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Random Stuff X: Something Weird


soyperson

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On 5/25/2017 at 3:36 AM, Chaos said:

At first this had me turn my head, but the answer to your question is simple: |x-1| and |1-x| are the same function. 

You see, when you integrate 1/x, the antiderivative is not ln(x), but actually ln|x|. That absolute value is important because now those two functions have the same domains. That is essential here because when you integrate 1/(x-1) you get ln|x-1|. 

But really, by looking ln(x-1) vs. ln(1-x) they are the same function except for some domain issues. The absolute values fix that. ln|x-1| contains those smaller domained functions. Basically those two functions are the two halves of ln|x-1|.

Of course, if you use Wolfram Alpha plotting you'll see that their graph of their real part (if you allow complex solutions) are exactly the same: http://www.wolframalpha.com/input/?i=y+%3D+ln(x-1) http://www.wolframalpha.com/input/?i=y+%3D+ln(1-x)

How would I apply this in a differential equation? For example, if I had the equation dx/ab = (1/(ac - y)) dy, the first step to solving would be to integrate both sides, which would result in x/ab + C = -ln|ac - y|. How would I deal with this absolute value when trying to isolate y?

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41 minutes ago, StrikerEZ said:

So, I was in the shower, and I came up with something.

Branderson. When you're too lazy to say Brandon Sanderson, but don't want to just say Brandon or Sanderson.

Well, one of my friends called him Brand Sand and Brando Sando once so it could be worse.

@Sunbirb those are great. My school's music office has the three headed giraffe one on their noticeboard, with the caption 'alien giraffe.'

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7 hours ago, Oversleep said:

I already posted it in Math & Science thread but I figure more people frequent this one...

main-qimg-5b0690e302a38cf2a8068158199e7a

...one hour later, when I should probably have been doing homework or something useful:

Spoiler

The best I can come up with is "no, you can't".

 

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I thought this photo of an apparently headless Cooper's Hawk would amuse y'all:

Spoiler

592bb23c1ec3f_IMG_0211_Coopers_Darken_and_Crop.thumb.jpg.e88bcf6cbc0d698956da73cbce301bfb.jpg

He was preening his back feathers; here's what he looks like normally XD

Spoiler

592bb242f0af2_IMG_0216_Coopers_Lighten_and_Crop.thumb.jpg.8dec07b6a9850ad141b5cca42d3cb82f.jpg

Random side note: @Mestiv, your sig has nailed down my city but failed to notice that my display name is now Sunbirb, not Sunbird. :P

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On 05/26/2017 at 4:43 PM, Silverblade5 said:

How would I apply this in a differential equation? For example, if I had the equation dx/ab = (1/(ac - y)) dy, the first step to solving would be to integrate both sides, which would result in x/ab + C = -ln|ac - y|. How would I deal with this absolute value when trying to isolate y?

My knowledge of calculus doesn't extend far, but I'll share what I've got:

The shortcut here is, if you have a point on f(x), "initial condition" is what my teacher called it, to plug in that (x,y) pair into the half-solved differential equation and see if the term inside the absolute value is positive or negative. If it is positive, you can drop the absolute value, because it isn't necessary to find the solution. If the absolute value ends up containing a negative value, multiply the interior of the absolute value by -1 and then drop the absolute value bars. It's the same sort of thing that you have to do with the square root when you're solving a differential equation whose result is a circle and you need to determine whether or not you need the top or bottom half of it.

If you don't have an initial condition, I can't help you too much. My course didn't cover that. Maybe define a piecewise function based on y being less than or greater than ac?

[\CrazyMathStuff]

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19 hours ago, Oversleep said:

I already posted it in Math & Science thread but I figure more people frequent this one...

main-qimg-5b0690e302a38cf2a8068158199e7a

12 hours ago, Exalted Dungeon Master said:

...one hour later, when I should probably have been doing homework or something useful:

  Reveal hidden contents

The best I can come up with is "no, you can't".

 

(I don't want to deal with that quoting into edits glitch, so double-post it is)

Brute-force algebra has failed me, so I've attacked this one with critical thinking, and here's what I've got so far:

(A, B, and C could be any of the fruits; the problem doesn't change whether A is apple or banana or pineapple, etc)

IF a solution exists with the specified domain restrictions:

A > B >= C

A > (B + C)

A > 6

A, B, and C are either all odd numbers or all even numbers

And, with a little help from an online graphing calculator, I can say that none of A, B, or C = 0.

That's all the time I've got right now.

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@Oversleep - That's 1 equation and 3 unknowns, which means you cannot get a unique algebraic solution. 
That doesn't mean it's insoluble, just that the solution won't be unique, and you can't use standard algebraic methods. 

In other words, it would take a loooot of time, or a really lucky guess. 

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9 minutes ago, Erunion said:

@Oversleep - That's 1 equation and 3 unknowns, which means you cannot get a unique algebraic solution. 
That doesn't mean it's insoluble, just that the solution won't be unique, and you can't use standard algebraic methods.

I know :D

The whole thing is a very elaborate trolling.

Spoiler

The "95% of people cannot solve it" is actually true.
One solution is
x = 437361267792869725786125260237139015281653755816161361862143‌7993378423467772036;
y = 68751317941299998271978115652254748254929799689719709962831‌37471637224634055579‌;
z = 154476802108746166441951315019919837485664325669565431700026‌63489825320203527799‌9
here is explanation

 

Edited by Oversleep
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