$$ { ${- x^{y}}$ } $$
$$ {} $$

About 10,000 results in 0.53 seconds.

$$ -x^y$$

What does -(-... mean in this equation?

...The author seems to be assuming that the unary minus has a higher precedence than exponentiation; i.e. that \( -x^y \) means \( (-x)^y \) rather than \( -(x^y) \) . Without any parentheses...
$$ - x^y$$

Why isn't $x^\frac{-1}{2}$ equal to $-\sqrt{x}$

...\( x^{-y} \) is \( \frac1{x^y} \) , not \( - x^y \) . Added: \( \sqrt[-2]{x} \) is the positive real number \( u \) such that \( u^{-2}=x \) , so it's \( u=\frac1{\sqrt x} \) .
$$ (y-x^2)$$

Exercise in Hartshorne

[email protected] In fact, I don't get it. It is possible that a prime ideal strictly contains another one. Imagine the second example. the map is onto and \( (y-x^2) \) is prime, but still there is another prime ideal \( (y-x^2,z-x^3) \) strictly containing it.
$$ (y-x^2)$$

Showing a point is generic

...How does one show that something is a generic point? I feel I may be missing some subtly. Vakil in his notes asks one to show that the prime ideal \( (y-x^2) \) is a generic point of \( V(y-x^2) \) ? That is we need to show that the closure of \( (y-x^2) \) is \( V(y-x^2) \) . But this seems trivial since \( V(y-x^2) \) is by definition the smallest closed subset containing \( y-x^2 \) . I am tired, though...
$$ y^{2}-x^{3}$$

Can we classify plane cubics, What are they?

...There are four qualitatively distinct pictures of the plane cubics. What are the polynomials corresponding to them? I know two of them have special names: nodal cubic and cuspidal cubic with polynomials \( y^{2}-x^{3}-x^{2} \) and \( y^{2}-x^{3} \) respectively. What are the others?
$$ \sqrt{9-x^2-y^2}$$

Why are the contour lines of a sphere more spread out near the equator, if a sphere is homogeneous?

...I realize that this question may not be a formal as you're used too, but I'd appreciate it if you'd bear with me. So, I have a conceptual question about the contour lines of a sphere (in this case \( \sqrt{9-x^2-y^2} \) , the top half of a sphere. It is not so hard to follow the actual computations. But the lines are concentrated towards the edges of the sphere along the x - y axis. Since the curvature of the sphere is the same at all points, this seems to imply that the sphere is 'steeper' in that region than t...
$$ y^2-x^3-x^2$$

Show that two quotient rings are not isomorphic

...Why need the characteristic zero in order to conclude that \( y^2-x^3-x^2 \) is irreducible? This holds over every field (and over every integral domain).
$$ z=1-x^2-y^2$$
$$ \sqrt{\left(x-x_{o}\right)^{2}+\left(y-y_{o}\right)^{2}}$$

Proving the set $S=\left\{ \left(x,y\right);ax+by

[email protected]астонвіллаолофмэллбэрг yes i know it is \( \sqrt{\left(x-x_{o}\right)^{2}+\left(y-y_{o}\right)^{2}} \) ...
$$ (x-y)^2(y-z)^2(z-x)^2$$

Let $(x, y, z)\in \Bbb R$, prove that $\big(\frac{2x-y}{x-y}\big)^2+\big(\frac{2y-z}{y-z}\big)^2 +\big(\frac{2z-x}{z-x}\big)^2 \geqslant 5$

...um_{cyc} y^4 z^2 -4 \sum_{cyc} x y z^4 +\\ 14 \sum_{cyc} xy^2 z^3 -10 \sum_{cyc} x y^3 z^2 -3x^2 y^2 z^2 \geq 0 \end{eqnarray*} This can be rearranged to \begin{eqnarray*} (2x-y)^2(y-z)^2(z-x)^2+(x-y)^2(2y-z)^2(z-x)^2+(x-y)^2(y-z)^2(2z-x)^2 \geq 5(x-y)^2(y-z)^2(z-x)^2 \end{eqnarray*} Now divide by \( (x-y)^2(y-z)^2(z-x)^2 \) and we have \begin{eqnarray*} \frac{(2x-y)^2}{(x-y)^2}+\frac{(2y-z)^2}{(y-z)^2}+\frac{(2z-x)^2}{(z-x)^2} \geq 5. \end{eqnarray*} Edit: I worked my way backwards from the result, using the "reduce" algebra package. I then guessed a square that would "consume" some of the terms & got lucky ! (2*x-y)...
$$ (x+y+1-x^2-y^2)/(1-x^2-y^2)$$

Calculating the surface area of sphere above a plane

...ok I get a very messy formula for the cross product: sqrt( \( (x+y+1-x^2-y^2)/(1-x^2-y^2) \) )... I take that this is right and I set x^2+y^2=r^2? the limits would then be r=0 to r=sqrt(3/4) and \theta from 0 to 2pi? how did you "guess" the answer so quickly when this is so complicated thought?
$$ \sqrt{\frac{a^2-x^2-y^2+x^2+y^2}{a^2-x^2-y^2}}=\frac{a}{\sqrt{a^2-x^2-y^2}}$$

Surface area of a sphere of radius a using double integral area formula

...Just add all the fractions in there. You will have \( \sqrt{\frac{a^2-x^2-y^2+x^2+y^2}{a^2-x^2-y^2}}=\frac{a}{\sqrt{a^2-x^2-y^2}} \) ...
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