Sentence examples for a computational formula for from inspiring English sources

If p ≡ 1 mod 4, then we cannot give a computational formula for the hybrid mean value in Theorem 2.

For a general odd number q ≥ 3, whether there exits a computational formula for ∑ h = 1 q ( h, q ) = 1 R q ( 2 h + 1 ) ⋅ S 1 ( 2 h, q ). is an open problem.

Now we are concerned about whether there exists a computational formula for the mean value ∑ ∗ χ mod q | ∑ a = 1 q χ ( a r + n a s ) | 2 k, (3).

For general integer q ≥ 3, whether there exists a computational formula for the hybrid mean value ∑ ′ χ mod q | ∑ a = 1 q χ ( a ) ⋅ e ( n a 2 q ) | 2 ⋅ | ∑ ′ a = 1 q − 1 χ ( a + a ¯ ) | 2. is an interesting open problem, where n is any integer with ( n, q ) = 1.

For general odd number q ≥ 3, whether there exits a computational formula for the hybrid mean value ∑ ′ m = 1 q ∑ ′ n = 1 q | K ( m, q ) | 2 ⋅ | K ( n, q ) | 2 ⋅ S 1 ( 2 m ⋅ n ¯, q ). is an open problem.

A computational formula for C was given by Kakwani et al. [27] as ( C=frac{2}{ Nmu}{displaystyle sum_{i=1}^N{y}_i{R}_i-1} ), where ( mu =frac{1}{N}{displaystyle sum_{i=1}^N{y}_i} ) is the mean of observed costs, N the sample size, y i observed costs, and R i the fractional rank defined according to Kakwani et al. as ( {R}_i=frac{i-1}{N}+frac{1}{2} ).

A computational formula of the new discrepancy is also given by the functional method.

A computational formula of the new discrepancy, by the functional method, is also given.

end{cases}displaystyle end{aligned} Xu and Wang [3] obtained a complex computational formula for (a_{k}=F_{k}^{3}).

They considered the infinite sum of cubes of reciprocal (F_{n}) and then obtained a complex computational formula for Biggl[ Biggl( sum_{k= n}^{infty} frac{1}{F_{k}^{3}} Biggr) ^{-1} Biggr].

It is natural to ask whether there exists an exact computational formula for (M_{n}(p)) when n is a positive integer and p is an odd prime?