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Problem 44 Hard Difficulty

Use a Maclaurin series in Table 1 to obtain the Maclaurin series for the given function.

$ f(x) = \left\{\begin{array} (\frac{x -\sin x}{x^3} \text{ if } x \not= 0 \\\\ \frac{1}{6} \text { if } x = 0 \end{array}\right. $


$\sum_{k=0}^{\infty} \frac{x^{2} k}{(2 k+3)}$


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Video Transcript

OK, so using Mecca lannisters in table want obtain the mechanism is forgiven function so X equals two ataman science over X cube Fxx now zero and one sixth. Even that's a cereal. Okay. Sold in the first expand son of ex. And if X equals two F miners, so expenses could be X minus. That's cute or through pictorial us. That's the five off of pictorial minus answers Simon to service Victorio and so on or through pictorial. And that's No. Zero, which is equal to so one north of material minus esque at square overcovered, factorial plus X to the four over. Senator Vittorio. So long which physicals too x to the power of two K. And here's to Kay plus three case from zero to infinity Took Mr Victorio actually and minus one to power of K. Let's check it out. So if Case zero and this is one, this is three in case one. This This man is one. This is Exc. That's where this is. This is private tutorial. So they scrapped end. Of course, we have the value at zero. So this is the case. One that at least none. Zero. Of course, if at zero, it becomes to one sixth check if X goes to zero. That's the limit coincides with our function battle as euro. So we're gonna find this is one or three Victoria, which is just one second. Yes, they agree with each other.

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