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seen Mar 4 '13 at 15:51

Dec
13
comment 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?
Dec
13
asked Averaging the values of $\cos x$ over one period
Oct
24
comment $f(t)f'(t)$ 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.
Oct
24
comment $f(t)f'(t)$ 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}
Oct
24
comment $f(t)f'(t)$ 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}
Oct
22
comment $f(t)f'(t)$ 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.
Oct
20
comment Uniqueness of the Interpolating Function
OK, I'm in chat but I don't see you there ...
Oct
20
comment 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)?
Oct
20
comment 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.
Oct
20
comment 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?
Oct
20
comment 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 $i = 0, 1, 2, 3, 4$?
Oct
20
asked Uniqueness of the Interpolating Function
Oct
20
comment 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.
Oct
20
comment 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.
Oct
20
comment 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.
Oct
20
comment 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).
Oct
20
comment 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?
Oct
20
comment 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?
Oct
20
comment 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.
Oct
20
comment 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 $t = 0$.