Unformatted text preview: =0 Hence
∞ cos(x) ∼
n=0 f (n) (0) n
x
n! 0x −1x2 0x3 1x4
+
+
+
+ ···
1!
2!
3!
4!
x2 x4 x6
= 1−
+
−
+ ···
2!
4!
6!
∞
x 2k
(−1)k
=
(2k )!
k=0
= 1+ A similar calculation shows that
x3 x5 x7
+
−
+ ···
3!
5!
7!
∞
x2k+1
=
(−1)k
(2k + 1)!
k=0 sin(x) ∼ x − The major problem that remains is to determine if
∞ (−1)k cos(x) =
k=0 and if ∞ (−1)k sin(x) =
k=0 4.5.1 x 2k
(2k )! x 2k +
?
(2k +)! Introduction to Taylor Polynomials
and Approximation Recall that if f (x) is diﬀerentiable at x = a, then if x ∼ a
=
f (x) − f (a)
f (a) ∼
=
x−a
Cross multiplying gives DE Math 128 f (a)(x − a) ∼ f (x) − f (a)
=
312 (B. Forrest)2 CHAPTER 4. Sequences and Series 4.5. Introduction to Taylor Series and ﬁnally that
f (x) ∼ f (a) + f (a)(x − a)
=
This leads us to deﬁne the linear approximation to f (x) at x = a to be the function
La (x) = f (a) + f (a)(x − a).
The geometrical signiﬁcance of the linear approximation is that it graph is the tangent line
to the graph of f (x) through the point (a, f (a)). The linear approximation has the following two important properties:
1. La (a) = f (a).
2. La (a) = f (a).
In fact, amongst all polynomials of degree at most 1, that is function...
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 Spring '09
 Math, Derivative, Power Series, Taylor Series, Sequences And Series, Taylor's theorem, B. Forrest

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