How do you graph #a_n=2(3)^(n1)#?
2 Answers
The sequence cannot be represented by a continuous graph as
Explanation:
It seems reasonable to assume
Hence, the sequence is discrete, where
Now consider:
To represent this sequence graphically one could present a chart showing the initial points
NB: The important point to realise here it that the sequence cannot be represented by a continuous graph as
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
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(x1)^2+(y2)^21)(100(x2)^2+(y6)^21)(100(x3)^2+(y18)^21)(100(x4)^2+(y54)^21) = 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
graph{(100(x1)^2+(y2)^21)(100(x2)^2+(y6)^21)(100(x3)^2+(y18)^21.2)(100(x4)^2+(y54)^21)(y2*3^(x1)) = 0 [4, 10, 4, 60]}
Note that the curve is not part of the graph of
It is possible to work around the vertical scale issue by plotting
graph{((x1)^2+(ylog(2))^20.01)((x2)^2+(ylog(6))^20.01)((x3)^2+(ylog(18))^20.01)((x4)^2+(ylog(54))^20.01)((x5)^2+(ylog(162))^20.01)((x6)^2+(ylog(486))^20.01)((x7)^2+(ylog(1458))^20.01)(ylog(2*3^(x1))) = 0 [2, 7.5, 1, 6]}
Here I have also included the graph of
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
So if your question was "What is the graph of
#{ (n, 2*3^(n1)) : n in ZZ_+ } = { (1, 2), (2, 6), (3, 18), (4, 54), (5, 162), (6, 486), (7, 1458),... } #