While finding the roots of f(x)=x2−4=0 using Newton - Raphson method, initial value of (x, x1 = 1). If the value obtained after first iteration is x2, find [100. x2], where [ ] is the greatest integer function.
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While finding the roots of f(x)=x2−4=0 using Newton - Raphson method, initial value of (x, x1 = 1). If the value obtained after first iteration is x2, find [100. x2], where [ ] is the greatest integer function.
The Newton-Raphson method uses an iterative process to approach one root of a function. The specific root that the process locates depends on the initial, arbitrarily chosen x-value.
xn+1=xn−f(xn)f′(xn)
Here, xn is the current known x-value, (f(xn)) represents the value of the function at xn, and f'(xn) is the derivative (slope) at xn. (xn+1) represents the next x-value that you are trying to find.
Here, we have f(x)=x2−2 and (xn = 1)\) f′(x)=2x ⇒f(xn)=f(1)=−1) (f′(xn)=f′(1)=2) ⇒x2=x1−f(x1)f′(x1) =1−−12=1.5, 100.x2=150 ([150] = 150)
The Newton-Raphson method uses an iterative process to approach one root of a function. The specific root that the process locates depends on the initial, arbitrarily chosen x-value.
xn+1=xn−f(xn)f′(xn)
Here, xn is the current known x-value, (f(xn)) represents the value of the function at xn, and f'(xn) is the derivative (slope) at xn. (xn+1) represents the next x-value that you are trying to find.
Here, we have f(x)=x2−2 and (xn = 1)\) f′(x)=2x ⇒f(xn)=f(1)=−1) (f′(xn)=f′(1)=2) ⇒x2=x1−f(x1)f′(x1) =1−−12=1.5, 100.x2=150 ([150] = 150)
