00:01
Okay, we're given that our error bound, ln of 1 .3 minus tn, sorry, excuse me, of 1 .3 is less than equal to 10 to the power of negative 4, at a is equal to 1.
00:23
We're asked to use our error bound to find the maximum value of n such that the given inequality is satisfied.
00:31
Okay, so we know that ln, the taylor polynomial of ln of x is equal to tn of x is equal to x minus 1, minus 1 over 2, x minus 1 to the power of 2 plus dot dot, dot, our general formula is negative 1 to the power of n minus 1 over n times x minus 1 to the power of n.
01:04
Okay, so what we can do is we can start out with n is equal to 1 test that if that doesn't work we go on to n is equal to 2 if that doesn't work we keep on going on until this inequality is satisfied okay, so if we have ln of 1 .3 minus t n or t1 of x that means we have t1 of x is equal to 1 of x at 1 .3 it's equal to this first term over here.
01:39
So we get 1 .3 minus 1.
01:42
That gives me 0 .3.
01:46
Okay, so we get 0 .3.
01:48
So ln of 1 .3 minus 0 .3 gives me, let's see.
01:58
Let's put this into my calculator.
02:02
And i get 3 .76 times 10 to the power of negative 2.
02:12
Is that less than or equal to 10 to we have negative 4? let's go on, so let's use ln of 1 .3, this is n is equal to 1.
02:26
Let's use n is equal to 2.
02:28
So this is t1 of 1 .3, 0 .3.
02:35
So i just need to add this additional term here.
02:39
That's minus 1 .3 minus 1 .3, 3, 3, 3 of 2, which is 0 .09 divided by 2.
02:47
Excuse me, 0 .045, so 2 of 1 .3 is equal to 0 .3 minus 0 .045, that gives me 0 .25.
03:06
Plugging that into here, 0 .255, subtracting ln of 1 .3 minus 0 .255.
03:15
I get 7 .36 times 10 to the power of 1, 2, 3, that's not less than or equal to 10 to the power of negative 4.
03:28
Okay, now for t3 of 1 .3, i just add to my t2 of 1 .3 term.
03:43
Just put at 3, so let's see, so we get negative 1 to the power of 3 minus 1, which is 2, and 1 .3 minus 1, which is 0 .3 to the power of 3 over 3.
04:02
This gives me 0 .2.
04:06
Okay...