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SashulF [63]
3 years ago
5

F(x)=x^3+3

Mathematics
1 answer:
lawyer [7]3 years ago
5 0

Answer:

  (a)   -3/4

  (b)  -0.75

  (c)  -0.75

Step-by-step explanation:

It's a bit hard to tell what constitutes an "iteration" when using the bisection method to approximate a polynomial root. For the purpose here, we'll say one iteration consists of ...

  • evaluating the function at the midpoint of the bracketing interval
  • choosing a smaller bracketing interval
  • identifying the x-value known to be closest to the solution

Thus, the result of the iteration consists of a bracketing interval and the choice of one of the interval's ends as the solution approximation.

__

(a) We observe that the graphs intersect in the interval (-1, 0). For the first iteration, we evaluate f(x)-g(x) at x=-1/2. This tells us the solution is in the interval (-1, -1/2). The x-value closest to the root is x=-1/2.

For the second iteration, we evaluate the function f(x)-g(x) at x=-3/4. This tells us the solution is in the interval (-1, -3/4). The x-value closest to the root is x=-3/4.

For the third iteration, we evaluate the function f(x)-g(x) at x=-7/8. This tells us the solution is in the interval (-7/8, -3/4). The x-value closest to the root is x=-3/4.

__

(b) The graph tells us the solution is approximately 0.7549. Rounded to 2 decimal places, the solution is approximately 0.75.

__

(c) The above solution found after 3 iterations rounded to 2 decimal places is exactly 0.75.

__

See the attached table for function values.

_____

<em>Comment on bisection iteration</em>

Since you cut the interval containing the root in half with each iteration, you gain approximately one decimal place for each 3 iterations. When the function value is very nearly zero at one of the interval endpoints, it can take many more iterations to achieve a better result.

Here, it takes 4 more iterations before an x-value becomes closer to the solution (x≈-97/128). And it takes one more iteration to move the end of the interval away from -3/4. After these 5 more iterations (8 total), the solution is known to lie in the interval (-97/128, -193/256). The corresponding solution approximation is -193/256. It is still only correct to 2 decimal places.

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Calculating:

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\theta = tan^{-1}\frac{y}{x}

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d) (−2√3,−2,3)

\rho = \sqrt{(-2\sqrt{3} )^{2}+(-2)^{2}+3^{2}} = 5

\phi = cos^{-1}(\frac{3}{5})

Since x < 0, use 2nd option:

\theta = \pi + tan^{-1}\frac{1}{\sqrt{3} }

\theta = \pi + \frac{\pi}{6}

\theta = \frac{7\pi}{6}

Cilindrical coordinate describes a 3 dimension space: 2 distances (r and z) and 1 angle (θ). To express cartesian coordinates into cilindrical:

r=\sqrt{x^{2}+y^{2}}

Angle θ is the same as spherical coordinate;

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r=\sqrt{4^{2}+2^{2}} = \sqrt{20}

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z = -4

b) (0, 8, 15)

r=\sqrt{0^{2}+8^{2}} = 8

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z = 15

c) (√2,1,1)

r=\sqrt{(\sqrt{2} )^{2}+1^{2}} = \sqrt{3}

\theta = \frac{\pi}{3}

z = 1

d) (−2√3,−2,3)

r=\sqrt{(-2\sqrt{3} )^{2}+(-2)^{2}} = 4

\theta = \frac{7\pi}{6}

z = 3

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