find the smallest positive root of the equation f(x)=2sinx+x-2=0 using the false positive method.
well, you know that
f(0) = -2
f(pi/2) = pi/2
So, start with those two values
a1 = 0, b1=pi/2
so, c1 = pi/2-(pi/2)(pi/2-0)/(pi/2-(-2)) = 0.88
f(c1)=f(0.88) = 0.421
Since f(c1) > 0,
a2 = a1, b2 = c1 = 0.421
I assume you can follow the method, as explained at
https://en.wikipedia.org/wiki/False_position_method
Come back if you get stuck, and show your calculations.
To find the smallest positive root of the equation f(x) = 2sin(x) + x - 2 = 0 using the false position method, follow these steps:
Step 1: Choose two initial points, x1 and x2, such that f(x1) < 0 and f(x2) > 0. Since we are looking for the smallest positive root, we can choose x1 = 0 and x2 = 1.
Step 2: Determine f(x1) and f(x2). Plug in the values of x1 and x2 into the equation f(x) = 2sin(x) + x - 2 and calculate the corresponding function values.
f(x1) = 2sin(0) + 0 - 2 = -2
f(x2) = 2sin(1) + 1 - 2 ≈ -0.1585
Step 3: Calculate the next approximation, x3. Use the false position formula:
x3 = x2 - (f(x2) * (x2 - x1)) / (f(x2) - f(x1))
In this case, x1 = 0, x2 = 1, f(x1) = -2, and f(x2) ≈ -0.1585. Plugging these values into the formula:
x3 = 1 - (-0.1585 * (1 - 0)) / (-0.1585 - (-2))
= 1 + (0.1585) / (1.8415)
≈ 1.0862
Step 4: Calculate f(x3) using the newly obtained x3.
f(x3) = 2sin(1.0862) + 1.0862 - 2
If f(x3) > 0, replace x2 with x3 and repeat steps 3 and 4 until you obtain a solution that's accurate enough (e.g., f(x3) ≈ 0).
Continue iterating until you converge to the smallest positive root of the equation f(x) = 2sin(x) + x - 2 = 0.