How the chain rule applies to integral equations.

In Calculus, one of the most basic things that can be solved for, is that a principal function receives a parameter, multiplies it by a multiplier, and then passes the product to a nested function, of which either the derivative or the integral can subsequently be found. But what needs to be done over the multiplier, is opposite for integration, from what it was for differentiation. The following two work-sheets illustrate:

PDF File for Desktop Computers

EPUB File for Mobile Devices

Please pardon the poor typesetting of the EPUB File. It’s the result of some compatibility issues (with EPUB readers which do not support EPUB3 that uses MathML.)

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Exploring the Discrete Sine Transform…

I can sometimes describe a way of using certain tools – such as in this case, one of the Discrete Cosine Transforms – which is correct in principle, but which has an underlying flaw, that needs to be corrected, from my first approximation of how it can be applied.

One of the things which I had said was possible was, to take a series of frequency-domain ‘equalizer settings’, which be at one per unit of frequency, not, at so many per octave, compute whichever DCT was relevant, such that the result had the lowest frequency as its first element, and then to apply that result as a convolution, in order finally to apply the computed equalizer to a signal.

One of the facts which I’m only realizing recently is, that if the DCT is computed in a one-sided way, the results are ‘completely non-ideal’, because it gives no control over what the phase-shifts will be, at any frequency! Similarly, such a one-sided convolution can also not be applied as the sinc function, because the amount of sine-wave output, in response to a cosine-wave input, will approach infinity, when the frequency is actually at the cutoff frequency.

What I have found instead is, that if such a cosine transform is mirrored around a centre-point, the amount of sine response, to an input cosine-wave, will cancel out and become zero, thus giving phase-shifts of zero.

But a result which some people might like is, to be able to apply controlled phase-shifts, differently for each frequency, such that those people specify a cosine as well as a sine component, for an assumed input cosine-wave.

The way to accomplish that is, to add-in the corresponding (normalized) sine-transform, of the series of phase-shifted response values, and to observe that the sine-transform will actually be zero at the centre-point. Then, the thing to do is, to apply the results negatively on the other side of the centre-point, which were to be applied positively on one side.


 

 

I have carried out a certain experiment with the Computer Algebra System named “wxMaxima”, in order first to observe what happens if a set of equal, discrete frequency-coefficients belonging to a series is summed. And then, I plotted the result of the definite integral, of the sine function, over a short interval. Just as with the sinc function, The integral of the cosine function was (sin(x) – sin(0)) / x, the definite integral of the sine function will be (1 – cos(x)) / x, and, Because the derivative of cos(x) is zero at (x = 0), the limit equation based on the divide by zero, will actually approach zero, and be well-behaved.


 

 

(Update 1/31/2021, 13h35: )

There is an underlying truth about Integral Equations in general, which people who studied Calculus 2 generally know, but, I have no right just to assume that any reader of my blog did so. There exist certain standard Integrals, which behave in the reverse way of how the standard Derivatives behave, just because ‘Integrals’ are ‘Antiderivatives’…

When one solves the Derivatives of certain trig functions repeatedly, one obtains the sequence:

sin(x) -> cos(x) -> -sin(x) -> -cos(x) -> sin(x)

Solving the Indefinite Integrals of the same trig functions yields the result:

sin(x) -> -cos(x) -> -sin(x) -> cos(x) -> sin(x)

Hence, the Indefinite Integral of sin(x) is in fact -cos(x), and:

( -(-cos(0)) = +1 )

(End of Update, 1/31/2021, 13h35.)

 

(Updated 2/04/2021, 17h10…)

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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 : )

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