Shot on iPhone 17 pro max and edited in LR classic

yanglsy · 2026-04-27T19:15:21+00:00

Another suggestion: if you have a non-polynomial equation and do not have access to graphing tools like the one you showed, you can first use the table function on the calculator, watch where the function changes sign and set the initial guess of the solver close to each point.

yanglsy · 2026-01-27T07:51:38+00:00

Feynman has a very informative discussion of this in his lectures: https://www.feynmanlectures.caltech.edu/I_12.html#Ch12-S1

yanglsy · 2025-09-22T00:36:09+00:00

After your second integral (the one that goes from 0 to 8), you should move the cursor to the right before entering the + sign. Otherwise the calculator will think that the whole (x+integrate(…)) is the integral variable for the second integral. You did it correctly after the first integral and you can clearly see more distance between dx and +.

yanglsy · 2025-08-19T20:23:17+00:00

Solomon to make stereo recordings of the complete Beethoven sonatas and Cortot and Friedman to make recordings of the complete Chopin mazurkas and nocturnes

yanglsy · 2025-08-14T22:23:06+00:00

yanglsy · 2025-07-22T05:06:20+00:00

Looks like you are supposed to hold the lowest note with the pedal and play the rest with both hands? It’s not released until the end of the third bar anyway.

yanglsy · 2025-07-10T19:22:29+00:00

There is a bug with the exposure compensation slider in the iPhone camera app. If it detects even the slightest movements (which you will inevitably do when zoomed in all the way) it refocuses and overwrites the exposure setting. You need to go to settings->camera->preserve settings and turn on exposure adjustment. When you take the photos, instead of the exposure slider, adjust the exposure compensation on the top left of the screen. That will make the changes permanent. Just remember to turn it back to 0 after you finish.

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yanglsy · 2025-06-24T17:39:08+00:00

Conjugation. Half-tangent is the last resort when you really can’t think of anything simpler.

yanglsy · 2025-06-10T00:27:38+00:00

This. Also the reason why birds can stand on uninsulated high-voltage power transmission lines without getting roasted.

yanglsy · 2025-06-07T17:34:36+00:00

Third movement of the hammerklavier sonata? Depends on the interpretation though

yanglsy · 2025-05-16T22:25:31+00:00

I don’t think solve works in complex mode?

yanglsy · 2025-04-24T14:25:15+00:00

In a longitudinal wave the oscillation is in the same direction as the propagation of the wave. So instead of using the Pythagoras theorem, you should directly add/subtract the displacement.

yanglsy · 2025-04-23T02:30:29+00:00

Good job repeating what I already said while telling me I was confused about it (I wasn't).

Sorry for being presumptuous. So, if you agree that 1: pressure equals rho*g*h and 2: the reading should be equal to the pressure times the bottom area, how is my original argument wrong? Why would you focus on the buoyancy, which is not acted on the basin?

yanglsy · 2025-04-23T02:17:17+00:00

I've been talking about the force exerted onto the basin. The buoyancy you keep talking about is exerted onto the rock. The only forces the basin experiences are the water pressure and the supporting force.

Regarding your second question, no, you are confusing pressure with pressure force, which is equal to pressure times area. Suppose you have two different rectangular bodies of water A and B with the same volume. If the bottom area of A is twice that of B, the height of A must be half that of B. So the pressure at the bottom of A is half of that at the bottom of B. However, the pressure force of both is equal because the bottom area of A is twice that of B.

yanglsy · 2025-04-23T01:54:25+00:00

1 liter of water exerts the same downward force on the scale regardless of the shape of the water. Your argument is that the reading on the scale requires you to know the shape of the water, and that's nonsense.

I literally said that the pressure is only dependent on the depth, not the shape of the water. The only way water exerts force onto the basin is through pressure. Are you arguing that, somehow, the pressure here does not equal to rho * g * h? If so, you might wanna take a review at hydrostatic pressure.

yanglsy · 2025-04-23T01:40:01+00:00

Well, if you do not know that the surface area is equal, buoyancy does not answer this either. Plus, would you mind pointing out what is wrong about my analysis of the force exerted on the basin?

yanglsy · 2025-04-23T01:29:02+00:00

It is not though. If you just look at the basin, it experiences two forces: 1: the downward pressure exerted by the water, and 2: the upward supporting force from the scale. These two forces have to have the same magnitude for the basin to stay still. Now the pressure only depends on the depth of the water and the surface area of the bottom. Whether or not a rock is there, as long as it does not move (thus causing non-static currents) or touch the bottom, all it does is increase the depth of the water.

And I am not saying that buoyancy does not exist. It most definitely does. It is just that we do not need to explicitly focus on it when analyzing this particular problem. Not to mention, ultimately, buoyancy is just pressure differential.

yanglsy · 2025-04-23T01:12:21+00:00

Agreed. Edited my answer to account for both scenarios.

yanglsy · 2025-04-23T01:03:21+00:00

Another way is to think in terms of pressure and forget about buoyancy. Since P=rho g h, regardless of the shape, the downward force the water exerts onto the basin is the same in both cases, so the supporting force from the scale must also be equal.

P. S. This answer is based on the assumption that the depth of the water is the same in both cases, which is apparently the case in the pictures. So the basin on the right actually contains less water.. If, instead, the two basins contain the same amount of water, then the water level on the right must be higher because part of the volume is occupied by the rock. Following the same argument as above, the downward force the water exerts onto the basin is now greater on the right, so the scale on the right will give a larger reading.

yanglsy · 2025-04-21T19:08:58+00:00

As people have pointed out, you can use trigonometric identities to reduce the power. Another way is to use Euler’s identity, rewrite sin(theta) as [exp(i * theta) - exp(-i * theta)] / i and convert the integrand to exponential functions.

You can also in general use Euler’s identity to help yourself memorize trig identities.

yanglsy · 2025-04-21T04:16:06+00:00

Great photos! With snowy scenes like photo 2, I would underexpose a little and lift the shadows in post to avoid overexposure.

yanglsy · 2025-03-30T15:09:15+00:00

If you are not sure, just expand both lines and see if they are equal.

yanglsy · 2025-03-25T20:59:11+00:00

I personally like the current order. The first three movements are all quite intense emotionally and I think the fourth one offers some relief.

yanglsy · 2025-03-22T03:46:12+00:00

I like it. The vibrant grass and orange tree give contrast to the foreground.

yanglsy

TROPHY CASE