Coming Soon!

[FrediFizzx]

If you post nonsense on this thread it will be very swiftly and mercilessly deleted!
This applies to everyone not just Gill! So, be careful of what you spew.
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[FrediFizzx]

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I haven't missed any mark. Both < f > and < g > are exactly equal to 1, but evidently f =/= g.

In which Universe the two functions f = 1 + x and g = 1 + sin(x) are the same function?

f is a straight line and g is a sinusoidal function. But they both average exactly equal to 1.

Please demonstrate how their averages are both exactly equal to 1 when I have clearly demonstrated otherwise. You have again missed the mark.

On the contrary, you have missed the entire Universe. Do the math, or look at the graph, to recognize that < f > and < g > are exactly equal to 1. No more demonstrations from me.

??? This looks like math to me.



Very clearly showing that you are wrong. You keep claiming no more demonstrations but you are missing the most important one for the simulation.

LOL! If there is some other kind of math I don't know about please get out your pencil and paper and demonstrate.
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[Joy Christian]

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That is not math. That is junk. You will have to throw it away to be liberated from it.

Any "calculator" that gives you wrong results is junk. Your "calculator" is giving you the wrong results.
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[local]

Fred, you use simple calculus. For example, take Joy's function 1+x. If we integrate that we get x + x^2/2. Now take limits of -n and n (don't worry, n = infinity is included). We get:

i = (n + n^2/2) - (-n + (-n2)^2/2)
i = 2n

Now divide by the span 2n to get the mean:

m = 2n/2n = 1

So the mean is exactly 1. Probably there are other ways to show this analytically.

[Joy Christian]

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I haven't missed any mark. Both < f > and < g > are exactly equal to 1, but evidently f =/= g.

In which Universe the two functions f = 1 + x and g = 1 + sin(x) are the same function?

f is a straight line and g is a sinusoidal function. But they both average exactly equal to 1.

Here is my original post:

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Here is another example:

f = 1 + (x/180), with -180 < x < +180 in degrees.

g = 1 + sin(x), with x in the same range.

So < f > = 1 = < g >, but f =/= g. QED.

Now let us look at the function (x/180). Because x ranges from -180 to +180 degrees, (x/180) clearly ranges from -1 to +1. There are thus as many negative points of (x/180) between -1 to 0 as there are positive points of (x/180) between 0 to +1. Therefore < (x/180) > = 0, exactly. Consequently, < f > = < 1 + (x/180) > = < 1 > + < (x/180) > = 1 + 0 = 1, exactly.

Next, look at the function sin(x). Since x ranges from -180 to +180 degrees, sin(x) clearly ranges from -1 to +1. Thus there are as many negative points of sin (x) between -1 to 0 as there are positive points of sin(x) between 0 to +1. Therefore < sin(x) > = 0, exactly. Consequently, < g > = < 1 + sin(x) > = < 1 > + < sin(x) > = 1 + 0 = 1, exactly.

Therefore, < f > = 1 and < g > = 1, exactly. Any "calculator" that does not give these results is junk and should be thrown away.
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[FrediFizzx]

@Joy Sorry, everything you have done is only true if you invoke infinity. We are NOT doing infinity with the simulation so your examples are the real junk. You are still missing the one example that you might be able to get to work.
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[local]

@Joy Sorry, everything you have done is only true if you invoke infinity. We are NOT doing infinity with the simulation so your examples are the real junk. You are still missing the one example that you might be able to get to work.
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In my demonstration above, you can choose n < infinity and the result is still 1. Here's another way to view it. We have 1 + x. For every value of x, there is a value -x that cancels it out when integrating from n to -n, leaving just 1. Joy also made that point.

[Joy Christian]

@Joy Sorry, everything you have done is only true if you invoke infinity. We are NOT doing infinity with the simulation so your examples are the real junk. You are still missing the one example that you might be able to get to work.

No. I have not invoked infinity. Did I mention the points have to be infinitely many? My argument works for any finite number of points. It is your "calculator" that has been tripping you off.
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[FrediFizzx]

@Joy Sorry, everything you have done is only true if you invoke infinity. We are NOT doing infinity with the simulation so your examples are the real junk. You are still missing the one example that you might be able to get to work.

No. I have not invoked infinity. Did I mention the points have to be infinitely many? My argument works for any finite number of points. It is your "calculator" that has been tripping you off.

Yes, you have invoked infinity and don't realize it. Your demonstrations are all infinite junk but there is still one demonstration that you might be able to get to work right.
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[FrediFizzx]

@Joy Sorry, everything you have done is only true if you invoke infinity. We are NOT doing infinity with the simulation so your examples are the real junk. You are still missing the one example that you might be able to get to work.
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In my demonstration above, you can choose n < infinity and the result is still 1. Here's another way to view it. We have 1 + x. For every value of x, there is a value -x that cancels it out when integrating from n to -n, leaving just 1. Joy also made that point.

I think I deleted your demonstration so you need to do it again so I can show you where it fails.
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[Joy Christian]

@Joy Sorry, everything you have done is only true if you invoke infinity. We are NOT doing infinity with the simulation so your examples are the real junk. You are still missing the one example that you might be able to get to work.

