{VERSION 4 0 "SUN SPARC SOLARIS" "4.0" } {USTYLETAB {CSTYLE "Maple Input" -1 0 "Courier" 0 1 255 0 0 1 0 1 0 0 1 0 0 0 0 1 }{CSTYLE "" -1 256 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 1 } {CSTYLE "" -1 257 "" 1 14 0 0 0 0 0 1 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 258 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 259 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 260 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 261 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 } {PSTYLE "Normal" -1 0 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "Maple Output" 0 11 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 }3 3 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "Maple Plot" 0 13 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 }3 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE " " 0 256 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 }3 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 0 257 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 }3 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 11 258 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 }1 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 0 259 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 } {PSTYLE "" 0 260 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }} {SECT 0 {PARA 11 "" 1 "" {TEXT -1 0 "" }{TEXT 256 0 "" }{TEXT 257 11 " MATH 2270-1" }}{PARA 256 "" 0 "" {TEXT -1 15 "MAPLE PROJECT 4" }} {PARA 257 "" 0 "" {TEXT -1 16 "November 1, 2000" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 258 23 "Part I: Body Mass Index " }}{PARA 0 "" 0 "" {TEXT -1 503 "1) In this first part of our proje ct we will use the method of least squares to find a power law which ( approximately) relates human weights to heights. Within the past few \+ years you might have noticed a series of newspaper and magazine artic les about the so-called body mass index, and its use in determining ri sk factors for overweight people. If you search the internet for \"bo dy mass index\" you will find many sites which let you compute your B. M.I., and which tell you a little bit about it. " }}{PARA 0 "" 0 "" {TEXT -1 1183 " A person's body mass index (B.M.I.) is computed b y dividing their weight by the square of their height, and then multip lying by a universal constant. (If you measure weight in kilograms, \+ and height in meters, this constant is the number one.) Thus, the prop onants of the B.M.I index seem to be assuming, or claiming, that for adults at equal risk levels (but different heights), weight should be proportional to the square of height. As we discussed in class, if p eople were to scale equally in all directions when they grew, weight w ould scale as the cube of height. That particular power law seems a \+ little high, since adults don't look like uniformly expanded versions \+ of babies; we seem to get relatively stretched out when we grow taller . One would expect the best predictive power to be somewhere between \+ 2 and 3. If the power is much larger than 2 then one could argue that the body mass index might need to be modified to reflect this fact. O f course, the sample heights and weights which we have collected may o r may not be representative of a healthy population, but let's proceed anyway. (In previous years the powers have come out between 2.35 an d 2.65.)" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 260 40 "Using logarithms to discover power laws:" }}{PARA 0 "" 0 "" {TEXT -1 333 " Here's how to use a least squares method to test whether \+ a power law can adequately describe a correlation between two differen t variables. Suppose we have some data points, \{[xi,yi]\}, and we exp ect a power law yi=C*(xi)^p to relate the yi's to the xi's. Here p \+ is the power, and we can call C the roportionality constant. " } {TEXT 259 205 "Then we expect the logarithms to satisfy ln(yi) = ln(C) + p*ln(xi). In other words, the ln-ln data should satisfy the equati on of a line having slope equal to p and vertical axis intercept equal to ln(C)." }{TEXT -1 277 " So we can take the least squares fit for \+ the ln-ln data, and then use the slope and intercept to deduce C and p in the power law: take the power p equal to the slope of the least sq uare fit to the ln-ln data, and take C to equal to the exponential of \+ its vertical intercept." }}{PARA 0 "" 0 "" {TEXT -1 261 " Here's \+ a carried-out example. By the way, it was constructed so that each yi is close to 2*(xi)^3. So when we go through the process described ab ove we should come out with something close to a cubic power law, with proportionality constant close to 2. \n" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 64 "with(linalg):with(plots):\n #we use these two comma nd libraries" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 133 "S:=[[1,2.2 ],[2,15.8],[3,55],[5,255],[6,430]];\n #A list of five data points [x i,yi], for which we seek\n #a power law relationship\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 147 "A1:=convert(S,matrix); \n #Conv ert S from a list data structure into\n #a matrix. We can manipulat e this matrix\n #using our linalg commands.