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Using Taylor expansion for limit solving

  1. Apr 26, 2012 #1
    Hi. I just want to ask: how can I realize that I need to do the 4th order taylor's expansion for solving a precise limit? e.g.

    [tex]\mathop {\lim }\limits_{x\to 0} \frac{{{e^x}-1-\frac{{{x^2}}}{2}+\sin x-2x}}{{1-\cos x-\frac{{{x^2}}}{2}}}[/tex]

    We need the 4th order of the expansion but how can I realize just seeing it?


    another question. Why is this true:
    [tex]\displaystyle\mathop {\lim }\limits_{x \to 0} \frac{{o(x)}}{x} = 0[/tex]?

    o(x) is the remainder (Landau notation)
    Thanks!
     
    Last edited: Apr 26, 2012
  2. jcsd
  3. Apr 26, 2012 #2

    Office_Shredder

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    Your tex isn't working for me so I can't answer your first question, but in general you need to use enough terms so that all larger terms just become zero in the limit. Try re doing the calculation using a 5th degree taylor polynomial, our a sixth degree one and you should get the same answer. With time it will become intuitive how many terms you will need fire this to happen
     
  4. Apr 26, 2012 #3
    Hmm, alright and what about the remainder o(x)?

    PD: I've fixed the LaTeX just when you were typing I guess


    Thank you1
     
  5. Apr 26, 2012 #4

    vela

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    Remainder of what?
     
  6. Apr 26, 2012 #5
    Of the Taylor polynomial
     
  7. Apr 26, 2012 #6
    Tried to solve the following but couldn't:

    [tex]\begin{array}{c}
    \mathop {\lim }\limits_{x\to 0} \frac{{\log (1+x)-x-\frac{{{x^2}}}{2}}}{{\tan x-\sin x}}=\frac{{-2x-{x^2}-\frac{{{x^3}}}{3}+{o_1}({x^3})}}{{2x+\frac{{{x^3}}}{3}+{o_2}({x^3})+{o_3}({x^3})}}\\
    =\frac{{\frac{{-2x-{x^2}-\frac{{{x^3}}}{3}+{o_1}({x^3})}}{{{x^3}}}}}{{\frac{{2x+\frac{{{x^3}}}{3}+{o_2}({x^3})+{o_3}({x^3})}}{{{x^3}}}}}\\
    =\frac{{\frac{{-2}}{{{x^2}}}-\frac{1}{x}-\frac{1}{3}+\overbrace {\frac{{{o_1}({x^3})}}{{{x^3}}}}^0}}{{\frac{2}{{{x^2}}}+\frac{1}{6}+\underbrace {\frac{{{o_2}({x^3})}}{{{x^3}}}}_0+\underbrace {\frac{{{o_3}({x^3})}}{{{x^3}}}}_0}}\\
    =?
    \end{array}[/tex]


    EDIT: I could solve it. It's -infinity.


    Why is this true:
    [tex]\displaystyle\mathop {\lim }\limits_{x \to 0} \frac{{o(x)}}{x} = 0[/tex]?

    o(x) is the remainder (Landau notation)
    Thanks!
     
    Last edited: Apr 26, 2012
  8. Apr 26, 2012 #7

    vela

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    It's true by definition of Landau notation, but that's probably not the answer you're looking for. When you write, say, o(x), you're talking about the terms that go to 0 faster than x does. In the quotient o(x)/x, there are two competing behaviors. The numerator going to 0 tends to drive the quotient toward 0, but the denominator going to 0 tends to cause the quotient to blow up. The behavior that dominates determines what happens. In this case, o(x), by definition, goes to 0 faster than x does, so o(x)/x goes to 0 in the limit that x goes to 0.

    Consider the limit
    $$\lim_{x \to 0} \frac{e^x-1}{x} = \lim_{x \to 0} \frac{\left(1+x+\frac{x^2}{2!}+\cdots\right) - 1}{x} = \lim_{x \to 0} \frac{x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots}{x} = \lim_{x \to 0} \frac{x + o(x)}{x}$$ The quadratic and higher-order terms go to 0 faster than x does, which is why you can replace them with o(x). If you skip the o-notation, though, you can see what happens is simply this:
    $$\lim_{x \to 0} \frac{x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots}{x} = \lim_{x \to 0} \frac{x + x\left(\frac{x}{2!} + \frac{x^2}{3!} + \cdots\right)}{x} = \lim_{x \to 0} \left[1 + \left(\frac{x}{2!} + \frac{x^2}{3!} + \cdots\right)\right]$$ All the terms that correspond to o(x)/x, the stuff inside the parentheses in the last limit, have a factor of x in them, so as x approaches 0, they vanish.
     
    Last edited: Apr 26, 2012
  9. Apr 26, 2012 #8
    But I'm not replacing the Taylor expansion by o(x) as you have just done. We prooved that the remainder of order n is $$o((x-a)^n))$$ when x->a. See how I solved the limit above your post. I just added the remainder to the Taylr polynomial (o(x))

    Could it be that, as the remainder tends to be zero, then the quotient o(x)/x is zero?
     
  10. Apr 26, 2012 #9

    vela

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    I didn't replace the Taylor expansion by o(x).
     
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