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  1. #1
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    square root using Taylor series

    As I am new to Taylor series, and im little confused, and in an urgent to expansion of sqrt
    I want Taylor expansion for


    where "p" is constant. and "c" is variable.
    and c>=1,

    I need it in urgent.can anyone please expand it for me.
    Last edited by BlackMamba; 27th August 2010 at 14:02.

    •   Alt6th June 2010, 18:14

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    Re: square root using Taylor series

    Taylor series at c=0 ? (you say c>=1)
    At c=1 ?
    At c= ?

    Try Wolfram Alpha with: taylor series sqrt((100 c + 21 - p)^2 - 8400 c) at c=0 to order 6

    http://www.wolframalpha.com/input/?i=taylor+series+sqrt%28%28100+c+%2B+21+-+p%29^2++-+8400+c%29+at+c%3D0+to+order+6



    •   Alt7th June 2010, 00:17

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    Re: square root using Taylor series

    Quote Originally Posted by _Eduardo_
    Taylor series at c=0 ? (you say c>=1)
    At c=1 ?
    At c= ?

    Try Wolfram Alpha with: taylor series sqrt((100 c + 21 - p)^2 - 8400 c) at c=0 to order 6

    http://www.wolframalpha.com/input/?i=taylor+series+sqrt%28%28100+c+%2B+21+-+p%29^2++-+8400+c%29+at+c%3D0+to+order+6
    Thank u so much.

    Can u exactly say what is the exact value for "c" I should take.



    •   Alt7th June 2010, 04:18

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    Re: square root using Taylor series

    How can I know? The exercise was given to you.

    The most common is at 0, but as was specified that c>= 0 perhaps may be at another point.



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    square root using Taylor series

    You should still be able to obtain an expansion around such an arbitrary c but I really advice you to rephrase your problem, perhaps start by replacing c with x and call this function f(x), consider the Taylor series expansion of the function f(x) around a constant c where c>=1)! Maybe the assumption that c>=1 is probably to put a constraint on your choice of p so that the result is always real or take sure the f is infinitely differentiable!!
    To evaluate Taylor Series, follow the Taylor series "equation" by taking the first, second, third and fourth, ... derivatives of f(x) with respect to x, take each derivative and when you're done replace x with the constant c, and multiply each i'th derivative by (x-c)^(i)/(factorial(i))!
    What you have now is a Taylor series expansion of of the function f(x) around a constant c>=1, x appears as a variable, c only as a constant, as well as p of course!



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