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Consider the nonlinear equation f(x)=0 which we know has a solution x in the interval [0,1] (e.g. by using the Intermediate Value Theorem). We would like to find a numerical approximation of the solution using the bisection method or the NewtonRaphson Method. Check all statements below that are true regarding the application of these two methods. Generally, we can expect that the approximation obtained with the Newton-Raphson method will be more accurate than the approximation made using the bisection method. The bisection method is an iterative method while the Newton-Raphson method is not. The Newton-Raphson method always shows quadratic convergence. The bisection method is guaranteed to converge while the Newton-Raphson method is not. The bisection method always shows linear convergence. In the context of a linear systems of equations Ax=b, we defined a condition number for the matrix A. We say that this linear system is ill-conditioned if the condition number of A is zero. the condition number of A is very large. the condition number of A is negative the condition number of A is close to1. the condition number of A is very small.. If a quantity y is expected to depend on a quantity x according to a power law, i.e. yaxb and an experiment or computation gives us pairs (xi,yi)(i=1,,N) then these data fall approximately on a straight line if the horizontal scale is and the vertical scale is The upper triangular nn matrix A has elements Aij=vi2+wj,ij, where v,wRn. Consider the following pseudo-code for constructing this matrix: - Input: arrays v and w with n elements of type float - Initialize A as empty n by n array - For i=1,n do - For j=i,n do - A[i,j]=v[i]v[i]+w[j] - end - end - Output: n by n array A What is the FLOP count for this algorithm? n2+3nn2+n21n221(n2+n) Suppose we would like to solve the linear system Ax=b, where: A=1.01.01.00.00.03.01.92.112.0,b=1.01.04.0 and we find the following numerical approximation for the solution x=1.0000230.9999710.000019 Write a script (Python code) named Quest 6 . py that computes and prints out the maximal relative error of the approximate solution x. Assume x is as given above; i.e. do not recompute x, and assume that you do not know the exact solution. In the space provided below, write out your answer: keep at least 4 decimal places in your answer and do NOT use scientific notation. For instance, if your answer is 4.8321568103 then enter: "0.0048". Consider the following system of linear equations: 1.2x1+0.1x21.1x3=0.10.9x11.1x2+0.2x3=0.30.3x1+2.1x20.9x3=1.2 Write a script (Python code) named Quest 7. py that computes and prints out an approximate solution to this system. You can use whatever method that you wish. In the space provided below, write out the first component of the final approximate solution. Keep at least 6 decimal places in your answer and do NOT use scientific notation. For instance, if your answer is 4.8321568103 then enter: " 0.004832 ". Consider the nonlinear equation: excos(x)x1=0 Write a script (Python code) called Quest8. py that uses the Newton-Raphson method to find the approximate solution to the given equation to a tolerance of less than 1010, with a starting guess of x(0)=1.0 In your script, print out the approximate error and residual on each iteration. In the space provided below, write out the final approximate solution. Keep at least 8 decimal places in your answer and do NOT use scientific notation. For instance, if your answer is 4.832156812103 then enter: "0.00483215". Consider the nonlinear equation f(x)=0 which we know has a solution x in the interval [0,1] (e.g. by using the Intermediate Value Theorem). We would like to find a numerical approximation of the solution using the bisection method or the NewtonRaphson Method. Check all statements below that are true regarding the application of these two methods. Generally, we can expect that the approximation obtained with the Newton-Raphson method will be more accurate than the approximation made using the bisection method. The bisection method is an iterative method while the Newton-Raphson method is not. The Newton-Raphson method always shows quadratic convergence. The bisection method is guaranteed to converge while the Newton-Raphson method is not. The bisection method always shows linear convergence. In the context of a linear systems of equations Ax=b, we defined a condition number for the matrix A. We say that this linear system is ill-conditioned if the condition number of A is zero. the condition number of A is very large. the condition number of A is negative the condition number of A is close to1. the condition number of A is very small.. If a quantity y is expected to depend on a quantity x according to a power law, i.e. yaxb and an experiment or computation gives us pairs (xi,yi)(i=1,,N) then these data fall approximately on a straight line if the horizontal scale is and the vertical scale is The upper triangular nn matrix A has elements Aij=vi2+wj,ij, where v,wRn. Consider the following pseudo-code for constructing this matrix: - Input: arrays v and w with n elements of type float - Initialize A as empty n by n array - For i=1,n do - For j=i,n do - A[i,j]=v[i]v[i]+w[j] - end - end - Output: n by n array A What is the FLOP count for this algorithm? n2+3nn2+n21n221(n2+n) Suppose we would like to solve the linear system Ax=b, where: A=1.01.01.00.00.03.01.92.112.0,b=1.01.04.0 and we find the following numerical approximation for the solution x=1.0000230.9999710.000019 Write a script (Python code) named Quest 6 . py that computes and prints out the maximal relative error of the approximate solution x. Assume x is as given above; i.e. do not recompute x, and assume that you do not know the exact solution. In the space provided below, write out your answer: keep at least 4 decimal places in your answer and do NOT use scientific notation. For instance, if your answer is 4.8321568103 then enter: "0.0048". Consider the following system of linear equations: 1.2x1+0.1x21.1x3=0.10.9x11.1x2+0.2x3=0.30.3x1+2.1x20.9x3=1.2 Write a script (Python code) named Quest 7. py that computes and prints out an approximate solution to this system. You can use whatever method that you wish. In the space provided below, write out the first component of the final approximate solution. Keep at least 6 decimal places in your answer and do NOT use scientific notation. For instance, if your answer is 4.8321568103 then enter: " 0.004832 ". Consider the nonlinear equation: excos(x)x1=0 Write a script (Python code) called Quest8. py that uses the Newton-Raphson method to find the approximate solution to the given equation to a tolerance of less than 1010, with a starting guess of x(0)=1.0 In your script, print out the approximate error and residual on each iteration. In the space provided below, write out the final approximate solution. Keep at least 8 decimal places in your answer and do NOT use scientific notation. For instance, if your answer is 4.832156812103 then enter: "0.00483215