taeto

It makes sense, because remdesivir has been useful against other coronaviruses, SARS so far as I recall, or was it Ebola. Also because the molecule is similar to adenosine and seems able to trick a viral RNA polymerase to try to build it into new RNA strings where adenosine would have belonged, and thus blocking the further production of the viral RNA. On the other hand, the link states that "Gilead says", and Gilead is a (the?) manufacturer of remdesivir in the US. The available information about the study says that the decrease in lethality among test patients treated with remdesivir was not statistically significant (the shorter recovery times presumably were?) compared to the patients treated with placebo. And an earlier study of the same drug did not produce any determination. So maybe more tests are needed.

Well, some of the dinosaurs. At 30,578 km/h even a collision with such a smallish size would pose concern. Since there apparently exists a good estimate of the mass of the asteroid that hit 65M years ago, is there also a good estimate of its velocity? After all the energy depends only linearly on mass, but quadratically on speed.

Maybe. But it is gone now and can no longer answer.

Darn it, you had to do that. I do not even click the video, and I am still going to hear Road to Nowhere inside my head for at least the rest of the day. I will take one hour off:

And Speed of the Sound of Loneliness https://www.youtube.com/watch?v=eFvenjll1Bk&list=RDSJPX2SpcCqw&index=3

John Prine is in heaven now, due to corona. https://www.youtube.com/watch?v=JKPDFQRmG_M Life changed in many ways during the past four weeks. This however is one news that gives me most sadness.

From usual definitions it would be \(\pi/\infty = 0\) and \(0 \cdot \infty = 0,\) so that means \((\pi/\infty)\cdot \infty = 0.\) But it seems you are trying to treat \(\infty\) as if it were a natural number, and that has only small chance or working out. I do not think that you need to introduce \(\infty\) in the first place anyway.

What do you mean by dividing by "infinity"? When extending the real numbers by new elements \(\{-\infty, +\infty\},\) and extending the usual arithmetic of real numbers to the new structure, usually you get \(0\) when you divide a real number by \(\infty\). Maybe your "infinity" is not the same as \(\infty\)?

Dima, your guess is that there is some value of \(a\) for which the value of \(y\) does not change if we look at all possible value of \(t\), is that right? My guess is that this is not true. Meaning that for any value of \(a\) that we choose, there will be at least two different values of \(t\) that produce different values of \(y\). Do you agree that my description of our guesses is correct? Now, since you are the one who says that there exists a specific value of \(a\) which wins the argument for your case, do you not think that you should have to present this particular value of \(a\)? Are you waiting for someone who is smarter than you to do so?

I will try to figure out bold math symbols. Simply typing everything in bold: S^2 = (ct)^2 - x^2 - y^2 makes \(S^2 = (ct)^2 - x^2 - y^2 \) so doesn't do anything. Using bm: \bm{S^2 = (ct)^2 - x^2 - y^2} produces \(\bm{S^2 = (ct)^2 - x^2 - y^2}\). With mathbf: \mathbf{S^2 = (ct)^2 - x^2 - y^2} \(\mathbf{S^2 = (ct)^2 - x^2 - y^2} \) it works almost like it should, except for the missing italics. There is a mathrm, so why not \mathit{\mathbf{S^2 = (ct)^2 - x^2 - y^2} }? \(\mathit{\mathbf{S^2 = (ct)^2 - x^2 - y^2} }\). Nope. Or maybe \mathbf{\mathit{S^2 = (ct)^2 - x^2 - y^2} }: \(\mathbf{\mathit{S^2 = (ct)^2 - x^2 - y^2} }\). Interesting. And the slow approach \boldsymbol{S}^2 = (ct)^2 - x^2 - y^2: \(\boldsymbol{S}^2 = (ct)^2 - x^2 - y^2\). But \boldsymbol{S^2 = (ct)^2 - x^2 - y^2} \(\boldsymbol{S^2 = (ct)^2 - x^2 - y^2}\) actually nearly does it. Neat. But now the = and - are bold too. So it must be \boldsymbol{S}^2 = (\boldsymbol{ct})^2 - \boldsymbol{x}^2 - \boldsymbol{y}^2: \(\boldsymbol{S}^2 = (\boldsymbol{ct})^2 - \boldsymbol{x}^2 - \boldsymbol{y}^2\). Tedious but alright. Maybe \bs: \bs{S}^2 = (\bs{ct})^2 - \bs{x}^2 - \bs{y}^2? \(\bs{S}^2 = (\bs{ct})^2 - \bs{x}^2 - \bs{y}^2\). Well, then not. Done with this.

