02 Product Notation

The notation \(\prod _{i=1}^ N p_ i\) denotes the product with \(N\) factors

\[ \prod _{i=1}^ N p_ i= p_1 p_2\cdots p_ N. \]

Question 01

\[ \prod _{i=1}^ M \frac{1}{\theta }= \]

Explanation Let’s break it down step by step:

• The symbol \(\prod\) denotes the product of a series of terms.

• The subscript \(i=1\) indicates that the product starts from \(i=1\).

• The superscript \(M\) indicates that the product goes up to \(M\).

• The term inside the product, \(\frac{1}{\theta}\), represents the value of each term in the product.

In closed form, this can be written as \((\frac{1}{\theta})^M\).

This is because when you multiply a number by itself multiple times, it is equivalent to raising that number to the power of the number of times you multiply it.

So, \(\prod _{i=1}^ M \frac{1}{\theta }\) is equivalent to \((\frac{1}{\theta})^M\)

To understand this notation better, let’s consider an example. Suppose we have the following values:

• \(M = 4\)

• \(\theta = 2\)

The expression \(\prod _{i=1}^ M \frac{1}{\theta }\) can be expanded as:

\[ \left(\frac{1}{\theta}\right) \times \left(\frac{1}{\theta}\right) \times \left(\frac{1}{\theta}\right) \times \left(\frac{1}{\theta}\right) \]

Substituting the values of \(\theta\) and \(M\):

\[ \left(\frac{1}{2}\right) \times \left(\frac{1}{2}\right) \times \left(\frac{1}{2}\right) \times \left(\frac{1}{2}\right) \]

Simplifying the expression:

\[ \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{16} \]

So, in this example, the value of \(\prod _{i=1}^ M \frac{1}{\theta }\) is \(\frac{1}{16}\) or \((\frac{1}{2})^4\)

Question 2

\[\prod _{k=1}^ K \frac{k}{k+1}=\]

Explanation This product notation represents the product of a sequence where each term is the ratio of \(k\) to \(k+1\).

If we expand this for a few terms, we get:

\[ \frac{1}{1+1} \times \frac{2}{2+1} \times \frac{3}{3+1} \times \ldots \times \frac{K}{K+1} \]

This simplifies to:

\[ \frac{1}{2} \times \frac{2}{3} \times \frac{3}{4} \times \ldots \times \frac{K}{K+1} \]

You can see that each term in the numerator cancels out with the previous term in the denominator, leaving us with:

\[ \frac{1}{K+1} \]

So, the closed form of \(\prod _{k=1}^ K \frac{k}{k+1}\) is \(\frac{1}{K+1}\).

Question 3

\[\ln \left(\prod _{k=1}^ K e^ k\right)=\]

Explanation The expression \(\ln \left(\prod _{k=1}^ K e^ k\right)\) involves both the natural logarithm function (denoted by \(\ln\)) and the product notation (denoted by \(\prod\)). Here’s what it means:

• The symbol \(\ln\) denotes the natural logarithm, which is the logarithm to the base \(e\).

• The symbol \(\prod\) denotes the product of a series of terms.

• The subscript \(k=1\) indicates that the product starts from \(k=1\).

• The superscript \(K\) indicates that the product goes up to \(K\).

• The term inside the product, \(e^k\), represents the value of each term in the product.

This product notation represents the product of a sequence where each term is \(e\) raised to the power of \(k\).

If we expand this for a few terms, we get:

\[ e^1 \times e^2 \times e^3 \times \ldots \times e^K \]

This simplifies to:

\[ e^{1+2+3+...+K} \]

Because of the properties of logarithms, specifically the property that \(\ln(ab) = \ln(a) + \ln(b)\), the logarithm of a product is the sum of the logarithms. So, we can rewrite the original expression as:

\[ \ln(e^1) + \ln(e^2) + \ln(e^3) + \ldots + \ln(e^K) \]

And since \(\ln(e^k) = k\), this simplifies to:

\[ 1 + 2 + 3 + \ldots + K \]

This is the sum of the first \(K\) positive integers, which has a closed form given by the formula \(\frac{K(K+1)}{2}\). e.g. if K=5, then 1+2+3+4+5 = 15 and \(\frac{5(5+1)}{2} = 15\)

So, the closed form of \(\ln \left(\prod _{k=1}^ K e^ k\right)\) is \(\frac{K(K+1)}{2}\).