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Homework Help: Differentiation help

  1. Feb 7, 2005 #1
    I don't understand how
    [tex]\frac{\partial{f}}{\partial{x_k}}(\vec{a}) = \lim_{\epsilon\rightarrow 0} \frac{f(a_1,...,a_k+\epsilon,...,a_n)-f(a_1,...,a_k,...,a_n)}{\epsilon}[/tex]
    can be equal to...
    [tex]\Delta_k f = f(a_1,...,a_k + \epsilon,...,a_n)-f(a_1,...,a_k,...,a_n) \approx \Delta x_k \frac{\partial{f}}{\partial{x_k}}[/tex]
    and why is it approximately?

    what does [tex] \Delta x_k [/tex] really repersent? graphiclly and with an example.

    why is

    [tex] df = \sum \frac{\partial{f}}{\partial{x_i}} dx_i [/tex]
    using [tex] dx_i [/tex] what does that graphicly represent? and what is Legendre transformation and what is its significance. Please try to explain (calc I-III, Linear algebra background)
     
    Last edited: Feb 7, 2005
  2. jcsd
  3. Feb 7, 2005 #2

    dextercioby

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    U messed up the tex code... :rolleyes:

    At the first,it's simply the partial derivative's definition,though improperly written,while at the second,it's simply the first term (linear in differentials) of the Taylor expansion of a function of multiple variables.

    Graphically,it's difficult to describe in words.Imagine a surface.The partial derivative wrt "x" at a point on the surface is nothing else but the tangent of the angle made by a tangent line in that point to the curve delimited by the Oxz plane and the surface and the Ox axis...Really disgusting... :yuck:

    Legendre transformations are fundamental in physics,yet a description of them cannot be given within a couple of lines.

    Daniel.
     
  4. Feb 7, 2005 #3
    I don't know what the Taylor expanision of a function of multiple variables means, can u elaborate? Still what is [tex]\Delta x_k [/tex] how do u calculate that?
     
  5. Feb 7, 2005 #4

    dextercioby

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    Then i'll have to advise you to read a calculus book on multiple variable calculus,where i'm sure you're gonna find the Taylor series explained much better than i'd succed if i were to try to.
    Those delta's are small variations,you do not calculate them by any mean,they're an analogus for the differentials.

    Da
     
  6. Feb 7, 2005 #5
    i see so is there an example u can give me that uses the total differential or where I'd need to? Such as in phyiscs...
     
  7. Feb 7, 2005 #6

    dextercioby

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    Thermodynamics:
    [tex] dS= \frac{dU}{T}+pdV+\mu dN+... [/tex]

    [tex] dS=(\frac{\partial S}{\partial U})_{V,N,...} dU +(\frac{\partial S}{\partial V})_{U,N,...} dV+(\frac{\partial S}{\partial N})_{U,V,...} dN+... [/tex]

    Daniel.
     
  8. Feb 8, 2005 #7

    Galileo

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    I`ll take the singe variable case, because the reasoning is similar.

    [tex]\frac{d}{dx}f(x)=\lim_{h \to 0} \frac{f(x+h)-f(x)}{h}=f'(x)[/tex]

    which means, that for any [itex]\epsilon>0[/itex] we can find a [itex]\delta[/itex], such that:
    [tex]|h|<\delta \Rightarrow \left|\frac{f(x+h)-f(x)}{h}-f'(x)\right|<\epsilon[/tex].

    So if we make h small enough, it will be close enough to f'(x). Therefore, for small h:
    [tex]\frac{f(x+h)-f(x)}{h} \approx f'(x)[/tex]
    or
    [tex]f(x+h)-f(x) \approx hf'(x)[/tex]

    The h in this case is what the [itex]\Delta x_k[/itex] represents in your multivariable case.
     
  9. Feb 8, 2005 #8
    i see good explaination! thanks
     
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