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  • one year ago

Quick little thing explaining how to get the gamma function integral @astrophysics

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  1. Empty
    • one year ago
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    Ok solve this integral, a is a constant with respect to x. $$\int_0^\infty e^{-ax}dx$$

  2. Astrophysics
    • one year ago
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    Wait wait do this

  3. Astrophysics
    • one year ago
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    |dw:1437992640744:dw| i saw this and wanted to do it

  4. Empty
    • one year ago
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    Yeah I know, trust me

  5. Empty
    • one year ago
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    lol

  6. Astrophysics
    • one year ago
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    by parts

  7. Astrophysics
    • one year ago
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    ok

  8. Astrophysics
    • one year ago
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    ima do it then

  9. Empty
    • one year ago
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    By parts is one way to do it, but this way is much cooler

  10. Empty
    • one year ago
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    Hahaha trust me you will be doing the same thing but cooler

  11. Empty
    • one year ago
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    don't you wanna be cool

  12. Empty
    • one year ago
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    all the kids are doing it

  13. Empty
    • one year ago
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    it's just one integral

  14. Empty
    • one year ago
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    one can't hurt

  15. Empty
    • one year ago
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    Woah this is OPEN STUDY not PEER PRESSURE ANSWER

  16. Empty
    • one year ago
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    c'mon what's the integral of e^-ax already hurrrrry

  17. Empty
    • one year ago
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    ty

  18. Astrophysics
    • one year ago
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    |dw:1437993094157:dw| I forgot to take the limit the first time ok, relax it's been a while since I've done an improper integral

  19. Empty
    • one year ago
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    haha yeah no problem this looks perfect!

  20. Astrophysics
    • one year ago
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    Jk I did one earlier...|dw:1437993189968:dw|

  21. Astrophysics
    • one year ago
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    |dw:1437993463484:dw|

  22. Empty
    • one year ago
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    Boom ok perfect now we have this nice looking thing: $$a^{-1} = \int_0^\infty e^{-ax}dx$$ This is where the magic happens, this is the cool way. We differentiate both sides WITH RESPECT TO A. This is weird if you're not used to it, so it'll be good practice. If you don't make a mistake doing this, then you will be officially better than me cause I messed this up the first time I saw it lol.

  23. Astrophysics
    • one year ago
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    I am better

  24. Empty
    • one year ago
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    lolol woahhhh champ

  25. Empty
    • one year ago
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    I was gonna give you a medal + fan but now I'm considering giving it to Jhan

  26. ParthKohli
    • one year ago
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    JUST DO IT.

  27. Jhannybean
    • one year ago
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    |dw:1437993589194:dw|

  28. Astrophysics
    • one year ago
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    I take that back, please I need my medals and fans ok leggo

  29. Empty
    • one year ago
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    lolol

  30. Astrophysics
    • one year ago
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    I tried putting the a over the t JB

  31. Empty
    • one year ago
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    Parth gets the medal

  32. Empty
    • one year ago
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    sry

  33. Empty
    • one year ago
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    I haven't got all night

  34. ParthKohli
    • one year ago
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    It's because I'm better than him.

  35. Empty
    • one year ago
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    I'm gonna pull a Michele_Laino on you if you don't hurry up

  36. Astrophysics
    • one year ago
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    Go do your discriminants and quit discriminating against me so I can do this

  37. Astrophysics
    • one year ago
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    |dw:1437993707615:dw|

  38. Astrophysics
    • one year ago
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    Differentiate both sides you say?

  39. Empty
    • one year ago
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    yeah with respect to a

  40. Empty
    • one year ago
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    and with respect to me, your sensei

  41. Astrophysics
    • one year ago
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    |dw:1437993761110:dw| I'm better

  42. ParthKohli
    • one year ago
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    FLAWLESS VICTORY

  43. Jhannybean
    • one year ago
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    Why take the limit when you can differentiate - his logic.

  44. Empty
    • one year ago
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    |dw:1437994060905:dw| You evaluated the left side, but not the right side and now Parth is enjoying his birthday wish pssssh

  45. ParthKohli
    • one year ago
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    Yes, my birthday wish was a free medal. <3

  46. Astrophysics
    • one year ago
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    That was the right side

  47. Astrophysics
    • one year ago
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    |dw:1437994295728:dw|

  48. Empty
    • one year ago
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    $$\frac{d}{da} \int_0^\infty e^{-ax}dx = \int_0^\infty \frac{\partial}{\partial a} (e^{-ax})dx$$

  49. Astrophysics
    • one year ago
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    Yeah ok I see what you're doing, you want me to do use leibniz rule pretty much and not just plug and chug haha

  50. Jhannybean
    • one year ago
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    defining a is indeed a constant and not just another variable

