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How, in general, do you expand an expression in powers of some variable?

  1. Feb 23, 2009 #1
    How, in general, do you "expand" an expression in powers of some variable?

    I'm studying mechanics right now out of the Landau-Lifschitz book, and he often talks about expanding expressions. But I honestly don't know how to do this, in general. I know about forming Taylor expansions of functions of one variable, but when things get to more than one variable, I start to get confused. Let me give you an example: Apparently, if we expand a function

    [tex]
    L(\vec{v}^2 + 2\vec{v} \cdot \vec{\epsilon} + \vec{\epsilon}^2)
    [/tex]

    "in powers of [tex]\vec{\epsilon}[/tex]," to first order, where [tex]\vec{\epsilon}[/tex] is some infinitesimal vector, we get

    [tex]
    L(\vec{v}^2) + \frac{\partial L}{\partial \vec{v}^2} 2 \vec{v} \cdot \vec{\epsilon}.
    [/tex]

    Could someone please explain why that is? You know, how to get that? And what's so special about expanding "in powers of [tex]\vec{\epsilon}[/tex]?" As best I can tell, we are letting [tex]\vec{\epsilon} = 0[/tex] when we do the expansion. So are we expanding [tex]L[/tex] about [tex]\vec{\epsilon} = 0[/tex]? If so, why are we taking a partial of [tex]L[/tex] with respect to [tex]\vec{v}^2[/tex]?

    I have tried to use the information provided at http://en.wikipedia.org/wiki/Taylor_Series to figure this out - particularly what's at the bottom of the page - but without much success.
     
  2. jcsd
  3. Feb 23, 2009 #2

    lurflurf

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    Re: How, in general, do you "expand" an expression in powers of some variable?

    a simple generalization easily seen by expanding by each variable in turn
    let x,h be vectors
    D the vector derivative so h.D is the directional derivative times h's magnitude
    f(x+h)=exp(h.D)f(x)=f(x)+[h.D]f(x)+[(h.D)^2]f(x)/2!+[(h.D)^3]f+...+[(h.D)^k]f(x)/k!+...
     
  4. Feb 24, 2009 #3

    HallsofIvy

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    Staff Emeritus
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    Re: How, in general, do you "expand" an expression in powers of some variable?

    If you are asking about how to get that and why it involves a derivative, "expanding in powers" is just the Taylor's series: The Taylor's series for function f(x), about x= a is
    [tex]f(a)+ \frac{df}{dx}(x-a)+ \frac{1}{2}\frac{d^2f}{dx^2}(x- a)^2+ \cdot\cdot\cdot[/tex]
    [tex]= \sum_{n=0}^\infty \frac{1}{n!}\frac{d^n f}{dx^n} (x- a)^n[/tex]
    Of course, if f is a function of several variables, that derivative just becomes the partial derivative with respect to the relevant variable.
     
  5. Dec 17, 2009 #4
    Re: How, in general, do you "expand" an expression in powers of some variable?

    I'am a little confused about this expantion too.

    [tex]
    L(\vec{v'}^2) = L(\vec{v}^2 + 2\vec{v} \cdot \vec{\epsilon} + \vec{\epsilon}^2)
    [/tex]

    At point [tex]a = \vec{v}'^2 = \vec{v}^2 + 2\vec{v} \cdot \vec{\epsilon} + \vec{\epsilon}^2[/tex] this expands to

    [tex]
    \begin{align*}
    L(\vec{v'}^2) &= L(\vec{v}^2) + \frac{\partial L}{\partial \vec{v'}^2} ( \vec{v}^2 + 2 \vec{v} \cdot \vec{\epsilon} + \vec{\epsilon}^2 - \vec{v}^2) + \cdots \\

    &= L(\vec{v}^2) + \frac{\partial L}{\partial \vec{v'}^2} ( 2 \vec{v} \cdot \vec{\epsilon} + \vec{\epsilon}^2) + \cdots\\

    &= L(\vec{v}^2) + \frac{\partial L}{\partial \vec{v'}^2} 2 \vec{v} \cdot \vec{\epsilon} + \frac{\partial L}{\partial \vec{v'}^2} \vec{\epsilon}^2 + \cdots
    \end{align*}
    [/tex]
    We leave just two terms and drop everything else, because then [tex]\vec{\epsilon}[/tex] approaches [tex]0[/tex] everything else is alot smaller.

