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{SECT 0 {SECT 0 {PARA 3 "" 0 "" {TEXT -1 16 "Approximating Pi" }}
{PARA 256 "" 0 "" {TEXT -1 232 "Let us show you a quick way using a fe
w commands to approximate pi.  We are going to find the Taylor series \+
expansion of a trig function.  Let us look at tan(pi/4)=1.  We are goi
ng to solve this for pi so we can get a taylor series:" }}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "solve(tan(pi/4)=x,pi);" }}}{PARA 0 
"" 0 "" {TEXT -1 0 "" }}{PARA 256 "" 0 "" {TEXT -1 394 "So, it seems t
his is what we want a Taylor series for.  To do this in Maple, we use \+
the command \"taylor(..)\".  Taylor takes three arguements.  The first
 is your function, the second is what the series will be centered at (
for our purposes, zero), and the third is the highest power you want t
he series to go to.  You can change this third parameter to add more t
erms to your series if you wish." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 28 "taylor(4*arctan(x), x=0, 8);" }}}{PARA 0 "" 0 "" {TEXT -1 0 "
" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 256 "" 0 "" {TEXT -1 646 "Once you have the series you wish to l
ook at, you can use it begin approximating pi.  We use a complicated s
tring of commands that starts with evalf(...) to numerically evaluate \+
our series.  The parameter this command takes is the expression to be \+
evaluated.  Our expression will be our series with a value substitued \+
for x.  What value should we substitute for x?  In our situation, we w
ant to use x = 1 since 4arctan(1)=pi.  We now have to use the mtaylor(
...) command.  This takes three commands:  your expression, the point \+
your series will be centered at (once again, zero for us), and the num
ber of terms we will be using to approximate pi." }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 47 "evalf(subs(x
=1, mtaylor(4*arctan(x), x=0, 8)));" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" 
}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }
{TEXT 264 331 "How close was your approximation to pi?  Can you make i
t closer?  Sure!  Increase the number of terms in your series and the \+
approximation will approach pi.  Try to do so on the lines below (you \+
can copy and paste to make this easier).  Compare your results to peop
le that used different expressions or a different number of terms:" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "
> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 
"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "
" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }
{TEXT 265 130 "For all real applications we really only need pi correc
t to about 10 places.  Try to get one of your approximations that accu
rate?" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "
" }{TEXT 266 296 "You can compare your answer to the actual value of p
i by by calculating a percent error or check the number of correct dig
its by subtracting your value from the actual value.  Try some of thes
e comparisons in Maple.  Remember the actual value of pi is built in a
nd is stored in the variable \"Pi\"." }}{PARA 0 "" 0 "" {TEXT -1 0 "" 
}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 50 "evalf(((Pi-(copy and paste
 result here))/Pi)*100);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{PARA 0 "
" 0 "" {TEXT -1 0 "" }}}{MARK "0 1 0" 6 }{VIEWOPTS 1 1 0 1 1 1803 1 1 
1 1 }{PAGENUMBERS 0 1 2 33 1 1 }