I can personally barely wait for a TAS of the SG-1000 version, once a suitable emulator is released (to my knowledge Kega Fusion and MEKA lack rerecording and frame advance).
I found the pipe in BitFS by BLJing and going out of bounds. Too bad the warp trigger isn't there. The thing I don't understand is how I got to the third tier of the level from BLJing at the second. Anyways, video here.
So at the end, is that thing that Mario ground-pounded on the pipe?
The fact that those functions approach gaussians as n->infinity can be readily seen from the fact that
(1+x2/n)-n -> exp(-x^2) as n-> Infinity.
This result isn't just true of 1/(1+x^2); it actually works for any function that is twice differentiable at the origin. This fact is the reason behind the Central Limit Theorem.
That's actually more relevant than I thought at first...
This is the member of the normalized family that attains the value 1 when x=0:
(1+pi*(Gamma(n-1/2))/Gamma(n)))2x2)-n
Gamma(n-1/2)/Gamma(n) is asymptotic to 1/sqrt(n), so in the limit this really does behave just like (1+pi*x2/n)-n, which does obviously approach e-pi*x^2 as n approaches infinity.
The normalized family itself, parameterized by b for each fixed n, is
b*(1 + pi*(Gamma(n-1/2))/Gamma(n)))2b2x2)-n
However, when I decided to make a neat animation, I instead substituted n=1/m and b=tan(a), so that constant steps in m or a would yield a smoother animation and cover the relevant range better.
The envelopes of the family are
±x-1(1+1/(2n-1))-nGamma(n)/(sqrt(pi)sqrt(2n-1)Gamma(n-1/2))
and the limit as n approaches infinity is
±x-1/sqrt(2e*pi)
which is also the envelope of the family of normalized Gaussians
b*e-pi*b^2*x^2
(I discovered this part before discovering that the family of Gaussians was actually the limit of the families of inverted polynomials, and this actually inspired me to see whethe it was, and fortunately this particular diagram did commute.)
I just learned about a couple of interesting families of functions that can be used to make Dirac delta distributions, and I was greatly aided by my intuition after plotting them out in Mathematica
As you may be aware, the function e-x^2, suitably normalized, can be made into a family of functions, by means of horizontal compression and vertical expansion by a factor of b, that all have integrals of 1 over the real line and converge to a Delta distribution as b->0; it turns out the envelope of the graphs of this family has the equation 1=2e*pi*x2y2.
Then I tried out the same thing with another obvious choice: 1/(1+x2); this time, however, the envelope turns out to be 1=2pi2x2y2, which is closer to the coordinate axes.
I thought that maybe a tweaking of that original base-function may work; one idea is to try to just raise the exponent on the x, but it turns out that the envelope of the family based on 1/(1+x2n) approaches 1=4x2y2, and the normalized family itself approaches one based on Heaviside step functions: H(x+1/2)-H(x-1/2).
Oddly Mathematica was unable to determine this analytically except when x>1/2, but a numerical plot made it unmistakable.
Another one turned out to be more fruitful, just raising that entire rational function to the power of n; then the envelopes of the normalized family turn out to reach their greatest extent in the limit as n->infinity, when the envelopes become exactly the same as for that Gaussian family, and the normalized family based on 1/(1+x2)nbecomes the normalized Gaussian family based on e-x^2; I'm thinking about releasing a .nb file that can be viewed in MathPlayer to show this sort of thing, because it is truly astounding. Maybe the normalized Gaussian family is the most "open" possible family of functions converging to a delta distribution, in the sense of having envelopes with the greatest distance from the origin.
Really? Because I got the impression that you did, considering you said this...
...eschewing my typical strategy of powerleveling in order to make the total play time shorter so that the VBM would fit on Microstorage. Also, what is needed is a verifiable save *file* like what the emulator produces when using the in-game save feature.
I got the impression that you at least knew the rudimentary steps for how to beat this game quickly and efficiently.
I got that inspiration from watching the TAS of the original Genesis game.
Be sure to check "Disable HTML in this post"
Also, to make those nice-looking images, try playing around with the mathematical typesetting in Wikipedia, which uses LaTeX internally; I looked at p4wn3r's post and he just uploaded the images to his Photobucket account and linked them from there.
Sorry, I conflated "Div" with "derivative" (of a multi-variable function)
Bobo the King wrote:
The last problem reminded me of one that I dreamed up a while ago.
1) Find a difference equation for q such that the sum
sum( f(qi, qi-qi-1, i) )
is minimized for an arbitrary function f. In the sum, i ranges from 1 to N and you may assume that q0 and qN are both given so that the problem is well-posed. (Hint: Take your partial derivatives carefully!)
