appear frequently enough to have this special name. They are easy to write using summation notation because of a simple fact:</p>
<div class="displaymath">
\begin{equation*}
(-1)^k=
\begin{cases}
1 \amp \text{if \(k\) is even}\\
-1 \amp \text{if \(k\) is odd}
\end{cases}
\end{equation*}
</div>
<p class="continuation">We can then write</p>
<div class="displaymath">
\begin{equation*}
\sum_{k=1}^6 (-1)^k k^2=-1+4-9+16-25+36=21 \text{.}
\end{equation*}
</div>
<p class="continuation">We can interchange the order of the signs by adding (or subtracting) one to the exponent of \(-1\text{:}\)</p>
<div class="displaymath">
\begin{equation*}
\sum_{k=1}^6 (-1)^{k+1} k^2=1-4+9-16+25-36=-21\text{.}
\end{equation*}
</div></section><section class="subsection" id="subsection-100"><h3 class="heading hide-type">
<span class="type">Subsection</span> <span class="codenumber">7.2.3</span> <span class="title">Manipulation with summation notation identities</span>
</h3>
<p id="p-1295">Summation notation is more that mere shorthand. There are rules of manipulation that are quite useful.</p>
<p id="p-1296">For any constant \(c\text{,}\)</p>
<div class="displaymath">
\begin{equation*}
\sum_{k=1}^n c=\underbrace{c+c+\cdots+c+c}_{n \textrm{ summands}}=nc
\end{equation*}
</div>
<p class="continuation">and</p>
<div class="displaymath">
\begin{equation*}
\sum_{k=1}^n ca_k
=ca_1+\cdots+ca_n
=c(a_1+\cdots+a_n)
=c\sum_{k=1}^n a_k
\end{equation*}
</div>
<p class="continuation">The process of the last equation is sometimes referred to as <em class="emphasis">pulling a constant across the summation sign</em>.</p>
<p id="p-1297">In addition,</p>
<div class="displaymath">
\begin{gather*}
\sum_{k=1}^n(a_k+b_k)= \sum_{k=1}^na_k+\sum_{k=1}^nb_k,\\
\sum_{k=1}^n(a_k-b_k)= \sum_{k=1}^na_k-\sum_{k=1}^nb_k
\end{gather*}
</div>
<p class="continuation">and</p>
<div class="displaymath">
\begin{equation*}
\sum_{k=0}^na_k= \sum_{k=t}^{n+t}a_{k-t}
\end{equation*}
</div>
<p class="continuation">The last identity is sometimes called <em class="emphasis">shifting the index of summation.</em></p>
<article class="example example-like" id="example-78"><a data-knowl="" class="id-ref example-knowl original" data-refid="hk-example-78"><h6 class="heading">
<span class="type">Example</span><span class="space"> </span><span class="codenumber">7.2.3</span><span class="period">.</span>
</h6></a></article><div class="hidden-content tex2jax_ignore" id="hk-example-78"><article class="example example-like"><p id="p-1298">Suppose we want to evaluate the sum of the first \(n\) positive integers. In other words, we want to find</p>
<div class="displaymath">
\begin{equation*}
s_n=1+2+\cdots+n=\sum_{k=1}^nk.
\end{equation*}
</div>
<p class="continuation">If we are convinced that \(\sum_{k=1}^n(2k-1)=1+3+5+\cdots+2n-1=n^2\text{,}\) then</p>
<div class="displaymath">
\begin{align*}
n^2\amp =\sum_{k=1}^n(2k-1)\\
\amp =\sum_{k=1}^n(2k)-\sum_{k=1}^n1\\
\amp =2\sum_{k=1}^nk-n\\
\amp =2s_n-n
\end{align*}
</div>
<p class="continuation">which yields</p>
<div class="displaymath">
\begin{equation*}
s_n=\frac12n(n+1).
