Homework Answers
Answer:
As per the newtons's divided difference formula, the approximate value of function at x can be calculated using the following formula,
Here we need to compute value at 1.21
Add Answer to:
The following divided difference table has been constructed from measured values of f(x): ху f(x) 0 0. 0.121249 0.5 0.0...
Problem 6: Interpolation, least squares, and finite difference Consider the following data table: 2 = 0...
Problem 6: Interpolation, least squares, and finite difference Consider the following data table: 2 = 0 0.2 0.4 0.6 f(x) = 2 2.018 2.104 2.306 A cubic spline interpolation over these data is constructed as a cubic polynomial interpolating over all 4 data points. None of the above. is constructed by piece-wise locally cubic polynomial interpolation that maintains continuity of the first and second derivatives at the data points is constructed by plece-wise locally cubic polynomial interpolation that has discontinuous...
Consider the following data table: 0 2i = 0.2 0.4 f(xi) = 2 2.018 2.104 2.306...
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...
Problem 2: Interpolation, least squares, and finite difference Consider the following data table: 0 2 0.2...
Problem 2: Interpolation, least squares, and finite difference Consider the following data table: 0 2 0.2 2.018 ti f(x) = 0.4 2.104 0.6 2.306 Using Lagrange linear interpolation between data points 22 = 0.2 and as = 0.4, the value obtained at z = 0.3 is (Chop after 2 decimal places) فو واقع = 2.06 2.09 2.03 None of the above. 2.05
You are given the table below. 16 20 4 8 12 X f(x) 12 2417 6...
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...
it is one question it just has many parts ere is only a single correct answer...
it
is one question it just has many parts
ere is only a single correct answer to all these questions. Question 1 Problem 1: Interpolation, least squares, and finite difference Consider the following data table: 0.2 0 2 2 2.018 2 f() = 3 0.4 2.104 0.6) L2.306) The linear Lagrange interpolator L1,1() used to linearly interpolate between data points x = 0.2 and x3 -0.4 is (Chop after 2 decimal places) 0 -5.00x+2.00 0 5.00x+2.00 None of the above....
A function f(x) has a period 2t and function values over one cycle is shown in the table below. x...
*Fourier Series Expansion
A function f(x) has a period 2t and function values over one cycle is shown in the table below. x0 0 30 60 90 120 150 180 210 240 270 300 330 f(x) 4.1 5.5 9.3 10.4 5.2 3.3 4.1 3.3 5.2 10.4 9.3 5.5 Given that Σ,21f(xr) sin(%) DE1f(x) sin(%)-Σ 21f(xr) sin(3%) 0.Find the Fourier series expansion up to the third harmonic. All calculation must be correct to 3 decimal places.
A function f(x) has a...
Problem 3: Interpolation, least squares, and finite difference Consider the following data table: 2 = f(x)...
Problem 3: Interpolation, least squares, and finite difference Consider the following data table: 2 = f(x) = 0.2 0.4 0.6 2 2.018 2.104 2.306 The quadratic Lagrange interpolator L2,0 (2) used to quadratically interpolate between data points 2 = 0.22 = 0.2. and 2'3 = 0.4 is (Chop after 2 decimal places) 25.05x^2-10.25x -5.00x^2+7.50x+2.00 12.50x 2-7.50x+1.00 5.00x^2-1.00 None of the above.
1/A/Consider the following.:f(x) = x2 − 4x − 1 Complete the table. (Round your answers to four decimal places.) x 0.9...
1/A/Consider the following.:f(x) = x2 − 4x − 1
Complete the table. (Round your answers to four decimal
places.)
x
0.9
0.99
0.999
1
1.001
1.01
1.1
f(x)
?
Use the result to estimate the limit. Use a graphing utility to
graph the function to confirm your result. (Round your answer to
four decimal places. If an answer does not exist, enter DNE.)
lim x→1 (x2 − 4x − 1) = ?
Can you please explain in clear details,step by...
Problem 3: Interpolation, least squares, and finite difference Consider the following data table: 2 0 0.2...
Problem 3: Interpolation, least squares, and finite difference Consider the following data table: 2 0 0.2 0.4 0.6 f(x) = 2 2.018 2.104 2.306 3 The quadratic Lagrange interpolator L2,0() used to quadratically interpolate between data points #1 = 0.12 = 0.2, and as = 0.4 is (Chop after 2 decimal places) None of the above -5.00x^2+7.50x+2.00 5.00x^2-1.00 12.50x^2-7.50x+1.00 25.05x^2-10.25%
A function, f. has the following table of values: 4 6 X 0 f(x) 2 23...
