Therefore if we let c1 f a as before then we will get

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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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