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Lettris is a curious tetris-clone game where all the bricks have the same square shape but different content. Each square carries a letter. To make squares disappear and save space for other squares you have to assemble English words (left, right, up, down) from the falling squares.
Boggle gives you 3 minutes to find as many words (3 letters or more) as you can in a grid of 16 letters. You can also try the grid of 16 letters. Letters must be adjacent and longer words score better. See if you can get into the grid Hall of Fame !
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taken over n-dimensional space Rn where i is the imaginary unit. Here f and g are real-valued smooth functions. The role of g is to ensure convergence; that is, g is a test function. The large real parameter k is considered in the limit as k → ∞.
The main idea of stationary phase methods relies on the cancellation of sinusoids with rapidly-varying phase. If many sinusoids have the same phase and they are added together, they will add constructively. If, however, these same sinusoids have phases which change rapidly as the frequency changes, they will add incoherently, varying between constructive and destructive addition at different times.
Consider a function
The phase term in this function, is "stationary" when
Solutions to this equation yield dominant frequencies for a given and . If we expand in a Taylor series about and neglect terms of order higher than ,
When is relatively large, even a small difference will generate rapid oscillations within the integral, leading to cancellation. Therefore we can extend the limits of integration beyond the limit for a Taylor expansion. If we double the real contribution from the positive frequencies of the transform to account for the negative frequencies,
This integrates to
The first major general statement of the principle involved is that the asymptotic behaviour of I(k) depends only on the critical points of f. If by choice of g the integral is localised to a region of space where f has no critical point, the resulting integral tends to 0 as the frequency of oscillations is taken to infinity. See for example Riemann-Lebesgue lemma.
The second statement is that when f is a Morse function, so that the singular points of f are non-degenerate and isolated, then the question can be reduced to the case n = 1. In fact, then, a choice of g can be made to split the integral into cases with just one critical point P in each. At that point, because the Hessian determinant at P is by assumption not 0, the Morse lemma applies. By a change of co-ordinates f may be replaced by
The value of j is given by the signature of the Hessian matrix of f at P. As for g, the essential case is that g is a product of bump functions of xi. Assuming now without loss of generality that P is the origin, take a smooth bump function h with value 1 on the interval [−1,1] and quickly tending to 0 outside it. Take
then Fubini's theorem reduces I(k) to a product of integrals over the real line like
with f(x) = x2 or −x2. The case with the minus sign is the complex conjugate of the case with the plus sign, so there is essentially one required asymptotic estimate.
In this way asymptotics can be found for oscillatory integrals for Morse functions. The degenerate case requires further techniques. See for example Airy function.
The essential statement is this one:
In fact by contour integration it can be shown that the main term on the right hand side of the equation is the value of the integral on the left hand side, extended over the range [−∞,∞]. Therefore it is the question of estimating away the integral over, say, [1,∞].
This is the model for all one-dimensional integrals I(k) with f having a single non-degenerate critical point at which f has second derivative > 0. In fact the model case has second derivative 2 at 0. In order to scale using k, observe that replacing k by ck where c is constant is the same as scaling x by √c. It follows that for general values of f″(0) > 0, the factor √(π/k) becomes
For f″(0) < 0 one uses the complex conjugate formula, as was mentioned before.