![Answers Given that table x 012 3 15. I 9=f(1) 12 14 15.]. f(3) =2 f.cx) = f (2) + f(3) = f(2) (-2) 3-2 = !!. t .(-4-10) (0-2)](http://img.homeworklib.com/questions/dfc84ac0-d59f-11ea-8375-7b1e168325b6.png?x-oss-process=image/resize,w_560)
Exercise 6: Given the table of the function f(x)-2" 2 X 0 3 2 f(x) 1 2 4 8 a) Write down the Newton polynomials P1(x), P2(x), Pa(x). b) Evaluate f(2.5) by using Pa(x). c) Obtain a bound for the errors E1(x), E2(x), Es(x) Exercise 7: Consider f(x)- In(x) use the following formula to answer the given questions '(x) +16-30f+16f,- 12h a) Derive the numerical differentiation formula using Taylor Series and find the truncation error b) Approximate f'(1.2) with h-0.05...
(3.2) Consider the data given in the following table 05 1 15 f(x) 0 2 0 6 1 2 20 (4) (a) Approximate f with a function of the form q (x) = kxm (4) (b) Approximate f with a function of the form g2(x) = be Which approximation between q and g2 1s more appropriate for the given data? Justify your (3) (c) answer < In, and a piecewise cubic polynomial Consider a set of points (I,) Such that...
Consider the following data table: 0 2i = 0.2 0.4 f(xi) = 2 2.018 2.104 2.306 0.6 0.2 and 23=0.4 is The linear Lagrange interpolator L1,1 (2) used to linearly interpolate between data points 12 (Chop after 2 decimal places) None of the above. -2.50x+0.20 -5.00x+2.00 -5.00x+2.00 5.00x-1.00 Consider the following data table: 2 Ti = 0 0.2 0.4 0.6 f(x) = 2.018 2.104 2.306 0.2 and 23 = 0.4, the value obtained at 2=0.3 is Using Lagrange linear interpolation...
You are given the table below. 16 20 4 8 12 X f(x) 12 2417 6 30 Use the table and n = 4 to estimate the following. Because the data is not monotone (only increasing or only decreasing), you should sketch a possible graph and draw the rectangles to ensure you are using the appropriate values for a lower estimate and an upper estimate. 20 f(x)dx lower estimate upper estimate Estimate the area of the region under the curve...
The following y vs. x data is given x y 1 2.25|3.7 15.1 4.256 13.7 15.1 The data is fit by a quadratic spline given by ( 2.85 + 1.4x, 1< x < 2.25 f(x) = { cx? + dx +e, 2.25 < x < 3.7 (fx2 + gx + h, 3.7 < x < 5.1 The value of d most nearly is
FITTING A QUADRATIC PARABOLA 8-11 Fit a parabola (7) to the points (x, y). Using MATLAB 10. hr] Worker's time on duty, y [sec] = His/her reaction time, (t, y) (1, 2.0), (2, 1.78), (3,1.90), (4, 2.35), (5, 2.70)
FITTING A QUADRATIC PARABOLA 8-11 Fit a parabola (7) to the points (x, y). Using MATLAB 10. hr] Worker's time on duty, y [sec] = His/her reaction time, (t, y) (1, 2.0), (2, 1.78), (3,1.90), (4, 2.35), (5, 2.70)
Consider the data: X- 1 Y- 6 3 14 5 7 2 20 9 11 10 18 13 15 26 22 (a) Calculate the correlation between X and Y. 0.7399 (b) What percent of the variation in Y can be attributed to X? (Round to a whole percent) 55 % (c) Obtain the equation of the regression line for these data y = X +
Assuming that the first spline is linear in quadratic spline interpolation, the number of unknowns in the splines for 10 data points is QUESTION 2 The following y vs. x data is given 1 225 3.7 5.1 4.25 68615.1 The data is fit by quadratic spline interpolants given by AX) - 2.85 +1.4x, 1 5X5 2.25 AXX2dx +2.25 SX5 3.7 A x) = fx2gx+, 3.7<*5.1 The value of d most nearly is QUESTION 3 The following y vs. x data...
Consider the following table of data points:
Using least squares fitting, find the polynomial Q(x) of degree
2 that fits the data points given in the table above. Approximate
f(0.3) using Q(0.3).
f(x) i Xi 0 0.000 1.00000 1 0.125 0.98450 2 0.250 0.93941 0.375 0.86882 4 0.500 0.77880 5 0.625 0.67663 6 0.750 0.56978 0.875 0.46504 8 1.000 0.36788
f(x) i Xi 0 0.000 1.00000 1 0.125 0.98450 2 0.250 0.93941 0.375 0.86882 4 0.500 0.77880 5 0.625 0.67663...
6. (20%) Consider two random variables X and Y with the joint PMF given in Table 2. Table 2: Joint PMF of X and Y Y =0Y 1 X 0 X 1 0 (a) (5%) Find the PMF of X and PMF of Y. (b) (5%) Find EX, EY, Var(X), Var(Y (c) (10%)Find the MMSE estimator of X given Y, (M) for both Y 0 and Y 1