00:01
In this problem we are asked to expand sign effects using taylor's polynomial up to degree of 5.
00:08
We know that taylor's expansion of a series is of the form like this and we can approximate a given function of effects around a point a using derivatives and actually higher order derivatives.
00:23
So since we are needing to expand it up to order of 5, we need f prime effect.
00:31
F, double prime of x, f triple prime of x, f4 of x, as well as f5 of x.
00:46
All right.
00:47
We know that f of x is sine of x, so the first derivative of that would be cosine of x.
00:52
Derrative of cosine would be negative sine of x.
00:55
Derrative of negative sine is negative cosine of x.
00:58
Therewith 2 of negative cosine is sine of x.
01:01
And it has a cyclic pattern, as you can see now, derivative of x.
01:05
Fifth there with it would again be cosine of x.
01:07
Now let's say we'll let this function around 0.
01:13
So 1a is 0.
01:18
We have f prime of 0 as 1.
01:22
F double prime of 0 is 0.
01:25
F triple prime of 0 is negative 1.
01:29
F4 of 0 is 0 since sine 0 is 0 and cosine 0 is 1.
01:34
So f5 of 0 5 derivative of f would also be 1.
01:42
All right.
01:43
Now we have pretty much everything that we need.
01:46
All we need to do is to take these values and then plug it in to this equation.
01:52
Then we can estimate, approximate, sine of x as sine of 0 plus f prime of 0 is 1.
02:01
So 1 times x minus 0 plus x plus f second derivative is 0, so that is 0, plus third derivative is negative 1, so that is negative 1 of x minus 0, cube divided by 3 factorial.
02:19
4th derivative is 0, so we have plus 0, plus 5th derivative is 1.
02:29
So we have 1 times x minus 0 to the 5 divide by 5 factorial.
02:34
We know that sign 0 is also 0, so we can write this one as x minus x cube over 3 factorial plus x to the 5 factorial.
02:50
From this, what we see is this...