# ganzewoort

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# 33 Comments

Dec13  comment  Averaging the values of $\cos x$ over one periodIt should be included twice because it belongs to the interval $[0,2\pi]$ (not $[0,2\pi)$), doesn't it?
Oct24  comment  $f(t)f'(t)$ where $f$ is part of a GaussianCORRECTION: edit time expired so I couldn't remove the $\frac{1}{4.5}$'s and the $\frac{1}{10}$ typo.
Oct24  comment  $f(t)f'(t)$ where $f$ is part of a Gaussianthen, 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}
Oct24  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}
Oct22  comment  $f(t)f'(t)$ where $f$ is part of a GaussianHow 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.
Oct20  comment  Uniqueness of the Interpolating FunctionOK, I'm in chat but I don't see you there ...
Oct20  comment  Uniqueness of the Interpolating FunctionHow do yo make Lagrangian polynomial periodic (I'm thinking in terms of physical signals where trigonometric functions are the usual choice)?
Oct20  comment  Periodic Function with Discrete ValuesSince 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.
Oct20  comment  Uniqueness of the Interpolating FunctionThis 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?
Oct20  comment  Uniqueness of the Interpolating FunctionSo, 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 $i = 0, 1, 2, 3, 4$?
Oct20  comment  Periodic Function with Discrete ValuesInstead 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.
Oct20  comment  Periodic Function with Discrete ValuesAll 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.
Oct20  comment  Periodic Function with Discrete ValuesNot 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.
Oct20  comment  Periodic Function with Discrete ValuesLike 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).
Oct20  comment  Periodic Function with Discrete ValuesSomething 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?
Oct20  comment  Periodic Function with Discrete ValuesAll 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?
Oct20  comment  Periodic Function with Discrete ValuesThat'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.
Oct20  comment  Periodic Function with Discrete ValuesWell, $f(t) = asin(t) - 10$ is differentiable and it, as well as its first derivative, does have a value at $t = 0$.
Oct20  comment  Periodic Function with Discrete ValuesThe 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?
Oct20  comment  $f(t)f'(t)$ 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 $[0,T]$?
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