\n" }}{PARA 11 "" 1 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 209 "A2:=map(ln , A1); \n #the map command will apply a given function,\n #in thi s case the natural logarithm, to each\n #entry of the specified matr ix. So the entries of\n #A2 will be the ln of entries in A1." }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 47 "A3:=map(evalf,A2); \n #ge t each decimal value" }}{PARA 11 "" 1 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 182 "rowdim(A3); \n #rowdim computes the number of rows in\n #a matrix. Of course, for this small matri x,\n #we know there are five rows. Your B.M.I. matrix\n #will be \+ much bigger\n" }{TEXT -1 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 361 "We use \+ A3 to construct the matrix \"A\" and right hand side b, for the least squares line fit. The first column of A will be the ln(x)-values, th e second column will be a vector of 1's and the vector y will be the c orresponding ln(y) values, since we want a least squares line fit (lny )=m(lnx) + b. Maple has commands to extract columns, augment matrices , etc." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 300 "col2:=vector(rowd im(A3),1); \n #this will be the second column, a vector of 1's\n \+ #for our least-squares line fit matrix, see e.g.\n #page 454 of the \+ text. This command creates a vector\n #with number of entries equal to the first argument\n #and makes each entry equal to the second a rgument\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 56 "A4:=delcols(A3 ,2..2); \n #remove the last column of A3," }}{PARA 11 "" 1 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 59 "A:=augment(A4,col 2); #This is our matrix for least squares" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 68 "b:=delcols(A3,1..1); #This is the right-hand side \+ for least squares" }}{PARA 11 "" 1 "" {TEXT -1 0 "" }}}{PARA 0 "" 0 " " {TEXT -1 75 "The next three commands find the least-squares solution , as in section 8.4." }}{PARA 11 "" 1 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 110 "ATA:=evalm(transpose(A)&*A);\nATb:=evalm (transpose(A)&*b);\nlinsolve(ATA,ATb);\n #solve the system (ATA)x=(A Tb)" }}{PARA 11 "" 1 "" {TEXT -1 0 "" }}}{PARA 258 "" 1 "" {TEXT -1 73 "Alternately, we could multiply the right-hand side by the inverse \+ matrix:" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "evalm(inverse(ATA )&*ATb);" }}{PARA 11 "" 1 "" {TEXT -1 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 103 "Actually, least squares for linear systems is a standard tool \+ so Maple has the command built right in:\n" }}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 212 "b1:=col(b,1); #I had to do this because Maple thou ght\n #b was a 5 by 1 matrix, not a vector. This command\n #makes b1 a vector, and so I can use it in the command\n #below. Don't as k me to explain, I can't!" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 62 "leastsqrs(A,b1); #returns the least square solution to Ax=b1." }} }{PARA 0 "" 0 "" {TEXT -1 144 "The first entry above is the least-squa res line slope (for the ln-ln data), the second point its intercept. \+ See visually how the line-fit went:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" } }{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 174 "lnlnplot:=pointplot(\{seq( [A3[i,1],A3[i,2]],i=1..rowdim(A3))\}):\n #those are the points from \+ the ln-ln data in\n #the A3 matrix. the index i ranges over all rows \n #in A3.\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 170 "line:=pl ot(2.959072398*t + .7589027908, t=0..2):\n #I used the mouse to past e in the coefficients\n #from my work above. Here t is standing for the\n #ln(x) variable." