Recent content by Billy Bob

The question is right. You are making an error when you say \pi^{-1}(\pi(x_1),\pi(x_2),\dots,\pi(x_k))=(\pi^{-1}\pi(x_1),\pi^{-1}\pi(x_2),\dots,\pi^{-1}\pi(x_k)). Try an example. \pi=(1 2 3),\sigma=(x_1,x_2,x_3,x_4,x_5)=(8 2 4 3 5), \pi^{-1}=(3 2 1).

Isn't this Wirtinger's Inequality?

Here's my way. Start over. (You won't use Abel or SBP.) You want to show |(1-x)f(x)-L| is small. Use a "trick" that L=(1-x)L/(1-x) and write 1/(1-x) as the geometric series. This enables you to express |(1-x)f(x)-L| as a single power series. Since a_n approaches L, we have |a_n -...

A= [cos x sin x -sin x cos x] A^2= [cos 2x sin 2x -sin 2x cos 2x] A^3= what would you guess! Yes, it's very simple, sorry. :smile: Now compute A^3, and if it doesn't match your guess, then your computation is probably wrong, not your guess!

Can't you guess the correct pattern just from this! :smile: Additional hint: A(x) corresponds to rotation through an angle of x.

What is the trig identity for cos(A+B)? For sin(A+B)? Also, your A^3 has two incorrect entries.

OK, I think I understand. I assume "integrable" means "Riemann integrable." Some incomplete ideas: Similar to your space filling curve idea, how about a line of irrational slope through the unit square, and whenever it hits the top or the right, it wraps around to the bottom or left...

Are you required to use ordinary Lebesgue measure for integration with respect to x and with respect to y? Or can you use, say, counting measure for the integration with respect to y?

An idea: An ordinary multivariable calculus textbook might have an example (I found one). It is in the same spirit as the standard examples of functions that don't have limit at the origin even though the limits along paths do exist. I can't yet figure out how the example was created...

Your answers are correct. The expected value of T1 is θ. However, any observed value of T1 may or may not be close to θ.

Use your intuition to decide what t should be. Draw an open disc of radius r centered at a. Draw an open disc of radius s centered at b. Draw c in the intersection. How far is c from the boundary of the first disc? How far is c from the boundary of the second disc? Take t to be the minimum...

Maybe for A^{2^{n-1}} but that doesn't help with A^5. In other words, multiply A^2 by A, not by A^2, to find A^3. (I know you know this -- your brain just went into overdrive and caused a blunder.)

Since 2 cos x sin x = sin 2x, a pattern is emerging, just not the one you thought.

0.6 Then try it with 1.7 Then try a generic t value.

It seems you took the root of the ratio. That's wrong. Don't combine the two tests. Use one or the other. The ratio test will work. (Your first step has a typo, but the second step has fixed it.) To finish it off, observe \frac{n^n}{(n+1)^n}=\frac{1}{\left( \frac{n+1}{n}...