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Hm, I didn't think of a proof without using calculus.
If you are allowed to use Σ 1/x^2 = pi²/6 without proof, then your argument works. But even if you aren't allowed to use that, you can still show that Σ {i=9 to infinity, i odd} 1/i^2 < 1/8, since by rounding each term of the sum up to the next power of 1/2:
1/9^2 + 1/11^2 + 1/13^2 + ...
<
4*1/8^2 + 8*1/16^2 + 16*1/32^2 + ... = 1/16 + 1/32 + 1/64 + ... = 1/8 = 0.125.
So R(N)<0.125*2=0.25.
Then filtering out the multiples of 1/2^2 twice, 1/3, 1/5^2 twice, 1/7^2 twice, 1/13 gives (1/2)(2/3)(23/25)(48/49)(12/13) = ~0.277, and 0.27-0.25=0.02. So S(N)>0.02N.
BrunoVisnadi wrote:
Here I'm considering N as the whole natural numbers set. Since (1/2)(2/3)(12/3)(23/25)(8/9) = ~0.25, and 0.25-0.17 = 0.08, there must be 0.08N, and thus, infinite solutions for f(n) = f(n-1) + 1 in N.
By the way, the (8/9) isn't necessary. In the equation 169A+1=27B, A can never be a multiple of 3, so you don't need to filter out a multiple of 1/3^2 in A. The only condition related to multiples of 3 is that B is not a multiple of 3, which (2/3) covers already. I also guess you meant (12/13) instead of (12/3).
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I held off replying for a couple days because I thought it was easy enough that someone else would have replied. Perhaps there would have been a response if the fact that it was an Olympiad question was omitted.
Suppose that there is a polynomial f(x) with at least one non-integer coefficient, deg(f)=n, g(x) is an integer polynomial, and S is a set of n+1 integers such that f(t)=g(t) for all t in S. Let h(x)=g(x)-f(x). Note that h(x) is not the zero polynomial since f and g cannot be the same. Now h(t)=0 for all t in S, so h has at least n+1 integer roots and so deg(h)≥n+1. Since deg(f)=n, then deg(g)≥n+1. So h is a polynomial with at least one non-integer coefficient, such that the coefficients of xn+1, xn+2, ... , are all integers.
Let s1, s2, ..., sn+1 be elements of S. Then h=(x-s1)(x-s2)...(x-sn+1)k(x), where k(x) is a polynomial of degree m=deg(h)-(n+1). Now k must have at least one non-integer coefficient; otherwise h is an integer polynomial. Let k=amxm+am-1xm-1+...+a1x+a0, and let i be the largest integer (0≤i≤m) such that ai is not an integer. Then the (n+1+i)th coefficient of h is equal to:
{coefficient of xn+1+i} (x-s1)(x-s2)...(x-sn+1)(amxm+...+aixi+...+a1x+a0)
=
ai - ai+1*sum{j=1 to n+1}(sj) + ai+2*sum{1≤j(1)<j(2)≤n+1}(sj(1)*sj(2)) - ...
+ (-1)m-i*am*sum{1≤j(1)<...<j(m-i)≤n+1}(sj(1)*...*sj(m-i)),
which is ai plus an integer, which is not an integer. This contradicts that the coefficients in h of xn+1, xn+2, ... , are all integers.
Therefore the answer to
Is it true that there is a polynomial f(x) with rational coefficients, but at least one non integer, which degree is n>0, a polynomial g(x) with only integer coefficients, and a set S with n+1 integer elements so that g(t) = f(t) to every t that belongs to S?
is no.
By the way, the statement that f(x) has rational coefficients is a red herring; the answer is no even if f(x) is allowed to have any real coefficients, as long as it has at least one non-integer coefficient.
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Pokota wrote:
So I've run into an interesting issue.
joypad.get(1) returns nil if there's only one controller in the core. Easy way to see this is to call it in the Gambatte core. Why is this interesting?
Config -> Controllers...
