Kuuki ga "Yomeru" Shinnyushain to Buaiso na Senpai no Hanashi (Web-comic) - Vol. 2 Ch. 7

[OmniscientOnion]

Thanks for the translation!

Bread. Bread is love, bread is life.

 

[Purplelibraryguy]

That was a very cute, but actually kind of crappy explanation he gave. And then they got buns together, but didn't get her computer any more memory. I know, I know, the point is they were cute together--it just niggles at me.

 

[Chiyamii]

This series is so cute

 

[RedInACup]

Thanks for the chapter but there is no way in hell in doing that math.

 

[potatofood]

@redinacup you just have to translate it to english lol, how hard can it be? /s

 

[nmyhv1]

I don't need maths at the end of a chapter, I dropped out of my Bachelors Degree for a reason.

 

[shutup]

Whoooolesome.

 

[mase_aga]

No, not the calculus again

 

[alannondarklord]

@Bistai Good job expanding the analogy out.

 

[berryoragami]

Yay bread, very cute chapter
Thanks for translating!

 

[shira952]

It's morning and I haven't get a wink of sleep but then a wild math appeared, I collapsed.

 

[tilkku]

Been a while since I did calculus, but here goes.
(i) First prove that Ψ is an isomorphism and infinitely differentiable. Isomorphism follows by writing Ψ(x)={{1,0},{-1,1}}*x and noting that said matrix is invertible. Differentiability follows by Ψ being linear. V is given by finding the preimage Ψ^-1(U), that is, the set of points x in ℝ² for which {1,1}*Ψ(x)<u. Simple matrix multiplication!
(ii) This is just a change of variables using Ψ. The earlier matrix is the Jacobian of Ψ, which you can plug into the usual formula.
(iii) Apply the equation of (ii) and note that the integrand is convolution. Write h(y2)=∫ f(y1)f(y2-y1)dy1, where y1 is integrated over ℝ. Denote the Fourier transform of h by H and recall that the transform of (1) convolution is multiplication and (2) exp(-x^2) is sqrt(pi)*exp(-pi^2*x^2). Therefore H(x)=pi*exp(-2*pi^2*x^2), which is easily inverse transformed into h(x)=sqrt(pi/2)*exp(-x^2/2). In all, ∫f(x1)f(x2)dx = ∫∫f(y1)f(y2-y1)dy1dy2 = ∫h(y2)dy2 = sqrt(pi/2) * ∫ f(y2/2)dy2. Substituting y2 ↦ y2*sqrt(2) yields the result.

 

[potatofood]

@tilkku lmao I'm pretty sure I barely passed my math analysis final but it's all good since the class is pass/fail anyways now

 

[MasterPannya]

You should be asking your IT for this.

 

[BCS]

MANSPLAINING

 

[smugcatto]

-

 

[Makinbaconpancakes]

@tilkku I read the pop quiz to be "translate this question into English", rather than solve it. Trick question?

I know enough to read out all the parts, but not solve. Never did enough proofs, at some point will go back to learn them to level up my algorithm skills

 

[Milanin]

No matter how I look at it, it's Anpan! Is that another favourite food??
Will she conquer his stomach to get to his heart? Is this what you're telling me?

 

[sad_historian]

Was he dreaming about bread, or Shino? 🤔🤔🤔

 

[DatOneGuy]

Let's talk about bread final-shaping!
I've been baking recently and one (albeit minor) slow down in my workflow is the amount of workbench space that I have.
There is a process (pre-shaping) where I need to push/pull/shape my dough to get a nice tight outer layer, naturally that requires space. After that process, I transfer the dough into a basket for the final shaping. But I can only fit two dough balls on my workbench at once so I have to walk over to my basket, bring it to my bench, and transfer one loaf, then push/pull/shape my second loaf.
If I had a larger workbench I could have both baskets and both loaves all together.
(That's what happens when you run out of memory on a process and have to leverage slow storage.)

There's a good reason why gamers like to stack up on RAM to increase memory. And why you have to be careful of how much data you are manipulating at once, chunking it may be faster than trying to do it in action.