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Taylor series?

  1. Mar 10, 2010 #1
    1. The problem statement, all variables and given/known data
    find a value of h such that for |x|<h implies sin(x)=x - x^3/6 + x^5/120 + R where |R|<10^(-4)

    2. Relevant equations

    3. The attempt at a solution
    it's tedious to type out my working but I've got h= (6!/10^4)^1/6 but I'm not sure about this...
  2. jcsd
  3. Mar 10, 2010 #2
    That certainly works since the difference in sin(x) and your T(x)with your x value is 9.16x10^-6
    How did you get this answer?
  4. Mar 10, 2010 #3
    I got that (mathmathmad's) as well but I chose x such that I could evaluate without using a calculator. It's not too difficult to be honest. What concerns me is that do we need to prove that sin(x) and its derivatives of all orders are continuous in the given domain or should we take it as given.
    Last edited: Mar 10, 2010
  5. Mar 10, 2010 #4
    sin(x)=x - x^3/6 + x^5/120 + R

    erm, I take |R| = |sin x - x + x^3/6 - x^5/120 |

    since R_n = f(x) - T_n (which the value of n I'm not sure of but I take n=5)

    and there's a formula which states there exists c in (x,0) such that :
    R_n = f^(n+1)(c)*x^(n+1)/(n+1)!

    since I take n as 5 then I evaluate |R|<10^(-4) blablabla
    get x^6 < 6!/10^4

    can we just get the 6th root of 6!/10^4 to evaluate x?

    so does h equal to (6!/10^4)^1/6 which is approximately 0.64499...?

    please correct me!
  6. Mar 10, 2010 #5
    I agree, Ive done it with n = 6 but the principles the same.
    You guys done the 3rd question?
    And if so, how?
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