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An Error Formula for Linearization (involving second Derivative)

  1. Aug 7, 2017 #1
    1. The problem statement, all variables and given/known data
    In textbook i was given formula to calculate error.

    I know that:
    E(t) = f(t)- L(x) = f(t) - f(a)- f'(a)(t- a) [L(x) is linear approximation]; [Lets call this Formula 1]

    I understand that, but that I have formula:
    E(x) = f''(s)/2 * (x-a)^2 [lets call this Formula 2]
    Here I was given Prove that this is correct, by going from formula 2 to formula 1.


    2. Relevant equations

    L(x) = f(a)+f'(a)(t- a)

    ex.jpg

    3. The attempt at a solution

    I can't connect this 2 formulas to each other logically and i dont want just tu remember (i want to understand it well)

    P.S. I will attach screenshot if I didnt explain everything.
     
  2. jcsd
  3. Aug 7, 2017 #2

    Charles Link

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    The error term is essentially the next term in the Taylor expansion. (## f(x)=f(a)=f'(a)(x-a)+\frac{1}{2} f''(a)(x-a)^2+...##). The first two terms in the Taylor expansion of the function gives you a straight line. If there is a quadratic term in computing the Taylor expansion of the function, that is the error (from the straight line being the exact result), the one difference being instead of computing the second derivative at ## x=a ##, they compute it at some point on the interval (a,x) for some ## s ## and ignore the higher order (3rd order, 4th order, etc.) terms. It should allow you to assign error bars to the straight line approximation where the second order term is ignored. Hopefully this explanation was helpful.
     
  4. Aug 7, 2017 #3

    Ray Vickson

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    The attachment seems clear enough: it is expressing the error of the "linear" approximation as an expression involving the second derivative. It derives everything in detail, so I cannot figure out what is your difficulty. However, an alternative explanation is the following.

    For an "analytic" function ##f(x)## we may express the result exactly as a series about the point ##x = a##:
    $$ f(x) = f(a) + f'(a) (x-1) + \underbrace{f^{\prime \prime}(a) (x-a)^2 /! + f^{\prime \prime \prime }(a) (x-a)^3/3! + \cdots}_{\text{infinite series involving all derivative orders}} $$
    that involves an infinite series of second, third, fourth, fifth ..... derivatives. So, the exact error in using the first two terms (constant + linear) consists of the infinite series that we are omitting. Estimating the error from the infinite series is not very practical, so we look for a simpler expression. Magically, it turns out that
    $$f^{\prime \prime}(a) (x-a)^2 /2! + f^{\prime \prime \prime }(a) (x-a)^3/3! + \cdots = f^{\prime \prime}(\xi) (x-1)^2/2!$$
    for some value ##\xi## that lies between ##a## and ##x##. Basically, that is what the attachment establishes, but does it in another way (and also does it for functions that are not "analytical" and so could not be expanded in an infinite series like one above).

    Usually we cannot figure out what is the actual value of ##\xi##, but we can usually develop error bounds by finding a reasonable estimate of the maximum of ##|f''(\xi)|## on the interval between ##a## and ##x##. That is, if ##M## is the maximum (or more) of ##|f''|## on the interval, then we know that the size of the error is does not exceed ##M (x-a)^2/2##. Sometimes we may not be able to determine an exact maximum of ##|f''|## on the interval, but can, instead, determine a larger value of ##M## than the maximum. We can just go ahead and use that ##M## in the error bound.
     
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