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UnkleRhaukus

  • 3 years ago

\[=\sum\limits_{r=1}^\infty\frac{(-1)^r}{1-2r}\] \[=1-\frac13+\frac15-\frac17+\dots\]

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  1. UnkleRhaukus
    • 3 years ago
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    \[\large=\frac\pi4\]

  2. UnkleRhaukus
    • 3 years ago
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    @experimentX

  3. experimentX
    • 3 years ago
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    http://tinyurl.com/d5x9bk9

  4. experimentX
    • 3 years ago
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    \[ \int_0^1{1 \over 1+x^2}dx = \arctan(1) = {\pi \over 4} \]

  5. experimentX
    • 3 years ago
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    http://www.wolframalpha.com/input/?i=expand+1%2F%281%2Bx^2%29

  6. experimentX
    • 3 years ago
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    \[ f(u) = \int_0^u {1\over 1+x^2}dx = \int_0^u(1-x^2 + x^4 - x^5 + ...)dx \\\] put 'u=1'

  7. UnkleRhaukus
    • 3 years ago
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    hmm

  8. sauravshakya
    • 3 years ago
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    |dw:1356251299233:dw|

  9. sauravshakya
    • 3 years ago
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    I guess he was trying to say that

  10. sauravshakya
    • 3 years ago
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    |dw:1356251680713:dw|

  11. sauravshakya
    • 3 years ago
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    So,|dw:1356251895939:dw|

  12. experimentX
    • 3 years ago
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    yep!! that's correct ...

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