Sigma and Pi Notation
Sigma and Pi Notation
Suppose we have a sequence like \(A=\{3,7,11,15,19,23,…, 203\}\), which is generated by the rule \(a_i=3+4\cdot i\), and we want to add up all the sequence elements, or perhaps multiply all the elements of \(A\) together. In many contexts, it would be unwieldy to write something like
$$S=3+7+11+15+19+23+…+203$$
$$or$$
$$P=3\cdot 7\cdot 11\cdot 15\cdot 19\cdot 23\cdot … \cdot 203.$$
Below, we describe a more compact way of writing lengthy sums and products.
With either \(\Sigma^n_{i=1}a_i\) or \(Pi^n_{i=1}a_i\), the subscript \(i=1\) indicates which element of the sequence we start with (in this case, \(i=1\) indicates that we start our sum or product with \(a_1\)). This \(i\) is called the index of the sum. The superscript \(n\) indicates where the sum or product stops (in the event there is no stopping point, then we would write \(\infty\) in the superscript. The subscript and superscript are sometimes referred to as limits of the sum.
Looking at the sequence we were given at the outset of this lesson, namely \(A=\{3,7,11,15,19,23,…, 203\}\), this sum is an arithmetic sequence with common difference \(c=4\). Thus, the rule for this sequence is \(a_i=3+4\cdot i\). Using this formula, we determine that \(203\) is the \(50\)th element of the sequence (since 203=3+4(50)). So, if I wanted to write the sum of all elements in \(A\), I can write
$$\sum^{50}_{i=1} (3+4\cdot i).$$
If I wanted to write the product of all elements of \(A\), I can write
$$\Pi^{50}_{i=1} (3+4\cdot i).$$
Expanding Sums and Products from \(\Sigma\) and \(\Pi\) Notation
When it comes to directly computing the value of a long sum or product, sigma and pi notation don’t really help us much (unless you have access to computer algebra software that allows you to type in sigma notation). So we need to practice writing out sums and products using sigma and pi notation in order to compute such sums and products.
Solution: Essentially, what this notation \(\Sigma^7_{i=2}2^i\) says is “plug in all \(i\)’s between \(2\) and \(7\) (including \(2\) and \(7\)) into the rule \(2^i\), and add up the results.”
This gives us
$$\begin{align}\Sigma^7_{i=2}2^i&=2^2+2^3+2^4+2^5+2^6+2^7\\&=252\end{align}$$
Solution: Similar to the last problem, what this notation \(\Pi^{23}_{i=20} (2i+1)\) says is “plug in all \(i\)’s between \(i=20\) and \(i=23\) (including \(20\) and \(23\)) into the rule \((2i+1)\), and then multiply the results.“
This gives us
$$\begin{align}\Pi^{23}_{i=20} (2i+1)&=(2(20)+1)\cdot (2(21)+1)\cdot (2(22)+1)\cdot (2(23)+1)\\&=41\cdot 43\cdot 45\cdot 47\\&=3728745\end{align}$$
Properties of Sigma and Pi Notation
In plain English, the first property states that you can “distribute a sum” or “break up a sum” over sums and differences.
For example, one can write the sum \(\sum^{27}_{i=1} 2^i+3i\) as
$$\sum^{27}_{i=1} 2^i+3i=\sum_{i=1}^{27} 2^i+\sum^{27}_{i=1} 3i$$
and then compute the value of each individual sum if so desired.
Essentially, what the property “For any \(m\) between \(i=1\) and \(i=n\), \(\sum_{i=1}^n a_i=\sum_{i=1}^m a_i+\sum^n_{i=m+1} a_i\)” says is that you can split up a sum over the values for which the index ranges if needed.
For example, if I wanted, I can break up the sum \(\sum^{10}_{i=1} 3^i\) into two separate sums so that the first starts at \(i=1\) and stops at \(i=4\), and the second starts where the other left off, namely at \(i=5\) and stops where the original stopped, i.e. at \(i=10\).
$$\sum^{10}_{i=1} 3^i = \sum^4_{i=1}3^i+\sum^{10}_{i=5} 3^i.$$
If one were to write out the entire original sum, this “breaking up of the sum” can be seen in action:
$$\sum^{10}_{i=1} 3^i=3^1+3^2+3^3+3^4+3^5+3^6+3^7+3^8+3^9+3^{10}=(3^1+3^2+3^3+3^4)+(3^5+3^6+3^7+3^8+3^9+3^{10})=\sum^4_{i=1}3^i+\sum^{10}_{i=5} 3^i$$
The third property, \(\sum^n_{i=1} c\cdot a_i = c\cdot \sum^n_{i=1} a_i\), is essentially a generalized distribution/factoring property for sums. It says that if every term of the sum is multiplied by the same number \(c\), then that number can be factored out in front of the sum. Somewhat conversely, multiplying the whole sigma sum by a number \(c\) results in distributing that \(c\) to every term being added (just FOILing/distributing as usual).
