Recent content by 6c 6f 76 65

Thank you for the reply! Sorry if I'm wasting your time, but I don't get what you mean by copy? (Is ⊕ the same as ×?) And ##A[X, Y]/(XY)## means, as far as I know, all polynomials, with coefficients from A, of variable X and Y. Mod XY means, XY = 0, so shouldn't the new ring be functions of only...

Homework Statement My textbook says that A[X, Y]/(XY) is a subring of A[X]⊕A[Y], but aren't they isomorphic? (A is any commutative ring) Homework Equations 1st Ismorphism Theorem The Attempt at a Solution I can construct the map φ: A[X, Y] → A[X]⊕A[Y] f(X)+g(Y)+h(X, Y)*X*Y → f(X)+g(Y), this...

Have you typed the the first equation correct? Because if not w_φ=0, and for the rest: remember that when you differentiate / integrate with respect to r you may have lost a function of θ.

Correct! Correct! Almost correct, what is the height of the "inner cylinder"? Well, you can make it a little simpler by saying the volume you're interested in finding is composed by: - Volume bounded by: x=1, x=2, y=0, y=\frac{1}{2} then get V_1=\pi\int_0^\frac{1}{2} (2^2-1^2)dy =...

Seems like you're heading in the right direction! As you creatively thought removing the cylinder is the key, now in order to compute the whole volume divide it into: - The volume of the "large thin cylinder", bounded by x=0, x=2, y=\frac{1}{2} (is the radius constant here?) - The volume of the...

Exactly!

Imagine seeing the rotated figure from above, it kinda looks like a volcano. Try to make a radius from the origin: one from the O to the yellow line, and one from O to the green line. How long will both of these radii be (as a function of x, then make the substitution y=\frac{1}{x})?

Think of it this way: You'll get the volume of the figure if you first find the volume under the purple line and the remove the volume between the purple line and the blue curve

I was looking for a more analytic expression like \sum_{i=1}^n i = \frac{n(n+1)}{2}. Maybe it's possible to find yet another connection to the Bernoulli numbers? But thank you nevertheless!

Good evening dearest physicians and mathematicians, I recently came across the so-called "Svein-Graham sum", and i wondered: is it possible to find a simple formula for evaluating it? \sum_{i=0}^k x\uparrow\uparrow i = \left .1+x+x^x+x^{x^x}+ ... +x^{x^{x^{x^{.^{.^{.^x}}}}}}\right \}k

I=\frac{1}{2}m R^2, thank you so much!

Thank you so much! Here's what I got: \frac{dK}{dt}=-\frac{B^2 R^4 \omega^2}{4 R_\Omega}=-\frac{B^2 R^2}{2 m R_\Omega}K, used that K=\frac{1}{2}m \omega^2 R^2 Solution for diff. equation: K=K_0 e^{-\frac{B^2 R^2}{2 m R_\Omega}t} \frac{1}{2}=\frac{K}{K_0}=e^{-\frac{B^2 R^2}{2 m R_\Omega}t}...

\frac{dKE}{dt}=-\frac{(\frac{B R^2 \omega}{2})^2}{R_\Omega}

Sorry, I don't quite know what you mean. The KE and the electrical power produced is linked via the angular velocity with the formula: W=\frac{(\frac{B R^2 \omega}{2})^2}{R_\Omega}t=\Delta KE

Open the parenthesis, then differentiate