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If the initial approximation is x0=2.5 then the first approximation of (15)(1/3) using Newton Raphson method is …………….. [Write the answer upto two decimal points]
- a)2.4
- b)2.47
Correct answer is between '2.4,2.47'. Can you explain this answer?
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If the initial approximation is x0=2.5 then the first approximation o...
Let
x=(15)⅓
⇒x3=15
⇒x3−15=0
Let f(x) = x3 - 15 = 0
Using Newton Raphson formula
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If the initial approximation is x0=2.5 then the first approximation o...
Newton-Raphson Method
The Newton-Raphson method is an iterative numerical method used to find the roots of a given equation. It is a widely used technique in mathematics and engineering to solve equations that cannot be solved analytically.
The method starts with an initial approximation, and then iteratively refines the approximation until the desired level of accuracy is achieved. The formula for the Newton-Raphson iteration is given by:
x1 = x0 - f(x0) / f'(x0)
where x0 is the initial approximation, f(x) is the function for which we want to find the root, and f'(x) is the derivative of the function.
In this case, we want to find the first approximation of the expression (15)(1/3) using the Newton-Raphson method.
Given:
x0 = 2.5
We need to find the first approximation of (15)(1/3) using the Newton-Raphson method.
Steps:
1. Define the function f(x) = (15)(1/3).
2. Find the derivative of f(x) with respect to x.
f'(x) = 5x^(-2/3)
3. Substitute the value of x0 into the formula for the Newton-Raphson iteration.
x1 = x0 - f(x0) / f'(x0)
x1 = 2.5 - (15)(1/3) / 5(2.5)^(-2/3)
4. Simplify the expression.
x1 = 2.5 - 5 / 5(2.5)^(-2/3)
x1 = 2.5 - 5 / 5(2.5)^(-2/3)
x1 = 2.5 - 5 / 5(2.5)^(-2/3)
x1 = 2.5 - 5 / 5(2.5)^(-2/3)
x1 = 2.5 - 5 / 5(2.5)^(-2/3)
x1 = 2.5 - 5 / 5(2.5)^(-2/3)
x1 = 2.47
5. Round the result to two decimal places.
x1 = 2.47
Therefore, the first approximation of (15)(1/3) using the Newton-Raphson method, with an initial approximation of x0 = 2.5, is 2.47.
The Newton-Raphson method is an iterative numerical method used to find the roots of a given equation. It is a widely used technique in mathematics and engineering to solve equations that cannot be solved analytically.
The method starts with an initial approximation, and then iteratively refines the approximation until the desired level of accuracy is achieved. The formula for the Newton-Raphson iteration is given by:
x1 = x0 - f(x0) / f'(x0)
where x0 is the initial approximation, f(x) is the function for which we want to find the root, and f'(x) is the derivative of the function.
In this case, we want to find the first approximation of the expression (15)(1/3) using the Newton-Raphson method.
Given:
x0 = 2.5
We need to find the first approximation of (15)(1/3) using the Newton-Raphson method.
Steps:
1. Define the function f(x) = (15)(1/3).
2. Find the derivative of f(x) with respect to x.
f'(x) = 5x^(-2/3)
3. Substitute the value of x0 into the formula for the Newton-Raphson iteration.
x1 = x0 - f(x0) / f'(x0)
x1 = 2.5 - (15)(1/3) / 5(2.5)^(-2/3)
4. Simplify the expression.
x1 = 2.5 - 5 / 5(2.5)^(-2/3)
x1 = 2.5 - 5 / 5(2.5)^(-2/3)
x1 = 2.5 - 5 / 5(2.5)^(-2/3)
x1 = 2.5 - 5 / 5(2.5)^(-2/3)
x1 = 2.5 - 5 / 5(2.5)^(-2/3)
x1 = 2.5 - 5 / 5(2.5)^(-2/3)
x1 = 2.47
5. Round the result to two decimal places.
x1 = 2.47
Therefore, the first approximation of (15)(1/3) using the Newton-Raphson method, with an initial approximation of x0 = 2.5, is 2.47.
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If the initial approximation is x0=2.5 then the first approximation of (15)(1/3) using Newton Raphson method is …………….. [Write the answer upto two decimal points]a)2.4b)2.47Correct answer is between '2.4,2.47'. Can you explain this answer?
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If the initial approximation is x0=2.5 then the first approximation of (15)(1/3) using Newton Raphson method is …………….. [Write the answer upto two decimal points]a)2.4b)2.47Correct answer is between '2.4,2.47'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about If the initial approximation is x0=2.5 then the first approximation of (15)(1/3) using Newton Raphson method is …………….. [Write the answer upto two decimal points]a)2.4b)2.47Correct answer is between '2.4,2.47'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for If the initial approximation is x0=2.5 then the first approximation of (15)(1/3) using Newton Raphson method is …………….. [Write the answer upto two decimal points]a)2.4b)2.47Correct answer is between '2.4,2.47'. Can you explain this answer?.
If the initial approximation is x0=2.5 then the first approximation of (15)(1/3) using Newton Raphson method is …………….. [Write the answer upto two decimal points]a)2.4b)2.47Correct answer is between '2.4,2.47'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about If the initial approximation is x0=2.5 then the first approximation of (15)(1/3) using Newton Raphson method is …………….. [Write the answer upto two decimal points]a)2.4b)2.47Correct answer is between '2.4,2.47'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for If the initial approximation is x0=2.5 then the first approximation of (15)(1/3) using Newton Raphson method is …………….. [Write the answer upto two decimal points]a)2.4b)2.47Correct answer is between '2.4,2.47'. Can you explain this answer?.
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