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Section 10.16 : Taylor Series
1. Use one of the Taylor Series derived in the notes to determine the Taylor Series for \(f\left( x \right) = \cos \left( {4x} \right)\) about \(x = 0\).
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Start SolutionThere really isn’t all that much to do here for this problem. We are working with cosine and want the Taylor series about \(x = 0\) and so we can use the Taylor series for cosine derived in the notes to get,
\[\cos \left( {4x} \right) = \sum\limits_{n = 0}^\infty {\frac{{{{\left( { - 1} \right)}^n}{{\left( {4x} \right)}^{2n}}}}{{\left( {2n} \right)!}}} \] Show Step 2Now, recall the basic “rules” for the form of the series answer. We don’t want anything out in front of the series and we want a single \(x\) with a single exponent on it.
In this case we don’t have anything out in front of the series to worry about so all we need to do is use the basic exponent rules on the 4\(x\) term to get,
\[\cos \left( {4x} \right) = \sum\limits_{n = 0}^\infty {\frac{{{{\left( { - 1} \right)}^n}{4^{2n}}{x^{2n}}}}{{\left( {2n} \right)!}}} = \require{bbox} \bbox[2pt,border:1px solid black]{{\sum\limits_{n = 0}^\infty {\frac{{{{\left( { - 1} \right)}^n}{{16}^n}{x^{2n}}}}{{\left( {2n} \right)!}}} }}\]