A function, f. has the following table of values: 4 6 X 0 f(x) 2 23 43 -2 - 1 -4 1 Approximate] пом. Approximate f(x)d.r using a Riemann sum with 2 rectangles and midpoints. 6 0 -2 4 O2 Question 4(5 points) ✓ Saved firmaland itsdaluritius Р w o 19 MacBou
Problem 6: Interpolation, least squares, and finite difference Consider the following data table: 2 = 0 0.2 0.4 0.6 f(x) = 2 2.018 2.104 2.306 A cubic spline interpolation over these data is constructed as a cubic polynomial interpolating over all 4 data points. None of the above. is constructed by piece-wise locally cubic polynomial interpolation that maintains continuity of the first and second derivatives at the data points is constructed by plece-wise locally cubic polynomial interpolation that has discontinuous...
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...
Problem 2: Interpolation, least squares, and finite difference Consider the following data table: 0 2 0.2 2.018 ti f(x) = 0.4 2.104 0.6 2.306 Using Lagrange linear interpolation between data points 22 = 0.2 and as = 0.4, the value obtained at z = 0.3 is (Chop after 2 decimal places) فو واقع = 2.06 2.09 2.03 None of the above. 2.05
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...
it
is one question it just has many parts
ere is only a single correct answer to all these questions. Question 1 Problem 1: Interpolation, least squares, and finite difference Consider the following data table: 0.2 0 2 2 2.018 2 f() = 3 0.4 2.104 0.6) L2.306) The linear Lagrange interpolator L1,1() used to linearly interpolate between data points x = 0.2 and x3 -0.4 is (Chop after 2 decimal places) 0 -5.00x+2.00 0 5.00x+2.00 None of the above....
*Fourier Series Expansion
A function f(x) has a period 2t and function values over one cycle is shown in the table below. x0 0 30 60 90 120 150 180 210 240 270 300 330 f(x) 4.1 5.5 9.3 10.4 5.2 3.3 4.1 3.3 5.2 10.4 9.3 5.5 Given that Σ,21f(xr) sin(%) DE1f(x) sin(%)-Σ 21f(xr) sin(3%) 0.Find the Fourier series expansion up to the third harmonic. All calculation must be correct to 3 decimal places.
A function f(x) has a...
Problem 3: Interpolation, least squares, and finite difference Consider the following data table: 2 = f(x) = 0.2 0.4 0.6 2 2.018 2.104 2.306 The quadratic Lagrange interpolator L2,0 (2) used to quadratically interpolate between data points 2 = 0.22 = 0.2. and 2'3 = 0.4 is (Chop after 2 decimal places) 25.05x^2-10.25x -5.00x^2+7.50x+2.00 12.50x 2-7.50x+1.00 5.00x^2-1.00 None of the above.
1/A/Consider the following.:f(x) = x2 − 4x − 1
Complete the table. (Round your answers to four decimal
places.)
x
0.9
0.99
0.999
1
1.001
1.01
1.1
f(x)
?
Use the result to estimate the limit. Use a graphing utility to
graph the function to confirm your result. (Round your answer to
four decimal places. If an answer does not exist, enter DNE.)
lim x→1 (x2 − 4x − 1) = ?
Can you please explain in clear details,step by...
Problem 3: Interpolation, least squares, and finite difference Consider the following data table: 2 0 0.2 0.4 0.6 f(x) = 2 2.018 2.104 2.306 3 The quadratic Lagrange interpolator L2,0() used to quadratically interpolate between data points #1 = 0.12 = 0.2, and as = 0.4 is (Chop after 2 decimal places) None of the above -5.00x^2+7.50x+2.00 5.00x^2-1.00 12.50x^2-7.50x+1.00 25.05x^2-10.25%
A function, f. has the following table of values: 4 6 X 0 f(x) 2 23 43 -2 - 1 -4 1 Approximate] пом. Approximate f(x)d.r using a Riemann sum with 2 rectangles and midpoints. 6 0 -2 4 O2 Question 4(5 points) ✓ Saved firmaland itsdaluritius Р w o 19 MacBou
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