For example, in the sum \(\sum_{i=1}^6 5\cdot 3^i\) we notice
$$\sum_{i=1}^6 5\cdot 3^i=5\cdot 3^1+5\cdot 3^2+5\cdot 3^3+5\cdot 3^4+5\cdot 3^5+5\cdot 3^6=5( 3^1+3^2+3^3+3^4+3^5+3^6)=5\sum_{i=1}^6 \cdot 3^i.$$
The above is really nothing more than factoring out a \(5\). One can read this string of equalities in reverse to see how distribution works with sigma notations.
What property 1 is saying is that products that are being multiplied can be “absorbed” into one big product, assuming that the product’s limits are the same (i.e. they both start and stop at the same values). Can’t absorb them into one sum as neatly otherwise. Reading the equation in the opposite direction, the property says that you can break products up over products.
This property comes as the result of being able to rearrange products of real numbers. This is illustrated (and the proof of this property is hinted at) in the example below.
$$\Pi^4_{i=1}2^i\cdot \Pi^{4}_{i=1} 3^i=(2^1\cdot 2^2\cdot 2^3\cdot 2^4)(3^1\cdot 3^2\cdot 3^3\cdot 3^4)=2^1\cdot 3^1 \cdot 2^2\cdot 3^2\cdot 2^3\cdot 3^3\cdot 2^4\cdot 3^4=\Pi^4_{i=1} 2^i\cdot 3^i$$
This property is interesting! We cannot just “Factor out” a constant multiple from a product because as a part of that product, that constant is getting multiplied repeatedly to itself (according to the number of terms that are being multiplied). This property is illustrated in the following example.
$$\begin{align}\Pi^5_{i=1}2\cdot 4^i&=(2\cdot 4^1)\cdot (2\cdot 4^2)\cdot (2\cdot 4^3)\cdot (2 \cdot 4^4)\cdot (2\cdot 4^5)\\&=2^5\cdot 4^1\cdot 4^2\cdot 4^3\cdot 4^4\cdot 4^5&=2^5\cdot \Pi^5_{i=1} 4^i\end{align}$$
The last equality was obtained by rearranging all the different factors, putting all five factors of \(2\) out front, and writing it as a fifth power.
Essentially, what the property “For any integer \(m\) between \(i=1\) and \(i=n\), \(\Pi^n_{i=1}a_i=\Pi^{m}_{i=1} a_i \cdot \Pi^n_{i=m+1} a_i\)” says is that you can split up a product over the values for which the index ranges if needed.
For example, if I wanted, I can break up the product \(\Pi^{10}_{i=1} 3^i\) into two separate products so that the first starts at \(i=1\) and stops at \(i=4\), and the second starts where the other left off, namely at \(i=5\) and stops where the original stopped, i.e. at \(i=10\).
$$\Pi^{10}_{i=1} 3^i = \Pi^4_{i=1}3^i\cdot \Pi^{10}_{i=5} 3^i.$$
If one were to write out the entire original sum, this “breaking up of the Product” can be seen in action:
$$\Pi^{10}_{i=1} 3^i=3^1\cdot 3^2\cdot 3^3\cdot 3^4\cdot 3^5\cdot 3^6\cdot 3^7\cdot 3^8\cdot 3^9\cdot 3^{10}=(3^1\cdot 3^2\cdot 3^3\cdot 3^4)\cdot(3^5\cdot 3^6\cdot 3^7\cdot 3^8\cdot 3^9\cdot 3^{10})=\Pi^4_{i=1}3^i\cdot \Pi^{10}_{i=5} 3^i$$
\(\sum_{i=1}^{10} 3=30\)
Note that in the given sum \(\sum_{i=1}^{10} 3\) there is nowhere to plug the \(i\)-values into… so don’t plug em in anywhere! Technically, what this sum is saying is “for \(i=1\) to \(i=10\) add \(3\) (to itself).” This gives us 10 of those \(3\)’s added together; in other words, \(10\cdot 3\) which gives the above answer.
