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Use zero- through third-order Taylor series expansions to predict f(2, 7) for f(x) = cos(3x) using a base point of x_0 = 2.5. For each approximation, list the following:

a. Approximate value of f(2, 7)
b. On what order of the step size h is the truncation error?
c. Compute the true percent relative errors, epsilon_t, for each approximation.

Respuesta :

Answer:

Following are the solution to this question:

Step-by-step explanation:

Given:

[tex]\bold{f(x) = cos(3x)}\\\to f'(x) = - 3 sin (3x)\\\to f''(x) = - 9 cos (3x)\\\to f'''(x) = 27 sin (3x)\\[/tex]

Similarly:

[tex]\bold{f(2.5)' = cos(6.75)}\\\to f'(2.5) = - 3 sin (6.75)\\\to f''(2.5) = - 9 cos (6.75)\\\to f'''(2.5) = 27 sin (2.5)\\[/tex]

calculating the order values [tex]\longrightarrow \cos(6.75)[/tex]:

In the 1st order:

[tex]\to cos(6.75) + (x-2.5) 3 sin (6.75)[/tex]

In the 2nd order:

[tex]\to cos(6.75) + [1- \frac{(x-2.5)^2}{2}9] + (x-2.5) 3 sin (6.75)[/tex]

In the 3rd order:

[tex]\to cos(6.75) + [1- \frac{9(x-2.5)^2}{2}] + 3 sin (6.75)(3(x-2.5) - \frac{27}{6} (1-2.5))[/tex]

In point a:

[tex]\boxed{\left\begin{array}{cccc}{Zero&First &Second&Third}\\0.993068457& 0.420690485&9.2635909&16.0135909\\\end{array}\right}[/tex]

In point b:

[tex]\boxed{\left\begin{array}{cccc}{First &Second&Third&Fourth}\\0.993068457& 0.420690485&9.2635909&16.0135909\\\end{array}\right}[/tex]

In point C:

At this point, data is missing.