Waldspurger formula over function fields
... Let K = k( D), where D ∈ A is square-free with non-zero even degree and the leading coefficients√ of D is not a square in F√ q (then ∞ is inert in K ). Let OK := A[ D] and Oc := A + c · A[ D] for each ideal c of A. Every character χ of Pic(Oc ) can be viewed as a Hecke character on K × \KA× via the ...
... Let K = k( D), where D ∈ A is square-free with non-zero even degree and the leading coefficients√ of D is not a square in F√ q (then ∞ is inert in K ). Let OK := A[ D] and Oc := A + c · A[ D] for each ideal c of A. Every character χ of Pic(Oc ) can be viewed as a Hecke character on K × \KA× via the ...
Partial Fraction Decomposition Examples
... The next step is to compare the numerators of the original question with the calculated one and solve for A, B, and C. Method #1 (difficult) is to simplify the RHS of the equation and compare coefficients thereby setting up three equations in three unknowns (e.g. all coefficients of x 2 terms will s ...
... The next step is to compare the numerators of the original question with the calculated one and solve for A, B, and C. Method #1 (difficult) is to simplify the RHS of the equation and compare coefficients thereby setting up three equations in three unknowns (e.g. all coefficients of x 2 terms will s ...
Semester 1 Final Exam Prompts
... I. During our study of quadratic and piecewise - absolute value functions in the first semester we studied manipulation of expressions and equations into different forms. Demonstrate your understanding of these different forms and the reason why you would make the conversion. ...
... I. During our study of quadratic and piecewise - absolute value functions in the first semester we studied manipulation of expressions and equations into different forms. Demonstrate your understanding of these different forms and the reason why you would make the conversion. ...
