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For example, the polynomial x 2 y 2 + 3x 3 + 4y has degree 4, the same degree as the term x 2 y 2. However, a polynomial in variables x and y, is a polynomial in x with coefficients which are polynomials in y, and also a polynomial in y with coefficients which are polynomials in x. The polynomial
Tunnell's theorem states that supposing n is a congruent number, if n is odd then 2A n = B n and if n is even then 2C n = D n. Conversely, if the Birch and Swinnerton-Dyer conjecture holds true for elliptic curves of the form y 2 = x 3 − n 2 x {\displaystyle y^{2}=x^{3}-n^{2}x} , these equalities are sufficient to conclude that n is a ...
In mathematics, an expansion of a product of sums expresses it as a sum of products by using the fact that multiplication distributes over addition. Expansion of a polynomial expression can be obtained by repeatedly replacing subexpressions that multiply two other subexpressions, at least one of which is an addition, by the equivalent sum of products, continuing until the expression becomes a ...
If y 2 = x 3 − x − 1, then the field C(x, y) is an elliptic function field. The element x is not uniquely determined; the field can also be regarded, for instance, as an extension of C(y). The algebraic curve corresponding to the function field is simply the set of points (x, y) in C 2 satisfying y 2 = x 3 − x − 1.
For example, y = x 2 fails the horizontal line test: it fails to be one-to-one. The inverse is the algebraic "function" x = ± y {\displaystyle x=\pm {\sqrt {y}}} . Another way to understand this, is that the set of branches of the polynomial equation defining our algebraic function is the graph of an algebraic curve .
Thus, the second partial derivative test indicates that f(x, y) has saddle points at (0, −1) and (1, −1) and has a local maximum at (,) since = <. At the remaining critical point (0, 0) the second derivative test is insufficient, and one must use higher order tests or other tools to determine the behavior of the function at this point.
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The two iterated integrals are therefore equal. On the other hand, since f xy (x,y) is continuous, the second iterated integral can be performed by first integrating over x and then afterwards over y. But then the iterated integral of f yx − f xy on [a,b] × [c,d] must vanish.