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Estimates of the remainder in Taylor's theorem

  1. Feb 7, 2015 #1
    Here is the exercise question;
    Use the general binomial series to get ##\sqrt{1.2}## up to 2 decimal points
    In the solution the ##R_1## was given as
    ##|R_1|\leq {\frac{1}{8}} {\frac{(0.2)^2}{2}}## But it doesn't say where this came from and comparing this with the estimate of remainder given in Taylor's theorem didn't help. Another thing that was frustrating was that the solutions for the end-of-chapter exercises was somewhat different what the book said for example they didn't include the n+1 factorial in the estimates of the remainder.
    Please help this is driving me crazy.
     
  2. jcsd
  3. Feb 7, 2015 #2

    Mark44

    Staff: Mentor

    What is f here? IOW, what is the function whose binomial series you are estimating?
    What does your book give for the formula of the remainder?
    Actually, I think they did. In what you showed above, what is n? What is n + 1?
    Please show us what you've done so far.
     
  4. Feb 8, 2015 #3
    The function is ##\sqrt{1+x}##, the question doesn't give you what the function is, it was a part of the process.
    The book says ##|R_n(x)|\leq {\frac{M_{n+1}}{(n+1)!}} |x|^{n+1}##
    The question I posted is not the one that didn't include the factorial, the one above has an additional ##{\frac{1}{2}}## and a couple others had this as well. I did the calculation for about 6 times, there still could have been a mistake but not likely.
     
  5. Feb 8, 2015 #4

    Mark44

    Staff: Mentor

    The "process" is just recognizing what the underlying function is, which in this case is f(x) = (1 + x)1/2.
    For R1 (where n = 1), the error is going to have a factor of 1/2! = 1/2. Is that what you're asking about?

    Also, since this is a problem involving a binomial series expansion, it might be helpful to write the first few terms of the expansion of (1 + x)1/2.
     
  6. Feb 9, 2015 #5
    No that's not what I'm asking about, if you do the calculation you'll see that it's different from the value here.
    The max is 1/4 and you divide it by 2! and multiply the whole thing again by 0.2 squared.
     
  7. Feb 9, 2015 #6

    Mark44

    Staff: Mentor

    OK, now I see what you are saying, and I agree with you. Here f(x) = (1 + x)1/2, so f'(x) = (1/2)(1 + x)-1/2, and f''(x) = (-1/4)(1 + x)-3/2. The maximum value of |f''(x)| occurs when x = 0, so |f''(x)| <= 1/4. This means that |R1| <= (1/4) * (1/2!) * (.2)2, same as what you're saying.

    Unless I'm missing something here, it looks like there's a typo in your book.
     
  8. Feb 10, 2015 #7
    Yeah looks like it. Thanks for the help :) really needed it
     
  9. Feb 10, 2015 #8

    Mark44

    Staff: Mentor

    It wouldn't hurt to ask your instructor about it.
     
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