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ch4aika [34]
4 years ago
12

Alculate two iterations of Newton's Method to approximate a zero of the function using the given initial guess. (Round your answ

ers to three decimal places.)
f(x) = x5 − 5, x1 = 1.6
Chemistry
2 answers:
Sav [38]3 years ago
7 0

Answer:

x_2=1.433

x_3=1.383

Explanation:

Hello,

In this case, for the Newton-Raphson method, we use the following equation in order to show the iterations <em>i</em>:

x_{i+1}=x_i-\frac{f(x_i)}{f'(x_i)}

Hence the derivative of the given function is:

f'(x)=5x^4

In such a way, the first iteration:

x_{1+1}=1.6-\frac{(1.6)^5-5}{5(1.6)^4}\\x_2=1.433

Then the second iteration:

x_{2+1}=1.433-\frac{(1.433)^5-5}{5(1.433)^4}\\x_3=1.383

Best regards.

nataly862011 [7]3 years ago
6 0

Answer:

x_{1} = 1.6, x_{2} = 1.433, x_{3} = 1.384

Explanation:

The expression for the approximation via Newton's Method has the following form:

x_{n+1} = x_{n} - \frac{f(x_{n})}{f'(x_{n})}

The function and its derivative are, respectively:

f(x_{n}) = x^{5}-5

f'(x_{n})= 5\cdot x ^{4}

After substituting the known variable, the Newton's expression is left as follows:

x_{n+1} =x_{n}- \frac{x_{n}^{5}-5}{5\cdot x_{n}^{4}}

The first two iterations are presented herein:

x_{2} = 1.6 - \frac{1.6^{5}-5}{5\cdot (1.6)^{4}}

x_{2} = 1.433

x_{3} = 1.433 - \frac{1.433^{5}-5}{5\cdot (1.433)^{4}}

x_{3} = 1.384

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