  51. Empty
    • one year ago
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    a is not a constant, it is just an independent variable from x

  52. Astrophysics
    • one year ago
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    |dw:1437994507765:dw| oOoOoOo

  53. Empty
    • one year ago
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    Constant with respect to x though, you could say that, also https://www.youtube.com/watch?v=Lcw6xBaCaXM

  54. Empty
    • one year ago
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    Boom good job, ok now we have something good

  55. Jhannybean
    • one year ago
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    Ah, I see.

  56. Empty
    • one year ago
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    \[a^{-1} = \int_0^\infty e^{-ax}dx\] \[a^{-2} = \int_0^\infty xe^{-ax}dx\] ...? Keep going do this 2 or 3 more times until you get the picture.

  57. Empty
    • one year ago
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    differentiate with respect to a I mean haha

  58. Astrophysics
    • one year ago
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    Integrate it, or keep taking derivatives?

  59. Empty
    • one year ago
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    keep taking derivatives :D

  60. Astrophysics
    • one year ago
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    K good lol

  61. Astrophysics
    • one year ago
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    |dw:1437994844847:dw|

  62. Empty
    • one year ago
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    noice

  63. Astrophysics
    • one year ago
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    I think we can write a series for this...

  64. Empty
    • one year ago
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    no don't!

  65. Empty
    • one year ago
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    just keep taking derivatives

  66. Jhannybean
    • one year ago
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    how do partials work again? \(a^{-2} = \int_0^\infty \color{red}{x}e^{-ax}dx\) work again? The highlighted red portion I mean...

  67. Astrophysics
    • one year ago
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    Ok I'll take one more lol

  68. Empty
    • one year ago
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    x is a constant when we're differentiating with respect to a @Jhannybean since a and x are independent of each other nothing happens

  69. Jhannybean
    • one year ago
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    Oh that's right,

  70. Jhannybean
    • one year ago
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    I kept reading \(\frac{ \partial}{\partial x}\) instead of \(\frac{\partial}{\partial a}\) and I was confusing myself. -_-

  71. Astrophysics
    • one year ago
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    Taking derivatives in integrals is amazing, I saw this technique once before, did not know you would treat it like a constant in the integral, awesome. |dw:1437995191725:dw|

  72. Astrophysics
    • one year ago
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    Oki, what's next

  73. Empty
    • one year ago
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    Do one more!

  74. Astrophysics
    • one year ago
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    |dw:1437995573741:dw|

  75. Astrophysics
    • one year ago
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    |dw:1437995632368:dw|

  76. Empty
    • one year ago
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    \[a^{-1} = \int_0^\infty e^{-ax}dx\] \[a^{-2} = \int_0^\infty xe^{-ax}dx\] \[6a^{-3} = \int_0^\infty x^2e^{-ax}dx\] \[24a^{-4} = \int_0^\infty x^3e^{-ax}dx\] ...?

  77. Empty
    • one year ago
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    What are we doing again

  78. Empty
    • one year ago
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    something something gamma is the continuous version of the factorial something something whatever I already forgot

  79. Astrophysics
    • one year ago
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    Ok I see, there's a factorial relationship, so there's a reoccurrence relationship.

  80. Empty
    • one year ago
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    Reoccurrence relationship?

  81. Empty
    • one year ago
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    \[a^{-1} = \int_0^\infty e^{-ax}dx\] \[a^{-2} = \int_0^\infty xe^{-ax}dx\] \[6a^{-3} = \int_0^\infty x^2e^{-ax}dx\] \[24a^{-4} = \int_0^\infty x^3e^{-ax}dx\] ...? This is the pattern, what's the general form?

  82. Astrophysics
    • one year ago
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    Haha recurrence*

  83. Empty
    • one year ago
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    No recurrence relation here either

  84. Empty
    • one year ago
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    Whoops I had been writing it wrong lol oh well don't look at what I write you wrote it yourself correctly anyways \[a^{-1} = \int_0^\infty e^{-ax}dx\] \[a^{-2} = \int_0^\infty xe^{-ax}dx\] \[2a^{-3} = \int_0^\infty x^2e^{-ax}dx\] \[6a^{-4} = \int_0^\infty x^3e^{-ax}dx\] \[24a^{-5} = \int_0^\infty x^4e^{-ax}dx\] ... \[\frac{n!}{a^{-???} }= \int_0^\infty ???dx\] Here's my hint, what the hell is this last thing!

