00:01
Okay, we're asked to find the maximum one polynomial of t, t2x for secant, for f of x is equal to secant h of x.
00:17
Okay, so we need to take the derivative two times, but we're also asked to find the maximum possible value of our error bound.
00:28
So we need to take the third derivative in order to bound our k.
00:33
Okay, so let's take the derivative of this three times.
00:40
So f prime of x, of sequence of x, is this negative tangents of x, secence of x, f double prime of x is equal to, well, negative 1 minus tangents squared of x, secants, h of x, plus tangents h squared of x, secant, h squared of x, secense h of h of h, well, 1 minus tangent squared of x is equal to sequence h squared of x.
01:14
So that says negative sequence cubed of x plus tangents square out of x sequence h of x.
01:24
And a third derivative is equal to, well, let's see.
01:33
We bring down the 3, so we have a negative 3...