By Idun Reiten, Sverre O. Smalø, Øyvind Solberg
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Extra resources for Algebras and Modules One
95 × 10−4 , so our estimate is a tad high, but certainly it is within the range of acceptable estimation. 4. Construct a linear interpolating polynomial to the function f (x) = x−1 using x0 = 21 and x1 = 1 as the nodes. What is the upper bound on the error over the interval [ 21 , 1], according to the error estimate? √ 5. Repeat the above for f (x) = x, using the interval [ 14 , 1]. Solution: The polynomial is p1 (x) = 1−x x − 1/4 (1) + (1/2) = (2x + 1)/3. 140625. 04. 6. Repeat the above for f (x) = x1/3 , using the interval [ 81 , 1].
What value do you now get for y8 ≈ y(1)? 827207570 4. 25 to compute approximate solution values for y = et−y , y(0) = −1. 7353256638? 5. 20. What value do you now get for y5 ≈ y(1)? 2, t0 = 0, and y0 = −1. 7945216786. 6. 125. What value do you now get for y8 ≈ y(1)? 7. 0625 to compute approximate solution values over the interval 0 ≤ t ≤ 1 for the initial value problem y = t − y, y(0) = 2, which has exact solution y(t) = 3e−t + t − 1. Plot your approximate solution as a function of t, and plot the error as a function of t.
Then we estimate the error using the suggested device for approximating the second derivative. 8960417333. Now, the error is bounded according to |Γ(x) − p1 (x)| ≤ 1 (x1 − x0 )2 max |(Γ(t)) | 8 where the maximum is taken over the interval [x0 , x1 ]. We don’t have a formula for Γ(x), so we can’t get one for the second derivative. 049... 917... 049... 00131... 95 × 10−4 , so our estimate is a tad high, but certainly it is within the range of acceptable estimation. 4. Construct a linear interpolating polynomial to the function f (x) = x−1 using x0 = 21 and x1 = 1 as the nodes.
Algebras and Modules One by Idun Reiten, Sverre O. Smalø, Øyvind Solberg