\end{equation*}
</div></article></div>
<article class="exercise exercise-like" id="exercise-50"><a data-knowl="" class="id-ref exercise-knowl original" data-refid="hk-exercise-50"><h6 class="heading">
<span class="type">Checkpoint</span><span class="space"> </span><span class="codenumber">7.2.4</span><span class="period">.</span>
</h6></a></article><div class="hidden-content tex2jax_ignore" id="hk-exercise-50"><article class="exercise exercise-like"><p id="p-1299">Evaluate the following sums</p>
<ul class="disc">
<li id="li-425"><p id="p-1300">\(\displaystyle \sum_{j=1}^5 2j\)</p></li>
<li id="li-426"><p id="p-1301">\(\displaystyle \sum_{k=1}^5 k(k-1)\)</p></li>
<li id="li-427"><p id="p-1302">\(\displaystyle \sum_{i=0}^4 2^i\)</p></li>
<li id="li-428"><p id="p-1303">\(\displaystyle \sum_{\ell=1}^3\ell^2+\frac1\ell\)</p></li>
<li id="li-429"><p id="p-1304">\(\displaystyle \sum_{j=1}^5 ij\)</p></li>
</ul>
<div class="solutions">
<a data-knowl="" class="id-ref solution-knowl original" data-refid="hk-solution-40" id="solution-40"><span class="type">Solution.</span> </a><div class="hidden-content tex2jax_ignore" id="hk-solution-40"><div class="solution solution-like"><ul id="p-1305" class="disc">
<li id="li-430"><p id="p-1306">\(\displaystyle 2+4+6+8+10=30\)</p></li>
<li id="li-431"><p id="p-1307">\(\displaystyle 2+6+12+20+30=70\)</p></li>
<li id="li-432"><p id="p-1308">\(\displaystyle 1+2+4+8+16=31\)</p></li>
<li id="li-433"><p id="p-1309">\(\displaystyle 1+1+4+\frac12+9+\frac13=15\frac56\)</p></li>
<li id="li-434"><p id="p-1310">\(\displaystyle i+2i+3i+4i+5i=15i\)</p></li>
</ul></div></div>
</div></article></div>
<article class="exercise exercise-like" id="exercise-51"><a data-knowl="" class="id-ref exercise-knowl original" data-refid="hk-exercise-51"><h6 class="heading">
<span class="type">Checkpoint</span><span class="space"> </span><span class="codenumber">7.2.5</span><span class="period">.</span>
</h6></a></article><div class="hidden-content tex2jax_ignore" id="hk-exercise-51"><article class="exercise exercise-like"><p id="p-1311">Evaluate \(\sum_{k=0}^n 2^k\) for \(n=1,2,3,4\text{.}\) Find a simple expression that works for any \(n\text{.}\) Justify your answer.</p>
<div class="solutions">
<a data-knowl="" class="id-ref solution-knowl original" data-refid="hk-solution-41" id="solution-41"><span class="type">Solution.</span> </a><div class="hidden-content tex2jax_ignore" id="hk-solution-41"><div class="solution solution-like">
<div class="displaymath" id="p-1312">
\begin{equation*}
\begin{array}{cc}
n \amp \sum_{k=0}^n 2^k\\
\hline
1 \amp 3\\
2 \amp 7\\
3\amp 15\\
4\amp 31
\end{array}
\end{equation*}
</div>
<p class="continuation">We note that each answer is one less than a power of \(2\text{.}\) This leads to the equation \[ \sum_{k=0}^n 2^k=2^{n+1}-1. \] To justify the equation we note that \[ 2(\sum_{k=0}^n 2^k)=\sum_{k=0}^n 2^{k+1}=\sum_{k=1}^{n+1} 2^k =\bigl(\sum_{k=0}^{n+1} 2^k\bigr) -1 \] so we get the sum for a given value on \(n\) by doubling the value for \(n-1\) and adding \(1\text{.}\)</p>
</div></div>
</div></article></div></section><section class="subsection" id="subsection-101"><h3 class="heading hide-type">
<span class="type">Subsection</span> <span class="codenumber">7.2.4</span> <span class="title">Double summations</span>
</h3>
<p id="p-1313">Sometimes it is useful to have a formula with two summations, each with its own index of summation. It is interpreted in the following way: \[ \sum_{i=1}^m \sum_{j=1}^n f(i,j)= \sum_{i=1}^m \bigl(\sum_{j=1}^n f(i,j)\bigr) \] and so, for example, \[ \sum_{i=1}^3 \sum_{j=1}^2 i^j=\sum_{i=1}^3 (i^1+i^2)=(1+1)+(2+4)+(3+9)=20 \]</p>
<p id="p-1314">Double summations are often used with matrices. Suppose we have a matrix</p>
<div class="displaymath">
\begin{equation*}
A=
\begin{bmatrix}
a_{1,1}\amp a_{1,2}\amp \cdots \amp a_{1,n}\\
a_{2,1}\amp a_{2,2}\amp \cdots \amp a_{2,n}\\
\vdots\amp \vdots\amp \vdots\amp \vdots\\
a_{m,1}\amp a_{m,2}\amp \cdots \amp a_{m,n}
\end{bmatrix}
\end{equation*}
</div>
<p class="continuation">Then \[ \sum_{j=1}^n a_{i,j} = a_{i,1}+a_{i,2}+\cdots+a_{i,n} \] which is the sum of the entries in the \(i\)-th row of the matrix. Similarly \[ \sum_{i=1}^m a_{i,j} = a_{1,j}+a_{2,j}+\cdots+a_{m,j} \] which is the sum of the entries in the \(j\)-th column of the matrix. We can now see that</p>
<div class="displaymath">
\begin{equation*}
\sum_{i=1}^m\sum_{j=1}^n a_{i,j}
=\sum_{j=1}^n\sum_{i=1}^m a_{i,j}
\end{equation*}
</div>
<p class="continuation">since each sides of the equation is equal to the sum of all of the entries in the matrix. This is called <em class="emphasis">interchanging the order of summation.</em></p>
<p id="p-1315">Another way of looking at these double summations \(\sum_{i=1}^m \sum_{j=1}^n f(i,j)\) is that it is the sum of \(f(i,j)\) as \(i\) takes on values from \(1\) to \(m\) and \(j\) takes on values from \(1\) to \(n\text{.}\) Clearly there are \(mn\) summands altogether.</p></section></section></div></main>
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