Recent content by hotvette

Thanks!

Not homework, just trying to understand a statement in the book. On page 158 in Fisher, the following statement is made: In these applications of the Residue Theorem, we often need to estimate the magnitude of the line integral of e^{iz} over the semicircle = Re^{i\theta}, \; 0 \le \theta \le...

I figured it must be simple. I too often miss the obvious. Thanks!

Not hw, just reading the textbook. In section 1.5, page 50, the book goes through an explanation that \sin(x+iy) is one-to-one if 0 \le x < \pi/2 and y \ge 0. At one point the book states that for 1 = -e^{-i x_1}\,e^{-i x_2}\,e^{y_1}\,e^{y_2} the absolute value of the left side is 1 and that...

Drat, I see it now. Thanks! \begin{align*} \frac{y^2-x^2}{(x^2+y^2)^2} &= \frac{y^2-x^2}{(x^2+y^2)^2} + g'(y) \\ \rule{0mm}{18pt} g'(y) & = 0 \\ \therefore v &= \frac{-y}{x^2+y^2} + C \end{align*}

Solution Attempt: \begin{align} \frac{\partial u}{\partial x} &= \frac{\partial v}{\partial y} = (x^2+y^2)^{-1} -x (x^2+y^2)^{-2} (2x) = (x^2+y^2)^{-1} - 2x^2 (x^2+y^2)^{-2} \\ \rule{0mm}{18pt} \frac{\partial u}{\partial y} &= -\frac{\partial v}{\partial x} = -x (x^2+y^2)^{-2} (2y) =...

The page is attached. The portfolio is hypothetical whose returns follow a normal distribution with mean 5.8% and SD of 6%. What I mean by given is that for a specified mean, SD, and number of periods, what is the distribution of returns after the n periods assuming mean and SD are constant...

I saw an interesting table in Asset Allocation (Roger Gibson) showing the distribution of portfolio annualized returns for a hypothetical portfolio with mean of 5.8% and standard deviation of 6%. It shows the return percentiles for various holding periods from 1 to 25 years. Can this...

Sure, that's just M. Ponens. Makes sense, thanks!

Really? So this means I can take any of the logic Rules and just replace anything with its negative and vice versa and it is still valid? Example of M. Ponens ##[p \land (p \implies q] \implies q## can be written as ##[\neg p \land (\neg p \implies \neg q] \implies \neg q##? The book makes no...

Is this as simple as letting ##p = \neg r## and ##q = \neg s##?

Book shows a proof where a conclusion is reached of: ##\neg r##. The next step says ##\neg r \lor \neg s## using the rule of disjunctive amplification. The rule of disjunctive amplification as I know it is ##p \implies p \lor q##. I don't see how from this you can also say ##\neg p \implies...

Thanks!

Not homework, just working odd numbered problems in the book. Sue has 24 each of n different colored beads. If 20 beads are selected (with repetition allowed) what is the value of n if there are 230,230 possible combinations. I view this as a problem of number of integer solutions to a linear...

I think I get it. For now I need to just use definitions and not try to inject my (flawed) sense of logic and intuition. Thanks for the explanations!