# graphing a sequence

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• Jan 2nd 2011, 10:07 AM
michaeljoannou1990
graphing a sequence
Hi guys, i have a problem with this past homework question i've come accross whilst studying for my exams:

Sketch the graphs of
f(x) = (1 + x)^(1/2) and g(x) = x on the same axes. Let the sequence

a
n
be defined by a0 = 0 and an+1 = &1 + an for any n # 0. Interpret the sequence an

graphically using your sketch.

Here is the solution:

Starting at
(an, an) on the graph of g and moving vertically to the the graph of f we reach the point

(
an, an+1). Moving horizontally back to the graph of g takes us to (an+1, an+1). This is represented

by the dotted zig-zag line.

Attachment 20308

I don't understand how this is the solution, am i plotting a graph of a against a n+1? any help would be much appreciated, thanks
• Jan 2nd 2011, 12:41 PM
pickslides
Here's a start, $\displaystyle (0,0),(0,1),(1,1),(1,f(1)),\dots = (0,g(0)),(g(0),f(0)),(f(0),g(f(0))),\dots$
• Jan 3rd 2011, 10:48 AM
michaeljoannou1990
i just don't understand the relationship between the sequence and the coordinates, if i didn't know the solution, i would just plot a0, a1, a2 etc. on the x axis and plot the corresponding value.
• Jan 3rd 2011, 11:07 AM
michaeljoannou1990
Right I think I understand this now so I'll leave how I understood it here for any other newbie members. The question asks to plot a sequence on the graph, we already have the function f(x) which defines the sequence, we just have to find the x-coordinates along the graph of f(x) which defines each term of the sequence. Because this is a sequence, each x-coordinate is the previous y/ f(x) coordinate. So to find this point we use the function g(x) which for any y-coordinate will an equal x-coordinate. Hence the zig zag is used to find a(n) along f(x). If anyone can see a mistake with this please explain.