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| visits | member for | 1 year, 7 months |
| seen | Mar 4 at 15:51 | |
| stats | profile views | 11 |
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Dec 13 |
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Averaging the values of $\cos x$ over one period It should be included twice because it belongs to the interval $[0,2\pi]$ (not $[0,2\pi)$), doesn't it? |
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Oct 24 |
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$f(t)f'(t)$ where $f$ is part of a Gaussian CORRECTION: edit time expired so I couldn't remove the $\frac{1}{4.5}$'s and the $\frac{1}{10}$ typo. |
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Oct 24 |
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$f(t)f'(t)$ where $f$ is part of a Gaussian then, every time the signal is applied (periodically or aperiodically) we get \begin{equation} \frac{1}{T} \int\limits_{0.5}^{5} f(t)f'(t)dt = \frac{1}{4.5} \frac{1}{2} \int\limits_{0.5}^{5} d (f(t))^2 = \frac{1}{10} (f(t))^2|_{0.5}^{5} = \end{equation} \begin{equation} \frac{1}{4.5} \frac{1}{2}(-e^{-\frac{(0.5t-1)^2}{2}} -e^{-\frac{(t-1)^2}{2}})^2|_{0.5}^{5} = \frac{1}{4.5} (-1.28763) < 0 \end{equation} |
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Oct 24 |
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$f(t)f'(t)$ where $f$ is part of a Gaussian @Gerry, is this better: Let the form of the signal which is applied periodically or aperiodically on the system be a part of a Gaussian, such as \begin{equation} f(t) = \begin{cases} \frac{1}{2}(-e^{-\frac{(0.5t-1)^2}{2}} -e^{-\frac{(t-1)^2}{2}})^2, & \mbox{for } t =\mbox{ $0.5<t<5$} \\ 0, & \mbox{for all other } t\mbox{ } \end{cases} \end{equation} |
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Oct 22 |
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$f(t)f'(t)$ where $f$ is part of a Gaussian How else can you express the fact that there will be a non-zero burst of that particular form, lasting for $\Delta t = 4.5$, starting from $t = 0$? The first burst will be, as said, from $t=0,5$ to $t = 5$. The next non-zero burst will occur from $t= 10.5$ to $t = 15$, the third from $t = 20.5$ to $t = 25$ and so on. |
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Oct 20 |
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Uniqueness of the Interpolating Function OK, I'm in chat but I don't see you there ... |
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Oct 20 |
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Uniqueness of the Interpolating Function How do yo make Lagrangian polynomial periodic (I'm thinking in terms of physical signals where trigonometric functions are the usual choice)? |
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Oct 20 |
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Periodic Function with Discrete Values Since the times civilized discussions are established in the modern world -- a couple of centuries or so ago. Down-voting and disappearing is an easy and mean way out when one feels that he/she should win the discussion at any rate. |
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Oct 20 |
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Uniqueness of the Interpolating Function This is the main point of my asking. If the interpolating function indeed is $f(t) = 5sin(t) - 10$ then it does have first derivative which is $5cos(t)$ and can be evaluated at the said 5 points. Also, from your answer I take it that there is no polynomial which would recover exactly the 5 points. All one can achieve through a polynomial is an approximation, right? |
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Oct 20 |
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Uniqueness of the Interpolating Function So, then, what is $f'(t)$ of that function? Will it differ from $f'(t)$ when it is defined as $f(t) = 5sin(t_i) - 5$ where $i = 0, 1, 2, 3, 4$? |
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Oct 20 |
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Periodic Function with Discrete Values Instead of down-voting you should propose another concrete function (other than $f(t) = 5sin(t) - 10$) which will recover the points in the above 5-point example. I may open a special separate question devoted to this uniqueness problem. |
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Oct 20 |
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Periodic Function with Discrete Values All right, can you propose another concrete function (other than $f(t) = 5sin(t) - 10$) which will recover the 5-point example? You didn't give any such function so far but only insisted that such function exists. |
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Oct 20 |
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Periodic Function with Discrete Values Not at all. I have the fixed 5 values of the discrete continuous function, that's a given, and it happens that $5sin(t) - 10$ is the only function which recovers all of them. Also, I don't see how a polynomial with constraints will express a physical signal. Somehow, it is usual to use trigonometric functions for that purpose. Thus, if we limit ourselves to trigonometric functions, $f(t) = 5sin(t) - 10$ appears to be unique as an interpolation function in the above 5-point case. |
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Oct 20 |
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Periodic Function with Discrete Values Like I said, this answer doesn't seem satisfactory because there is, in fact, a unique strong additional constraint on $f$, namely, the specific function $f(t) = 5sin(t) - 10$ which also accounts for the periodicity which a polynomial doesn't (the example with the 5-point discrete periodic function is had in mind). |
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Oct 20 |
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Periodic Function with Discrete Values Something has to be done so that one can carry out the discussions comfortably without limiting them to non-extended discussions, as well as avoiding the chat. What is the usual solution in such a case? |
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Oct 20 |
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Periodic Function with Discrete Values All I need to do is calculate the average value of $f(t)f'(t)$ for the discrete points (5 in this case) within the period. How is the result using a polynomial (accounting for periodicity too) going to differ, if at all, from the above result with the sine function? |
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Oct 20 |
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Periodic Function with Discrete Values That's true but suppose you have arranged the matters to have $f(t_0) = -10, f(t_1) = -5, f(t_2) = -10, f(t_3) = -15$ and $f(t_4) = -10$ and all the points in between to be zero (where $t_0$ and $t_4$ are the beginning and the end of the period), then $5sin(t) - 10$ will be the possible interpolation (can't think of any other function). The argument expressed in the question applies to this function as well. |
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Oct 20 |
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Periodic Function with Discrete Values Well, $f(t) = asin(t) - 10$ is differentiable and it, as well as its first derivative, does have a value at $t = 0$. |
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Oct 20 |
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Periodic Function with Discrete Values The way I understand it is as follows: if there is one non-zero point of the function and its non-zero first derivative, it is enough to carry out the summation (which is in fact integration) of $f(t)f'(t)$ over the entire interval. All the other, infinite in number products within the period, are zero. In the case at hand, it is undeniable that $f(t)f'(t) < 0$. How is that fact accounted for if the integral over the period is taken to be zero? |
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Oct 20 |
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$f(t)f'(t)$ where $f$ is part of a Gaussian $f(t)$ is, indeed, periodic. It is with a discrete-math tag because it resembles the earlier discussed discrete periodic functions. Notice, the integral over the entire $[0,T]$ period is zero and yet there is a section within $[0,T]$ where the integral isn't zero. Why should one ignore that fact when carryin out integration over the entire $[0,T]$? |
