In this problem, you will investigate the error in the nth degree Taylor approximation to ln(x+1) about 0 for various values of n.
(a)Let E_1=ln(x+1)−P_1(x)=ln(x+1)−(x). Using a calculator or computer, graph E1 for −0.1≤x≤0.1, and notice what shape the graph is. Then use the Error Bound for Taylor
polynomials to find a formula for the maximum error,as a function of x, in this case: |E_1(x)|≤?.
(b)Let E_2=ln(x+1)−P_2(x)=ln(x+1)−(x−x^2/2). Choose a suitable range and graph E_2 for −0.1≤x≤0.1. Again, notice what shape the graph of E_2 is. Then use the Error Bound for Taylor polynomials to find a formula for the maximum error, as a function of x, in this case: |E_2(x)|≤?.
To investigate the error in the nth degree Taylor approximation to ln(x+1) around 0, we need to compare the actual function with the Taylor polynomial approximation.
(a) For the first degree Taylor approximation, the polynomial is P_1(x) = x. So, the error function is E_1(x) = ln(x+1) - x.
To graph E_1 for -0.1 ≤ x ≤ 0.1, you can follow these steps:
1. Open a graphing calculator or computer graphing software.
2. Set up the graphing range from -0.1 to 0.1 on the x-axis.
3. Plot E_1(x) = ln(x+1) - x on the y-axis.
4. The resulting graph will show the shape of E_1 for the given range.
By graphing E_1, you should observe that it resembles an oscillating curve.
To find a formula for the maximum error using the Error Bound for Taylor polynomials, we have the formula |E_1(x)| ≤ (M * |x-a|^(n+1)) / (n+1)!, where M is the maximum value of the (n+1)th derivative on the interval [a, x].
Since we are working with first-degree approximation, n = 1, and the interval is [-0.1, 0.1], we plug these values into the formula. The maximum value of the second derivative ln(x+1) is found to be M = 1/4.
|E_1(x)| ≤ (1/4 * |x - 0|^(1+1)) / (1+1)! = 1/4 |x^2 / 2|
Thus, the formula for the maximum error in this case is |E_1(x)| ≤ |x^2 / 8|
(b) For the second degree Taylor approximation, the polynomial is P_2(x) = x - x^2/2. So, the error function is E_2(x) = ln(x+1) - (x - x^2/2).
To graph E_2 for -0.1 ≤ x ≤ 0.1, you can follow the same steps as in part (a).
By graphing E_2, you should observe that it resembles a parabolic curve.
Using the Error Bound for Taylor polynomials, we have the formula |E_2(x)| ≤ (M * |x-a|^(n+1)) / (n+1)!, where M is the maximum value of the (n+1)th derivative on the interval [a, x].
Since we are working with second-degree approximation, n = 2, and the interval is [-0.1, 0.1], we plug these values into the formula. The maximum value of the third derivative ln(x+1) is found to be M = 1/8.
|E_2(x)| ≤ (1/8 * |x - 0|^(2+1)) / (2+1)! = 1/8 |x^3 / 6|
Thus, the formula for the maximum error in this case is |E_2(x)| ≤ |x^3 / 48|