your bounds for (i) doesn't seem correct. If we are using (Z_i+1) then we would stop at z_3, since when i = 2, Z_(i+1) = Z_(3), However we don't have x_3, y_3. So It would make more sense to bound i to be less than or equal to 1, not 2. when i = 1, Z_(i+1) = Z_2. Which would also be our final Z. if you set some variable k = i + 1, then you can bound k to being less than or equal to 2.
why do we let z_0 to be =0? (at 3:13)
I assumed Z0 = 1 instead of 0, and got the right answer. i don't really know why but it works.
Very helpful video thank you
Glad it was helpful…
Loves from Turkey
Thank you a lot..
Thanks so much. What a nice and wonderful Job....am so blessed. Thanks so much. You have my subscription. Notify me always with all.
You are most welcome and thank you too…
Awesome video! Thank you!
You are most welcome!
Nice
your bounds for (i) doesn't seem correct. If we are using (Z_i+1) then we would stop at z_3, since when i = 2, Z_(i+1) = Z_(3), However we don't have x_3, y_3. So It would make more sense to bound i to be less than or equal to 1, not 2. when i = 1, Z_(i+1) = Z_2. Which would also be our final Z. if you set some variable k = i + 1, then you can bound k to being less than or equal to 2.
Ma'am books name please
you teach very well but little slow,i watch at 2x but it is still slow.
and you donot covers cubic splines.
The method and everything is easy but the formula 😵
Tu yaha kya kr rha hai?