If it doesn't explicitly say "Player 1", "Player 2", etc., then you cannot get controller state by joypad.get(1) or such. In that case, you need to use joypad.get(). joypad.get() can also get Player 1 (or 2 or ...) input, but the key name is prefixed with "P1 " (or "P2 " or ...).
In any case, key names are as shown in Config -> Controllers... .
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The idea is, we need to prove that there are infinitely many (A,B) satisfying 169A+1=27B where A,B are square-free and 13 does not divide A and 3 does not divide B. Even though we don't specifically know in general if a number is square-free, we can determine that infinitely many such (A,B) exist because there are "not enough multiples of k^2 to cover everything". For example, the probability that a randomly selected "large" number is square-free is (3/4)(8/9)(24/25)... = product{p prime} (1-1/p^2) = 6/pi^2.
In the case of this problem, the probability that a "large" (A,B) satisfying 169A+1=27B also satisfies the other conditions is P = (2/4)(2/3)(23/25)(47/49)(119/121)(12/13) * product{p prime >= 17} (1-2/p^2). It suffices simply to show that P>0, although I think the value of P is closer to 0.26 or so. For a rigorous proof, it is better to state it as how I defined S(N) (number of M between 0 and N-1 for which A=27M+23 and B=169M+144 satisfies the conditions). It is sufficient to consider first the M for which these are satisfied (accounting for (2/4)(2/3)(23/25)(12/13)N, or about 0.28N of the numbers)
1) 2^2=4 divides neither 27M+23 nor 169M+144.
2) 3 does not divide 169M+144.
3) 5^2=25 divides neither 27M+23 nor 169M+144.
4) 13 does not divide 27M+23.
and then subtract these possible cases, regardless of overlap (which can be bounded by 0.22N)
5) k^2 divides neither 27M+23 nor 169M+144 for all odd k>=7.
ceiling(x) is the smallest integer greater than or equal to x. I guess there are different terms in other languages, but I learned it as "floor" (greatest integer less than or equal to x) and "ceiling".
R(N), which is an upper bound for number of M that do not satisfy condition 5, is the sum of all 2*ceiling(N/k^2) over all odd k between 7 and sqrt(169N+144). We know that ceiling(N/k^2) <= (N/k^2) + 1, so we can bound using
R(N) <= N*( sum{i=3 to infinity} 2/(2i+1)^2 ) + 2*sqrt(169N+144)/2
The reason why the starting k for R(N) is chosen to be 7 is because this is the first point where the bound for R(N) is smaller than the estimate for the other conditions (the bound for that summation of square reciprocals is smaller for a larger starting k). Of course, if we made the starting k for R(N) to be much higher, then we can get a more accurate estimate for S(N), but this method proves that S(N)>0.06N, which is good enough for this problem.
By the way, you don't even need to figure out any specific (A,B) for this method, even though (77,482) gives the n=13013 that Amaraticando stated.
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BrunoVisnadi wrote:
Being n's prime factorization = (p1)^q1 * (p2)^q2 * (p3)^q3 (...) * (pk)^(qk).
And being f(x) = q1*p1^(q1-1) * q2*p2^(q2-1) * q3*p3^(q3-1) (...) * qk*pk^(qk-1).
Prove that there are infinite natural numbers n so that:
f(n) = f(n-1) + 1
This problem is difficult. I have a method to do it but the latter part feels like something from an advanced number theory textbook. I looked for an ingenious trick (it is an Olympiad question after all) but I couldn't find one. The proof I have is as follows:
Note: A square-free integer is one which is a product of distinct primes; that is, every prime divisor occurs to degree 1 in the prime factorization.
We have the following facts:
@ f(x*y)=f(x)*f(y) if gcd(x,y)=1.
@ f(p)=1 if p is prime.
@ f(13^2)=26.
@ f(3^3)=27.
Therefore:
@ f(N)=1 if N is square-free.
@ f(A*13^2)=26 if A is square-free and 13 does not divide A.
@ f(B*3^3)=27 if B is square-free and 3 does not divide B.