\(\sum_{i=1}^{10} 3=3+3+3+3+3+3+3+3+3+3=10\cdot 3=30\)
Plug all \(i\)-values \(i=1,2,3,4,5\) into the rule \(2i-3\) and add the results!
\(\sum_{i=1}^{5} (2i-3)=15\)
\(\begin{align}\sum_{i=1}^{5} (2i-3)&=(2(1)-3)+(2(2)-3)+(2(3)-3)+(2(4)-3)+(2(5)-3)\\&=(-1)+(1)+(3)+(5)+(7)\\&=15\end{align}\)
Plug all \(i\)-values \(i=3,4,5,6,7,8\) into the rule \(i^2\) and add up the results!
\(\sum_{i=3}^8 (i^2)=199\)
\(\begin{align}\sum_{i=3}^8 (i^2)&=(3)^2+(4)^2+(5)^2+(6)^2+(7)^2+(8)^2\\&=9+16+25+36+49+64\\&=199\end{align}\)
Use one of the Sigma Notation Properties! Makes your life a lot easier!
\(\sum_{i=1}^6 (3i-2) +\sum_{i=1}^6 (-2i+2)=21\)
$$\begin{align}\sum_{i=1}^6 (3i-2) +\sum_{i=1}^6 (-2i+2)&=\sum_{i=1}^6 ((3i-2)+(-2i+2))&\ &\ \\ &=\sum_{i=1}^6 i &\ &\text{combining like-terms}\\&=1+2+3+4+5+6&\ & \\&=21&\ &\ \end{align}$$
Here we are not finding the sum, but just expanding the sum so its written out in “long form.” The process is the same though. Plug in all numbers \(i=5,6,7,8,9,10,…,(n-1),n\) and then “add them up” by putting a plus-sign between them.
\(5^5+5^6+5^7+5^8+5^9+…+5^{n-1}+5^{n}\)
(Note: you don’t need to write terms like, \(5^{n-2}\) or \(5^{n-1}\) at the end of the sum to represent the third-to-last and second-to-last summand, respectively. I just chose to add them. It IS necessary to have the \(5^n\) term (i.e. the last term of the sum) , as it indicates the end of the sum. )
Plug \(i=1,2,3,4,5,…,8,9\) into the rule \(a_i=i\) and multiply the results!
\(\Pi^{9}_{i=1} i=362880\)
\(\Pi^{9}_{i=1} i=1\cdot 2\cdot 3\cdot 4\cdot 5\cdot 6\cdot 7\cdot 8\cdot 9=362880.\)
plugging in \(i=1,2,3,…\) into the rule \(a_i=i\) essentially does nothing with the \(i\) being plugged in, so we end up multiplying the numbers \(1\) through \(9\).
Plug in all \(i=1,2,3,4,…,9\) into the rule \(5i\) and multiply the results!
\(\Pi^{9}_{i=1} 5i=708750000000\)
\(\Pi^{9}_{i=1} 5i=(5(1))\cdot (5(2))\cdot (5(3))\cdot (5(4))\cdot … \cdot (5(8))\cdot (5(9))\).
Turn the product of products into a single product with a single \(\Pi\) using one of the \(\Pi\) notation properties. This will simplify your work considerably.
$$\begin{align}\Pi^5_{i=2} 2^i\cdot \Pi^5_{i=2}\left(\frac{1}{3}^i\right)&=\Pi^{5}_{i=2} 2^i\cdot \left(\frac{1}{3}\right)^i& & \\&=\Pi^{5}_{i=2} \left(\frac{2}{3}\right)^i& & \text{because } 2^i\cdot \left(\frac{1}{3}\right)^i= 2^i\cdot \left(\frac{1}{3^i}\right)=\left(\frac{2}{3}\right)^i\\&=\left(\frac{2}{3}\right)^2\cdot \left(\frac{2}{3}\right)^3\cdot \left(\frac{2}{3}\right)^4\cdot \left(\frac{2}{3}\right)^5& & \\&=\frac{2^2\cdot 2^3\cdot 2^4\cdot 2^5}{3^2\cdot 3^3\cdot 3^4\cdot 3^5}& & \\&=\frac{2^{14}}{3^{14}}& & \end{align}$$