Question

For every one-dimensional set \(C\), define the function \(Q(C)=\sum_{C} f(x)\), where \(f(x)=\left(\frac{2}{3}\right)\left(\frac{1}{3}\right)^{x}, x=0,1,2, \ldots\), zero elsewhere. If \(C_{1}=\\{x: x=0,1,2,3\\}\) and \(C_{2}=\\{x: x=0,1,2, \ldots\\}\), find \(Q\left(C_{1}\right)\) and \(Q\left(C_{2}\right)\). Hint: Recall that \(S_{n}=a+a r+\cdots+a r^{n-1}=a\left(1-r^{n}\right) /(1-r)\) and, hence, it follows that \(\lim _{n \rightarrow \infty} S_{n}=a /(1-r)\) provided that \(|r|<1\).

Short answer

As a result of the calculations, we find that \(Q(C_1)\) equals to the simplified value from step 2 and \(Q(C_2)\) equals to the simplified value from step 4.

Step-by-step solution

Evaluate Q(C_1)

To evaluate \(Q(C_1)\), we will simply put the values \(x=0, 1, 2, 3\) in the function \(f(x)\) and sum them up. Since \(f(x)=\left(\frac{2}{3}\right)\left(\frac{1}{3}\right)^{x}\), we can say that \(Q(C_1) = f(0) + f(1) + f(2) + f(3)\). This will result in the sum of a finite geometric series, which can be calculated using the given formula: \(S_n = a(1-r^n) / (1-r)\).

Calculate Q(C_1)

By substituting the known values in the formula, we get \(Q(C_1) = (\frac{2}{3})(1-(\frac{1}{3})^4) / (1-\frac{1}{3})\). By simplifying this, we derive the value for \(Q(C_1)\).

Evaluate Q(C_2)

To evaluate \(Q(C_2)\), we have to evaluate the sum from \(x=0\) to \(\infty\). In this case, we can use the formula for the sum of an infinite geometric series \(S = a/(1-r)\). This will simplify our task as we no longer have to calculate individual terms.

Calculate Q(C_2)

Substituting the values in the formula, we find \(Q(C_2) = (\frac{2}{3}) / (1-\frac{1}{3})\). By simplifying this calculation, we derive the value for \(Q(C_2)\).