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Calculus II - Taylor Series - Error Bounds

  1. Aug 17, 2011 #1
    1. The problem statement, all variables and given/known data

    Hi,

    I'm really struggling with trying to come up with the error bound when doing taylor series problems

    Use the reaminder term to estimate the absolute error in approximating the following quantitites with the nth-order Taylor Polynomial cnetered at 0. Estimates are not unique.

    e^.25; n=4

    2. Relevant equations

    If n is a fixed integer, suppose there exists a number M such that |f^(n+1)(c)|<=M for all c between a and x inclusive. The remainder in nth-order Taylor polynomial for f centered at a satisfies

    |Rn(x)|=|f(x)-Pn(x)|<= M*|x-a|^(n+1)/(n+1)!

    3. The attempt at a solution

    So apparently if we let f(x)=e^x, then f^(5)(x) = e^x, which I agree with fully

    But I don't see how e^.25 is bounded by 2 and this is what we set M equal to. e^.25 is about 1.284025417...

    I understood the previous similar problem that I did with sin(.3); n=4, because the nth derivative of sin(x) is either +/- sin(x) or +/- cos(x) which will always be <= 1 so this makes sense to use for a value of M, M=1 in this problem...

    but for the one that I'm stuck on I don't see why we set M=2. The nth derivative of e^x is always going to be e^x which doesn't have a max value and increases without bound...

    I hope it makes sense as to were I'm getting confused. Thanks for any help.
     
  2. jcsd
  3. Aug 17, 2011 #2

    micromass

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    So basically, you're confused why they picked M=2, right?

    Well, we need to pick M such that

    [itex]|f^{(5)}(x)|\leq M[/itex]

    for all x between 0 and 0.25.

    So we need to pick M such that

    [itex]|e^x|\leq M[/itex]

    for all x between 0 and 0.25.

    We can choose M=2 since indeed [itex]|e^x|<2[/itex] for all such x.
     
  4. Aug 17, 2011 #3
    then why not just go with like 1.3 or something sense

    |e^x|<1.3
    if x is between 0 and .25
     
  5. Aug 17, 2011 #4

    micromass

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    That"s also possible.
     
  6. Aug 17, 2011 #5
    so M can be any number at all? as long as
    |f^(n+1)(c)|<=M
    is true?
    Won't this different M value change the value of
    |Rn(x)|=|f(x)-Pn(x)|<= M*|x-a|^(n+1)/(n+1)!

    which i guess is ok because of the
    <=
    sign, so what's the best value to chose for M then in a any given situation, the smallest value that it can be or does it not really matter. Like I guess the smallest value you could use is e^.25 itself for M for this problem
     
  7. Aug 17, 2011 #6

    micromass

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    Well, you could choose M=100000000 for all I care, it would be correct. HOWEVER: choosing such a large bound, will result in a bad estimation. If you choose a tighter bound like 2, then the estimation will be better. An even tighter bound would be 1.3 and this would be an even better estimation.
    They chose 2 here because it seems to work with such a bound.
     
  8. Aug 17, 2011 #7
    What exactly do you mean by work, just like the statement is true about |f^(n+1)(c)|<=M, I'm just trying to fully understand this, and so like I guess if Taylor series appeared on my examination my professor would have to actually check the final answer himself because like you said you can chose anything for M as long as the statement is true, |f^(n+1)(c)|<=M, and there's not just one answer for the rough estimation of M*|x-a|^(n+1)/(n+1)! that you would put at the end of the problem by plugging it into your calculator, that would really suck, and ya that makes sense, to use a tighter bound so that way the estimate is better
     
  9. Aug 17, 2011 #8

    micromass

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    Well, there are indeed multiple answers available. Like it says is the OP

    So if your professor asks this questions, then in principle M=10000000 should be correct. However, he probably ask something more specific.
     
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