We want to prove that there are infinitely many A,B which satisfy the above and such that A*13^2+1=B*3^3 (or in other words 169A+1=27B).
Consider solutions (A,B) to 169A+1=27B. One solution is (A,B)=(23,144). Then for every integer M, (27M+23,169M+144) is a solution.
So we want to prove that there are infinitely many M>=0 which satisfy the following (condition (#)):
@ 27M+23 is square-free and 13 does not divide it, and
@ 169M+144 is square-free and 3 does not divide it.
We may reword condition (#) as:
1) 2^2=4 divides neither 27M+23 nor 169M+144.
2) 3 does not divide 169M+144.
3) 5^2=25 divides neither 27M+23 nor 169M+144.
4) 13 does not divide 27M+23.
5) k^2 divides neither 27M+23 nor 169M+144 for all odd k>=7.
Let N>10^6 be a number divisible by 2*3*5*13. Let S(N) be the number of M between 0 and N-1 that satisfy condition (#). We want to show that, as N grows without bound, then so does S(N). Then
S(N)>= N*(2/4)*(2/3)*(23/25)*(12/13) - R(N)
where R(N) is the sum of all 2*ceiling(N/k^2) over all odd k between 7 and sqrt(169N+144). Note that gcd(27M+23,169M+144)=1. The (2/4)*(2/3)*(23/25)*(12/13) comes from the conditions 1 to 4 and the R(N) from condition 5. Now:
R(N) <= N*( sum{i=3 to infinity} 2/(2i+1)^2 ) + 2*sqrt(169N+144)/2
Note that sum{i=3 to infinity} 2/(2i+1)^2 is less than the integral from x=2 to infinity of 2/(2x+1)^2 dx, which is 1/5=0.2 . Furthermore, since N>10^6, sqrt(169N+144)<sqrt(400N)=20*sqrt(N)<20*(N/10^3)=0.02N . Therefore R(N)<0.22N, and so
S(N) >= 0.28N - R(N) > 0.28N - 0.22N = 0.06N .
So then as N goes to infinity, so does S(N). QED
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BrunoVisnadi wrote:
Being n's prime factorization = (p1)^q1 + (p2)^q2 + (p3)^q3 (...) + (pk)^(qk).
And being f(x) = q1*p1^(q1-1) + q2*p2^(q2-1) + q3*p3^(q3-1) (...) + qk*pk^(qk-1).
Prove that there are infinite natural numbers n so that:
f(n) = f(n-1) + 1
Just to be clear.
Do you mean n = (p1)^q1 * (p2)^q2 * (p3)^q3 * ... * (pk)^qk ? That is normally how a prime factorization is written out.
If so, is it correct that f(n) = q1*p1^(q1-1) + q2*p2^(q2-1) + q3*p3^(q3-1) + ... + qk*pk^(qk-1) ?
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If you have already made savestates at or near the end of the movie before stopping, you can just load them after starting to play the movie back. Must be in "read+write" (i.e. not "read-only") mode in order to resume recording from that savestate. There is also a way to set a button to switch between "read-only" and "read+write" mode: Config -> Hotkeys... -> Movies -> Toggle read-only.
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fsvgm777 wrote:
I can do squares just fine in AviSynth, oddly enough:
So maybe it's on a per system basis?
It appears that the ANSI extended codes are possible, and squares (exponent to the 2) is one of them, as are most European-language Latin alphabet forms. The many other Unicode characters are not possible, though.
And yes, this is regarding the standard Subtitle() function.
Edit: By the way, AviSynth only sees files in the ANSI character set, not in UTF-8 or anything else.
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I see.
I had based my post on what Hetfield90 said in his second-last post:
Hetfield90 wrote:
but I think that unfairly puts a "bad ending" label on the Z-saber run when distinguishing between the two categories(as it is most commonly called when referring to both runs as "100%") when the Z-saber run is the only run that collects 100% of the items
and that every known RTAcategoryorganization does not have any category that is a "Zero alive" variant of the well-known ones (Zero being alive in low% is merely the natural result of going for low%). Having seen Hetfield90 say that it unfairly puts a "bad ending" label on the Z-saber run, I thought it was a bigger issue than it actually was.
