This OFFICIAL MIT blog is probably not designed for late-night delirious whims of posting. Oh well.
So, it's late (3:00 AM) and I just finished most of a Diff Eq. PSET (I started at 10:45 PM, it's been a long day).
Here's when you know it's time to go to bed.
Roommate Jon: "Did you get Part II, 4) d?"
Me: "No, solving for k was impossible, not even Mathematica could do it."
Jon: "I know, the same thing happened to me. I think I figured out how to do it though."
Me: "How?"
Jon: "Here, look what I did. . . "
Me: ". . . what!?!"
Jon: "Wait, you can't even solve this! I hate math."
Me: "I'm going to bed."
Comments (Closed after 30 days to reduce spam)
(that's Chinese...probably not the best place for this heh~)
(meaning first in forums)
the comic's from the Tech right?
Posted by: jinziling on February 22, 2008
Posted by: Alexander on February 22, 2008
Posted by: Alexander on February 22, 2008
Yea.....now i'm going to sleep to....!
PLEASE POST THE SOLUTION TOO!
Posted by: Roshini on February 22, 2008
Posted by: Tanmay on February 22, 2008
Posted by: Roshini on February 22, 2008
it's 4:30 PM in my timezone, so apparently my brain's working alright
Posted by: jinziling on February 22, 2008
Posted by: Tanmay on February 22, 2008
Posted by: 0 on February 22, 2008
Posted by: Anonymous on February 22, 2008
2. Hahaha your response to the nonsensical method is probably the funniest part of the whole thing.
"...what?!"
OH and did you know? "?!" has its own name! I heard it was called an interrobang.
Posted by: Shamarah on February 22, 2008
(I emailed you the solution snively, post it if you want to!)
Posted by: Khaled Saad ('12 hopeful) on February 22, 2008
Posted by: 0 on February 22, 2008
Posted by: Khaled Saad ('12 hopeful) on February 22, 2008
Posted by: Khaled Saad ('12 hopeful) on February 22, 2008
Posted by: Libin Daniel on February 22, 2008
Posted by: Libin Daniel on February 22, 2008
Posted by: Lauren on February 22, 2008
Posted by: Libin Daniel on February 22, 2008
Posted by: Shruthi on February 22, 2008
Posted by: AnonSpoiler on February 22, 2008
MIT ~ Brains.
Caltech ~ Brains.
MIT = Massachusetts Institute of Technology.
Caltech = California Institute of Technology.
MIT-> Good with science/technology.
Caltech-> Good with science/technology.
Germany-> Good with science/technology.
German-> Language of Germany (good with science/technology).
mit = German word for 'with.'
cit = not the German word for 'with.'
Caltech = not the German word for 'with.'
MIT -> "German word for 'with'" Brains.
Caltech -> "Not the German word for 'with'" Brains -> Not with Brains -> Without Brains.
q.e.d.
Questions? Comments? Can I help clarify? :D
Posted by: AwayfromHome on February 22, 2008
Posted by: steph on February 22, 2008
Posted by: kevb on February 22, 2008
Posted by: sara '11 on February 22, 2008
Posted by: sara '11 on February 22, 2008
And yes, although it looks like an XKCD comic (it was indeed parodied), Sara was right. It was the comic CalTech hacked my CPW with. I like to have fun with it. In fact, it's a greeting now between me and some friends.
"Math."
"Math."
"LOL."
In answer to some of the other questions (mainly as to why this looks nothing like a differential equation), that's because there was a whole lot of work done until this point and this is the boiled down version of an initial value problem. Do you want to know what the whole problem is?
Part II #4
Thanks for all of the help, now I may actually get this right when I turn it in this afternoon.
Posted by: Snively on February 22, 2008
I am asking this because I might be missing some important theory taught at the classes of MIT( Sigh....How I wish to be there).
Posted by: Libin Daniel on February 22, 2008
If only math worked like Jon thinks it does at 3:00 am, it would make my life much easier...
Posted by: Suzanne on February 22, 2008
Click the link in the subtitle. . .
Posted by: Snively on February 22, 2008
If you just want the real values of k, you can plot both sides and see where they intersect:
k = .5
k = -.76
Of course, if you replaced the k on the right with a T, and wanted K(T), I can't see how you'd do that.
I'm not that good with math, so there's probably just something I'm not seeing.
Posted by: Alex on February 22, 2008
The question only uses T(1/2) = Ln(2)/k
The trick is to figure out dx/dt and then solve for x. Use your known value of x to set up dy/dt and solve for y, and then use everything you know so far to set up dz/dt solve for z.
Hint: For part 4, when Kr is at a maximum, that's an initial value problem with t=2ln(2) and dy/dt = 0.
Oh, and this may help in general:
http://scitation.aip.org/journals/doc/AJPIAS-ft/vol_71/iss_7/684_1.html
Posted by: Snively on February 22, 2008
In d) How comes you equated and got the result that you have blogged. Throw some light man on it man.
Posted by: Libin Daniel on February 22, 2008
Awesome post, loved the YouTube video!
Posted by: 0 on February 22, 2008
Posted by: Snively on February 22, 2008
Posted by: Libin Daniel on February 22, 2008
k=0.5 or k=-0.765
Posted by: mike on February 22, 2008
Posted by: 0 on February 22, 2008
k= .5
k= 0
DID ANY OF YOU BOTHER TO CHECK YOUR OWN ANSWERS!!!
