Pad 6:39 AM * 88%.. a d2l.lonestar.edu Homework Assignments 2414 Calulus II MATH
ID: 2887122 • Letter: P
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Pad 6:39 AM * 88%.. a d2l.lonestar.edu Homework Assignments 2414 Calulus II MATH-2414 63011 approproate the function - Google Search x+1 2. Find the nth Taylor polynomial centered at c for the functions: a. f(x)--n 3, c 1 3. Approximate the function at the given value of x using the polynomials found above: a. f(x) - ex.x b. f(x)- c. f(x)-2.x-1 d, f(x)-Vx, x-3.5 Use Taylor's Theorem to obtain an upper bound for the error of the approximation. Then calculate the exact value of the error 4. (0.4) arctan(0.4)0.4 5. Determine the degree of the Maclaurin polynomial required for the error in the approximation of the function at the indicated value of x to be less than 0.001 0.6 9.10: 44, 46, 48 10.1: New instructions for problems 4, 5, 8: a) Graph the path of the particle using the parametric equations given and the parametric interval b) Write the parametric equations in rectangular form 20, 22, 24, 28, 33 2, 8, 10, 14, 26, 28, 30, 32 (an improper integral - caution), 34 10.2:Explanation / Answer
According to Taylor's theorem,
f(x)= f(0)+xf'(0)+(x2/2!)f''(0)+(x3/3!)f''(0)+.......
Now, for f(x)=arctan(x),
f'(x)=1/(1+x2). Therefore, f'(0)=1.
f''(x)=-2x/(1+x2)2. Therefore, f''(0)=0
Similarly, we can find f'''(0)=2 and so on....
Substituting these values in the above series, we get
f(x)=0+x.1+(x2/2!).0+(x3/3!).2+.......
Now, taking the approximate value of the series upto fourth term (with third order derivative, as stated in the question), we get
arctan(0.4)=0.4+0.43/3=0.421333 (upto 6th decimal place)
Actual value of arctan(0.4)=0.380506 (upto 6th decimal place)
Therefore, the exact value of the error is 0.421333-0.380506=0.040827 (upto 6th decimal place).
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