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Consider the following equation:
[tex]3^x=2^{-x}+4[/tex]
If the equation is solved using successive approximation, select the value of [tex]x[/tex] after the following number of iterations.
1 iteration: [tex]x=[/tex] $\square$
2 iteration: [tex]x=[/tex] $\square$
3 iteration: [tex]x=[/tex] $\square$
Answer (1)
We are given the equation 3 x = 2 − x + 4 and will solve it using successive approximation with Newton-Raphson method.
The Newton-Raphson method uses the formula x n + 1 = x n − f ′ ( x n ) f ( x n ) , where f ( x ) = 3 x − 2 − x − 4 and f ′ ( x ) = 3 x ln ( 3 ) + 2 − x ln ( 2 ) .
After 1 iteration, x ≈ 1.4118 .
After 2 iterations, x ≈ 1.3480 .
After 3 iterations, x ≈ 1.3308 .
Therefore, the values of x after 1, 2, and 3 iterations are approximately 1.4118 , 1.3480 , 1.3308 .
Explanation
Understanding the Problem We are given the equation 3 x = 2 − x + 4 and asked to solve it using successive approximation. This means we will make an initial guess for x and then iteratively refine our guess using the Newton-Raphson method. The Newton-Raphson method uses the formula x n + 1 = x n − f ′ ( x n ) f ( x n ) , where f ( x ) = 3 x − 2 − x − 4 and f ′ ( x ) is its derivative.
Finding the Derivative and Initial Guess First, we need to find the derivative of f ( x ) .
f ( x ) = 3 x − 2 − x − 4 f ′ ( x ) = 3 x ln ( 3 ) + 2 − x ln ( 2 ) We will start with an initial guess of x 0 = 1 .
Iteration 1 Now we perform the iterations: Iteration 1: x 1 = x 0 − f ′ ( x 0 ) f ( x 0 ) f ( 1 ) = 3 1 − 2 − 1 − 4 = 3 − 0.5 − 4 = − 1.5 f ′ ( 1 ) = 3 1 ln ( 3 ) + 2 − 1 ln ( 2 ) = 3 ln ( 3 ) + 0.5 ln ( 2 ) ≈ 3 ( 1.0986 ) + 0.5 ( 0.6931 ) ≈ 3.2958 + 0.34655 ≈ 3.64235 x 1 = 1 − 3.64235 − 1.5 ≈ 1 + 0.4118 ≈ 1.4118
Iteration 2 Iteration 2: x 2 = x 1 − f ′ ( x 1 ) f ( x 1 ) f ( 1.4118 ) = 3 1.4118 − 2 − 1.4118 − 4 ≈ 4.7215 − 0.3741 − 4 ≈ 0.3474 f ′ ( 1.4118 ) = 3 1.4118 ln ( 3 ) + 2 − 1.4118 ln ( 2 ) ≈ 4.7215 ln ( 3 ) + 0.3741 ln ( 2 ) ≈ 4.7215 ( 1.0986 ) + 0.3741 ( 0.6931 ) ≈ 5.1870 + 0.2593 ≈ 5.4463 x 2 = 1.4118 − 5.4463 0.3474 ≈ 1.4118 − 0.0638 ≈ 1.3480
Iteration 3 Iteration 3: x 3 = x 2 − f ′ ( x 2 ) f ( x 2 ) f ( 1.3480 ) = 3 1.3480 − 2 − 1.3480 − 4 ≈ 4.4806 − 0.3912 − 4 ≈ 0.0894 f ′ ( 1.3480 ) = 3 1.3480 ln ( 3 ) + 2 − 1.3480 ln ( 2 ) ≈ 4.4806 ln ( 3 ) + 0.3912 ln ( 2 ) ≈ 4.4806 ( 1.0986 ) + 0.3912 ( 0.6931 ) ≈ 4.9224 + 0.2709 ≈ 5.1933 x 3 = 1.3480 − 5.1933 0.0894 ≈ 1.3480 − 0.0172 ≈ 1.3308
Final Results After 1 iteration, x ≈ 1.4118 .
After 2 iterations, x ≈ 1.3480 .
After 3 iterations, x ≈ 1.3308 .
Examples
Successive approximation, like the Newton-Raphson method used here, is crucial in many real-world applications where finding exact solutions is impossible. For instance, in engineering, it helps optimize designs by iteratively refining parameters to meet specific criteria, such as minimizing material usage or maximizing efficiency. Similarly, in finance, it's used to determine the fair value of complex financial instruments by iteratively solving equations that lack closed-form solutions. This iterative approach allows professionals to achieve increasingly accurate results, making it an indispensable tool in various fields.
