**Unformatted text preview: **0. Hence
T3,0 (x) = 1 + 0(x − 0) + −1
0
(x − 0)2 + (x − 0)3
2!
3! x2
2
= T2,0 (x) = 1− We also have that
T4,0 (x) = 1 + 0(x − 0) +
= 1− DE Math 128 −1
0
1
(x − 0)2 + (x − 0)3 + (x − 0)4
2!
3!
4! x2 x4
+
2
24
319 (B. Forrest)2 4.5. Introduction to Taylor Series CHAPTER 4. Sequences and Series and
T5,0 (x) = 1 + 0(x − 0) + 0
1
0
−1
(x − 0)2 + (x − 0)3 + (x − 0)4 + (x − 0)5
2!
3!
4!
5! x2 x4
+
2
24
= T4,0 (x)
= 1− An important observation to make is that not all of these polynomials are distinct. In
fact, T0,0 (x) = T1,0 (x), T2,0 (x) = T3,0 (x) and T4,0 (x) = T5,0 (x). In general, this equality of
diﬀerent order Taylor polynomials happens when one of the derivatives is 0 at x = a. (In
this example at x = 0.) This can be easily seen by observing that for any n
Tn+1,a (x) = Tn,a (x) + f (n+1) (a)
(x − a)n+1
(n + 1)! so if f (n+1) (a) = 0, we get Tn+1,a (x) = Tn,a (x).
The diagram below shows cos(x) and its Taylor polynomials up to degree 5. You will notice
that there are only four distinct graphs. In the next example, we wi...

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