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{SECT 0 {EXCHG {PARA 18 "" 0 "" {TEXT -1 32 "Solving a Transendental E
quation" }}{PARA 256 "" 0 "" {TEXT -1 19 "Original Version:  " }{TEXT 
256 15 "Earl C. Swallow" }{TEXT -1 10 "   7/22/97" }}{PARA 257 "" 0 "
" {TEXT -1 38 "Revised and Annotated:  ECS   10/11/99" }}{PARA 0 "" 0 
"" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 115 "This worksheet is an \+
example of solving a (rather simple) transendental equation which has \+
no closed-form solution." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "
" 0 "" {TEXT -1 13 "The equation " }{TEXT 258 17 "sin(x) = x*cos(x)" }
{TEXT -1 77 " gives the positions of diffraction maxima for single-sli
t diffraction (here " }{TEXT 259 1 "x" }{TEXT -1 93 " is the diffracti
on angle).  It herefore occurs is acoustics, optics, and quantum mecha
nics (" }{TEXT 257 4 "i.e." }{TEXT -1 89 ", wherever there are waves).
  It has no closed-form solution.  By dividing both sides by " }{TEXT 
260 6 "cos(x)" }{TEXT -1 24 ", it can be turned into " }{TEXT 261 10 "
tan(x) = x" }{TEXT -1 33 ", which is a bit more convenient." }}{PARA 
0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 
0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart
;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "assume(x, real);" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "eq1 := sin(x) = x*cos(x);" }
}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "eq2 := tan(x) = x;" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "with(plots):" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 49 "plot([x,tan(x)], x=0..25, y=0..25, \+
discont=true);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 61 "Let's look at t
he first couple of intersections more closely." }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 47 "plot([x,tan(x)], x=0..8, y=0..8, discont=true)
;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 61 "So we can read from the grap
h that the first 3 solutions are " }{TEXT 265 7 "roughly" }{TEXT -1 
75 " x1 = 0, x2 = 4.46, x3 = 7.68 .  Thinking about tan(x), it has asy
mptotes (" }{TEXT 263 4 "i.e." }{TEXT -1 62 ", it goes off to infinity
) at 0.5*Pi, 1.5*Pi, 2.5*Pi, 3.5*Pi, " }{TEXT 262 3 "etc" }{TEXT -1 
89 ".  Recognizing that x2 is a little less than 1.5*Pi and x3 is a li
ttle less than 2.5*Pi, " }{TEXT 264 4 "etc." }{TEXT -1 70 ", we see th
at successive solutioons will be between N*Pi and (N+1)*Pi." }}{PARA 
0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 
0 "" {TEXT -1 28 "So now go for the solutions." }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "
x2 := fsolve(eq1, x, 1*Pi..2*Pi);" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 13 "evalf(x2/Pi);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 6 "\"/1.5;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 41 "So this is just
 a bit less than 1.5*Pi.  " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 
"" 0 "" {TEXT 267 44 "Now you get the next few.  Also try it with " }
{TEXT 266 3 "eq2" }{TEXT 268 22 " and see what happens." }}}{EXCHG 
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