Taylor polynomials.

is called the Taylor polynomial of degree of the function at .

If , this polynomial is called the Mac-Laurin polynomial of degree of .

For any natural number , we have , thus .

Thus the MacLaurin series of is:

The Taylor polynomial of order at 0 of is therefore:

This theorem gives an estimation of the error when approximating a function by its Taylor polynomial of order .

- With the settings of Thm 6.3, suppose that there exist two positive numbers
and
such that for any
between
and
, the following inequality is verified:

Then, Lagrange's remainder verifies the inequality:

- If all these conditions hold for every natural number , then Taylor series converges and its sum is equal to .

Therefore and .

We have:

As all values of the cosine function are between and 1, the remainder verifies the inequality:

As , for all real , the MacLaurin series of the cosine function is convergent for every real number .

In Fig. 2 is a Maple display for MacLaurin polynomials of the cosine function (blue), of order 2 (yellow curve),4 (green curve) and 6 (red curve). Pay attention to the fact it seems that the larger the degree of the approximation, the better the approximation.

As , for all real , the MacLaurin series of the cosine function is convergent for every real number .

By the same way we can prove that all the MacLaurin series in 5.2 converge towards the given functions.

In Fig. 3 is a Maple display for MacLaurin polynomials of the sine function (in blue), of order 1 (in green),3 (in red) and 5 (in yellow).

Finally, we give in Fig. 4 a mathplot display for MacLaurin polynomials of the function such that of order 1 (curve ),2 (curve ) and 3 (curve ).

Studying the convergence domain of a Taylor series is a non trivial issue. Let us see a couple of examples. Take . The function is defined over . Its first MacLaurin polynomials are given as follows:

The MacLaurin series of the function is a geometric series whose ratio is equal to , thus it is convergent iff . It has no meaning (at least for us now) out of this interval. This situation is illustrated in Figure 5.

We can find a first partial explanation for this situation: at -1, the function is not defined. As the domain of convergence of the MacLaurin series is an interval centered at 0, it cannot be larger than .

Take now . This function is defined over the whole of .The first partial sums of its MacLaurin series are as follows:

The series is geometric and its ratio is equal to . Thus as previously the series is convergent iff . The graphs of the function and of these partial sums are displayed in Figure 6. The function is defined over the whole of . There is no opportunity to use an argument as the one used above for . We will wait until we learn the course in Complex Variables to have a more convincing argument (http://ndp.jct.ac.il/tutorials/complex/node51.html).

Noah Dana-Picard 2007-12-28