2^(2(.5)) = 1 + 1/(2(.5))
2^(1)=2 and 1 + 1/1 =2
2 = 2
2^(2(.5)) = 1 + 1/(2(0))
2^0=1 and 1 + 0 = 1
1 = 1
nothing else works
Posted by: MIT math wiz on February 22, 2008
for the bottom one
2^(2(.5)) = 1 + 1/(2(0))
2^0=1 and 1 + 0 = 1
1 = 1
this is supposed to be
2^(2(0)) = 1 + 1/(2(0))
2^0=1 and 1 + 0 = 1
1 = 1
Posted by: 0 on February 22, 2008
Posted by: Snively on February 22, 2008
Posted by: MIT math wiz on February 22, 2008
didn't know you can't divide by zero =0
Posted by: MIT math wiz on February 22, 2008
Posted by: Will on February 22, 2008
I'm actually kind of happy that I managed to solve that
Posted by: Harish on February 22, 2008
Nice blog btw.
Posted by: 0 on February 22, 2008
let 2^(2k) = a
therefore 2k = ln(a)/ln(2)
then substitute into the original equation...and manipulate it untill you get
a^a = 2a
(btw at one point you are going to have to take e^ of both sides.....post it snively!!)
The answer you get is a=2....then substitute back into 2^(2k) = a and find k....it turns out to be 0.5!
Posted by: Khaled Saad on February 22, 2008
Posted by: Khaled on February 22, 2008
However, I think finding the second root (k=-0.765), a numerical approach would have to be taken. A good method would be graphing the rhs and lhs and finding the intersection points, as Tanmay did.
Posted by: Andrew '12 on February 22, 2008
Posted by: math is coooolll on February 22, 2008
Set it to zero and use Newton's Method!
My Calc teacher would be so proud.
Posted by: PinxiSimitu on February 22, 2008
Posted by: Tanmay on February 22, 2008
Posted by: Lielainee on February 22, 2008
"Part II contains more challenging and novel problems. They will be graded with care
(Complain if they are not!) and contribute the bulk of the Homework grade."
LOL!
Posted by: Anion on February 23, 2008
The calculations for #4 was a beast. All that work to get "ln 2" at the end. How anticlimatic.
Posted by: Oasis on February 23, 2008
Posted by: Oasis on February 23, 2008
Posted by: Shauna on February 23, 2008
Posted by: Rachel '12 on February 23, 2008
I disagree, when an answer comes out to something simple like that, I think back to my calc teacher's words of wisdom,
"It's too beautiful to be wrong."
Posted by: 0 on February 23, 2008
It was still a pain in the butt though. :D
Posted by: Paul on February 23, 2008
Posted by: Eric '12 Hopeful on February 23, 2008
So have come to a suitable solution( method indeed) yet?
Posted by: Libin Daniel on February 23, 2008
Posted by: Hawkins on February 23, 2008
Posted by: Rahul Jain on February 23, 2008
Posted by: Twilight Bob on February 23, 2008
Posted by: Max 12 on February 24, 2008
Or maybe it's just the math.
Posted by: 0 on February 24, 2008
Posted by: baggy on February 24, 2008
Don't be hatin' on Newton's Method.
It's Legitimate.
Just say that 0= 2^2k-1-1/(2k)
So then F(k)= 2^2k-1-1/(2k).
And then K2= k1 - F(k1)/F'(k1)
So just pick any old number for k1 to find k2.
And then, K3= k2 - F(k2)/F'(k2)
So just do that infinity times and you get the right answer.
It sooooo simple why would you just want to use that complicated algebra and substitution mumbo jumbo.
Newton is where it's at.
Posted by: PinxiSimitu on February 24, 2008
Posted by: Suggestion on February 24, 2008
Posted by: Carolina on February 24, 2008
Could you please tell me the tentative dates by which the RA's can hear from the MIT admissions office?
Thank you!
Posted by: 0 on February 24, 2008
I'd place decisions at 11:48 AM EST on March 22nd
Posted by: Snively on February 24, 2008
Sum from k = 1 to infinity of k / (5 ^ k)
Posted by: Fun Math Challenge on February 25, 2008
Posted by: Libin Daniel on February 25, 2008
Here's my logic:
Decisions always come out mid-march. They're also always released online on a Saturday afternoon (at least in my experience). That narrows it down to the 15th or the 22nd. Two years ago they released on Pi Day (March 14th). Last year they tried to release that early but couldn't pull it off because of the larger number of applicants. They have even more applicants this year, so March 15th looks unreasonable. March 22nd looks pretty reasonable though.
As for the 11:48 AM, decisions are officially released online at 12:00 PM EST, but they actually sneak them on about 12 minutes prior. Haven't figured out why yet, but it's a handy little bit of info.
Sooooo, that's my thinking. This could be completely wrong so don't quote me on it (although I'm an admissions blogger I don't actually know the inner workings and secret stuff from the admissions office), but it'd be safe to guesstimate.
Posted by: Snively on February 25, 2008
that really helped a lot... solved it
I know there must be some way other than compute it numerically... especially when there is a rational number...
Posted by: Mgccl on February 26, 2008
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