Lesson 28a - Approximating Taylor Series

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Unformatted text preview: ion. Example 2 Show that the Maclaurin Series of Sin(x) converges to Sin(x) for every possible x: Using the remainder our limit, we want to make sure that f ( k 1) ( z )( x a ) k 1 lim 0 k (k 1)! f ( k 1) ( z )( x) k 1 lim k (k 1)! We know that the k+1’th derivative evaluated at z will be one of the following functions: sin(z), cos(z), ‐sin(z), ‐cos(z). No matter which trig function we have we get: ‐1 ≤ f(k+1)(z) ≤ 1 ( x) k 1 f ( k 1) ( z )( x) k 1 ( x) k 1 lim lim This means that we can write: lim k ( k 1)! k k ( k 1)! (k 1)! Again, we know that the both limits will go to 0, so by squeeze theroem we have...
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