An affirmation of a concept that exists in Calculus 2, the Integral of (1/x).

There are certain concepts in Calculus 2, which introduces definite and indefinite integrals, that are taught to College and University Students, and which are actually considered to be basic information in Higher Math. One of them is, that the integral of (1/x) is the natural logarithm of (x).

Yet, some people just like to go around and dispute such things, much as the concept is popular, that (2+2) does not equal (4). And so, what I have just done is to ignore the obvious fact, that people who studied Calculus at a much higher level than I have, have found an analytical proof, and to ask the question:

‘What would happen if the integrals of simple power functions were given, that have powers slightly more-negative and slightly more-positive than (-1), in relation to this accepted answer, the natural logarithm of (x)?’ The accepted answer should always fall between those two curves, even if some plausible arbitrary constant is added to each power-function integral, such as one which sets all the functions to equal zero, when the parameter equals one. Not only that, but it’s easy for me to plot some functions. And so, the following two worksheets have resulted:

Testing the Integral of (1/x) – EPUB File for Mobile Devices

Testing the Integral of (1/x) – PDF File for Desktop and Laptop Computers

Further, I’d just like to remind the reader, that a function can easily be defined that follows a continuous line, except at one parameter-value, at which it has a different value, such that the neighbouring intervals in the domain of said function do not include this endpoint, in either case. The only question which remains is, whether that function is a correct answer to a question. And, because such functions are possible, the answer depends on additional information, to the idea that there are exceptions to how this function is to be computed.

(Update 1/26/2020, 20h20 : )

Continue reading An affirmation of a concept that exists in Calculus 2, the Integral of (1/x).

An observation about computing the Simpson’s Sum, as a numerical approximation of an Integral.

One of the subjects which I posted about before, is the fact that in practical computing, often there is an advantage to using numerical approximations, over trying to compute Algebraically exact solutions. And one place where this happens, is with integrals. The advantage can be somewhere, between achieving greater simplicity, and making a solution possible. Therefore, in Undergraduate Education, emphasis is placed in Calculus 2 courses already, on not just computing certain integrals Algebraically, but also, on understanding which numerical approximations will give good results.

One numerical approximation that ‘gives good results’, is The so-called Simpson’s Sum. What it does, is to perform a weighted summation of the most-recent 3 data-points, into an accumulated result. And the version of it which I was taught, sought to place a weight of 2/3 on the Midpoint, as well as to place a weight of 1/3 on the Trap Sum.

In general, the direction in which the Midpoint is inaccurate, will be opposite to the direction in which the Trap Sum is inaccurate. I.e., if the curve is concave-up, then the Midpoint will tend to underestimate the area under it, while the Trap Sum will tend to overestimate. And in general, with integration, consistent under- or over-estimation, is more problematic, than random under- or over-estimation would be.

But casual inspection of the link above will reveal, that this is not the only way to weight the three data-points. In effect, it’s also possible to place a weight of 1/2 on the Midpoint, plus a weight of 1/2 on the Trap Sum.

And so a logical question to ask would be, ‘Which method of summation is best?’

The answer has to do, mainly, with whether the stream of data-points is ‘critically sampled’ or not. When I was taking ‘Cal 2′, the Professor drew curves, which were gradual, and the frequencies of which, if the data-points did form a stream, would have been below half the Nyquist Frequency. With those curves, the Midpoint was visibly more-accurate, than the Trap Function.

But one parameter for how Electrical Engineers have been designing circuits more recently, not only states that an analog signal is to be sampled in the time domain, but that the stream of samples which results will be critically sampled. This means considerable signal-energy, all the way up to the Nyquist Frequency. What will result is effectively, that all 3 data-points will be as if random, with great variance between even those 3.

Under those conditions, there is a slight advantage to computing the average, between the Midpoint, and the Trap Sum.

(Update 12/08/2018, 20h25 : )

I’ve just prepared a small ‘wxMaxima’ work-sheet, to demonstrate the accuracy of the 2/3 + 1/3 method, applied to a sine-wave at 1/2 Nyquist Frequency:

Worksheet

Worksheet in EPUB2 for Phones

(Update 1/6/2019, 13h55 : )

Continue reading An observation about computing the Simpson’s Sum, as a numerical approximation of an Integral.