00:04
In this question, you also use tarrus here is to show that this expression is second order in h.
00:09
Well, basically, we're looking at h very small, obviously.
00:13
And what we need to do is actually to write x plus h, right, equals i'm righted as fx, right, and plus h times the derivative of f, right, x.
00:26
And then, of course, plus the second order i'm going to write as well, right, h squared, right? and the second word derivative, right, like this.
00:35
And similarly, i can write if x plus 2h, right, and i can read it as f used tally series, fx plus 2h, right, and f prime x, and then plus, now this 2h squared is actually get 2h squared, right, and if double prime, right, and of course these x here, and this x here as well.
00:59
Now you substitute everything in, you'll find the expression left hand is given by fx, by the right hand side you will find minus three fx.
01:08
But you see that there's fx here, right? and there's four fx here.
01:12
See they can't completely cancel each other.
01:13
We don't know what about this first term anymore, right? so how about the linear term? the inimity would have four, uh, four h f prime x.
01:24
Basically, right? that's what you would actually, uh, get, right? so you get a four h f prime.
01:32
Right, x, basically, right? and then you have to, but you have to get minus of this, you get actually, instead of getting there, you get actually one, right? so you get out 2 here.
01:45
And then, how about the squared, the edge of square term, you would have four of this, you get 2 h squared, and this you get 2 h squared as well...