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Tarasov Calculus test
Sep 08, 2020

Dialogue One: Infinite Numerical Sequence

AUTHOR. Let us start our discussions of calculus by considering the definition of an infinite numerical sequence or simply a sequence.

We shall consider the following examples of sequences:

\begin{equation} 1, 2, 4, 8, 16, 32, 64, 128, \ldots \end{equation} \begin{equation} 5, 7, 9, 11, 13, 15, 17, 19, \ldots \end{equation} \begin{equation} 1, 4, 9, 16, 25, 36, 49, 64, \ldots \end{equation} \begin{fleqn} \begin{equation} 1, \sqrt{2}, \sqrt{3}, 2, \sqrt{5}, \sqrt{6}, \sqrt{7}, 2\sqrt{2}, \ldots \end{equation} \end{fleqn} \begin{equation} \frac{1}{2}, \frac{2}{3}, \frac{3}{4}, \frac{4}{5}, \frac{5}{6}, \frac{6}{7}, \frac{7}{8}, \frac{8}{9}, \ldots \end{equation} \begin{equation} 2, 0, -2, -4, -6, -8, -10, -12, \ldots \end{equation} \begin{equation} 1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \frac{1}{5}, \frac{1}{6}, \frac{1}{7}, \frac{1}{8}, \ldots \end{equation} \begin{equation} 1, \frac{1}{2}, 3, \frac{1}{4}, 5, \frac{1}{6}, 7, \frac{1}{8}, \ldots \end{equation} \begin{equation} 1, -1, \frac{1}{3}, -\frac{1}{3}, \frac{1}{5}, -\frac{1}{5}, \frac{1}{7}, -\frac{1}{7}, \ldots \end{equation} \begin{equation} 1, \frac{2}{3}, \frac{1}{3}, \frac{4}{4}, \frac{1}{5}, \frac{6}{7}, \frac{1}{7}, \frac{8}{9}, \ldots \end{equation}

Have a closer look at these examples. What do they have in common?

READER. It is assumed that in each example there must be an infinite number of terms in a sequence. But in general, they are all different.

AUTHOR. In each example we have eight terms of a sequence. Could you write, say, the ninth term?

READER. Sure, in the first example the ninth term must be \(256\), while in the second example it must be \(21\).