Open yasmensarhan27 opened 8 months ago
It appears to me that you did the 2nD task @tirthbha.
Your results for the root of tan is 90 degrees or 1.59 rad. and this value is a minima that your function is stuck in.
I had this value before but I started with first guess of pi/2 +0.0001 to avoid this problem so it has found the root at pi which is reasonable.
for the bisection, I've chosen X0 and X1 as pi/2+0.0001 and 3pi/2 - 0.0001 to avoid being stuck in the (n+1) pi/2 values
@laserlab what do you think prof.??
choosing the right start condition helps. That’s why I showed the animated plot to easily check on it
It appears to me that you did the 2nD task @tirthbha.
Your results for the root of tan is 90 degrees or 1.59 rad. and this value is a minima that your function is stuck in.
I had this value before but I started with first guess of pi/2 +0.0001 to avoid this problem so it has found the root at pi which is reasonable.
for the bisection, I've chosen X0 and X1 as pi/2+0.0001 and 3pi/2 - 0.0001 to avoid being stuck in the (n+1) pi/2 values
@laserlab what do you think prof.??