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Consider the function π of π₯ equals cos of π₯.
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Find the Taylor series expansion of π of π₯ equals cos of π₯ at π₯ equals π and write the Taylor series expansion of π of π₯ in sigma notation.
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And we recall that the Taylor series expansion of a function about π₯ equals π is given by π of π₯ equals π of π plus π prime of π over one factorial times π₯ minus π plus π double prime of π over two factorial times π₯ minus π all squared and so on.
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In this question, our function is cos of π₯.
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And weβre looking to find the Taylor series expansion at π₯ equals π.
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So weβre going to let π be equal to π.
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And so we see that π of π₯ here is equal to π of π plus π prime of π over one factorial times π₯ minus π and so on.
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Now, we can quite easily evaluate π of π.
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We would substitute π into the function cos of π₯.
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But what about π prime of π, π double prime of π, and so on?
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Well, weβre going to differentiate our function with respect to π₯.
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We might recall the first derivative of cos of π₯ to be negative sin of π₯.
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Then to get the second derivative, we differentiate negative sin π₯ with respect to π₯ and we get negative cos π₯.
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Differentiating once more, we find that π triple prime of π₯ is sin π₯.
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And in fact, weβll repeat this process one more time because when we differentiate sin π₯, we get cos π₯ again.
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And you might notice, we have a cycle.
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The fifth derivative of π will be negative sin π₯, the sixth derivative will be negative cos π₯, and so on.
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And letβs use all of these to evaluate π of π, π prime of π, π double prime of π, and so on and look for a pattern.
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π of π is cos π, which is negative one.
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π prime of π is negative sin π, which is zero.
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π double prime of π is negative cos π, which is one.
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And π triple prime of π is sin π, which is once again zero.
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It, of course, follows that the fourth derivative of π will be cos of π again, which is negative one.
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Letβs substitute all of these back into our Taylor series expansion.
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Now of course, every other term is going to be equal to zero as weβve seen.
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And so we find the Taylor series expansion to be negative one plus a half times π₯ minus π all squared minus one over 24 times π₯ minus π to the fourth power plus one over 720 times π₯ minus π to the sixth power and so on.
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The second part of this question asks us to write the Taylor series expansion of π of π₯ in sigma notation.
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And so letβs see if thereβs a way we can find a pattern.
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We saw that π prime of π was equal to zero, while π triple prime of π is equal to zero.
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And extending the pattern, we would see that the fifth derivative, the seventh derivative, and so on would also be equal to zero.
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And so our derivative alternate between negative one, one, negative one, and so on.
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Similarly, our denominators are factorials of ascending even numbers each time this even number is the same as the exponent.
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So letβs define that even number as two π.
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Then the denominator is two π factorial and the exponent of π₯ minus π is two π.
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To achieve alternating powers of negative one, we write negative one times π plus one.
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That means when π is zero, negative one to the power of one is negative one, when π is one, negative one to the power of two gives us one, and so on.
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And so we find that π of π₯ equals the sum of negative one to the power of π plus one times π₯ minus π to the power of two π over two π factorial for values of π between zero and infinity.