# How do you graph a_n=2(3)^(n-1)?

Nov 23, 2016

${a}_{i} = {a}_{\text{i-1}} \times 3 , {a}_{1} = 2$
The sequence cannot be represented by a continuous graph as ${a}_{n}$ is only defined $\forall n \in \mathbb{N}$

#### Explanation:

${a}_{n} = 2 {\left(3\right)}^{n - 1}$

It seems reasonable to assume $n \in \mathbb{N}$

Hence, the sequence is discrete, where ${a}_{n}$ exists only $\forall n \in \mathbb{N}$

Now consider:

${a}_{1} = 2 \cdot {3}^{0} = 2$
${a}_{2} = 2 \cdot {3}^{1} = 6$
${a}_{3} = 2 \cdot {3}^{2} = 18$

${a}_{i} = {a}_{\text{i-1}} \times 3$

To represent this sequence graphically one could present a chart showing the initial points ${a}_{i}$ $\left(i \in \mathbb{N}\right)$ or a similar histogram of such values.

NB: The important point to realise here it that the sequence cannot be represented by a continuous graph as ${a}_{n}$ is only defined $\forall n \in \mathbb{N}$

Nov 25, 2016

See explanation...

#### Explanation:

Assuming that this formula describes the general term of a sequence, it would normally only be used for positive integer values of $n$, so as a function is only defined on the positive integers.

Note that the terms are also given by the recursive formulation:

$\left\{\begin{matrix}{a}_{1} = 2 \\ {a}_{n + 1} = 3 {a}_{n}\end{matrix}\right.$

So we can start to list the points which could be graphed:

$\left(1 , 2\right) , \left(2 , 6\right) , \left(3 , 18\right) , \left(4 , 54\right) , \left(5 , 162\right) , \left(6 , 486\right) , \left(7 , 1458\right) , \ldots$

Let us try plotting the first few points:

graph{(100(x-1)^2+(y-2)^2-1)(100(x-2)^2+(y-6)^2-1)(100(x-3)^2+(y-18)^2-1)(100(x-4)^2+(y-54)^2-1) = 0 [-4, 10, -4, 60]}

Since the function is exponential, it quickly goes off any linear vertical axis, resulting in at least a couple of problems:

• You can only see a few points.

• It is not clear whether the function has a vertical asymptote.

We can supplement the graph with the graph of $f \left(x\right) = 2 \cdot {3}^{x}$ to indicate the trend of the points:

graph{(100(x-1)^2+(y-2)^2-1)(100(x-2)^2+(y-6)^2-1)(100(x-3)^2+(y-18)^2-1.2)(100(x-4)^2+(y-54)^2-1)(y-2*3^(x-1)) = 0 [-4, 10, -4, 60]}

Note that the curve is not part of the graph of ${a}_{n} = 2 \cdot {3}^{n - 1}$, but it clarifies the pattern of the isolated points.

It is possible to work around the vertical scale issue by plotting $\log \left({a}_{n}\right)$ instead of ${a}_{n}$...

graph{((x-1)^2+(y-log(2))^2-0.01)((x-2)^2+(y-log(6))^2-0.01)((x-3)^2+(y-log(18))^2-0.01)((x-4)^2+(y-log(54))^2-0.01)((x-5)^2+(y-log(162))^2-0.01)((x-6)^2+(y-log(486))^2-0.01)((x-7)^2+(y-log(1458))^2-0.01)(y-log(2*3^(x-1))) = 0 [-2, 7.5, -1, 6]}

Here I have also included the graph of $\log \left(f \left(x\right)\right) = \log \left(2 \cdot {3}^{x - 1}\right)$, making the linear trend of the points obvious.

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Footnote

In a technical sense, the set:

$\left\{\left(n , {a}_{n}\right) : n \in {\mathbb{Z}}_{+}\right\} = \left\{\begin{matrix}1 & 2 \\ 2 & 6 \\ 3 & 18 \\ 4 & 54 \\ 5 & 162 \\ 6 & 486 \\ 7 & 1458 \\ \ldots\end{matrix}\right\}$

is the graph of the function.

Given any function $f : A \to B$ the graph of $f$ is the subset of $A \times B$ consisting of ordered pairs $\left\{\left(a , f \left(a\right)\right) : a \in A\right\}$

So if your question was "What is the graph of ${a}_{n} = 2 \cdot {3}^{n - 1}$?" then the answer could be expressed as:

$\left\{\left(n , 2 \cdot {3}^{n - 1}\right) : n \in {\mathbb{Z}}_{+}\right\} = \left\{\begin{matrix}1 & 2 \\ 2 & 6 \\ 3 & 18 \\ 4 & 54 \\ 5 & 162 \\ 6 & 486 \\ 7 & 1458 \\ \ldots\end{matrix}\right\}$