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This is a continuation of the question asked in second-order divided differenc.

Suppose the periodic function $f(t)$ has only one value, $f(t_0) = -10$ within the period $[0,T]$, all other values of the function being zero within that period. We can interpolate $f(t)$ by $asin(t) - 10$ which for $t_0 = 0$ will give the value $-10$. Now, if we accept the interpolation then $f'(t) = acos(t)$ and therefore for $t_0=0$ we will have $f'(t_0) = acos(0) = a$. Thus, at $t_0 =0$ we will obtain $f(t_0)f'(t_0) = -10a$. At the end of the period the value of $f(t)f'(t)$ will be also $-10a$. Therefore, the average value of $f(t)f'(t)$ for the period is $\frac{2(-10a)}{2} = -10a$. Do you agree with this?

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If I understand your question correctly, you have an $f: [0,T] \to \mathbb{R}$, $$ f(t) = \begin{cases} -10 &\text{if } t=t_0 \\ 0 &\text{otherwise} \end{cases} $$ and are asking if for the value of $$ \frac{1}{T}\int_0^T f(t)f'(t) dt \text{ .} $$

That question doesn't make a lot of sense then, because $f$ is not differentable, hence $f'(t)$ isn't even defined. And if you go on to define $f'(t)$ somehow, the integral will be $0$, because the integrand is zero everywhere except at $t_0$.

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    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?2012-10-20
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    No, if $g$ is zero on $A$ except at finitely many points, then $\int_A g(t)dt = 0$. Or you do actually envision $f$ to be a *measure*? I.e. do you have a measure $\mu$ with $\mu([a,b]) = f(b) - f(a)$ and want to compute $\int_A f(t)d\mu(t)$? In that case, $\int_A f(t)d\mu(t) = \int_A f(t)f'(t)dt$ *only* if $f$ is differentiable, which your $f$ is not.2012-10-20
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    Well, $f(t) = asin(t) - 10$ is differentiable and it, as well as its first derivative, does have a value at $t = 0$.2012-10-20
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    So? Pick any other interpolation, and you'll get a different result. In fact, I bet that you can get *every* possible result just by picking a suitable interpolation for $f$. So what's the point?2012-10-20
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    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.2012-10-20
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    Pick the value of $f$ at any $n$ points. There always is a polynomial of degree $n$ which coincides with $f$ at those $n$ points. If you add additional conditions, e.g. about $f'$, you'll need to pick a polynomial with a larger degree, but again, you *will* find a polynomial which fulfills *all* of your conditions. I really don't know what you're trying to do there, but you need to assume some *pretty* strong property of $f$ to be able to make assumptions about $f'$ purely from samples values of $f$. Limiting the *bandwidth* of $f$ seems like it might work, but I didn't check...2012-10-20
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    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?2012-10-20
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    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?2012-10-20
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    Because you can choose *arbitrary* values of $f'(t)$, and *always* find an interpolating polynomial, so you can get *any* answer you want. Which makes the whole thing completely devoid of any meaning. I'll say it one last time: *You cannot deduce anythinga bout $f'$ from samples of $f$ without some strong additional constraint on $f$. Nothing. Nada. Zilch*.2012-10-20
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    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).2012-10-20
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    Sure, yeah, if you *know* that $f(t) = a\sin(t) - b$, you can do that. But you never stated that you *know* that, you basically said "I'm going to just pick $a\sin(t) - b$ 'cause it seems easy to do". There's a *huge* difference there. And btw, you can *easily* find polynomial interpolations which make $f$ periodic - just add constraints $p(0) = f(T)$....2012-10-20
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    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.2012-10-20
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    This is wrong, I told you why *multiple* times, even showed you *how* to find other functions which recover them. Quite frankly, I don't think you're really listening, and it seems that you're *not* actually interested in an answer but rather in arguing that you are right and I'm wrong. That's certainly your prerogative, but I chose not to be part of it any longer.2012-10-20
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    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.2012-10-20
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    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.2012-10-20
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    ganzewoort: *Instead of down-voting you should*... They SHOULD? Since when @fgp *should* do anything? Please come back to Earth. (And for that matter, how do you know who downvoted your question?)2012-10-20
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    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.2012-10-20