Question

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

Answer 1

`a_i = a_"i-1" xx 3, a_1 = 2`
The sequence cannot be represented by a continuous graph as `a_n` is only defined `forall n in NN`

Explanation:

`a_n = 2(3)^(n-1)`

It seems reasonable to assume `n in NN`

Hence, the sequence is discrete, where `a_n` exists only `forall n in NN`

Now consider:

`a_1 = 2*3^0 = 2`
`a_2 = 2* 3^1 = 6`
`a_3 = 2* 3^2 = 18`

`a_i = a_"i-1" xx 3`

To represent this sequence graphically one could present a chart showing the initial points `a_i` `( i in NN)` 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 NN`

Answer 2

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:

`{ (a_1 = 2), (a_(n+1) = 3a_n) :}`

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

`(1, 2), (2, 6), (3, 18), (4, 54), (5, 162), (6, 486), (7, 1458),...`

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(x) = 2*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*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(a_n)` 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(f(x)) = log(2*3^(x-1))`, making the linear trend of the points obvious.

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Footnote

In a technical sense, the set:

`{ (n, a_n) : n in ZZ_+ } = { (1, 2), (2, 6), (3, 18), (4, 54), (5, 162), (6, 486), (7, 1458),... }`

is the graph of the function.

Given any function `f:A->B` the graph of `f` is the subset of `AxxB` consisting of ordered pairs `{ (a, f(a)) : a in A }`

So if your question was "What is the graph of `a_n = 2*3^(n-1)`?" then the answer could be expressed as:

`{ (n, 2*3^(n-1)) : n in ZZ_+ } = { (1, 2), (2, 6), (3, 18), (4, 54), (5, 162), (6, 486), (7, 1458),... }`