# Apply Newton-Raphson method to find the solutions to

Apply Newton-Raphson method to find the solutions to the equation $$x^3-5x=0$$ starting with an initial guess of $$x_0 = 1$$.

While using Newton Raphson method, the value doesn't converge to a specific number. Rather, every iteration either gives $$1$$ or $$-1$$. Why does this happen? Any graphical interpretation for this?

• All you have to do is draw a sketch. There is a similar question here, which has some pictures. – TonyK May 4 '19 at 12:16

For the graph of $$f(x)=x^3-5x$$, we have $$f'(x)=3x^2-5$$ $$f(1)=-4$$ $$f'(-1)=4$$ $$f'(1)=-2=f'(-1)$$

The tangent line at $$x=1$$ is $$y+4=-2(x-1)$$ which is $$y=-2x-2$$.

The tangent line at $$x=-1$$ is $$y-4=-2(x+1)$$ which is $$y=-2x+2$$

Geometrically, what has happened is you are trapped in the following cycle. Starting from $$(1,-4)$$, by traveling along the tangent, we intercept the $$x$$-axis at $$x=-1$$. From $$(-1,4)$$, by trveling along the tangent, we intercept the $$x$$-axis at $$x=1$$.

• $f'(1)=f'(-1)=-2$. Isn't it? – pi-π May 4 '19 at 12:39
• yes, that's a typo, thanks. – Siong Thye Goh May 4 '19 at 12:40

The basic reason for this is that the graph of $$f(x) = x^3 -5x$$ has some symmetry. Specifically, $$f(x)$$ is an odd function.

The intuitive visualization of the Newton-Raphson method is that you are finding the tangent line to the curve at the initial guess point, and then finding the x-intercept of this line. You are then just repeating this, using the x-intercept you found as the next ‘guess’ point.

The first tangent line we find is given by $$-2x -2$$, which intersects the x-axis at $$x=-1$$, and because of the symmetry of the graph (the slope at $$-x$$ is the same as the slope at $$x$$), the tangent line here will intersect back at $$x=1$$. So we are stuck in a never-ending loop.