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Use Linear Approximation to Find Margin of Error

  1. Oct 2, 2012 #1
    1. The problem statement, all variables and given/known data
    For x near 0, local linearization gives the following equation.
    e^x ≈ 1 + x

    Estimate to one decimal place the magnitude of the error for −1 ≤ x ≤ 1.
    2. Relevant equations

    3. The attempt at a solution
    I'm no exactly sure what to do here to be honest, but what I thought I'd do is try to work backwards. Generally I'd be given a margin of error to be accurate within, but instead I am given the values that are accurate within the margin of error(i assume). So what I did was take -1 & 1 and stick it in |e^x - 1+x|. I got 1/e and e-2. I took the difference of the two and got .3504. I'm pretty sure this is wrong I think I messed up when I took 1 & 1 for |e^x - 1+x|.
     
  2. jcsd
  3. Oct 2, 2012 #2

    Mark44

    Staff: Mentor

    Have you studied power series, and particularly Maclaurin and Taylor series? If so, look up the Taylor Remainder Theorem.
     
  4. Oct 2, 2012 #3
    Unfortunately I haven't covered teh topic of series yet. I've heard of them though, but I'm sure that's not good enough :P

    ----Edit-----
    I took a look at the Taylor Remainder Theorem, and I understand the concept well enough to give a simple explanation, but I still don't feel comfortable using it. Just to make sure that I do understand your suggestion correctly, if I tried to estimate the Remainder using the form q<=f^(k+1)(x)<=Q (k being an arbitrary value I assign for accuracy) that should give me the correct solution right?
     
    Last edited: Oct 3, 2012
  5. Oct 2, 2012 #4

    Ray Vickson

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    Homework Helper

    Why would you take the difference? When x = +1 the error is e -2 ≈ 0.718, and when x = -1 the error is exp(-1)-1+1 = exp(-1) ≈ 0.368. The difference of these two numbers has no meaning at all in this question.

    RGV
     
  6. Oct 3, 2012 #5
    Well I was never sure what to do in the first place, but since the site requires me to attempt something that's what I did. Can't really explain it. :tongue2: I'm still a bit confused on what to do with 2 margin of errors. I feel like .7 is the margin of error, but I'm not sure why.
     
  7. Oct 3, 2012 #6

    Ray Vickson

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    Where do you get 2 margins of error? I just see one, and that is the one that guarantees |exp(x) - 1 - x| ≤ margin for all x in (-1,1).

    A more serious consideration is that *maybe* you are not supposed to be able to compute exp(x) at all yet (so cannot do the computations above), but are somehow expected to be able to place an upper bound on the error anyway. I don't know if that is what the questioner wants.

    RGV
     
  8. Oct 3, 2012 #7
    Ahhhh i see what I was doing wrong. I went ahead a graphed it and got a better visualization of what you are saying. Out of the two results you got (.718, and .368) .718 fits the inequality |exp(x) - 1 - x| ≤ margin. So .368≤ margin (margin=.718) works wit that. Is that right?
     
    Last edited: Oct 3, 2012
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