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "display(\{lnlnplot,line\});" }}{PARA 13 "" 1 "" {TEXT -1 0 "" }}} {PARA 0 "" 0 "" {TEXT -1 174 "Pretty good fit. The slope of this line will be our experimental power, its intercept will be the ln of our p roportionality constant. Now work backwards to get these values" }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 112 "C:=exp(.7589027908); #propo rtionality constant\np:=2.959072398; \n #power. I used the mouse t o paste these in." }}{PARA 11 "" 1 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 203 "Notice, the power came out close to 3 and the pr oportionality constant came out close to 2, as expected from the origi nal choice of points. Now see how our power law works for the origina l (xi,yi) data:" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "realplot:=pointp lot(S):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "powerplot:=plot( C*x^p,x=0..6):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "display( \{realplot,powerplot\});" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 259 "" 0 "" {TEXT -1 133 "1a) Your job: carry out the same sort of an alysis for the height-weight data which we have accumulated. (See the set S just below.)" }}{PARA 260 "" 0 "" {TEXT -1 106 "1b) What does \+ your power law predict for the weight of typical 5 foot, 5.5 foot, 6 f oot, 6.5 foot people?" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 538 " Here is the height- weight data which you have p rovided to me. Thanks to all who contributed. The first number of ea ch pair is a height, in inches, the second number is a weight, in poun ds. Heights vary from smallish baby (17 inches) to tall person (77 in ches). Weights vary from 6 to 320 pounds. Find a least squares line \+ fit for the corresponding ln-ln data, then work backwards to find a po wer law. Show graphs of the least squares line fit to the ln-ln data, and of the power law to the original data. Finally, answer 1b. " }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 871 "S:=[[64,118],[75,151],[63,1 28],[63,130],[66,125],[77,320],[71,180],\n[40,37.5],[42,41.5],[43,41], [43,43],[44,42.5],[44,38.5],[44,40.5],[45,55],\n[45,46.5],[46,47.5],[4 7,54.5],[48,70.5],[48,43.5],[49,49],[50,50.5],\n[50,58.5],[50,51],[52, 73.5],[52,65.5],[53,49],[53,71],[53,73],[53,63.5],\n[54,72],[55,101.5] ,[56,106.5],[56,138],[56,88],[57,72],[58,82.5],[58,80.5],\n[59,87],[60 ,94.5],[60,95.5],[60,102.5],[62,101],[63,134],[63,133.5],[64,135],\n[6 9,175],[72,180],[74,205],[17,6.0+15/16],[21,8.0+11/16],[19,6.0+15/16], \n[69,145],[66,130],[72,170],[73,168],[65,115],[64,145],[69,195],[67,1 70],\n[73,180],[64,120],[64,155],[62,140],[67,110],[67.25,204],[76.25, 199],[66.5,124],\n[75,228],[21,8.0+15/16],[44,38],[61.5,92],[44,43],[3 9,36],[22,8.0+2/16],[49,55],\n[56,80],[70,167],[73,166],[52,54],[54,65 ],[52.5,65],[52,74],[50,50],\n[50,49],[50,55],[53,52],[52.5,72],[65,12 0],[71.5,183]];" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 261 29 "Part 2: Inner Product Sp aces:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 431 " We will explore inner product function spaces. This part of the \+ project is like examples 4-7, 9-10 in Appendix B.1, except we will us e the x-interval from [-1..1] instead of [0..1]. It is more standard \+ to do so, and this is also how we did the examples in class. In order that your time is used efficiently, part of your work in this section is to check your answers to the homework from appendix B.1: #9, 14, 1 5, 17, 28. " }}{PARA 0 "" 0 "" {TEXT -1 197 " Even though we talk ed about Fourier Series in class, and even though they are fun to comp ute and look at in Maple, we will postpone doing so until Math 2280, w hen we will use them extensively." }}{PARA 0 "" 0 "" {TEXT -1 1 " " }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart:" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 273 " Here's the inner pr oduct we will use. We'll call it ``dot'' in honor of the dot product \+ of R^n, to which this inner product is very similar indeed. With the \+ dot product we can therefore define the magnitude of vectors, as well as distance and angle between vectors." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 57 "dot:=(f,g)->int(f(x)*g( x),x=-1..1);\n #our inner product" }}{PARA 11 "" 1 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 65 "dot(x->x,x->x^3);\n #th e dot product of f(x)=x with g(x)=x^3\n " }}{PARA 11 "" 1 "" {TEXT -1 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 30 "2a) Check B.1 #9, using \"dot \"" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 64 "mag:=f->sqrt(dot(f,f)); \n \+ #computes the magnitude of a vector " }}{PARA 11 "" 1 "" {TEXT -1 0 " " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 54 "mag(x->x^2);\n #the ma gnitude of the function f(x)=x^2" }}{PARA 11 "" 1 "" {TEXT -1 0 "" }}} {PARA 0 "" 0 "" {TEXT -1 31 "2b) Check B.1 #14, using \"mag\"" }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 76 "dist:=(f,g)->mag(f-g); \n \+ #computes the ``distance'' between two functions" }}}{EXCHG {PARA 0 " > " 0 "" {MPLTEXT 1 0 109 "dist(x->x^2,x->x^2+.01*x);\n #these funct ions are \"close\" so the distance\n #between them should be small. \+ " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 98 "cosangle:=(f,g)->(dot(f ,g)/(mag(f)*mag(g)));\n #computes the \"cos of the angle\" between f unctions" }}{PARA 11 "" 1 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 81 "fangle:=(f,g)->evalf(arccos(cosangle(f,g)));\n #c omputes angle between functions" }}{PARA 11 "" 1 "" {TEXT -1 0 "" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 127 "fangle(x->x^2,x->x^2+.01*x) ;\n #these vectors should almost be parallel, so the\n #angle betw een them should be close to zero" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 65 "2d) Use the commands above to check your answers to B.1 #15, 17." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 " " 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 85 "2e) Use the dot p roduct and the Gram-Schmidt algorithm to check your anser to B.1#28" } }{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 279 "Now we will do a projection problem like we did in class. We will try to find the closest degree 3 polynomial to f(x)=sin(x), using our inner product space distance to measure \"c lose\". (See below.) First, let's get names for our standard basis p olynomials in this subspace W: " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 40 "P0:=x->1;P1:=x->x;P2:=x->x^2;P3:=x->x^3;" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 70 "Here's one way to get an \+ orthogonal basis from these four polynomials:" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 41 "Q0:=P0; #first orthogonal basis element" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 157 "proj0:=f->(dot(f,Q0)/mag(Q0 )^2)*Q0;\nQ1:=P1 - proj0(P1); \n #second orthogonal basis element is obtained from\n #P1 by subtracting off its projection onto P0" }} {PARA 11 "" 1 "" {TEXT -1 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 62 "Notice \+ that P1 was already orthogonal to P0=Q0, so also Q1=P1." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 190 "proj1:=f->proj0(f) + (dot(f,Q1)/ma g(Q1)^2)*Q1;\nQ2:=P2 - proj1(P2);\n #third orthogonal basis element \+ is obtained from\n #P2 by subtracting off its projection onto the sp an\n #of Q0 and Q1" }}}{EXCHG {PARA 11 "" 1 "" {TEXT -1 0 "" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 183 "proj2:=f->proj1(f) + (dot(f ,Q2)/mag(Q2)^2)*Q2;\nQ3:=P3-proj2(P3);\n #final orthog basis element is obtained from\n #P3 by subtracting off its projection onto the\n #span of Q0,Q1,Q2" }}{PARA 11 "" 1 "" {TEXT -1 0 "" }}{PARA 11 "" 1 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 51 "Q0(x); Q1(x);Q2(x);Q3(x);\n #what polys did we make?" }}{PARA 11 "" 1 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 99 "dot(Q0,Q1); dot(Q0,Q2);dot(Q0,Q3);\ndot(Q1,Q2); dot(Q1,Q3);\ndot(Q2,Q3);\n #thes e should all be zero!" }}{PARA 11 "" 1 "" {TEXT -1 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 471 "The orthogonal poly nomials which you have been constructing are called the Legendre polyn omials. Maple knows about them. To see the first few you can load th e `orthopoly' package. By the way, if you define different weighted i nner products you get different (famous to experts) orthogonal polynom ial families, you can read about some of them on the help windows, sta rting at `orthopoly'. Orthogonal polynomials are used in approximatio n problems, as you might expect." }}{PARA 0 "" 0 "" {TEXT -1 1 " " }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "with(orthopoly):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 99 "P(0,x);P(1,x);P(2,x);P(3,x);P(4,x); P(5,x);\n #these should look familiar, after what you just did!" }} {PARA 11 "" 1 "" {TEXT -1 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 223 "2f) Fi nd the projection of f(x) = sin(x) onto the subspace spanned by \{1, x , x^2, x^3\}, using our inner product, the orthogonal basis \{Q0, Q1, \+ Q2, Q3\}, and the projection formula you know which works with orthog onal bases." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 " " }}{PARA 0 "" 0 "" {TEXT -1 123 "2g) Show that the the polynomial yo u get in (2f) is closer to sin(x) than the degree 3 Taylor polynomial \+ p(x) = x - x^3/6." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 235 "2h) Plot, in one picture, the graph of sin(x), its projection from 2f), and the Taylor poly from 2g . You can find the commands you will need on the Oct 27 notes from cl ass, available on your internet browser, at our class Maple page." }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "with(plots):" }}}{PARA 0 "" 0 "" {TEXT -1 2 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 " " {TEXT -1 0 "" }}}{MARK "105 0" 0 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 } {PAGENUMBERS 0 1 2 33 1 1 }