    So we get
    [tex]
    L(\vec{v}^2) + \frac{\partial L}{\partial \vec{v'}^2} 2 \vec{v} \cdot \vec{\epsilon}
    [/tex]
    The remaining question is how to get?
    [tex]
    \partial \vec{v'}^2 = \partial \vec{v}^2
    [/tex]
     
  6. Dec 17, 2009 #5

    Landau

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    Re: How, in general, do you "expand" an expression in powers of some variable?

    In principle this is just Taylor's theorem in several variables. In physics this is used extensively, but I also don't always directly see how it is done. So I decided to write it out in a more formal way.

    We have a function of one variable [tex]L:\mathbb{R}\to\mathbb{R}: x\mapsto L(x)[/tex], and the inner product, a function of n variables, [tex]g:\mathbb{R}^n\to\mathbb{R}:\vec{x}\mapsto \vec{x}\cdot\vec{x}=\vec{x}^2.[/tex]

    We are interested in the composition [tex]L\circ g:\mathbb{R}^n\to\mathbb{R}[/tex], evaluated at [tex]\vec{v}+\vec{\epsilon}[/tex], since this is indeed [tex](L\circ g)(\vec{v}+\vec{\epsilon})=L((\vec{v}+\vec{\epsilon})^2)=L(\vec{v}^2+2\vec{v}\cdot\vec{\epsilon}+\vec{\epsilon}^2)[/tex]. We're going to appoximate this assuming epsilon is small.

    Taylor's theorem from Wikipedia says [tex]T(\vec{x}) = f(\vec{a}) + (\vec{x} - \vec{a})^T\mathrm{D} f(\vec{a}) + \frac{1}{2!} (\vec{x} - \vec{a})^T \,\{\mathrm{D}^2 f(\vec{a})\}\,(\vec{x} - \vec{a}) + \cdots[/tex], for the Taylor expansion of f around the point a.

    Substituting [tex]a=v, h=\epsilon, x=a+h=v+\epsilon[/tex], we get [tex]T(\vec{v}+\vec{\epsilon}) = f(\vec{v}) + (\vec{\epsilon})^T\mathrm{D} f(\vec{v}) + \frac{1}{2!} (\vec{\epsilon})^T \,\{\mathrm{D}^2 f(\vec{v})\}\vec{\epsilon} + \cdots[/tex].

    Now, the assumption that epsilon is small leads to the ignoring of powers of epsilon greater than one, i.e. all terms from [tex]\frac{1}{2!} (\vec{\epsilon})^T \,\{\mathrm{D}^2 f(\vec{v})\}\vec{\epsilon} + \cdots[/tex] on are ignored because of the occurence of at least two epsilon factors (here under the disguise of [tex](\vec{\epsilon})^T[/tex] and [tex]\vec{\epsilon}[/tex]).

    Now we compute this for [tex]f=L\circ g[/tex]:
    [tex]T(\vec{v}+\vec{\epsilon}) = (L\circ g)(\vec{v}) + (\vec{\epsilon})^T\mathrm{D} (L\circ g)(\vec{v})[/tex], using the chain rule in several dimensions: [tex]D(L\circ g)(x)=DL(g(x))\circ Dg(x)[/tex]. Since [tex]L[/tex] is just a function of one variable, this is equal to the product of the number [tex]L'(g(x))[/tex], i.e. the derivative of L evaluated at the point g(x), with the vector [tex]Dg(x)[/tex].

    Now, since [tex]g(\vec{x})=x_1^2+x_2^2+...+x_n^2[/tex], we have

    [tex]Dg(\vec{x})=\left(\frac{\partial g}{\partial x_1},\frac{\partial g}{\partial x_2},...,\frac{\partial g}{\partial x_n}\right)=(2x_1,2x_1,...,2x_n)[/tex].

    So
    [tex]T(\vec{v}+\vec{\epsilon}) = (L\circ g)(\vec{v}) + (\vec{\epsilon})^T\mathrm{D} (L\circ g)(\vec{v})[/tex]

    [tex]=L(\vec{v}^2)+(\vec{\epsilon})^TL'(\vec{v}^2)Dg(\vec{v})[/tex]
    [tex]=L(\vec{v}^2)+L'(\vec{v}^2)(\epsilon_1,\epsilon_2,...,\epsilon_n)\cdot (2v_1,2v_1,...,2v_n)[/tex]
    [tex]={L(\vec{v}^2)+L'(\vec{v}^2)2\vec{\epsilon}\cdot\vec{v}[/tex].