2) Introduce a symbol (if you haven't already done so) that will make your solution to (1) instantly familiar to any physicist or mathematician with a background in calculus of variations.
3) Apply your solution to the function
f=2qi - (qi-qi-1)^2
subject to the boundary conditions q0=0, qN=0, and N=11. What physical problem is this analogous to?
I'm guessing here that you meant sum(f(qi,qi-qi-1),i)...
1) The partial derivative with respect to qi of this sum is f1(qi,qi-qi-1)+f2(qi,qi-qi-1)-f2(qi+1,qi+1-qi) where 0<i<N, while the derivative with respect to qN is f1(qN,qN-qN-1)+f2(qN,qN-qN-1) and the derivative with respect to q0 is -f2(q1,q1-q0); here fn is the partial derivative with respect to the nth variable.
Setting all of those partial derivatives equal to 0 and then adding the equations yields sum(f1(qi,qi-qi-1),i)=0; however I don't know how to derive a difference equation here...
Maybe it would be better to take the Div of the aforementioned Grad, to get the Laplacian...
Second partial in qi for 0<i<N is f1,1(qi,qi-qi-1)+f1,2(qi,qi-qi-1)+f2,1(qi,qi-qi-1)+f2,2(qi,qi-qi-1)+f2,2(qi+1,qi+1-qi), second partial in qN is f1,1(qN,qN-qN-1)+f1,2(qN,qN-qN-1)+f2,1(qN,qN-qN-1)+f2,2(qN,qN-qN-1), and second partial in q0 is f2,2(q1,q1-q0), so the Laplacian, assuming that all of the respective functions are sufficiently well-behaved, is sum(f1,1(qi,qi-qi-1)+2f1,2(qi,qi-qi-1)+2f2,2(qi,qi-qi-1),i)
but I'm still not sure how to get any relations among the qi themselves...
2) I'll try to take a variational derivative here...
For a test function F, a perturbation is given by sum(f((qi,qi-qi-1)+E*F((qi,qi-qi-1),i) for some E>0; its derivative with respect to E is sum(F((qi,qi-qi-1),i) and this is also the derivative when E=0, so I think the variational derivative with respect to f is N...I'm not sure how this helps with the problem.
3) To make it easier, f(x,y)=2x-y2, so f1(x,y)=2 and f2(x,y)=-2y; then for 0<i<11, the partial of that sum with respect to qi is 2-2(qi-qi-1)+2(qi+1-qi), which is simplified to 2qi+1-4qi+2qi-1+2, the partial with respect to q0 is 2q1-2q0, and the partial with respect to q11 is 2-2q11+2q10. Setting all of these partials equal to 0 and adding equations yields 22=0, which is impossible; the Laplacian is 44, although I'm not sure of its significance.
It's a classic, but sadly it only runs on 32-bit or 16-bit Windows, because 64-bit Windows lacks a compatibility mode for 16-bit applications; however, the Tile World engine can be used instead if necessary.
This site contains links to numerous resources related to this game: http://www.telusplanet.net/public/nfield/ChipChallenge/chip.htm
Let a1 through an be real numbers such that:
What is the maximum possible value of a1a2 + a2a3 + ... + ana1?
This is a Lagrange optimization problem, with the first two equations as constraints.
If A=(a1,a2,...,an), g(A)=Sum(ai,i,1,n), h(A)=Sum(ai2,i,1,n)-1, and f(A)=Sum(aiai+1,i,1,n-1)+a1an
The system to solve is g(A)=0, h(A)=0, Grad(f(A))-r*Grad(g(A))-s*Grad(h(A))=0.
The ith element of Grad(f(A)) is ai-1+ai+1 for 1<i<n; the first element is a2+an, and the nth element is a1+an-1.
The ith element of Grad(g(A)) is 1, and the ith element of Grad(h(A)) is 2ai.
Therefore, for 1<i<n, we have ai-1+ai+1=r+2s*ai, and similarly cyclic equations for the cases i=1 and i=n; adding them all up yields
2Sum(ai,i,1,n)=nr+2ns*Sum(ai,i,1,n), and from the first constraint, 0=nr, so r=0.
Then we have, for i<1<n, ai-1+ai+1=2s*ai, and similarly cyclic equations for the cases i=1 and i=n.
Now multiply both sides by ai and then add all of these equations together to get 2f(A)=2s*Sum(ai2,i,1,n), and from the second constraint we have f(A)=s.
If instead both sides were multiplied by ai-1 or a similarly appropriate term for the i=1 and i=n cases, and then all terms were added together, the second constraint would yield 1+f(A)=2s*f(A), from which f(A)=1/(2s-1).