To promote hydroxychloroquine as a remedy against coVid-19 does not seem stupid at all, provided you own a lot of stocks in hydroxychloroquine manufacturing companies. Same if you replace "hydroxychloroquine" by "magic beans" in the previous sentence.

Why did you tell me that you know that \(a\) is not equal to \(\pi/6?\) I have to understand this first.

Sure, it depends on the theory you want to engage. A unit line segment has volume zero in the Euclidean plane, and measure one on the real line, and it has cardinality \(2^{\aleph_0}\) as a set of points in the real plane.

If I will not reach the end, then never. If I will, then, at whichever time that will be. So "vanishing", I get it now. At the end, if I reach it, I would have vanished to zero. But I am not sure that your image tells us much about places that are "infinitely small".

I only know Euclidean geometry and its various extensions that are locally Euclidean. The dimension of a point is zero here, and if you go to measure theory, then its size in terms of volume is zero as well.

I understand my own diagram in the way that it was constructed: there is an angle \(t\) that can be chosen freely, and there is an angle \(\alpha,\) or \(a\) as you call it now, which can also be chosen freely, except of course their sum \(t+\alpha\) must be less than \(\pi/2.\) There is nothing there about \(t\) or \(a\) being constant. Then the diagram contains an expression for the resulting value of the \(y\)-intercept of interest. So if you now say that your \(a\) is a constant, and you also say that it is not \(\pi/6\) (which is a constant), then how do you know that \(a\) is not \(\pi/6?\) Is it because you already know what \(a\) is? Can you explain how you know this, and what you think is the actual value of \(a,\) as a fraction of \(\pi\) or in decimal expansion?

I actually think that was the point.

On the other hand, if you look inside a black hole (don't go too close), then I suppose that mathematically, it has a point of size zero in its center which contains all its mass. Somehow the exact opposite of a photon, if you will. Whether this is absolutely a correct description or not of the reality or not, it could suggest why popular depictions of the big bang start from a very similar picture.

To say that something is "infinitely" small is certainly an unfortunate mode of description. And meaningless unless it really means that its size is zero. If we were to say that the mass of a photon is "infinitely small", we would make ourselves easy targets of ridicule. However, the mass of a photon is in reality not "infinitely small", rather it is identically zero. If we take "infinitely small", as a manner of speaking, to mean "of size zero", then we can all have a night of peaceful sleep. Except of course those who insist that nothing can be "infinitely small".

Assuming you worked it out correctly, this might be another version of the same identity. I have not checked it. I suppose that you have no additional questions? If we let \(t = \pi/4\) and \(a = \pi/6,\) then \(\sin t = \frac{\sqrt{2}}{2}\) and \(\sin a = \frac{1}{2}.\) Can you solve to find \(y\) for that case?

I do not understand any of that. But do you think that what you have now is useful to you, or have you got any additional questions? Just to make sure. Once you decide on assigning some value to \(t,\) like \(\pi/6, \pi/4, \pi/3\) or any other value, then we can calculate \(\sin t\) and \(\cos t\) just in the same way that we are used to?

I do not understand what you mean. If you set \(t = \pi /6\) in your equation, then we can (if we are lucky) solve the equation and find a solution \(a = a(\pi/6).\) If we set \(t = \pi/3\) we may also find a solution \(a = a(\pi/3)\) to the same equation. It is extremely unlikely that we will discover that \(a(\pi/6) = a(\pi/3)\) holds. I worry when you say that \(\sin(a)\) and \(\cos(a)\) are "unknown" and "can be defined". Once we have \(a,\) then normally they should be completely known and determined by calculation of the function values of \(\sin\) and \(\cos\) with argument \(a.\)

If this answers your question, then fine. All I was saying is that if you keep your \(a\) constant, then the point \((0,y)\) depends on the value of \(t.\)

It seems we agree on this, or at least something very close to it. To simplify more, it could be useful to introduce (in your notation) b = Pi/2 - a. Either way this is a function of t which is not constant. As t grows, the point (0,y) moves upwards on the y-axis.

I cannot find it either. Anyway, it means what it means 😂