CC's Degenerate Shitpost Hole @chaoticcreaturehideaway - Tumblr Blog

Honestly, would anyone say you haaaaave to let a law student observe your client meeting, for their educations?

Sometimes I have law student ducklings following me around. This video actually inspires me to be more careful about asking when it’s okay that they observe; I tend to not have much time on court days so I just set the ducklings to “FOLLOW” and skip past it.

I know it’s caused some discomfort in the past. I guess I tended to think of it as unimportant discomfort? That’s on me; it’s so easy in criminal justice to see that your clients are suffering at every stage of the case and forget that doing something like paying attention to minor discomforts can actually make a difference. And I love having interns to teach.

But I also have blindness in that I’m someone who always lets med students in. I tend to view it as a positive, in that they’re my captive audience and I can do bits on them until someone laughs. They also get to hear about how doctors have disappointed me.

curlicuecal

Oooh, I was so braced at the beginning of that video but I loved where he went with it.

(And I am someone who also always lets students into appointments--as someone who teaches this stuff it's definitely all familiar territory to me. But nobody should ever feel obliged.)

girlnephew

ryanthedemiguy

I don't think this is possible????

physicallyimprobable

Hello Ryan I am here to help. So the first step is pretty easy: Three cheeseburgers are worth 18, so each one is worth 6. If these are dollars, that's a steal!

From the second equation we get that cheeseburger plus fries-squared is five. Subtracting cheeseburger, which is six, from both sides, we get that fries-squared is negative-one. Math fans will know that there are two solutions to this; either fries are the "imaginary unit" 𝒾 or they are its negative, -𝒾. We'll do the rest of the problem with 𝒾, keeping in mind that at the end we should also take the complex conjugates as solutions.

Finally, we have that cup to the power of fries, minus cup, equals three. Replacing fries with 𝒾, and moving a cup to the other side, we get that cup-to-the-𝒾 is equal to cup-plus-three.

Now, the weird part about this is the cup-to-the-i. The problem with this is that complex exponentiation is technically not a thing. That is to say, there is no one function which is mathematically equal to "input-to-the-power-of-𝒾". In fact, there are infinitely many such functions.

Fortunately, due to reasons that take about six pages to explain (trust me I've done it), there is one particular function that many people have agreed is "the most reasonable one". This is not a mathematical notion, but a human preference. Seeing as this question was presumably written by a human, I am comfortable with using this function.

So, what function is this? Well, given a complex number r∠θ written in polar form (if you don't know what that means don't worry), where -π < θ ≤ π, then (r∠θ)^𝒾 = e^(-θ)∠ln(r).

Applying this to our problem a value r∠θ will be a possible solution for cup if e^(-θ)∠ln(r) = r∠θ + 3. Splitting this into real and imaginary parts, we get two equations: e^(-θ) cos(ln(r)) = r cos(θ) + 3 and e^(-θ) sin(ln(r)) = r sin(θ). We can graph these equations on Desmos:

The possible values of cup are the intersections between the red, green, and purple. There are infinitely many of these which have an angle of around -π/3, and there are two weirdos: One which is a complex number very close to -2.98, and one which is somewhere around -25. The possible values for cup are all of these infinitely many solutions, and also all of their complex conjugates.