| Thread Closed |
Differentiation help... |
Share Thread | Thread Tools |
| Feb7-05, 05:57 PM | #1 |
|
|
Differentiation help...
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) |
| PhysOrg.com |
science news on PhysOrg.com >> Hong Kong launches first electric taxis >> Morocco to harness the wind in energy hunt >> Galaxy's Ring of Fire |
| Feb7-05, 06:07 PM | #2 |
|
|
U messed up the tex code...
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... Legendre transformations are fundamental in physics,yet a description of them cannot be given within a couple of lines. Daniel. |
| Feb7-05, 06:41 PM | #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?
|
| Feb7-05, 06:47 PM | #4 |
|
|
Differentiation help...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 |
| Feb7-05, 08:02 PM | #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...
|
| Feb7-05, 08:12 PM | #6 |
|
|
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. |
| Feb8-05, 03:51 AM | #7 |
|
Recognitions:
|
[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. |
| Feb8-05, 11:18 AM | #8 |
|
|
i see good explaination! thanks
|
| Thread Closed |
| Thread Tools | |
Similar Threads for: Differentiation help...
|
||||
| Thread | Forum | Replies | ||
| For f(x) = x^n, I have to prove f'(x) = 1/n x^{n-1} | General Math | 18 | ||
| differentiation of tan^-1 | Calculus & Beyond Homework | 7 | ||
| Differentiation | Calculus & Beyond Homework | 2 | ||
| Differentiation | Introductory Physics Homework | 4 | ||
| differentiation | Differential Equations | 1 | ||