The latter. More generally, if ##f: \mathbb{R} \to \mathbb{R}## is not a polynomial but still a smooth function, then ##x_0## is called a root of order ##n \ge 1## if all coefficients of orders strictly smaller than ##n## vanish in a Taylor expansion of ##f## around ##x_0## while the ##n##th order coefficient itself is non-zero.Raghav Gupta said:Does a double root means to find two roots or a root which is repeated two times?
I agree with you and don't see any double roots here.Raghav Gupta said:I checked analytically with my calculator and this equation has 1 real root and 2 complex roots.
So how to proceed?
But suppose we don't know analytically before and we apply this formula,Krylov said:The latter. More generally, if ##f: \mathbb{R} \to \mathbb{R}## is not a polynomial but still a smooth function, then ##x_0## is called a root of order ##n \ge 1## if all coefficients of orders strictly smaller than ##n## vanish in a Taylor expansion of ##f## around ##x_0## while the ##n##th order coefficient itself is non-zero.
I agree with you and don't see any double roots here.
When you start "Newton" close to a root that is not simple, you are going to run into trouble because ##f'(x_0) = 0##. I don't understand why you would put a "2" in front of the quotient, this is not going to help. "Newton" can only be used when ##f## is regular at ##x_0##. Incidentally, you got a sign incorrect, it should beRaghav Gupta said:But suppose we don't know analytically before and we apply this formula,
##x_{n+1}=x_n +2 \frac{f(x_n)}{f'(x_n)}##
Since multiplicity is 2.
What we can infer from that?
Eventually it will also give a root.
How many iterations to do and what initial guess value to take
Raghav Gupta said:But suppose we don't know analytically before and we apply this formula,
## x_{n+1}=x_n +2 \frac{f(x_n)}{f'(x_n)} ##
Since multiplicity is 2.
What we can infer from that?
Eventually it will also give a root.
How many iterations to do and what initial guess value to take?
I used Wolfram Alpha to graph the function.Raghav Gupta said:Homework Statement
Find a double root of the equation ## f(x)= x^3+15.87x+24.34 ## correct to 3 decimal places.
Homework Equations
By Newton Raphson's Formula,
## x_{n+1}=x_n + \frac{f(x_n)}{f'(x_n)} ##
##f'(x)=3x^2+15.87 ##
The Attempt at a Solution
Does a double root means to find two roots or a root which is repeated two times?
I checked analytically with my calculator and this equation has 1 real root and 2 complex roots.
So how to proceed?
SammyS said:I used Wolfram Alpha to graph the function.
Clearly there is only one real root. Of course this is apparent if one considers the first derivative, which is positive definite.
Changing the sign of the linear term gives ##\ x^3-15.87x+24.34 \ ##, which still has only one real root, but does have a minimum of ≈ 0.006 .
Finally, subtracting 0.006 gives the function ##\ g(x)=x^3-15.87x+24.334 \ ##. It appears that g(x) has two roots, one of which is a double root.
Maybe there are two typos in the posted problem.
The concept of Double root of equation with Newton Raphson is a numerical method used to find the roots of a given equation. It is based on the Newton-Raphson iterative method, which uses an initial guess to approximate the root and then refines the guess until a desired level of accuracy is achieved.
The Newton Raphson method works by using the derivative of the equation to find the slope of the curve at a given point. This slope is then used to determine the next approximation for the root. The process is repeated until the desired level of accuracy is achieved or until the root is found.
One advantage of using Newton Raphson method for finding double roots is that it is a fast and efficient method. It also has a high convergence rate, meaning it can find the root with a small number of iterations. Additionally, it can handle complex equations and multiple roots.
The Newton Raphson method may not always converge to the root or may converge to a wrong root if the initial guess is not close enough to the actual root. It also requires knowledge of the derivative of the equation, which may not always be easy to obtain. Furthermore, it may not work well for equations with multiple roots that are close together.
The algorithm for finding double roots using Newton Raphson involves repeatedly using the formula xn+1 = xn - f(xn)/f'(xn) until the desired level of accuracy is achieved. This process is continued until the root is found or the maximum number of iterations is reached. The initial guess and the derivative of the equation are required inputs for the algorithm.