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 Nov 17 revised Using Runge-Kutta-Fehlberg 4-5 for higher dimension systems added 58 characters in body Nov 17 asked Using Runge-Kutta-Fehlberg 4-5 for higher dimension systems Nov 13 comment Show $\frac {1}{2\pi i}\int_\gamma\frac {f'(z)}{f(z)}$ is sum of poles and zeroes times their order Could $\int \frac{f'}{f} = \log f$ be of help here ? Nov 11 comment How to solve the Integral $\int_{-\infty}^\infty [\tanh(\frac{x+a/2}{b})-\tanh(\frac{x-a/2}{b})]e^{ikx} dx$? Also, perhaps Fourier transform or some integral transform method? Nov 11 comment How to solve the Integral $\int_{-\infty}^\infty [\tanh(\frac{x+a/2}{b})-\tanh(\frac{x-a/2}{b})]e^{ikx} dx$? Maybe differentiate wrt to $a$. Just a thought to try... Nov 7 revised Not every holomorphic function $f$ can be written as $f(z)=e^{g(z)}$ edited body Nov 6 accepted How to write down, think about, and evaluate a simple Lebesgue integral Nov 4 asked How to write down, think about, and evaluate a simple Lebesgue integral Nov 3 comment What does it mean for two functions to be orthogonal? I've been expecting that comment. I can mentally grasp the idea of orthogonality in higher dimensions. Visualisation is tricky of course. I think my original confusion was in accepting functions as vectors, but once i intuitively pictured these functions as infinitely many component values along the x axis i can by extension mentally accept orthogonality of function vectors too. It's all a bit mental and vague, but helps to grasp the maths. Nov 3 accepted What does it mean for two functions to be orthogonal? Nov 3 comment What does it mean for two functions to be orthogonal? @Thomas Andrews so one interpretation is that two orthogonal functions have zero linear relationship. Nov 3 revised What does it mean for two functions to be orthogonal? added 111 characters in body Nov 3 comment What does it mean for two functions to be orthogonal? Thanks, nice property to point out. Although, I was hoping to avoid referring back to the inner product (the integral - which is related to the area). I was thinking more of properties of $f$ and $g$ which didn't rely on the inner product. Much like perpendicularity can be viewed geometrically without referring back to the scalar product. If that makes sense. Nov 3 revised What does it mean for two functions to be orthogonal? deleted 138 characters in body Nov 3 comment What is the geometric meaning of the inner product of two functions? So when we compute the "angle" between two functions, $\cos\theta=\frac{\langle f,g\rangle}{\|f\|\|g\|}$, we don't really know what "angle" means any more ? Nov 3 asked What does it mean for two functions to be orthogonal? Oct 30 awarded Popular Question Oct 26 comment How to think of a function as a vector? I finally found something that made intuitive sense to me linking the traditionally taught idea of a vector to these vectors as functions. I could never quite understand where the integral product $\langle\cdot,\cdot\rangle=\int_0^1 f(x)g(x)dx$ came from. I was always told it was just defined that way, but the lecturers never explained why this might be the case... eng.fsu.edu/~dommelen/quantum/style_a/funcvec.html. Seems that a suitably well-behaved function defined over a finite interval $[0,1]$ can represent an infinite dimensional vector. Makes sense now. Oct 26 comment Sketch $\{z^2|\text{Re}(z)>0\}$. Having troubles with finding what points are or aren't in the graph. @Omnomnomnom I'd not considered the image/codomain difference before, so thanks for your comment. However, it was not clear to me. The OP mentions graph in the title which means to me the set of ordered pairs $(z,z^2)$, both elements being complex numbers. But yes, the sketching statement clears that up. Apologies. Oct 26 comment Sketch $\{z^2|\text{Re}(z)>0\}$. Having troubles with finding what points are or aren't in the graph. So what you actually want to do is plot the image of the function, i.e. the codomain.