Hi, David.

You could, of course, just take any list of 5 million primes, and write a simple program (or spreadsheet) to calculate the gaps.

Alternatively, the list of gaps is in OEIC here, and has a link to Vojtech Strnad, First 100000 terms [First 10000 terms from N. J. A. Sloane].

That doesn’t include the primes themselves, but you could correlate it with a list of primes.

prime number. ( or a very large available prime such as one millionth prime ). all the google and any

other internet search i have tried jumps over this direct listing of primes and their gaps so i am stopped.

a few examples are: 2 1, 3 2,5 2,7 4 and on up to the largest available.

can you help with this? thank you. ]]>

Hi, Anthony.

I have explained this several times. Quoting myself from a comment above, “Finding that 0/0 = 0 is not a contradiction, because the whole point is that it could be **any number** (and therefore can’t be defined to be any one of them).”

The problem is not that 0/0 **can’t be 0**, but that it **can be anything** with equal justification; defining it as any one value leads to contradictions (particularly with the calculus sense of indeterminacy, where it would mess up the concept of limits).

But when you suggest n/0 = 0, that breaks many rules; it makes no sense. (See Why Can’t You Divide by Zero?) And the fact that n*0 = 0 is not a made-up rule; that can be proved from other rules.

]]>One problem with this approach is that, since the cosine is an even function, you are losing the **sign** of the angle — it’s as if you were averaging, not the angles themselves, but their **absolute values**. So if you “average” the angles +1 and -1 degree, your answer will be 1 degree, not 0 degrees.

This explains your 0.5 degree limit; if your numbers are equally spaced from -1 to +1 degree, you are really “averaging” angles from 0 to 1 degree, so half a degree sounds about right. You are, to be more precise, finding the center of mass of the arc between those angles, and taking the inverse cosine of that, which I find to be 0.577 degrees, if my calculation is correct.

Your average might make a little sense if the angles are all positive; but it would depend, as I have said, on what you are using it for.

]]>Instead, what I came up with was to average the cosine of angles, = (cos(theta1) + cos(theta2) + …)/N, and then taking the arccos(). I never really gave it much thought, but this way it feels more natural, because cos(theta) is a continuous function defined for all reals, making the average well-defined and single-valued (whatever range of theta you have — I think, though never proved)…

Namely: cos(x) = 1 – x² + … for small angles |x|, so averaging 359 deg with 0 and 1,

acos((cos(1*pi/180)+cos(0)+cos(359*pi/180))/3)*180/pi = 0.8 degrees

which is almost zero (better than 120 degrees).

The same works at the other side, at 180 degrees:

acos((cos(pi)+cos(pi-pi/180)+cos(pi+pi/180))/3)*180/pi = 179.2 degrees

The difference to the real value of the average angle decreases as the number of samples increase:

if you consider 1000 angles from -1 degree to 1 degree, and average the cosine, and then take the arccos, you get 0.5 deg… In MATLAB:

acos(mean(cos(linspace(-1,1,1000).*pi/180)))*180/pi = 0.5 degrees

Two caveats: the 0.5 degrees accuracy seems to be a “fixed point” (larger and larger N converge to this error, and it can be smaller than this — I tested it numerically)… The second is that we must take care that is always bounded in -1 to 1.

]]>Hi, Sam.

You said, “I tried to find some equations of **parabola** for my presentation in the topic **hyperbola** itself.”

Are you asking about the **hyperbola**, or the **parabola**? Assuming the former, what about this post is insufficient? And if you meant what you said, and are talking about equations for the parabola, what are a, b, and c there?

Please don’t reply here, but through our Ask a Question link, which is where specific questions like this belong.

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