Recent content by anuttarasammyak

We can easily integrate your definition integral and get $$ y=\ln x - \ln 1 = \ln x $$ z does not appear there. So we know yoiur definition gives the condition $$ \ln z =1 $$ to the last equation in post #3 which is a general solution

Not only laser but light in general do not lose energy in vacuum. Laser beam go in one direction. A beam receiver get the energy without the loss. Light, for an example sun light, goes all the direction whose energy density drops by r^-2 law. Total r^2-surface surrounding the light source get...

Say $$x=z^y$$, then $$y=\frac{\ln x}{\ln z}$$. Is it enough ? Graphs for some z values.

Yes. For an example, let p, q, r be prime numbers. There are infinite number of prime numbers. Infinite number of sets $$\{p^1,p^2,p^3,...\}$$ are disjoint infinite subsets of $$\{1,2,3,...\}$$. Infinite number of sets $$\{p^{q^1},p^{q^2},p^{q^3},....\}$$ are disjoint infinite subsets of...

These solution wave functions must be zero outside the box. They are not simple sinusoids functions. Their momentum wave functions do not give simple two opposite values. Ref. https://en.wikipedia.org/wiki/Particle_in_a_box

Such a tower is considered as the space elevator

$$(x-x_1)(x-x_2)=(x-\frac{x_1+x_2}{2})^2-(\frac{x_1+x_2}{2})^2+x_1x_2=(x-\frac{x_1+x_2}{2})^2-(\frac{x_1-x_2}{2})^2$$ or $$(x-x_1)(x-x_2)=(x-\frac{x_1+x_2}{2}-\frac{x_1-x_2}{2})(x-\frac{x_1+x_2}{2}+\frac{x_1-x_2}{2})=(x-\frac{x_1+x_2}{2})^2-(\frac{x_1-x_2}{2})^2$$ Apply the same method to y...

https://www.feynmanlectures.caltech.edu/II_13.html 13-6 shows excellent thought on current wire which seems same to your condensor case.

In case : ##\frac{1m}{v} >>## time resolution width of simultaneity ##>>\frac{1m}{V_{bullet}},##time for information processing and shooting action The players can make use of simultaneity shooting rule : In case the first succeeds, the second shoot within the time resolution to disuturb the...

Using this optimal case, Plyaer 1 Left would find a strategy which he ties with Player 2 Right. Say Player 2 shoots at this optimal distance in case he has found no shooting of Player 1. Player 1 shoots a tiny bit before the optimal distance so that Player 2 does not recognaize it due to finite...

I would estimate it using post #5 $$p_1=0.381966... , p_2= 0.618033...$$ where 1 is left and 2 is right. Firing at the same time in that optimal case Probabilities of Player 1 wins : ##p_1(1-p_2)=p_1^2## Player 2 wins : ##p_2(1-p_1)=p_2^2## No contest : ##1-p_1^2-p_2^2## Relative...

$$p_1(r)+p_2(r)=1$$ gives optimal firing position r for both the players where ##p_i(r)## is the probability of success of i-th player at position r. In this particular case $$1-r+1-r^2=1$$ $$r=\frac{\sqrt{5}-1}{2}$$ $$p_1=\frac{3-\sqrt{5}}{2}$$ $$p_2=\frac{\sqrt{5}-1}{2}$$ Success probalities...

$$1-x^2=(1-x)(1+x)>1-x$$ I know that the right player has advantage. I do not think the left has a strategy to win with more than 50% probability.

I have learned expansion of $$\cos(A+B),\cos(A-B),\sin(A+B),\sin(A-B)$$ in high school math. Let w=0, d=0 in the formula we get 2 which corresponds to width of a synthesized single slit. Let w=1, d=0, we get 3 which again corresponds to width of a single slit. We deduce that width of a slit is...

These are called arithmetic operations. I do not find a direct relation to partitioning or decomposing a set into disjoint subsets which I suspect your main subject is. Results of arithmetic operations to natural number except division by 0 which is not defined belong to 0 and plus-minus...