TokenMonkey Posted February 5, 2011 Posted February 5, 2011 Hi all, I hope this is the right forum to post this question in, rather than one of the subforums, as it deals with notation, not actual mathematics. I've attached a pic of the problem I'm looking at: In Eq. (31), an implicit ODE is given; note how the last two derivatives are superscripted with a summation/capital sigma. What does this mean? I've Googled as best I can, but no luck. The only clue I have is that one of the terms superscripted with the sigma is a matrix of coefficients. There are more examples of this notation later on in the attachment. Any thoughts on this? Even a pointer in the right direction would be helpful. Thanks, TM
timo Posted February 5, 2011 Posted February 5, 2011 From a quick look: I've not seen the notation before, but if [math]\vec f(T, \vec x)[/math] is a vector-valued function with components [math]f_i[/math] depending on the real-valued variable T and the vector-valued [math]\vec x[/math] which has components [math]x_j[/math], then [math] \frac{d f_i}{dT} = \frac{\partial f_i}{\partial T} + \sum_j \frac{\partial f_i}{\partial x_j} \frac{\partial x_j}{\partial T}[/math] for all i. I'd guess that's what the equation means (note that the terms [math]\frac{\partial f_i}{\partial x_j}[/math] can be considered forming a matrix with indices i and j).
TokenMonkey Posted February 7, 2011 Author Posted February 7, 2011 From a quick look: I've not seen the notation before, but if [math]\vec f(T, \vec x)[/math] is a vector-valued function with components [math]f_i[/math] depending on the real-valued variable T and the vector-valued [math]\vec x[/math] which has components [math]x_j[/math], then [math] \frac{d f_i}{dT} = \frac{\partial f_i}{\partial T} + \sum_j \frac{\partial f_i}{\partial x_j} \frac{\partial x_j}{\partial T}[/math] for all i. I'd guess that's what the equation means (note that the terms [math]\frac{\partial f_i}{\partial x_j}[/math] can be considered forming a matrix with indices i and j). Thanks, Timo. I guess that that's what it must mean. Still, it's a weird, weird way of expressing a simple concept.
timo Posted February 7, 2011 Posted February 7, 2011 Oh, and I made a little typo. It's [math] \frac{d f_i}{dT} = \frac{\partial f_i}{\partial T} + \sum_j \frac{\partial f_i}{\partial x_j} \frac{d x_j}{d T}[/math], not [math] \frac{d f_i}{dT} = \frac{\partial f_i}{\partial T} + \sum_j \frac{\partial f_i}{\partial x_j} \frac{\partial x_j}{\partial T}[/math] (i.e. a total derivative in the last term, not a partial one). But I hope that typo was obvious when comparing to the image you linked (or at least insignificant).
TokenMonkey Posted February 7, 2011 Author Posted February 7, 2011 Oh, and I made a little typo. It's [math] \frac{d f_i}{dT} = \frac{\partial f_i}{\partial T} + \sum_j \frac{\partial f_i}{\partial x_j} \frac{d x_j}{d T}[/math], not [math] \frac{d f_i}{dT} = \frac{\partial f_i}{\partial T} + \sum_j \frac{\partial f_i}{\partial x_j} \frac{\partial x_j}{\partial T}[/math] (i.e. a total derivative in the last term, not a partial one). But I hope that typo was obvious when comparing to the image you linked (or at least insignificant). Haha, I did notice, but I wasn't about to go nitpicking with someone who was trying help me.
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