No. I have not invoked infinity. Did I mention the points have to be infinitely many? My argument works for any finite number of points. It is your "calculator" that has been tripping you off.

Yes, you have invoked infinity and don't realize it. Your demonstrations are all infinite junk but there is still one demonstration that you might be able to get to work right.

Show me where did I invoke infinity in my argument above. I didn't.
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[FrediFizzx]

The only way that you can get the average of "f" to equal the average of "g' in your scheme that I put in Mathematica is by taking the trials to infinity.

But..., guess what??? When you do that then f = g.
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[local]

Fred, you are deleting posts arbitrarily. What are you afraid of? I think you have suffered some kind of nervous breakdown. Therefore, I'll just say "sayonara", and if you come to your senses, send me an email.

[FrediFizzx]

Fred, you are deleting posts arbitrarily. What are you afraid of? I think you have suffered some kind of nervous breakdown. Therefore, I'll just say "sayonara", and if you come to your senses, send me an email.

I am NOT deleting posts arbitrarily. If I think they contain nonsense for this thread then they are gone. I did in fact warn everyone! Which I will do again.
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[FrediFizzx]

If you post nonsense on this thread it will be very swiftly and mercilessly deleted!
This applies to everyone not just Gill! So, be careful of what you spew.
.

[FrediFizzx]

It's quite simple. Don't post nonsense and your post won't be deleted. I have yet to see any reasonable arguments against what Mathematica says. I know some of you are smarter than what you are portraying. So, get with it! Let's see some no-nonsense posts. I can think of at least one example that you might be able to get it to work.
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[FrediFizzx]

It's quite simple. Don't post nonsense and your post won't be deleted. I have yet to see any reasonable arguments against what Mathematica says. I know some of you are smarter than what you are portraying. So, get with it! Let's see some no-nonsense posts. I can think of at least one example that you might be able to get it to work.

@Joy I figured out your mistake. You are in fact assuming infinite sampling for x. When done like a simulation like we are doing we have,



That depends on the number of trials and will actually vary from run to run. If you were able to do infinite trials, you would get your answers. But that is NOT what we are doing.
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[Gordon Watson]

It's quite simple. Don't post nonsense and your post won't be deleted. I have yet to see any reasonable arguments against what Mathematica says. I know some of you are smarter than what you are portraying. So, get with it! Let's see some no-nonsense posts. I can think of at least one example that you might be able to get it to work.
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Fred,
I posted a no-nonsense comment. It supported Mathematica, but not your interpretation. As I recall, you responded by asking a new question that my post had answered: but without quoting me. When I sought to find my original comment it had been deleted. I suggest: if you are looking for no-nonsense comments, please reduce the supposedly nonsensical replies by restating the questions. I accept that you deleted a later perhaps-too-cranky post of mine.

I, one of many, am trying to help!

All the best; Gordon
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[FrediFizzx]

It's quite simple. Don't post nonsense and your post won't be deleted. I have yet to see any reasonable arguments against what Mathematica says. I know some of you are smarter than what you are portraying. So, get with it! Let's see some no-nonsense posts. I can think of at least one example that you might be able to get it to work.
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Fred,
I posted a no-nonsense comment. It supported Mathematica, but not your interpretation. As I recall, you responded by asking a new question that my post had answered: but without quoting me. When I sought to find my original comment it had been deleted. I suggest: if you are looking for no-nonsense comments, please reduce the supposedly nonsensical replies by restating the questions. I accept that you deleted a later perhaps-too-cranky post of mine.

I, one of many, am trying to help!

All the best; Gordon

Well, thanks for trying to help but I'm probably going to delete this post also after you have had a chance to read it since it doesn't really directly address the issues. I hope everyone realizes now what Joy's and other's mistakes were. Even though finite intervals, there was infinite sampling of x in-between. Whereas that is NOT what happens in a simulation with a finite number of trials. There is no infinite sampling of x in-between the intervals.
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[Gordon Watson]

It's quite simple. Don't post nonsense and your post won't be deleted. I have yet to see any reasonable arguments against what Mathematica says. I know some of you are smarter than what you are portraying. So, get with it! Let's see some no-nonsense posts. I can think of at least one example that you might be able to get it to work.
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Fred,
I posted a no-nonsense comment. It supported Mathematica, but not your interpretation. As I recall, you responded by asking a new question that my post had answered: but without quoting me. When I sought to find my original comment it had been deleted. I suggest: if you are looking for no-nonsense comments, please reduce the supposedly nonsensical replies by restating the questions. I accept that you deleted a later perhaps-too-cranky post of mine.

I, one of many, am trying to help!

All the best; Gordon

Well, thanks for trying to help but I'm probably going to delete this post also after you have had a chance to read it since it doesn't really directly address the issues. I hope everyone realizes now what Joy's and other's mistakes were. Even though finite intervals, there was infinite sampling of x in-between. Whereas that is NOT what happens in a simulation with a finite number of trials. There is no infinite sampling of x in-between the intervals.
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Fred, as you increase the number of trials, N, so you will approach the infinite case in the limit. Gordon
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