# Tag Info

0

You can solve this problem very quickly. From $\partial_{q_n} L=0$, we have $$q_n(\lambda_1,\lambda_2) = e^{e_n - 1 + \lambda_1 + \lambda_2 n/N}.$$ We know that $q_n$ has to sum to 1 (the first constraint), so we can eliminate $\lambda_1$: $$q_n(\lambda_1,\lambda_2) = \frac{e^{e_n - 1 + \lambda_1 + \lambda_2 n/N}}{\sum_{n'} e^{e_{n'} - 1 + \lambda_1 + \... 0 I think for a 2m dimensional vector ( a,b,c,d.........z) you can say that ( -z, -x,......... a) is orthogonal where half are negative of their original values However for 2m + 1 dimensional vector you can't straightforwardly say the same ( if you want a non zero orthogonal vector ). But since it is a guess you can keep a zero in the middle i think 1 I assume we know how to simulate a Poisson point process with constant intensity in an interval (e.g. by considering partial sums of i.i.d. exponential variables.) That allows you to simulate a standard Poisson point process in a rectangle [a,b] \times [0,d] by simulating an intensity d Poisson process in [a,b] and assigning the resulting points \{... 0 Newton series is similar to the Taylor series, except we use delta operators instead of derivatives:$$f(x) = \sum_{k=0}^\infty \binom{x-a}k \Delta^k f\left (a\right)$$The binomial coefficients work similar to factorials. We can expand the deltas so to have:$$f(x)=\sum_{m=0}^{\infty} \binom {x}m \sum_{k=0}^m\binom mk(-1)^{m-k}f(k) The later formula is ...

4

To attack this more theoretically: if A and B have a common eigenvector, then that must lie in $\ker Q$ and obviously $[Q,A]$ and $[Q,B]$ act trivially on this guy. Thus, we can take the perp to this eigenvector and ask the same question about the leftover matrices. Thus, we should assume that $A$ and $B$ have no common eigenvectors; in this case, you get ...

3

The forward and backward finite differences and the derivative lower the degree of a polynomial by one. This property underlies the construction of series expansions of polynomials and, therefore, analytic functions with appropriate convergence properties in terms of diverse polynomial sequences, in particular, Sheffer sequences. In the terminology of the ...

3

Short answer from trying random matrix inputs: it is not true for $A = \begin{bmatrix}0.25634 & 0.417943& 0.696104 \\ 0.417943 & 0.0327021& 0.921007 \\ 0.696104 & 0.921007 & 0.0685762 \end{bmatrix}$ and \$ B = \begin{bmatrix}1.64721 & 0.29856 & 0.455064 \\ 0.29856 & 0.448668 & 1.56321 \\ 0.455064 & 1....

Top 50 recent answers are included