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Series[f[x] {x, 0, 10}] ,
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8.21 Obtain a representation of x5 in powers of x 2.
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Series[x5, {x, 2, 5}] //Normal 32 + 80( 2 + x) + 80( 2 + x)2 + 40( 2 + x)3 + 10( 2 + x)4 + ( 2 + x)5 %//TraditionalForm ( x 2)5 + 10 ( x 2)4 + 40 ( x 2)3 + 80 ( x 2)2 + 80 ( x 2) + 32
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8.22 Construct a Taylor polynomial of degree 5 about a = 1 for the function approximate 3 / 2 .
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p[x_]= Series[ x, {x,1,5}] / /Normal
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2 3 4 5 1 + 1 ( 1 + x) 1( 1 + x) + 1 ( 1 + x) 5 ( 1 + x) + 7 ( 1 + x) 2 8 16 128 256
approx = p[3/2] //N 1.22498 exact = 1.22474 Abs[% %%] 0.000230715
This is the absolute error of the approximation.
3/ 2 //N
8.23 Let f(x) = sin x and compute the Maclaurin polynomials of degrees 7, 9, and 11. Then plot f(x) and the three polynomials on one set of axes, 0 x 2 , and observe their behavior.
SOLUTION
f[x_]= Sin[x]; p7[x_]= Series[f[x], {x, 0, 7}] //Normal; p9[x_]= Series[f[x], {x, 0, 9}] //Normal; p11[x_]= Series[f[x], {x, 0, 11}] //Normal;
Differential Calculus
Plot[{f[x], p7[x], p9[x], p11[x]}, {x, 0, 2 }, PlotStyle {Thickness[.01], Thickness[.001], Thickness[.001], Thickness[.001]}]
3 2 1 1 1 2 3 4 The higher the degree of the polynomial, the better the polynomial approximates f(x) = sin x. 2 3 4 5 6
8.24 Let f(x) = sin x and compute the Maclaurin polynomial of degree 11. Construct an error function and compute its value from x = 0 to x = 1 in increments of 0.1. Place the results in the form of a table and comment on the values of the error as x gets further from 0.
SOLUTION
f[x_] = Sin[x]; p11[x_] = Normal[Series[f[x], {x, 0, 11}]]; error[x_] = Abs[f[x] p11[x]]; ; errorvalues = Table[{x, error[x]}, {x, 0, 6, 1.}] TableForm[errorvalues, TableSpacing {1,5}, TableHeadings {None, {"x"," error[x]"}}]
x 0. 1. 2. 3. 4. 5. 6. error[x] 0. 1.59828 10 10 1.29086 10 6 0.000245414 0.0100021 0.174693 1.78084
As x gets further from 0, the error gets larger.
8.25 Let f(x) = sin x. Construct the Maclaurin polynomials of degrees 1, 3, 5, 7, and 9 and compute their value at x = 1. Determine the error in the approximations and express in a tabular form.
SOLUTION
f[x_] = Sin[x] ; exactvalue = f[1] ; value[n_] Normal[Series[f[x], {x, 0, n}]] /. x 1 data = Table[{n, N[value[n]], N[exactvalue], N[Abs[value[n] exactvalue]]}, {n, 1, 9, 2}];
Differential Calculus
TableForm[data, TableSpacing {1,5}, TableHeadings {None, {"n","
p(1) ","
f(1)","
Error"}}]
n 1 3 5 7 9
p(1) 1. 0.833333 0.841667 0.841468 0.841471
f(1) 0.841471 0.841471 0.841471 0.841471 0.841471
Error 0.158529 0.00813765 0.000195682 2.73084 10 6 2.48923 10 8 As n gets larger, the error gets smaller.
8.26 Let f(x) = ln x and compute the Taylor polynomials about a = 1 of degrees 5, 10, and 15. Then plot f(x) and the three polynomials on one set of axes, 1 x 2.
SOLUTION
f[x_] = Log[x]; p5[x_] = Series[f[x], {x, 1, 5}]//Normal; p10[x_] = Series[f[x], {x, 1, 10}]//Normal; p15[x_] = Series[f[x], {x, 1, 15}]//Normal; Plot[{f[x], p5[x], p10[x], p15[x]}, {x, 1, 2}, PlotStyle {Thickness[.01], Thickness[.001], Thickness[.001], Thickness[.001]}]
8.27 Let f(x) = ln x and construct the Taylor polynomial of degree 5 about a = 1. Construct an error function and compute its value from x = 1 to x = 2 in increments of 0.1. Place the results in the form of a table and comment on the values of the error as x gets further from 1.
SOLUTION
f[x_] = Log[x]; p5[x_] = Normal[Series[f[x], {x, 1, 5}]]; error[x_] = Abs[f[x] p5[x]]; errorvalues = Table[{x, error[x]}, {x, 1, 2, .1}];
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