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4.1 Taylor series

One series you have encountered before is Taylor’s series,

f(x) ={ \mathop{∑ }}_{n=0}^{∞}{f}^{(n)}(a){{(x − a)}^{n}\over n!} ,
(4.1)

where {f}^{(n)}(x) is the nth derivative of f. An example is the Taylor series of the cosine around x = 0 (i.e., a = 0),

\eqalignno{ & &\qquad &\mathop{cos}\nolimits (0) & = 1, & & & & & & \cr \mathop{cos}\nolimits '(x) & = −\mathop{sin}\nolimits (x), & &\mathop{cos}\nolimits '(0) & = 0, & & & & & & \cr {\mathop{cos}\nolimits }^{(2)}(x) & = −\mathop{cos}\nolimits (x), & &{\mathop{cos}\nolimits }^{(2)}(0) & = −1, & & \text{(4.2)} \cr {\mathop{cos}\nolimits }^{(3)}(x) & =\mathop{ sin}\nolimits (x), & &{\mathop{cos}\nolimits }^{(3)}(0) & = 0, & & & & & & \cr {\mathop{cos}\nolimits }^{(4)}(x) & =\mathop{ cos}\nolimits (x), & &{\mathop{cos}\nolimits }^{(4)}(0) & = 1. & & & & & & }

Notice that after four steps we are back where we started. We have thus found (using m = 2n in (4.1)) )

\mathop{cos}\nolimits x ={ \mathop{∑ }}_{m=0}^{∞}{{(−1)}^{m}\over (2m)!} {x}^{2m},
(4.3)

Question: Show that

\mathop{sin}\nolimits x ={ \mathop{∑ }}_{m=0}^{∞} {{(−1)}^{m}\over (2m + 1)!}{x}^{2m+1}.
(4.4)

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