00:01
In this question, we want to find the taylor series for the function f of x equals x to the one -third power, where c equals 8.
00:07
So let's remember the formula for a taylor series.
00:10
We have f of c plus f prime of c over 1 factorial times the quantity of x minus c plus f double prime at c over 2 factorial times x minus c being squared plus f triple prime at c over 3 factorial times x minus c being cubed.
00:40
Now in this case, my c is 8, so we're going to have f of 8 plus f prime of 8 over 1 factorial times the quantity of x minus 8 plus f double prime of 8 over 2 factorial times the quantity of x minus 8 being squared plus f triple prime at 8 over 3 factorial times the quantity of x minus 8 being cubed.
01:15
And i should note that there is plus dot dot dot on the end of this.
01:18
This does go on forever and ever.
01:20
So we're going to need to get some derivatives of f of x equals x to the one -third power and evaluate them at 8.
01:28
So if f of x is x to the one -third power, first of all, my first derivative, f prime of x, would be one -third x to the negative two -thirds power.
01:42
Then my second derivative, f double prime of x, would be negative two -ninths x being raised to the negative five -thirds power.
01:54
And then my third derivative, f triple prime of x, is 10 over 27 times x to the negative eight -thirds power.
02:06
So now we're going to evaluate each of these at x equals 8.
02:10
So let's see, f of 8, that's 8 to the one -third power, and 8 to the one -third power should be the cube root of 8, which should be 2.
02:21
Indeed it is.
02:23
And then i want f prime of 8.
02:27
So let's get this.
02:28
It's going to be one -third times 8 to the negative two -thirds.
02:32
So we have one -third times 8 being raised to the negative two -thirds power.
02:41
And if i convert that into a fraction, i'm getting one -twelfth...