Strokes Theorem

My name is a dirty Calculus reference. Yes, math is an interest. Tech, nerdy references, deadpan one-liners, and things about life too (that apply to math).

[dacuala]

numpy

rhymes with lumpy

rhymes with dumb pie

Remaining time: 10 hours

[cauchyfeelings]

YOU HEARD IT HERE FOLKS @python3-official HAS SPOKEN

[stochastic-serenity]

Took a break from my women in maths drawing series to draw my cat

I’m thinking about drawing Sophie Germain next

[javascript-official]

I just fucked a platonic solid (dodecahedron for those curious ;). ) call that friends with benefits

[thememedaddy]

[javascript-official]

Women are objects

Men are objects

Non-binary people are objects

Everything is an object in JavaScrip becouse of its object oriented paradigm based on class constructors.

[icaseybaybee]

[alwaystiredshark]

More of my art @alwaystiredshark

Support me on: Kofi | Instagram | Youtube

[viralfrog]

[bagel-of-masyaf]

rarely laugh out loud at math memes but this got me

[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.

[sharkgirldick]

They were right, 99% of people can't solve it.

[peelerofpotatoes]

i've actually been working on some formulae to give all possible solutions to complex exponentiation problems recently, so here's my take on this:

let the value of the glass = z, for z ∈ ℂ:

z^(±i) = 3+z

let z = r·e^(iθ) for r,θ ∈ ℝ, -π < θ ≤ π

∴ z = r·e^i(θ+2πn) for all n ∈ ℤ

∴ (r·e^i(θ+2πn))^(±i) = 3+r·e^i(θ+2πn)

distribute powers (apologies for the use of ∓):

r^(±i)·e^(∓(θ+2πn)) = 3+r·e^i(θ+2πn)

convert to the same base:

e^i(±ln(r))·e^(∓(θ+2πn)) = 3+r·e^i(θ+2πn)

split into real and imaginary components:

re: cos(±ln(r))·e^(∓(θ+2πn)) = 3+r·cos(θ)

im: sin(±ln(r))·e^(∓(θ+2πn)) = r·sin(θ)

in effect, all this changes is the restriction on the domain of theta to be between -pi and pi, so you can just ignore that constraint.

[inthefallofasparrow]

[listacole]

@oddcabinboy

[1kit]

@anremithrl

[anremithrl]

*sigh*

I have a rant to go on, hold on.

[anremithrl]

A lot of these answers have the right idea. I have a lot of pedantic problems with the phrasing and what's going on here.

The first problem is there's no super clear indication of what ring we're working on.

For example another valid solution I might accept would simply be "we are working in a ring mod in some characteristic prime p where p mod 4 is 1." There are other ways. In general any ring in which -1 is a square, we have good algebraic extensions. In fact I'd allow any ring where there's a good argument for why exponentiation works, and allow people to extend it where necessary.

For example, in mod 29, we can do the same thing and get fries equal to 12 or 17.

On the point of 'infinitely many functions that can represent input to the power of'... this depends. There are, for example, infinitely many possible ways of doing input multiplied by other input, but in the end we choose one out of wanting it to correspond nicely with some properties. Input to the power of has a lot of functions that satisfy the properties we usually want, but we ordinarily choose additional properties with good reasons for them so that there is only one. I mean yeah at some point it's mathematical convention but so are the properties we care about for, say, addition. And then yeah this chosen function, which is no more a choice of convention than choosing to work in ℂ is not injective on ℂ.

(To this sentiment, 'addition' of ordinals violates the usual sense in which we define addition, even if it satisfied the general intuition of addition, while there are alternative ways to add do not satisfy this intuitive notion, (see symmetric difference in the natural ring on the power set) .)

In general I feel that if we're being pedantic the problem is not sufficiently well defined to admit solution.

Now let's say we do actually restrict ourselves to ℂ, which is a reasonable thing to do for an arbitrary thing. Then most of this works.

And yeah that kind of makes this inane but I just had this immediate feeling of annoyance at the phrasing above. This in fact goes fairly well to my point about the how cool the Lambert W function in some other post.

[strokes-theorem]

Tldr; werk that complex analysis. But like in life, in math, it's important to define your boundaries (or initial conditions). Makes it easier to identify what works and what doesn't.

[claimhte]

does anybody remember like maybe 10 or so yrs ago somebody's math friend died and they posted the late friend's books. I'm like 90% sure one of them was linear algebra? it was a really funny book

[ominous-signs]

Official terrifying sign

[anxietyproblem]

[thememedaddy]

[noelements-setempty]

Fun mathematical fact: you can divide by 0!

[dragonsampersanddragons]

One of my favorite things about working with hyperbolic 3-space is that i treat its boundary as the 1-point compactification complex plane, and then the symmetry group of the space is the mobius transforms on C with infinity. So .... I can divide by zero quite easily: any number divided by zero gives infinity. What i can never ever do is multiply by zero.

[noelements-setempty]

That's great, but it's even easier to divide by 0!

[strokes-theorem]

Still waiting on whether or not it's 0 factorial or an excited 0.

[rautadim]

i think every textbook should have this

From Information Theory, Inference, and Learning Algorithms (McKay)

[javascript-official]

Being into math is dangerous because you accidentally wonder "hmm I wonder if you can take an exponent and generalize instead of a number to a 2x2 matrix" and before you know it you watched 3 hours of lectures and learned Lie algebra and have come to realize it's one of the most beautiful pieces of mathematics ever constructed, don't do math kids, it's dangerous shit.