TokenMonkey Posted February 5, 2011 Share 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 Link to comment Share on other sites More sharing options...
timo Posted February 5, 2011 Share 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). Link to comment Share on other sites More sharing options...
TokenMonkey Posted February 7, 2011 Author Share 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. Link to comment Share on other sites More sharing options...
timo Posted February 7, 2011 Share 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). Link to comment Share on other sites More sharing options...
TokenMonkey Posted February 7, 2011 Author Share 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. Link to comment Share on other sites More sharing options...
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