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Homework Help: 3 questions in tailor series

  1. Dec 28, 2007 #1
    i added a file with the questions
    and the solutions of the book


    question 1 is:
    calculate "e" with an error less than 10^-9??
    i started the solving of this problem
    by finding the expretion of Rn for e^X
    we need to chhose a point C which is located between X and A
    i know that X equals to 1
    but why in the solution they tell that A=0???

    question 2 is:

    find cos 9 with an error less than 10^-5 (9 is degrees not radians)
    here after findinf the expretion of Rn
    where did i use to put "c"
    because they solve this problem without c
    i dont know why??

    question 3:
    i dint understand the way they solved it
    because they didnt use any tailor series here
    where is the function that i need to build a tailor series from??
  2. jcsd
  3. Dec 29, 2007 #2


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    Homework Helper

    a is zero because you are using a Maclaurin series for e^x, that is, a Taylor series for the function expanded about x = a = 0.

    It seems from your note that the problem is asking you to find the number of terms in the Taylor polynomial that would be required to find e^x to a precision of 10^-9 on the interval [0,1]. This will let you find e^1 = e to the desired precision.

    You will not need to choose c: it is just a point on the interval for which we are guaranteeing that the size of the remainder for the Taylor polynomial will be less than the selected precision. The numerator in the remainder term needs to be the maximum value that the nth derivative of e^x has on [0,1], which will in fact be e^1. But since we're supposed to be computing a value for e, we have to pretend that we don't already know it, so we set that maximum value to 3, which we're pretty sure is bigger than e...

    That leaves you to solve 3/(n+1)! < 10^-9 , which doesn't work algebraically and has to be done by one or another sophisticated "trial-and-error" method.

    Once again, they are using a Maclaurin series for cos(x), so a=0. The Taylor remainder term needs a largest value for the nth derivative on some interval around x = 0. But since all the derivatives of cos(x) are +/-sin(x) or +/-cos(x), the largest value is guaranteed to be one. So in practice, we don't worry about a value for c: we just put in a "1" in the remainder term and solve for the number of terms as you show in your notes.

    Actually, they did work out a Taylor series. The problem is apparently to find sqrt(5) using a Maclaurin series based on (1+x)^p (for some reason, they used something like a cursive-L for the exponent, which I'm not going to attempt to reproduce).

    So they said: let's factor 5^(1/2) as (4+1)^(1/2) =
    2 · [ (1 + {1/4})]^(1/2) and work out the general Taylor series for a binomial (1+x) raised to any real power (which they do on the next line).

    They then set up the remainder term for this series, with the exponent now set to p = (1/2) and the interval going out to
    x = 1/4 ; the nth derivative isn't hard to find, but solving the inequality to find the number of terms required to meet a precision of 5·10^-3
    ( which they write as (10^-4)/2 ) isn't going to get done algebraically, so they must have done a "trial-and-error" solution. They find that terms out to n=4 are needed and proceed to make the calculation to obtain an estimate for sqrt(5).
    Last edited: Dec 29, 2007
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