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

2 Answers
Nov 23, 2016

Answer:

#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#

Nov 25, 2016

Answer:

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.

#color(white)()#
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),... } #