If "best ending, max%" is all right with everyone, then I agree to it.
I apologize for my poorly-worded previous post.
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I think "max%" or "max items" should be included in the category name for the other run as well.
Also, keep in mind that Hetfield90 comes from the Mega Man RTA (speedrun) community where keeping Zero alive in Mega Man X3 as a notion by itself has never been (and still is not) taken seriously in any way, shape or form. Thus, calling the ending where Zero is alive the "good/best ending" may be offensive to some people. If it is an issue, perhaps "Zero alive" or something like that would be a better category name. Might add "Press Disposer route" as well.
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The function a() should be defined separately in its own block. Even if for nothing else other than to make the code easier to understand.
Also, all functions must be defined with the "function" keyword. For example:
function a()
-- this happens when clicking on the display:
menuscreen=0 -- menu gets closed
PositionY = PositionY + (Y-YBefore) -- enables dragging the display around
PositionX = PositionX + (X-XBefore) -- "
end
Furthermore, when passing a function in a parameter list, just pass the name, with no parentheses.
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Recently, there have been even fewer views on videos tagged "TAS" on nicovideo. As you can see, 30th place has less than 10000 views, when in the past you'd have to have well over 10000 views to have any chance of making the list. I wonder what is responsible for the downturn in views.
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According to ASCII tables, 0x26 is '&'.
Which makes sense, since the ROM names are MARIO&LUIGIJA88J and MARIO&LUIGIUA88E, respectively.
Edit: Same as https://github.com/TASVideos/BizHawk/issues/490 . Apparently it has been fixed in builds since Aug 31.
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Fortranm wrote:
I thought MMX3 couldn't be more glitchy than MMX1.
Well, even though MMX3 is glitched in one place, I think it would take a lot to be more glitchy than MMX1. Neon jumping (jumping in air by firing buster and jumping while in air dash) feels so normal that I don't even count it as a real glitch. :)
PikachuMan wrote:
I am requesting an encode that uses the MSU-1 hack.
Oh, first time I heard of this audio hack. It doesn't seem to work in BizHawk though. I guess it only works in higan?
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Jor02 wrote:
(and too how to fill gameshark codes)
BizHawk does not support gameshark (unless you mean Game Genie). You need to decode the gameshark code, and then fill out the form carefully.
I never used cheats in BizHawk so that's pretty much all I can say.
Translation:
----
I'm amazed at such a great improvement. I don't understand the numerical values for the names very well so I left it alone, but I didn't think to change them to equipment values. A wonderful piece of work.
----
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It looks like you are using the VBA-Next core. Does the same graphical problem happen with the mGBA core? To switch to mGBA, use Config -> Cores -> GBA with mGBA, and then reboot core.
mGBA discussion thread is here: http://tasvideos.org/forum/viewtopic.php?t=16165
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On Windows, there is a bug where a minimized BizHawk window cannot be restored by clicking on it while a video dump using FFmpeg writer (from Record AVI/WAV) is running and not paused.
1) Open ROM
2) Record AVI/WAV
3) FFmpeg writer
4) Any format (I tested mp4, WebM, Ogg, FLV and Uncompressed AVI)
5) Unpause emulation
6) Minimize window
7) Attempt to restore window by clicking on it on the toolbar
The window will fail to restore and an error noise will play. This is more obvious when "Run in background" is checked (in Config -> Customize...).
OS is Windows 7 Home Premium.
Note that the window can be restored if you click on it and then press the Enter key.
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Question about the framerate.
BizHawk says that the framerate is 59.2928626fps. With this TAS being 8462 frames, this gives a time of 2:22.715. However, the site parser uses 60.0fps, giving a time of 2:21.03. Which is correct?
So maybe it's on a per system basis?