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gaurangnaware

  • 3 years ago

integration of "e" raise to power "x" square

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  1. lgbasallote
    • 3 years ago
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    \[\huge \int e^{x^2}dx\] ???

  2. lgbasallote
    • 3 years ago
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    is that your question?

  3. gaurangnaware
    • 3 years ago
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    yes

  4. lgbasallote
    • 3 years ago
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    ..this isn't solveable by natural means...

  5. gaurangnaware
    • 3 years ago
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    well how do i slove it?

  6. lgbasallote
    • 3 years ago
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    numerical analysis was the term if i remember it right

  7. gaurangnaware
    • 3 years ago
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    erf...?

  8. lgbasallote
    • 3 years ago
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    yup. those kinds of functions

  9. lgbasallote
    • 3 years ago
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    only mathematicians solve this. Because they love solving imaginary problems.

  10. gaurangnaware
    • 3 years ago
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    well i am solving an ode and the question is if the given y satisfies it and it contains.. inte^x^2...

  11. lgbasallote
    • 3 years ago
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    they give these stuffs in ode now? are you by any chance from a premier university?

  12. gaurangnaware
    • 3 years ago
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    haha well Ph.D.. i have forgotten the basics.. was just revising

  13. Algebraic!
    • 3 years ago
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    http://www.wolframalpha.com/input/?i=error+function&dataset=&asynchronous=false&equal=Submit

  14. mukushla
    • 3 years ago
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    http://www.wolframalpha.com/input/?i=integrate+e^%28x^2%29 imaginary error function !!

  15. lgbasallote
    • 3 years ago
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    I am a PhD myself. However, I'm from Liberal Arts...

  16. Algebraic!
    • 3 years ago
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    @mukushla jinx

  17. lgbasallote
    • 3 years ago
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    i think @gaurangnaware wants to know how to get that erf function....

  18. mukushla
    • 3 years ago
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    lol Algebraic

  19. myko
    • 3 years ago
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    use series

  20. gaurangnaware
    • 3 years ago
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    Thanx Mukushla.. go it!

  21. mukushla
    • 3 years ago
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    no problem :)

  22. myko
    • 3 years ago
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    series expantion of ex2 is: \[e ^{x ^{2}}=1+\frac{x ^{2}}{1!}+\frac{x ^{4}}{2!}+\frac{x ^{6}}{3!}+....\] this is a uniformly convergent series, so you can integrate it term by term. So just integrate each term to get your answer.

  23. myko
    • 3 years ago
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    @gaurangnaware

  24. myko
    • 3 years ago
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    @lgbasallote

  25. heedcom
    • 3 years ago
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    or use polar

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