Mathematics

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Dec 15, 2014 19:28
:)
Dec 15, 2014 19:26
Saw it now :P
Dec 15, 2014 19:26
Yes
Dec 15, 2014 19:26
@Venus
Jul 2, 2014 16:09
Recently, I think it was asked by juanethron, I am not sure
Jul 2, 2014 16:08
That one was recently discussed on SE
Jun 21, 2014 11:18
@Chris'ssis: Sure. Thanks a lot. :)
Jun 21, 2014 11:17
@Chris'ssis: Please let me know if you solve this. :)

Even after a bounty on this one, nobody answered it.
Jun 21, 2014 11:14
@Chris'ssis: Can you please look at this one too: http://math.stackexchange.com/questions/818494/integral-int-0-infty-frac-sqrt-sqrt-alpha2x2-alpha-exp-big-bet

I am scratching my head from days but I am unable to come up with anything. :(
Jun 21, 2014 10:56
Sure, I will. :)
Jun 21, 2014 10:55
cute? lol but I will try this. :3
Jun 21, 2014 10:54
That result looks scary. :3
Jun 21, 2014 10:54
What the..? :O
Jun 21, 2014 10:52
Its fun to work on them but I am usually never able to reach any answer. :P
Jun 21, 2014 10:51
Very nice. :)
Jun 21, 2014 10:50
Woah! :O
Jun 21, 2014 10:50
@Chris'ssis: From where do you get these kind of integrals? They are crazy hard. :3
Jun 21, 2014 10:49
@Sawarnik: Ok. :)
Jun 21, 2014 09:30
Can somebody please solve this: http://math.stackexchange.com/questions/818494/integral-int-0-infty-frac-sqrt-sqrt-alpha2x2-alpha-exp-big-bet

I even set a bounty on it but nobody answered. :(
Jun 21, 2014 06:56
So if you integrate (1-x^2/2)/x^3 from x=2/pi to 1, you get 0.50799>1/2
Jun 21, 2014 06:56
You can use the fact that cos(x)>1-x^2/2
Jun 21, 2014 06:55
I think I see now
Jun 21, 2014 06:55
Hey @Sawarnik
Jun 21, 2014 06:24
@Sawarnik: Oops. the lower limit is cos(1)(1-2/pi). :P So I guess the problem still needs some work. I will have to think about it.
Jun 21, 2014 06:14
Yes, I am looking at that. It looks strange.
Jun 21, 2014 06:11
What problem?
Jun 21, 2014 06:09
lol
Jun 21, 2014 06:08
Not sure, maybe BITS Pilani or IIT Jodhpur.
Jun 21, 2014 06:07
No, I will be in a few days. :)
Jun 21, 2014 06:06
:)
Jun 21, 2014 06:04
?
Jun 21, 2014 06:04
@Sawarnik: Did you post this: integralsandseries.prophpbb.com/…
Jun 12, 2014 19:18
BTW, by any chance, does the other method involves the use of differentiation under the integral symbol? Just guessing...
Jun 12, 2014 19:17
@Chris'ssis: Yep, looks like it, thank you! :)
Jun 12, 2014 19:13
@Chris'ssis: I don't think I will be able to figure out a reason, dilogarithms is completely a new thing to me. Can you please give a hint? :)
Jun 12, 2014 19:11
The identity gives: $$\frac{\pi^2}{3}-\frac{1}{2}$$
Jun 12, 2014 19:10
Erm...it doesn't match the answer which the identity gives.
Jun 12, 2014 19:10
@Chris'ssis: But the imaginary part still remains.
Jun 12, 2014 19:09
@r9m: Ah ok and you too post great solutions on MSE, I like reading them. :)
Jun 12, 2014 19:07
@r9m: Integrals and Series? Never saw you there and thanks. :)
Jun 12, 2014 19:02
I guess you got admission in an early age, it really makes me feel stupid for taking so much time to get into a college :P
Jun 12, 2014 18:58
or what are you currently studying there? :)
Jun 12, 2014 18:58
What do you learn at CMI? I am really curious about it. :)
Jun 12, 2014 18:56
I didn't even know about CMI or ISI in my first attempt and this year, I forgot about the CMI exam. :P
Jun 12, 2014 18:52
I tried for ISI this year but didn't get selected, I am probably going to BITS-Pilani this year but I wanted to do a degree in Math from CMI or ISI
Jun 12, 2014 18:51
I wish I could study at CMI or ISI :(
Jun 12, 2014 18:50
No, I passed out of high school last year and will take admission in some college this year. :)
Jun 12, 2014 18:49
That's great!
Jun 12, 2014 18:49
Wow!
Jun 12, 2014 18:48
Do you mind telling me which one? :)