  85. Astrophysics
    • one year ago
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    (n-1)!

  86. Empty
    • one year ago
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    Stop looking at what it should be and look at what it actually is, forget the gamma function for a minute

  87. Empty
    • one year ago
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    Just find the truth don't worry about being right right now

  88. Empty
    • one year ago
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    idk check it: Here's what you KNOW \[2a^{-3} = \int_0^\infty x^2e^{-ax}dx\] try out your false formula to find out how to fix it

  89. Astrophysics
    • one year ago
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    Yeah I tried, that didn't work..one sec let me go over it again..6 am xD

  90. Empty
    • one year ago
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    Pythagoras didn't have anyone to check his work I think on the whole a^2+b^2 thing lol so you're at least fortunate :P

  91. Astrophysics
    • one year ago
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    \[\frac{ n! }{ a^{-(n+1)} } = \int\limits_{0}^{\infty} x^{n+1} e^{-ax} dx\]

  92. Astrophysics
    • one year ago
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    I tried by plugging in integers haha

  93. Empty
    • one year ago
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    Known: \[\frac{2!}{a^3}= \int_0^\infty x^2e^{-ax}dx\] Yours: \[\frac{ n! }{ a^{-(n+1)} } = \int\limits_{0}^{\infty} x^{n+1} e^{-ax} dx\] Plugged in n=2 \[\frac{ 2! }{ a^{-3} } = \int\limits_{0}^{\infty} x^{3} e^{-ax} dx\] Same or not?

  94. Astrophysics
    • one year ago
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    x^n but the known should be a^-(...) ye?

  95. Empty
    • one year ago
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    ???

  96. Empty
    • one year ago
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    Sorry I don't know what you're saying X_X

  97. Astrophysics
    • one year ago
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    Oh no I was just looking at when you said \[\frac{n!}{a^{-???} }= \int\limits_0^\infty ???dx\]

  98. Astrophysics
    • one year ago
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    \[\frac{ n! }{ a^{-(n+1)} } = \int\limits_{0}^{\infty} x^n e^{-ax} dx\]

  99. Empty
    • one year ago
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    better but why are you just going with my mistake?

  100. Empty
    • one year ago
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    like negative in the denominator is like "double dividing"

  101. Astrophysics
    • one year ago
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    Haha, I get scared so I thought it was right

  102. Astrophysics
    • one year ago
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    That works out

  103. Astrophysics
    • one year ago
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    Tried n = 2

  104. anonymous
    • one year ago
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    My gawd this is a long question

  105. Empty
    • one year ago
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    Hahaha ok that's what I like to hear, everything leading to this point doesn't really matter now, it's like the ladder we can kick out from under us and if you like you can pretend we were never wrong on the path to this. @isaac4321 yeah we should probably have flushed halfway through for courtesy :(

  106. anonymous
    • one year ago
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    It's all good at least your gettin edumacated

  107. Empty
    • one year ago
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    tru

  108. Astrophysics
    • one year ago
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    |dw:1437998982910:dw|

  109. anonymous
    • one year ago
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    Um ok

  110. Astrophysics
    • one year ago
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    Wow, empty, this happens every time spend so much time on one thing..then...yeah it doesn't matter xD

  111. Astrophysics
    • one year ago
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    You're like a real prof

  112. anonymous
    • one year ago
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    Who is that

  113. Astrophysics
    • one year ago
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    google your name sir

  114. anonymous
    • one year ago
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    Ok I thought it was him

  115. Empty
    • one year ago
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    ok so now you can derive the gamma function, you're on your own, you can think more about why this works

  116. anonymous
    • one year ago
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    👍

  117. Empty
    • one year ago
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    It should be really easy

  118. Empty
    • one year ago
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    and if it isn't, wellllll think about it until you do

  119. Astrophysics
    • one year ago
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    Ok so I don't really know what exactly the gamma function is, I mean is it actually the thing I posted above, I thought that was just a problem that was a gamma function haha..well that's cool, learned so much

  120. anonymous
    • one year ago
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    Obviously it is very easy 😉

  121. Astrophysics
    • one year ago
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    |dw:1437999236907:dw|

  122. Empty
    • one year ago
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    Yeah, the gamma function is basically defined as \[\Gamma(n)=(n-1)!\] for positive integers but extends further.

  123. Empty
    • one year ago
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    do you see where the n-1 is coming into play now?

  124. Astrophysics
    • one year ago
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    Yeah I think so

  125. Empty
    • one year ago
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    ok

  126. Empty
    • one year ago
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    |dw:1437999792138:dw|

  127. Astrophysics
    • one year ago
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    lmao

  128. Astrophysics
    • one year ago
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    Howd you get so many medals

  129. Empty
    • one year ago
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    it's a thumbs up not a pile of poop for the record

  130. Astrophysics
    • one year ago
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    Why do people always observe you teaching me

  131. Empty
    • one year ago
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    they're creepers

  132. Empty
    • one year ago
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    creepin'