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**Unformatted text preview: **t we outlined before. We want p(a) = f (a), but
p(a) = c0 + c1 (a − a) + c2 (a − a)2 + c3 (a − a)3 = c0 , so we only have to let c0 = f (a).
We can diﬀerentiate to get
p (x) = c1 + 2c2 (x − a) + 3c3 (x − a)2
so that
p (a) = c1 + 2c2 (a − a) + 3c3 (a − a)2 = c1 .
Therefore, if we let c1 = f (a) as before, then we will get p (a) = f (a).
Diﬀerentiating again gives
p (x) = 2c2 + 3(2)c3 (x − a).
Therefore,
p (a) = 2c2 + 3(2)c3 (a − a) = 2c2 .
Just as before, if we let c2 = f (a)
,
2 we will get
p (a) = f (a). Finally, observe that
p (x) = 3(2)c3 = 3(2)(1)c3 = 3!c3
for all x, so if we want
p (a) = 3!c3 = f (a),
then we need only let c3 = f (a )
.
3! It follows that if
p(x) = f (a) + f (a)(x − a) + f (a)
f (a)
(x − a)2 +
(x − a)3 ,
2
3! then
1. p(a) = f (a),
2. p (a) = f (a),
3. p (a) = f (a),
4. p (a) = f (a).
DE Math 128 317 (B. Forrest)2 4.5. Introduction to Taylor Series CHAPTER 4. Sequences and Series In this case, we call p(x) the third degree Taylor polynomial centered at x = a and denote...

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