00:01
So in this problem, we're going to use jacoby's method with an initial approximation of 0.
00:17
Right, this is x1, x2.
00:23
And work this until we get agreement to four decimal places.
00:31
So we get agreement within 0 .001.
00:40
Okay, so we saw the first equation for x1 so x1 for the next iteration will be add the point five x2 to the other side so 1 plus 0 .5 x2 divided by the 4 and a half so you get x1 on by itself okay then down here for x2 for the next iteration so we move the x1 to the other side so i have my 1 minus 1 minus x1 divided by the negative 3 and a half and all the negatives will cancel each other out right negative or negative is positive okay so here we go iteration 1 zero zero this gives me x1 is well x2 is 0 this is 1 over 4 and a half which is in my calculator, one divided by, that's 0 .2222.
02:12
And x2, well, x1 is zero, so this is one over three and a half.
02:23
So that's 0 .2857.
02:29
Okay, so iteration two, i use what i had up here, right, 0 .222, 0 .282 .0 .2 .0 .285.
02:40
So now x1 i simply plug in what i have up here, right? so i have one plus a half of x2.
02:59
So this is one plus 0 .5 times 0 .2857 over 4 .5.
03:14
Okay.
03:15
So 0 .2857 times 0 .5 plus 1 divided by 4 .5.
03:38
This is 0 .25396, which is 4 .0 when it rounds off.
03:47
The next 2 is 1 plus x1 over 3 .5.
03:53
So this is 1 plus 0 .222 over 3 .5.
04:01
Okay, so 1 .222 divided by 3 .5 in my calculator gives me 0 .3492.
04:19
Okay, so then iteration 3.
04:26
We're using 0 .2540 and 0 .3492.
04:33
Okay, so x1 is 1 plus a half times.
04:41
Going back to original formulas plus our x2 0 .3492 over 4 .5.
04:56
So 0 .3492 times 0 .5 plus 1 divided by 4 .5.
05:06
This is 0 .2610.
05:11
Getting there, aren't we? and x2 is 1 plus x1 0 .2540 over 3 .5 .2 divided by 3 .5.
05:31
This is 0 .3583.
05:38
Okay.
05:40
We just keep going, don't we? iteration 4.
05:43
And by this time we're using 0 .2610 and 0 .358.
05:49
So then x1 is 1 plus 0 .5 times 0 .3583 over 4 .5.
06:05
3583 times .5 divided by 4 .5.
06:14
This is 0 .2620 and x2.
06:22
1 plus x1, 0 .2610 over 3 .2610 over 3 .5.
06:29
Half so 1 .261 divided by 3 .5 is 0 .3603.
06:43
Okay so now these two are point 0 .01 apart but these two not there yet.
07:00
We're getting close though.
07:02
This is 0 .002 isn't it? okay so we got one more iteration here.
07:09
I iteration 5.
07:13
So i have 0 .2620 and 0 .3603.
07:22
So x1 is 1 plus 1 half times 0 .3583 over 4 .5.
07:35
So 0 .3583 times 0 .5 plus 1 divided by 4 .5.
07:45
That's 0 .2620.
07:50
So that matches.
07:54
And then x2 is going to be 1 plus 0 .2620 over 3 .5...