- #1

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Let G(s) be the moving average of all previous values of f(s).

G(s) and F(s) intersect at multiple points. Is it possible to prove that the intersections happen periodically?

- #1

- 75

- 4

Let G(s) be the moving average of all previous values of f(s).

G(s) and F(s) intersect at multiple points. Is it possible to prove that the intersections happen periodically?

- #2

pwsnafu

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Could you explain what the 1 over infinity is supposed to be?

- #3

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That's the summation term from 1 to infinity. The software garbled it.

- #4

mfb

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What do you mean by "moving average over all previous values", and does f(s) mean F(s)?

Do you have a proof that they intersect at all? With complex numbers this is not trivial.

- #5

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I just graphed the functions. If you dont know what a moving average is then look it up.

What do you mean by "moving average over all previous values", and does f(s) mean F(s)?

Do you have a proof that they intersect at all? With complex numbers this is not trivial.

As I suspected this problem seems very difficult or impossible to solve.

- #6

mfb

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I know what a moving average is, but the definition is not unique (average over which range?), and your description doesn't make sense.If you dont know what a moving average is then look it up.

Did you plot both the real and imaginary part?

Currently the problem is not even well-defined.

- #7

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I plotted both the real and imaginary parts.I know what a moving average is, but the definition is not unique (average over which range?), and your description doesn't make sense.

Did you plot both the real and imaginary part?

Currently the problem is not even well-defined.

The range is over the entire function, or if your software does not support that, then try from the origin to S.

- #8

mfb

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That doesn't make sense either.The range is over the entire function

How did you define G(s), can you show the formula?

And they had intersections at the same point?I plotted both the real and imaginary parts.

- #9

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I don't have the formula handy. You just average all the previous points.That doesn't make sense either.

How did you define G(s), can you show the formula?And they had intersections at the same point?

They intersected at different points. I want to know if either or both of the graphs are periodic.

Im not too concerned about the specifics. If you can answer anything related to this question then that would be interesting. I am just asking if its always possible to prove whether a function is periodic.

- #10

mfb

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That would work as description if there would be a finite set of those points, but you don't have that. Average from 0 to s? Is your function defined for negative s? Make some limit procedure to "average" from -infinity to s? Something else?You just average all the previous points.

Which means the functions do not intersect. Only their real and imaginary parts.They intersected at different points.

A clear no. You can construct functions that are periodic if and only if some turing machine halts, and there is no algorithm that can determine this for all turing machines (halting problem). Those function definitions will look weird, however (like "f(x)=sin(x) if this turing machine halts, f(x)=x if it does not").I am just asking if its always possible to prove whether a function is periodic.

- #11

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YesThat would work as description if there would be a finite set of those points, but you don't have that. Average from 0 to s?

YesIs your function defined for negative s?

Is that a proof? Are you sure they never intersect?Which means the functions do not intersect. Only their real and imaginary parts.A clear no. You can construct functions that are periodic if and only if some turing machine halts, and there is no algorithm that can determine this for all turing machines (halting problem). Those function definitions will look weird, however (like "f(x)=sin(x) if this turing machine halts, f(x)=x if it does not").

You seem to be claiming I claimed they always intersect at different points. No, I said they intersected at some points and the parts intersected at others.

Anyway you basically answered my questions to a good enough level with your last paragraph.

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- #12

mfb

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No, but you claimed they would intersect, and I don't see evidence for that so far.Are you sure they never intersect?

An intersection of two functions requires that the function values are identical - for complex numbers, this means both real and imaginary part have to be identical.G(s) and F(s) intersect at multiple points.

- #13

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I really don't see how that helps. If you have any more questions about the function let me know, but it seems pretty well defined for me. Cant you just use the moving average feature in matlab? I really don't see where the ambiguity was, I am able to graph the function fine.No, but you claimed they would intersect, and I don't see evidence for that so far.An intersection of two functions requires that the function values are identical - for complex numbers, this means both real and imaginary part have to be identical.

- #14

mfb

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The range of definition and the range you average over.I really don't see where the ambiguity was, I am able to graph the function fine.

In addition, the first post had the (probably wrong, but certainly not backed by evidence) claim that the functions intersect.

- #15

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So basically, the example problem is well defined and unanswerable.

- #16

jbriggs444

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Nothing in this thread has indicated that the example problem (as clarified in #11) is unanswerable. Only that it is unanswered and that it might be unanswerable.So basically, the example problem is well defined and unanswerable.

- #17

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So its unanswerable, as far as anyone knows.Nothing in this thread has indicated that the example problem (as clarified in #11) is unanswerable. Only that it is unanswered and that it might be unanswerable.

- #18

jbriggs444

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And it's answerable, as far as anyone knows. Might be answerable, might not be. We do not know. Trying to spin it one way or the other does not change that.So its unanswerable, as far as anyone knows.

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- #19

Mark44

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If you dont know what a moving average is then look it up.

People give help here out of the kindness of their hearts -- none of us gets paid for this. Also, I guarantee that @mfb knows what a moving average is, but that's not what he asked:

Your question was unclear from the start, and it took 11 posts to figure out what you were asking. It's not the responsibility of responders to go "look it up" -- it's your responsibility to ask a question that makes sense, and with as much information as you can provide.mfb said:What do you mean by "moving average over all previous values"?

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