Search Results for ""

Mergelyan's theorem can be stated as follows (Krantz 1999). Let K subset= C be compact and suppose C^*\K has only finitely many connected components. If f in C(K) is ...

The decimal expansion of the natural logarithm of 10 is given by ln10=2.302585092994045684... (1) (OEIS A002392). It is also given by the BBP-type formulas ln10 = (2) = ...

Pickover's sequence gives the starting positions in the decimal expansion of pi (ignoring the leading 3) in which the first n digits of e occur (counting the leading 2). So, ...

Closed forms are known for the sums of reciprocals of even-indexed Lucas numbers P_L^((e)) = sum_(n=1)^(infty)1/(L_(2n)) (1) = sum_(n=1)^(infty)1/(phi^(2n)+phi^(-2n)) (2) = ...

By analogy with the sinc function, define the sinhc function by sinhc(z)={(sinhz)/z for z!=0; 1 for z=0. (1) Since sinhx/x is not a cardinal function, the "analogy" with the ...

By analogy with the tanc function, define the tanhc function by tanhc(z)={(tanhz)/z for z!=0; 1 for z=0. (1) It has derivative (dtanhc(z))/(dz)=(sech^2z)/z-(tanhz)/(z^2). (2) ...

In each of the ten cases with zero or unit mass, the finite part of the scalar 3-loop tetrahedral vacuum Feynman diagram reduces to four-letter "words" that represent ...

An apodization function A(x)=1, (1) having instrument function I(k)=2asinc(2pika). (2) The peak of I(k) is 2a. The full width at half maximum of I(k) can found by setting ...

The apodization function A(x)=1-(x^2)/(a^2). (1) Its full width at half maximum is sqrt(2)a. Its instrument function is I(k) = 2asqrt(2pi)(J_(3/2)(2pika))/((2pika)^(3/2)) (2) ...

Andrica's conjecture states that, for p_n the nth prime number, the inequality A_n=sqrt(p_(n+1))-sqrt(p_n)<1 holds, where the discrete function A_n is plotted above. The ...