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/sci/ - Science & Math

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>> No.11657930 [View]
File: 108 KB, 566x720, __hong_meiling_touhou_drawn_by_uu_uu_zan__792bc027999ad3a93937b93d33bb52ee.jpg [View same] [iqdb] [saucenao] [google]
11657930

>>11657893
Very nice, thanks.

>> No.11639417 [View]
File: 108 KB, 566x720, __hong_meiling_touhou_drawn_by_uu_uu_zan__792bc027999ad3a93937b93d33bb52ee.jpg [View same] [iqdb] [saucenao] [google]
11639417

>>11639353
>40 dollars for a gigantic stack of loose paper
>imagine being ripped off into paying 40 dollars on fucking paper
I bought a 500 leaves stack of decent A4 white paper for 5 (five) dollars.
By printing each book page on half an A4, I can get 2000 pages out of it. If I'm a faggot and insist on printing shit properly, that's still a solid thousand pages.

>> No.11632494 [DELETED]  [View]
File: 108 KB, 566x720, __hong_meiling_touhou_drawn_by_uu_uu_zan__792bc027999ad3a93937b93d33bb52ee.jpg [View same] [iqdb] [saucenao] [google]
11632494

>>11630523
Diurnal emissions.
>>11632164
From staring at the equation I can guess that maximization occurs at [math](-2, 2)[/math].
So 0.
>>11632366
Kind of sorta. This is all very finicky so I'll try to explain it as best as I can.
We consider the good old [math]f(x) = f'(x)[/math], with solution [math]ae^x[/math].
Setting [math]y = -x[/math], you might think you get [math]f(y) = f'(y)[/math]. However, [math]f(y) = ae^{-y}[/math], so [math]\frac{df}{dy} = - ae^{-y}[/math], and the equation becomes [math]f(y) = - \frac{df}{dy}(y)[/math].
In your case, the only differentiation is square, so it probably works as is without you concerning yourself with "does [math]f'(y)= \frac{df}{dy}[/math] or [math]\frac{df}{dx} (y) [/math]" or any similar issues.
Otherwise it might just be a matter of flipping the signs for odd differentials.

The moral of the story is that notation can get fucking confusing sometimes.

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