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Taylor series remainder

  1. Dec 13, 2014 #1
    How is the Taylor remainder of a series (with given Taylor expansion) expressed if you want to make a calculation with known error? e.g. if I want to calculate π to, say, 12 decimal places using the previously-derived result π=4*arctan(1) and the Taylor series for arctan(x), how will I work out how many terms I need (or, imagine that number of decimal places is high enough that trial and error is not efficient)?

    The problem is that equating the error (e.g. 5*10^(-13) in my example above) to the integral form of the remainder leaves us an expression that can be numerically (but certainly not analytically) solved, whereas I am looking for a method that works (even approximately) by hand and handheld calculator.
     
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  3. Dec 14, 2014 #2

    mfb

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    It is always possible as you have to calculate every term individually.

    You can find an analytic upper limit I think.
     
  4. Dec 14, 2014 #3

    FactChecker

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    The remainder after expanding n terms of a Taylor series around point a is equal to ((x-a)n/n!)f(n)(x') for some x' in (a,x). So you can bound the error at x if you can bound the n'th derivative.
     
  5. Dec 14, 2014 #4
    How would this work, for, e.g. my example of π=4*arctan(1)? It seems I would have to take the nth derivative of arctan(x) - which doesn't exactly suggest itself easily to me - even then, I'm not sure what would have to be done with the remaining 3 variables (x, x', n) to find an upper bound for n?
     
  6. Dec 14, 2014 #5

    mfb

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    Yes. It might give ugly expressions, but no one said calculating the 12th decimal digit that way was easy without a computer.
    x is where you want to evaluate the series (at 1), a the point your taylor series is based on (0), x' can be anything between 0 and 1, you have to find an upper bound that is valid in the whole range. n is just the derivative you have.
     
    Last edited: Dec 14, 2014
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