Question

Use zeroth- through fourth-order Taylor series expansions to predict f(5) for 𝑓(π‘₯) = 25π‘₯^4 βˆ’ 6π‘₯^3 + 7π‘₯^2 βˆ’ 88π‘₯ Use a base point at x = 1, and compute the true percentage error for each approximation. (Round the final estimated values to the nearest whole numbers and the error values to two decimal places.) Enter the desired values in the spaces provided in Canvas. Problem 7 (20 points) Write a MATLAB program or Excel macro that will solve for sin(x) using the infinite series below. The program should have 2 inputs: the value of x and the stopping criterion (i.e. maximum desired true fractional error). Loop and add terms until the true fractional error is less than the stopping criterion. Report the approximate value of sin(x) and the number of terms needed. 𝑠𝑖𝑛(π‘₯) = π‘₯ βˆ’ π‘₯^3/3! + π‘₯^5/5! βˆ’ ... = 𝑛 = 0 to ∞ βˆ‘ (βˆ’1)^𝑛π‘₯^(2𝑛+1)/(2𝑛+1)! Order Value Error Ξ΅|% 0 ? ? 1 ? ? 2 ? 84.98 3 ? ? 4 14610 ?

          Use zeroth- through fourth-order Taylor series expansions to predict f(5) for
𝑓(π‘₯) = 25π‘₯^4 βˆ’ 6π‘₯^3 + 7π‘₯^2 βˆ’ 88π‘₯
Use a base point at x = 1, and compute the true percentage error for each
approximation. (Round the final estimated values to the nearest whole numbers and the
error values to two decimal places.) Enter the desired values in the spaces provided in
Canvas.
Problem 7 (20 points)
Write a MATLAB program or Excel macro that will solve for sin(x) using the infinite
series below. The program should have 2 inputs: the value of x and the stopping
criterion (i.e. maximum desired true fractional error). Loop and add terms until the true
fractional error is less than the stopping criterion. Report the approximate value of sin(x)
and the number of terms needed.
𝑠𝑖𝑛(π‘₯) = π‘₯ βˆ’ π‘₯^3/3! + π‘₯^5/5! βˆ’ ... = 𝑛 = 0 to ∞ 
βˆ‘ (βˆ’1)^𝑛π‘₯^(2𝑛+1)/(2𝑛+1)!
Order Value Error Ξ΅|%
0 ? ?
1 ? ?
2 ? 84.98
3 ? ?
4 14610 ?
        
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Added by Jesse E.

Computer Science and Information Technology
Computer Science and Information Technology
Trishna Knowledge Systems 2018 Edition
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Use zeroth- through fourth-order Taylor series expansions to predict f(5) for 𝑓(π‘₯) = 25π‘₯^4 βˆ’ 6π‘₯^3 + 7π‘₯^2 βˆ’ 88π‘₯ Use a base point at x = 1, and compute the true percentage error for each approximation. (Round the final estimated values to the nearest whole numbers and the error values to two decimal places.) Enter the desired values in the spaces provided in Canvas. Problem 7 (20 points) Write a MATLAB program or Excel macro that will solve for sin(x) using the infinite series below. The program should have 2 inputs: the value of x and the stopping criterion (i.e. maximum desired true fractional error). Loop and add terms until the true fractional error is less than the stopping criterion. Report the approximate value of sin(x) and the number of terms needed. 𝑠𝑖𝑛(π‘₯) = π‘₯ βˆ’ π‘₯^3/3! + π‘₯^5/5! βˆ’ ... = 𝑛 = 0 to ∞ βˆ‘ (βˆ’1)^𝑛π‘₯^(2𝑛+1)/(2𝑛+1)! Order Value Error Ξ΅|% 0 ? ? 1 ? ? 2 ? 84.98 3 ? ? 4 14610 ?
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Transcript

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00:02 We are going to use the function f of x equals natural logarithm of 1 plus x and the remainder term to estimate the absolute error in approximating the natural logarithm of 1 .05 with the third degree taylor polynomial centered at zero.
00:21 So we consider the function f of x equals natural logarithm of 1 plus x and so if we want to approximate natural logarithm of 1 .05, we see that this is the natural logarithm of 1 plus 0 .05 and looking at the definition of the function f this is exactly equal to f at 0 .05.
00:50 So to approximate this number we get to approximate the function f at 0 .05.
00:59 That's what we're going to do.
01:03 Then to find the taylor polynomial of degree three centered at zero we need the first three derivative of the function f and we also need the fourth derivative to evaluate the remainder.
01:18 So let's see the first derivative of f is one over one plus x which is the same as one plus x to the negative one.
01:29 The second derivative is derivative of this expression is negative one plus x to the negative two times the derivative of one plus x which is one.
01:43 So the third, sorry here, the third derivative at x is derivative of this expression is two times one plus x to the negative three times the derivative of one plus x which is one...
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