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{SECT 0 {EXCHG {PARA 256 "" 0 "" {TEXT 261 38 "Taylor and Maclaurin Se
ries\nApril 2006" }}{PARA 257 "" 0 "" {TEXT -1 32 "revised March 2007,
 October 2008" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{SECT 
1 {PARA 3 "" 0 "" {TEXT -1 4 "The " }{TEXT 262 6 "taylor" }{TEXT -1 9 
" and the " }{TEXT 263 6 "series" }{TEXT -1 9 " commands" }}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart;" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 64 "You can display the first few terms of a Taylor series wi
th the " }{TEXT 256 6 "taylor" }{TEXT -1 107 " command.  To do so, you
 need to communicate the function and the point where the expansion ta
kes place.  \n" }{TEXT 257 10 "Examples: " }{TEXT -1 8 " Expand " }
{XPPEDIT 18 0 "f(x) = exp(x);" "6#/-%\"fG6#%\"xG-%$expG6#F'" }{TEXT 
-1 5 " at  " }{XPPEDIT 18 0 "x = 0;" "6#/%\"xG\"\"!" }{TEXT -1 14 "  a
nd expand  " }{XPPEDIT 18 0 "f(x) = 1/x;" "6#/-%\"fG6#%\"xG*&\"\"\"F)F
'!\"\"" }{TEXT -1 6 "  at  " }{XPPEDIT 18 0 "x = 1;" "6#/%\"xG\"\"\"" 
}{TEXT -1 5 "  :  " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "taylo
r(exp(x),x=0);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "taylor(1/
x,x=1);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 163 "The number of terms c
an be controlled with a third optional entry.  Note, that the number i
ndicates the order of the approximation, and not the number of terms. \+
 \n" }{TEXT 264 8 "Example:" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 66 "tayl
or(sin(x),x=0,8);\ntaylor(sin(x),x=0,9);\ntaylor(sin(x),x=0,10);" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 4 "The " }{TEXT 258 6 "series" }{TEXT 
-1 39 " command accomplishes the same task.  \n" }{TEXT 265 8 "Example
:" }{TEXT -1 22 "  Expand the function " }{XPPEDIT 18 0 "f(x) = ln(x);
" "6#/-%\"fG6#%\"xG-%#lnG6#F'" }{TEXT -1 4 " at " }{XPPEDIT 18 0 "x = \+
2;" "6#/%\"xG\"\"#" }{TEXT -1 1 "." }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 
39 "series(ln(x),x=2);\nseries(ln(x),x=2,8);" }}}{EXCHG {PARA 0 "" 0 "
" {TEXT -1 72 "The series command lets you even find an expansion when
 taylor fails.  \n" }{TEXT 266 8 "Example:" }{TEXT -1 6 "  Let " }
{XPPEDIT 18 0 "f(x) = 1/sin(x);" "6#/-%\"fG6#%\"xG*&\"\"\"F)-%$sinG6#F
'!\"\"" }{TEXT -1 3 " = " }{XPPEDIT 18 0 "csc(x);" "6#-%$cscG6#%\"xG" 
}{TEXT -1 25 ".  A Taylor expansion at " }{XPPEDIT 18 0 "x = 0;" "6#/%
\"xG\"\"!" }{TEXT -1 22 " is impossible, since " }{XPPEDIT 18 0 "f(0);
" "6#-%\"fG6#\"\"!" }{TEXT -1 14 " is undefined." }}{PARA 0 "> " 0 "" 
{MPLTEXT 1 0 21 "taylor(1/sin(x),x=0);" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 48 "However, we can find a series for this function:" }}
{PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "series(1/sin(x),x=0,10);" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 63 "Let us check the respective graphs
 of f and its approximation:." }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 55 "y \+
:=x^(-1)+1/6*x+7/360*x^3+31/15120*x^5+127/604800*x^7;" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 76 "plot([1/sin(x),y],x=-4..4,view=[-4.
.4,-4..4],discont=true,color=[red,blue]);" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 59 "The series \+
approximates the cosecant at x=0 by approaching " }{XPPEDIT 18 0 "infi
nity;" "6#%)infinityG" }{TEXT -1 18 " at the same pace." }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{SECT 1 {PARA 3 "" 0 "" {TEXT 
-1 42 "Taylor coefficients and Taylor polynomials" }}{EXCHG {PARA 0 ">
 " 0 "" {MPLTEXT 1 0 8 "restart;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 267 
46 "Throughout this part we shall use the function" }{TEXT -1 3 "   " 
}{XPPEDIT 18 0 "f(x) = exp(-x/2)*sin(x);" "6#/-%\"fG6#%\"xG*&-%$expG6#
,$*&F'\"\"\"\"\"#!\"\"F0F.-%$sinG6#F'F." }{TEXT -1 2 "  " }{TEXT 268 
61 "as an example, and we begin by defining it once and for all.\n" }}
{PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "f := x -> sin(x)*exp(-x/2);\n" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 60 "A particular series coefficient ca
n be calculated using the " }{TEXT 259 8 "coeftayl" }{TEXT -1 55 " com
mand.  For example the fifth series coefficient is " }}{PARA 0 "> " 0 
"" {MPLTEXT 1 0 21 "coeftayl(f(x),x=0,5);" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 48 "This is mathematically equivalent to computing  " }
{XPPEDIT 18 0 "`@@`(D,5)(f)(0)/5!;" "6#*&---%#@@G6$%\"DG\"\"&6#%\"fG6#
\"\"!\"\"\"-%*factorialG6#F*!\"\"" }{TEXT -1 65 "  and of course we ge
t the same result when using this statement." }}{PARA 0 "> " 0 "" 
{MPLTEXT 1 0 16 "(D@@5)(f)(0)/5!;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 
145 "I am not certain which of the two commands (coeftayl or direct de
rivative calculation) is numerically more efficient.  Let us check our
 result.  " }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "taylor(f(x),x=0);" }}
}{EXCHG {PARA 0 "" 0 "" {TEXT -1 47 "We see that  our number is the co
efficient for " }{XPPEDIT 18 0 "x^5;" "6#*$%\"xG\"\"&" }{TEXT -1 15 ",
 as expected. " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 19 "We now turn to the " }{TEXT 260 
35 "construction of Taylor polynomials." }{TEXT -1 426 "  The simplest
 way is to accomplish this task is to obtain the first few terms of th
e series with the taylor or the series command, and then apply the pop
ular copy-and-paste technique.  A more elegant way is to calculate the
 Taylor coefficients in a loop, and then assemble the polynomial as a \+
sum.  A fairly generic model follows; recall that the function has alr
eady been defined, and we don't need to duplicate that step.. " }
{MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 200 "N := 8
:    # degree of the Taylor polynomial\nc := 0:    # point of expansio
n\n# begin the construction\nfor k from 0 to N do\n    a[k] := coeftay
l( f(x),x=c, k):\nod:\np := x -> sum( a[n]*(x-c)^n,n=0..N):\n" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 29 "Yes, the polynomial is there." }}
{PARA 0 "> " 0 "" {MPLTEXT 1 0 5 "p(x);" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 87 "Let's look at the graph of the function along with its 8t
h degree Taylor approximation." }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 62 "p
lot([f(x),p(x)],x=-5..5,color=[red,blue],view=[-5..5,-4..6]);" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}}{MARK "0 1 0" 32 }
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