00:01
We will find the nth taylor polynomial of the function 1 over 1 plus x at x equals 0 and compute p4 of 0 .1 and compare the result with f at 0 .1.
00:16
Okay so we have the function f of x equal 1 over 1 plus x which can be written as 1 plus x to the negative 1.
00:28
So we find the derivatives of f to generalize the formula for the nth derivative.
00:37
We get the first derivative is, using this expression, is negative 1 times 1 plus x to the negative 1 minus 1 is negative 2 times derivative of the base 1 plus x, but that derivative is 1.
00:52
So the chain rule gives us a multiplication by 1, so we are not going to say it anymore.
01:00
We know we got to use or apply the chain rule, but it's going to be a multiplication by 1, which doesn't change anything.
01:09
Okay, so we get this, which is, in fact, negative 1 plus x to the negative 2, or negative 1 over 1 plus x squared.
01:21
But we are going to use this expression to find the next derivative.
01:26
The second derivative of f is plus 2.
01:32
1 plus x to the negative 2 times 1 which is derivative of the base and that is 2 over 1 plus x squared let's find the third derivative the third derivative is using this expression here sorry i made a mistake here because we got to subtract one from the exponent we have over here that is negative 3 sorry okay so now we negative 3 times 2 1 plus x to the negative 3 minus 1 is negative 4 times the derivative of the base which is 1 and we get negative 3 times 2 over 1 plus x to the 4th we recognize the numerator is negative 3 factorial okay now let's do one more the 4th derivative of x is doing use of this expression is plus 4 times 3 times 2 times 1 plus x to the negative 4 minus 1 is negative 5 and that is 4 times 2 over 1 plus x to the 5th and now we can generalize any derivative of f for any k in the natural numbers or even okay let's say that first we have that the kth derivative of f at x is so we see that first of all we have a sign that is alternating negative here positive here negative here positive here and so the sign depends on the order of the derivative for example the first derivative is negative the second is positive, the third is negative.
03:50
So we can say it's negative 1 to the kth, because when k is even, we will get a positive derivative.
03:58
So, for example, 2, which is an even order of differentiation, we get a positive derivative.
04:07
And when we have 1, for example, we have a negative derivative.
04:10
So it's negative 1 to the case.
04:12
Then the numerator is factorial of the order of the derivative.
04:16
Effectively we have indeed three times two for the third derivative two times one for the second four times three times two for the fourth derivative so is k factorial over one plus x and now the power or the exponent of one plus x is one plus the order of the derivative that is k plus one and that's for k in the natural numbers.
04:48
But if we put zero here, which corresponds to the function itself, we'll get negative 1 to the 0 is 1, 0 factorial is 1, and 1 plus x to the 0 plus 1 is 1 plus 1, that is 1 plus x.
05:03
We will get 1 over 1 plus x, which is just a function we have.
05:08
That is, this can be written for any integer, the k greater than or equal to zero.
05:24
That is, we can include the function itself in this formula, and that's a good thing, because we don't need to separate that when we write the taylor polynomial.
05:36
Then the nth taylor polynomial is p .n...