I don't even know what a Maclaurin series is, but it shore do look pretty. Maybe I'm in the wrong community. At least I'm not the guy who wrote an essay in Latin for the free-response portion of the Calc AP exam.
A side note: it looks like the numerator in the summation is just (2n - 1)!.
However, I don't think this is correct. The original function is asymptotic at x = -13, and I just don't see that reflected in the Maclaurin series.
I think for this general type of function, the Maclaurin series is easily given by 13^(1.5 - n) multiplied by -1/2 choose (n - 2). (That is, the binomial coefficient (-1/2, n - 2).) You may want to verify with some type of graphing software for small n (3 or 4) and close to x = 0.
Now that I think about it, I may have spoken too quickly. First, I think your answer is very close (it's close to the explicit form of the binomial coefficients). Second, I don't think Taylor series ever handle asymptotes well, so that wasn't exactly a good reason for me to discredit your series.
Comments 3
Reply
However, I don't think this is correct. The original function is asymptotic at x = -13, and I just don't see that reflected in the Maclaurin series.
I think for this general type of function, the Maclaurin series is easily given by 13^(1.5 - n) multiplied by -1/2 choose (n - 2). (That is, the binomial coefficient (-1/2, n - 2).) You may want to verify with some type of graphing software for small n (3 or 4) and close to x = 0.
Reply
Reply
Leave a comment