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Quiz: calculating a definite integral
01-03-2014, 01:29 PM
Post: #36
RE: Quiz: calculating a definite integral
(01-03-2014 12:47 PM)Thomas Klemm Wrote:  [quote='Bunuel66' pid='2007' dateline='1388749354']
From a strict mathematical standpoint, I'm a bit puzzled as the Taylor expansion used seems a bit unjustified

This is not the Taylor-series expansion:
1-x log(x)+\frac{1}{2} x^2 log^2(x)-\frac{1}{6} x^3 log^3(x)+\frac{1}{24} x^4 log^4(x)-\frac{1}{120} x^5 log^5(x)+\frac{1}{720} x^6 log^6(x)+O(x^7)
The coefficients are not constant but dependent on \(x\).
Great, but then what is the justification for this expansion if it is not based on a Taylor one?
It looks a lot like the expansion of exp(u) with u replaced by xlog(x). My view is that it is the Taylor expansion of exp(u) around 0 with u taking the value given by xlog(x), which is numerically correct. But I still miss the justification for getting an equality when integrating on both sides. This equality would have been justified if we had built the Taylor expansion properly.
Just as an intuition, if we accept the above serie, as we have an identity for any value of x and as they are both continuous, it is probably feasible to show the identity of the integral by going back to a Riemann sum.

This is interesting because it leads to a more general approach like: if f(x) admits a Taylor expansion S(x) then f[u(x)] admits the expansion S[u(x)] and under some conditions to be datailed the equality holds for integrating on both sides. Which is much simpler than the computation of the nth derivative of f[u(x)].

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RE: Quiz: calculating a definite integral - Bunuel66 - 01-03-2014 01:29 PM

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