    The only question is, what do L&L mean by [tex]\frac{\partial L}{\partial \vec{v}^2}[/tex]? [tex]L[/tex] is just a function of one variable, so it doesn't really make sense to talk about partial derivatives. My notation [tex]L'(\vec{v}^2)[/tex] is unambiguous: THE derivative of L, evualated at [tex]\vec{v}^2[/tex].
     
  7. Dec 18, 2009 #6
    Re: How, in general, do you "expand" an expression in powers of some variable?

    I think your way is too difficult in this case. Why do you expand in several variables?

    I found the answer and will correct my previous post soon.
     
  8. Dec 18, 2009 #7
    Re: How, in general, do you "expand" an expression in powers of some variable?

    [tex]
    L(x) = L(\vec{v'}^2) = L(\vec{v}^2 + 2\vec{v} \cdot \vec{\epsilon} + \vec{\epsilon}^2)
    [/tex]

    At point [tex]a = \vec{v}^2[/tex] this expands to

    [tex]
    \begin{align*}
    L(x) &= L(a) + \frac{d L(a)}{d x} ( x - a) + \cdots \\

    &= L(\vec{v}^2) + \frac{d L(\vec{v}^2)}{d x} ( \vec{v}^2 + 2 \vec{v} \cdot \vec{\epsilon} + \vec{\epsilon}^2 - \vec{v}^2) + \cdots \\

    &= L(\vec{v}^2) + \frac{d L(\vec{v}^2)}{d x} ( 2 \vec{v} \cdot \vec{\epsilon} + \vec{\epsilon}^2) + \cdots\\

    &= L(\vec{v}^2) + \frac{d L(\vec{v}^2)}{d x} 2 \vec{v} \cdot \vec{\epsilon} + \frac{d L(\vec{v}^2)}{d x} \vec{\epsilon}^2 + \cdots
    \end{align*}
    [/tex]
    We leave just two terms and drop everything else, because then [tex]\vec{\epsilon}[/tex] approaches [tex]0[/tex] everything else is alot smaller.

    [tex]d x = d(\vec{v}^2 + 2\vec{v} \cdot \vec{\epsilon} + \vec{\epsilon}^2) = 2 d(\vec{v}) + 2 \vec{\epsilon} d(\vec{v}) = (1+\vec{\epsilon}) 2 d(\vec{v}) = (1+\vec{\epsilon}) d(\vec{v}^2)[/tex]

    Because [tex]\vec{\epsilon}[/tex] approaches [tex]0[/tex] and is alot smaller then [tex]1[/tex] we drop it. So

    [tex]d x = d(\vec{v}^2)[/tex]

    And finally
    [tex]
    L(\vec{v}^2) + \frac{d L(\vec{v}^2)}{d \vec{v}^2} 2 \vec{v} \cdot \vec{\epsilon}
    [/tex]
     
  9. Dec 19, 2009 #8

    Landau

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    Re: How, in general, do you "expand" an expression in powers of some variable?

    I don't think it is difficult, I just thoroughly explained what I was doing and using.
    How can you not expand in several variables, given that we are considering n-dimensional vectors [tex]\vec{v}[/tex] (of course, in classical mechanics, n=3)?
    I'm a bit confused about your notation. What does the expression [tex]\frac{d L(\vec{v}^2)}{d x}[/tex] mean? Just [tex]L'(\vec{v}^2)[/tex], the derivative of L evaluated at the point [tex]\vec{v}^2[/tex]? My problem with this notation is the use of d/dx; it suggests you are differentiating 'with respect to x' (whatever that means for a function of one variable), but you earlier defined x as v'^2.
    Same here:
    what does the expression [tex]\frac{d L(\vec{v}^2)}{d \vec{v}^2}[/tex] mean?

    Maybe my problem is that following mathematics courses has made me unconfortable with this kind of 'physics approach', of just calculating things like [tex]d(\vec{v}^2)[/tex], and substituting the result in the 'denominator of the derivative'.
     
  10. Dec 20, 2009 #9
    Re: How, in general, do you "expand" an expression in powers of some variable?

    But [tex]\vec{v'}^2[/tex] is scalar.

    Oh, of course, you are right, this notation is strange. Now I'm confused too :)
     
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