Therefore, s=1/(2s-1), so 2s2-s-1=0, so (2s+1)(s-1)=0, so s=-1/2 or 1; because f(A)=s at the critical points, the global maximum must be 1 (and the global minimum must be -1/2), because the intersection of the constraints is a smooth manifold.
The preceding logic only works if n>2, however, because only then are the terms in f(A) distinct and only then is the intersection of the constraints a smooth manifold (of dimension n-2): For n=2 f(A)=-1 at both points of the intersection (which is not a smooth manifold), for n=1 the intersection of constraints is empty, and for n=0 the second constraint is impossible while the first is trivial.
making a save for someone else
I lack the tenacity to actually do a TAS, let alone a 100% TAS
and indeed, although I normally stick with the defaults, that special one-character name Z will definitely help with the text speed
but I wonder...when you do the New Game +, what exactly gets retained? I know that the cards do, and some of the items do (based on the description in a FAQ of the conditions for getting the Teeny Bikini)...
If you just want to generate a verifiable save, then the movie that makes the save doesn't need to be optimized. I think FreshFeeling just feels that an optimized movie would be worth watching.
I just got an idea...maybe I'll do an ordinary run, send off the VBM and save-file (and savestate, because I remembered that the New Game + run can start from one), and then dump the AVI and chop it up to turn into a Let's Play on YouTube.
It is possible to have an infinite chain of elemental self-containment (or even an uncountably-large branching structure, like a set A={A,A}, which can be expanded as {{A,A},{A,A}} and so on); the paradox that you may have thought of is considering "the set of all sets that are not elements of themselves" because it is an element of itself if and only if it is not an element of itself, and this consideration is what leads out of naïve set theory into such concepts as "proper classes."
Now I just thought of an interesting computational exercise, which I shall attempt to solve because although it is interesting, it's not that hard for someone with a Calculus II background...
Find all linear real-valued functions of real variables with the property that the uniform lamina bounded by the graphs of that function and y=x^2 has a center of mass on the y-axis; then for any c>0, find all such functions for which the center of mass of this lamina is (0,c).
These functions are of the form ax+b, and the points of intersection are where x^2-ax-b=0, or A=(a-sqrt(a^2+4b))/2 and B=(a+sqrt(a^2+4b))/2 (so b>-a^2/4); then having the center of mass on the y-axis means that the x-coordinate is 0, so the numerator of the center-of-mass expression must be 0, so
Int(bx+ax^2-x^3,x,A,B)=0
The antiderivative is bx^2/2+ax^3/3-x^4/4, so the integral (thanks to Wolfram Alpha) is
a*(a^2+4b)^(3/2)/12, which is 0 when a=0 or b=-a^2/4; however only the former case is tenable, because the latter (as intimated above) would lead to a line tangent to the graph of x^2 at (a/2,a^2/4) (leading to an area of 0 and an x-coordinate of 0/0), so all of these functions are constant.
Then we can re-parameterize the boundaries (y=x^2 and y=b) in terms of functions of y, as -sqrt(y) and sqrt(y), ranging from 0 to b, and then the equation for the y-coordinate of the center of mass can be rearranged into
Int(2y^(3/2),y,0,b)=c*Int(2sqrt(y),y,0,b)
which becomes
(4/5)b^(5/2)=c*(4/3)b^(3/2)
so b=(5/3)c.
As an extension, let's consider the set of linear functions for which the center of mass lies on the same vertical line as the midpoint of the linear boundary of this lamina; it turns out that the center of mass *always* lies on the same vertical line as the midpoint of the boundary of the lamina.
For the next part, using an alternate formula for the center of mass, we have
c=(1/2)Int(b^2+2abx+a^2*x^2-x^4,x,A,B)/Int(b+ax-x^2,x,A,B)=2a^2/5+3b/5, or equivalently, b=5c/3-2a^2/3; because b>-a^2/4, this means 5c/3>(2/3-1/4)a^2, so 5c/3>5a^2/12, so c>a^2/4, which because a may be any real number means c can be any positive real.
Then the linear functions are of the form ax-2a^2/3+5c/3; in terms of any desired center of mass (C,c), C=a/2, so a=2C and the functions are the form 2Cx-8C^2/3+5c/3.
Interestingly this expression still makes sense if c<0, so let's, as an example, try to find the center of mass of the infinite lamina bounded by x^2 and x-1; in this case, C=1/2 and c=-1/5.
However, integrating from -r to r in the expression for x-coordinate and letting r approach infinity yields -1, while instead using the limits -r+1/2 and r+1/2 yields 1/2 (even without letting r approach infinity), so this is obviously inconsistent; even worse, the expression for y-coordinate approaches positive infinity.
What is the maximum possible value of a1a2 + a2a3 + ... + ana1?