In addition to writing new facts for February 29, the twins also need to add another image in the middle of their flip-book anim
ation. The animation sequence is of Dorothy's house from the Wizard of Oz as it is being carried over the rainbow by a tornado. The house in the February 28 drawing has been rotated to create the house in the March 1 drawing. Carlos believes that he can get from the February 28 drawing to the March 1 drawing by reflecting the February 28 drawing, and then reflecting it again.
A. What convinces you that the February 29 image is a reflection of the February 28 image about the given line of reflection?
B. What convinces you that the March 1 image is a reflection of the February 29 image about the given line of reflection?
C. What convinces you that the two reflections together complete a rotation between the February 28 and March 1 images?
A. What convinces you that the February 29 image is a reflection of the February 28 image about the given line of reflection?
B. What convinces you that the March 1 image is a reflection of the February 29 image about the given line of reflection?
C. What convinces you that the two reflections together complete a rotation between the February 28 and March 1 images?
1 answer:
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(a) See the first attachment for a graph. This graphing calculator displays roots to 3 decimal places. (The third attachment shows a different graphing calculator and 10 significant digits.)
(b) In the table of the first attachment, the column headed by g(x) gives iterations of Newton's Method. (For Newton's method, it is convenient to let the calculator's derivative function compute the derivative f'(x) of the function f(x). We have defined g(x) = x - f(x)/f'(x).) The result of the 3rd iteration is ...
... x ≈ 3.0473167
(c) The function h(x₁, x₂) computes iterations using the secant method. The results for three iterations of that method are shown below the table in the attachment. The result of the 3rd iteration is ...
... x ≈ 3.2291234
(d) The function h(x, x+0.01) computes the modified secant method as required by the problem statement. The result of the 3rd iteration is ...
... x ≈ 3.0477377
(e) Using <em>Mathematica</em>, the roots are found to be as shown in the second attachment. The highest root is about ...
... x ≈ 3.0466805180
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<em>Comment on these methods</em>
Newton's method can have convergence problems if the starting point is not sufficiently close to the root. A graphing calculator that gives a 3-digit approximation (or better) can help avoid this issue. For the calculator used here, the output of "g(x)" is computed even as the input is typed, so one can simply copy the function output to the input to get a 12-significant digit approximation of the root as fast as you can type it.
The "modified" secant method is a variation of the secant method that does not require two values of the function to start with. Instead, it uses a value of x that is "close" to the one given. For our purpose here, we can use the same h(x1, x2) for both methods, with a different x2 for the modified method.
We have defined h(x1, x2) = x1 - f(x1)(f(x1)-f(x2))/(x1 -x2).
