A Taylor polynomial for (lnx)/x

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Question from Dave, a student:dave1

I have a series problem that I cannot solve. The problem asks for you to compute a Taylor polynomial Tn(x) for f(x) = (lnx)/x. I calculated this poly out to T5(x) and attempted to use this to identify a pattern and create a series in order to calculate Tn(x). However, the coefficients on the numerator out to F5prime(x) are as follows: 1, -3, 11, -50, 274... Ok, so the negative is an easy fix -> (-1)^n-1. But the other coefficients are stumping me. I can't see any sort of pattern there and I've tried every trick I know. Is there another way to go about this?

I also attempted to solve this as a Maclaurin geometric series using f'(x) and then integrating the series. I find the derivative as f'(x) = 1-lnx/x^2 = (x^2/1-lnx)^-1 = 1/x^2(1/1-lnx)^-1. This can then be incorporated as a geometric series (1/1-x) = ∑x^n)

But I get stuck when I try to incorporate 1/x^2 into the summation of ∑(lnx)^-n. How would I reconcile the x^2 with my lnx and bring that into the summation?

Thanks!

Dave,

There is no surprise that you might run into trouble with your approach: It is not possible to expand your function about x = 0 -- as x approaches 0 from the positive side, f(x) goes to -(infinity). You must therefore choose another value of x to expand about. The usual choice is x = 1, in which case the Taylor polynomial for ln x in terms of x-1 is very simple to write out. (It will be a series in powers of x-1.) Alternatively, you can look it up in tables. It will represent ln x for all x between 0 and 2. To get the expansion for f(x) = (ln x)/x, I would write out the expansion for 1/x about x-1 (it's just the derivative of ln x), then multiply the two series together.

Chris

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