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Where can I get help with David Bachman's A Geometric Approach to Differential Forms?

  1. Aug 31, 2015 #1

    nearc

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    this starts as a calculus question, but springs into where i can get help with david bachman's A GEOMETRIC APPROACH TO DIFFERENTIAL FORMS second edition.

    looking at paul's notes cheat sheets http://tutorial.math.lamar.edu/cheat_table.aspx we have


    ##
    \int \frac{1}{\sqrt{a^{2}-x^{2}}} = sin^{-1}(\frac{x}{a})+c
    ##

    but this is different than wolfram http://www.wolframalpha.com/input/?i=integral&a=*C.integral-_*Calculator.dflt-&f2=1/sqrt(a^2-x^2)&f=Integral.integrand_1/sqrt(a^2-x^2)&a=*FVarOpt.1-_**-.***Integral.rangestart-.*Integral.rangeend--.**Integral.variable---.*--

    however, all i really want to know is this correct?

    ## \int \frac{1}{\sqrt{1-a^{2}-x^{2}}} = sin^{-1}(\frac{x}{\sqrt{1-a^{2}}})+c ##
     
    Last edited: Aug 31, 2015
  2. jcsd
  3. Aug 31, 2015 #2

    DEvens

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    To get your TeX stuff to work right, put two # signs before and two after.

    To compare the two different results for the integrals, consider a triangle with sides x, a, and ##\sqrt{a^2 - x^2}##. What angle is indicated by the ##\sin^{-1}()## version and what angle by the ##\tan^{-1}()## version?
     
  4. Aug 31, 2015 #3

    nearc

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    thanks, latex fixed, now i need to ponder the triangle approach
     
  5. Aug 31, 2015 #4

    mathwonk

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    Re your second question, there was here years ago, a thread devoted to reading BACHMAN'S BOOK, and featuring the participation of the author. Perhaps it is still accessible.
     
  6. Sep 1, 2015 #5

    nearc

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    thanks, i think that was for first edition but i'm not sure
     
  7. Sep 9, 2015 #6
    The integral with 1 - a2 - x2 under the square root sign is just the same basic integral as the original one with just a2 - x2 under the the square too sign, if you substitute the expression 1 - a2 for the expression a2.

    Of course, in any expression, anything under a square root sign is required to be non-negative. In the first example, should that be an "a" on the RHS, or perhaps a |a| ? (An "a" alone could be either positive or negative.)

    The easiest way to check if an indefinite integration is correct is to check whether the putative integral can be differentiated to arrive at the integrand.
     
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