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 Mar19 awarded Curious Jul25 awarded Popular Question Jul25 awarded Teacher May4 comment Solving a large system of non-linear equations @Jamie, Sorry for the late reply. Yes, FindRoot is a numerical approach. And yes, the Conjugate-Gradient approach would work as you say. It's not the best resource but this wiki link, en.wikipedia.org/wiki/Conjugate_gradient_method, is a good place to start if you're thinking of going that route. Apr26 answered Solving a large system of non-linear equations Mar26 asked How to find discretezation error of NDSolve Oct30 awarded Scholar Oct30 accepted Issues FindRootPlot command Oct30 comment Issues FindRootPlot command Crap, I feel like such an idiot. Thanks for clearing this up. Oct30 asked Issues FindRootPlot command Oct25 comment Getting increased accuracy for roots of determinant Please find a more detailed version of the question here: scicomp.stackexchange.com/questions/8873/… Oct21 comment Getting increased accuracy for roots of determinant @Mr.Wizard, I've tried to keep the numbers to a high degree of precision in the matrix (around 50 digits) before attempting this Singular Value List thing. I'll check on this again, and get back to you. To be honest, I've read the documentation about precision and accuracy and working precision lots of times without clearly understanding it or how to implement it. Oct21 comment Getting increased accuracy for roots of determinant @ssch, I believe I understand the examples in the link provided. However I cannot get it to work with my code. I tried f[x_?NumericQ] := Last[SingularValueList[ N[a /. \[Kappa]\[CapitalOmega] -> x, Tolerance -> 0], 0]] . And it gives me a plot with horizontal line through the origin. Oct21 asked Getting increased accuracy for roots of determinant Sep24 awarded Supporter Sep2 asked Issue with Coefficient command Aug13 awarded Nice Question Nov13 comment Small Issue with Chebyshev Derivative Appoximation Hello again. So I should have asked this earlier, but why does numerical recipes approximate functions as (Sum c_i T_i(y) - 1/2*c_0 where i=0, i=m-1)? When I approximated the function above I did not need to subtract off half the value of the c_0 constant. Sorry if this is a really basic question. Is it simply because I am doing integrals and they are actually doing the sum over a grid of values? Is that the significance of the m-1 in the sum? Oct26 comment Small Issue with Chebyshev Derivative Appoximation Thanks you very much J.M. I'm not that familiar with the Clenshaw algorithm, but I will check the references you put up, and work through your code. This is so very helpful. Oct25 asked Small Issue with Chebyshev Derivative Appoximation