Is there a particular point on the sine curve where if you make this your initial guess, the root jumping continues forever? (Except the tops of the curves at pi/2, 3pi/2, etc where the tangent is parallel to the x-axis)
@TimpBizkit I do not think that would happen. It would eventually converge on some root, because they repeated throughout the [-infinity,infinity] domain.
For Newton Raphson method, effectively you are poking upwards (or downwards) from the x-axis (f(x)=0 line) to the function (graph curve), then drawing a tangent back to the x-axis, then moving vertically again until you reach the function graph curve again, and then drawing another tangent. Depending on the initial guess and the function in question, sometimes this method can accidentally cause you to skip away from finding a root, (a point where the function graph meets the x-axis.)
@tshiong87 Points of inflection are where the second derivative is zero (stationary points the first derivative is zero), so differentiate x^3 + cos(x) twice. Once: 3x^2 - sin(x). Twice: 6x - cos(x).
Now solve 6x - cos(x) = 0 using Bisection or Newton Raphson method (I recommend Bisection) (sketch the graph 6x - cos(x) and guess near where there is likely a zero.) I know this is 6 months old but no one else appears to have chimed in so I thought I'd help.
Is there a particular point on the sine curve where if you make this your initial guess, the root jumping continues forever? (Except the tops of the curves at pi/2, 3pi/2, etc where the tangent is parallel to the x-axis)
TimpBizkit 1 year ago
@TimpBizkit I do not think that would happen. It would eventually converge on some root, because they repeated throughout the [-infinity,infinity] domain.
numericalmethodsguy 1 year ago
For Newton Raphson method, effectively you are poking upwards (or downwards) from the x-axis (f(x)=0 line) to the function (graph curve), then drawing a tangent back to the x-axis, then moving vertically again until you reach the function graph curve again, and then drawing another tangent. Depending on the initial guess and the function in question, sometimes this method can accidentally cause you to skip away from finding a root, (a point where the function graph meets the x-axis.)
TimpBizkit 1 year ago
Sir, can u help me give me hints on finding coordinates of the inflection point of the curve f(x)= x^3 + cos x by Numerical method?
Ans:( 0.164419, 0.990958)
What initial number i should guess? thank ;)
tshiong87 1 year ago
@tshiong87 Points of inflection are where the second derivative is zero (stationary points the first derivative is zero), so differentiate x^3 + cos(x) twice. Once: 3x^2 - sin(x). Twice: 6x - cos(x).
Now solve 6x - cos(x) = 0 using Bisection or Newton Raphson method (I recommend Bisection) (sketch the graph 6x - cos(x) and guess near where there is likely a zero.) I know this is 6 months old but no one else appears to have chimed in so I thought I'd help.
TimpBizkit 1 year ago
great explanation!!!
metalking0206 1 year ago