Recent content by D H

Do that and you should get zero, and you will get zero if you do the integration correctly. They are orthogonal functions, after all. What you need to integrate are ##\left(Y_1^0\right)^2## and ##Y_1^{\pm1}\,\left(Y_1^{\pm1}\right)^\ast##. You are integrating over the surface of the unit...

Your ##f(x+y)=f(x)f(y)-[f(x)+f(y)]-1/2## isn't consistent with the problem statement. If one takes ##x=0, y=1##, then ##f(0+1)=f(1)=f(0)f(1)-(f(0)+f(1))-\frac12##, and since ##f(1)=2## is a given, this means ##f(0)=\frac92##. However, the problem statement says ##f(2*0) = f(0)= f^2(0) -...

The equation ##\tan\frac\theta 2 = \sqrt{\frac{1-e}{1+e}} \tan\frac\phi 2## pertains only if ##0\le e<1##. Here, ##\theta## and ##\phi## are the eccentric and true anomalies. In the case that ##e=1##, one of two (or both) of total energy and total angular momentum is zero. Angular momentum...

Any logical expression can also be simplified to NAND only, NOR only, and if one allows ##c\to\text{false}##, also by implication only as ##c\to\text{false}## is NOT c. That expression can be write as ##((A\vee \neg B) \vee (\neg C \vee D))##, and now one can dispense with the parentheses...

An interval scale means both more and less than what you wrote. It does not mean "with respect to 0 °C". An interval scale instead means that the ten degree change from -5 °C to +5 °C has the same meaning as does the ten degree change from 10 °C to 20 °C. The zero point is arbitrary in an...

That is an awful diagram. It is unsourced and it is just wrong. While estimates of the Sun's effective temperature (ideal blackbody temperature) do vary, the variance is not that much, from 5772 to 5780 kelvins. That is well above 5250 °C, or 5523 kelvins.

The one solution that can easily be ascertained from the graph is ##y=y=0##. This is a false solution. There are four points that represent true solutions where neither ##x## nor ##y## is zero.

No! It refers to ##P\to Q##. You are still misunderstanding modus ponens and modus tollens, the intent of which are to create new knowledge. Suppose you know ##P\to Q## is true, and nothing else. That alone says nothing about ##P## or ##Q##. Now you are given that ##P## is false. Since ##P \to...

Just piling on: Please refrain from doing this. Instead keep the 0.75 as ##3/4##. Doing so will benefit you in the long run. Plus, this is a sign that you are relying on your calculator far too much. Keeping this to rational factors, this is better expressed as ##\frac{27}{64}m^3 -...

There's no need to solve for ##t_1## and ##t_2## as the problem statement only asks for ##h##. The OP has already solved for ##T##, except the OP plugged in numbers as soon as possible. Keeping it symbolic, T = \sqrt{\frac{2H}g}. Similarly, the time to fall from a height ##H## to a height ##h##...

I previously gave an incorrect explanation for the subtraction of half of a day in the second pass at estimating solar noon. Here is a correct explanation. The line prior to the calculation of newt (the calculation of solNooNOffset) can be merged with the line you are questioning as var newt...

Now I see the problem. Modus ponens is not about the truth table of ##[(p \to q) \land p ] \to q##. It is instead about what one can deduce about ##q##, namely that one can deduce that ##q## must be true if both ##p \to q## and ##p## are true. If either of (or both of) ##p \to q## or ##p## is...

That is not the case. Modern ponens is anything but a tautology. There is only one pair of ##(p,q)## values that lead to ##[(p \to q) \land p ] \to q## being true, and that is when both ##p \to q## and ##p## are true. What is a tautology is $$[(p \to q) \land q ] \to q$$.

That is correct. Sometimes you do get what you pay for when you use a free site, which is nothing. It's a bit less than a week; I get 4.83 days. The correct version of Kepler's third law is ##a^3\omega^2 = G(M+m)##, as opposed to ##GM## which is what one obtains ignoring the mass of the...

On the other hand, assuming constant acceleration all the way out to the Moon is unreasonable. This assumption leads to 2.4592 hours. Perhaps the calculator is using a non-standard value of ##g## to arrive at 2.4595 hours. It shouldn't be arriving at that value at all; the assumptions leading to...