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Orlov [11]
3 years ago
6

Evaluate the infinite geometric series 0.79 + 0.079 + 0.0079 + 0.00079 + 0.000079. Express your answer as a fraction with intege

r numerator and denominator.
Mathematics
1 answer:
tamaranim1 [39]3 years ago
3 0

Answer:

The fraction is 79/99

Step-by-step explanation:

We can express the given infinite geometric series as follows:

\frac{79}{10^{2} }+\frac{79}{10^{3} }+\frac{79}{79^{4} }+\frac{79}{10^{5} }+...

Which is an infinite geometric series with r=\frac{1}{10^{2} } that converges, then its sum is:

S=\frac{a}{1-r}

Where a is the first factor of the serie: a=\frac{79}{10^{2} }

Replacing values:

S=\frac{\frac{79}{10^{2} } }{1-\frac{1}{10^{2} } }  \\S=\frac{\frac{79}{10^{2} } }{\frac{99}{10^{2} } }\\S=\frac{79}{99}

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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.)

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(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 ...

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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).

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