Interesting consequence of differentation of inverse functions

[Nootropic]

BREAKTHROUGH!!!!!!!!!!!!!!!!!!!!!!!!

 

So while browsing through my calculus book in the power series section, I noticed a an integral involving y = x^x, with limits x = 0 and x = 1. I saw that this integral had a series representation of ((-1)^n-1)(n^n). This is of course interested me, knowing that someone had previously pioneered the taylor expansion of this function. My first thought was that this was an expansion about x =1, and attempted to reconstruct the original expression and then differentiate that to obtain the taylor series, but alas, that did not work (not quite sure why...probably made a mistake) and then I attempted to calculate the derivatives myself (for the umpteenth time), and got so sick of being tedious by the [massively long] fifth derivative that I just settled with the sequence 1,1,2,3 and 8, which contains evaluation up to the fourth derivative. I then turned to the computer, which was able to evaluate at least one more derivative (the fifth derivative is 10 at x = 1) and anything more than that was just too long for any derivative generator I found. So I remembered earlier I had seen, in another thread, an encyclopedia of integer sequences, so I figured, why not, and I entered the sequence up to ten, and the second sequence I found was exactly what I was looking for. How exciting!

 

The On-Line Encyclopedia of Integer Sequences

 

These numbers (the nth derivative of x^x evaluated at x = 1), are apparently called Lehmer-Comtet numbers, and I haven't been able to find much about them, but it begs for a investigation. Summer project! They certainly make creating a taylor series much easier, and much more aesthetic than my franken-series expansion. They have an interesting pattern of oscillation past 54. Hm...

[Nootropic]

So now I'm relatively confused. I can't remember where I heard about this, but I decided to check out what is called "Sophmore's dream" Sophomore's dream - Wikipedia, the free encyclopedia. Why it's called that, I do not know...but anyhow, this was the original series that Lambus had posted. What I'm confused about is how can you give a mclaurin series expansion for the function when it's first derivative is not even defined at x =0. Likewise, I saw a mclaurin series expansion for exp(x)/x in a textbook (which was then integrated), but exp(x)/x isn't defined at x = 0 either. However, I was check this out on my calculator, and I did notice that it appears that dividing the series for exp(x) by x did seem to converge to the function, though a proof is beyond me.

[Guest Lambus]

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[Nootropic]

So I have some correspondence with a professor and I asked him the same question and his reply certainly clears up a lot, at least for me. But here is his email verbatim:

 

You want to find int(exp(x)/x) dx. Now exp(x) is well defined and analytic at

0. You can find its maclaurin series which is sum (x^k)/k!. Now divide by x

(this has nothing to do with convergence at all) and you get sum (x^(k-1)/k!,

k = 0..infinity which you can write as 1/x + sum x^(k-1)/k!, k = 1..infinity.

Again this has nothing to do with convergence. It is simply splitting a sum

into two parts. But the second summand is convergent for all x. Thus it can be represented by a function, say g(x). So the problem becomes int 1/x +g(x) dx, which is the integral of a finite sum. You can split this integral of a finite sum into two parts and get the answer you have in Stewart.

 

Of course the Maclaurin series of exp(x)/x does not exist.

 

It would certainly seem to justify the integral of x^x represented as a the integral of the of the series representation of y = e^(x ln(x))

[Nootropic]

 

[math]f(x) = \frac{1}{x} \cdot e^x = \frac{1}{x} \cdot \sum_{n=0}^{\infty} \frac{e(x - 1)^n}{n!} = \frac{1}{x} + 1 + \frac{x}{2} + \frac{x^2}{6} + \frac{x^3}{24} + ... \frac{x^{n-1}}{n!} + ...[/math]

 

[math]f(x) = \frac{e^x}{x} = \frac{1}{x} + 1 + \frac{x}{2} + \frac{x^2}{6} + \frac{x^3}{24} + ... \frac{x^{n-1}}{n!} + ...[/math]

 

[math]f(x) = \frac{e^x}{x} = \sum_{n=0}^{\infty} \frac{x^{n-1}}{n!} \; \; \; x>0[/math]

 

[math]f(x) = \frac{e^x}{x} = \frac{1}{x} \cdot \sum_{n=0}^{\infty} \frac{e(x - 1)^n}{n!} = \sum_{n=0}^{\infty} \frac{x^{n-1}}{n!} \; \; \; x>0[/math]

 

 

[math]I(x) = \int_0^1 f(x) dx = \sum_{n=0}^{\infty} \int_0^1 \frac{x^{n-1}}{n!} dx = \sum_{n=0}^{\infty} \frac{1}{n \cdot n!} \; \; \; n>0[/math]

 

[math]I(x) = \sum_{n=0}^{\infty} \frac{1}{n \cdot n!} \; \; \; n>0[/math]

 

Negative, I cannot get this function to converge at 0.

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'A way to fix the problem with discontinuities in the taylor series'

 

Write an expansion for exp(x)/x about x = 1:

[math]g(x) = e + \frac{e(x - 1)^2}{2} - \frac{e(x - 1)^3}{3} + \frac{3e(x - 1)^4}{8} - ... + e(x - 1)^n \text{???} - ... [/math]

 

The expression becomes:

[math]g(x) = \sum_{n=0}^{\infty} e(x - 1)^n \text{???}[/math]

[math]h(x) = \sum_{n=1}^{1} \frac{x^{n-1}}{n!} = 1[/math]

 

Insert h(x) into expression g(x):

[math]f(x) = g(h(x)) = \sum_{n=0}^{\infty} e(1 - 1)^n \text{???} = \sum_{n=0}^{\infty} e(0)^n \text{???} = \text{Indeterminate}[/math]

 

Taylor Series:

[math]f(x) = \frac{e^x}{x} \neq \text{Indeterminate}[/math]

 

Reference:

Taylor series - Wikipedia, the free encyclopedia

Sophomore's_dream

Sophomore's Dream -- from Wolfram MathWorld

 

The integral for exp(x)/x does not converge at x = 0, but not because of where the indexed sum begins, because ln(x) has an infinite discontinuity at x = 0. However, the sum that is separate from ln(x) certainly converges. The taylor series expansion for exp(x)/x is incorrect; you have to divide the entires series by x. If you want to expand a composite function into a series, you have to insert the the entire series into the other series. There is a nice example at wikipedia. Taylor series - Wikipedia, the free encyclopedia

[Guest Lambus]

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[Nootropic]

Technically it's not a taylor series expansion, but sum splitting, as exp(x)/x does not have a taylor series expansion about x = 0. This can be done by by dividing any series about x = a for exp(x) by x, and then integrating. Actually, I can't recall what I thought was incorrect, though I'm not sure what you're trying to achieve by by by using composite functions.