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 19h comment Solving $\exp\bigg(\frac{2+\pi i}{4} \bigg)$ To solve an equation you need an equality sign and one unknown variable such as $x$. You probably mean simplify this expression, or similar. Wording is useful, especially in exams, as it tells you what you need to do. Took me a while to realise that many moons ago :-) Nov 25 comment is my answer correct? derivative of logarithmic functions $y=-1/\log (x)$ Nov 19 comment Using Runge-Kutta-Fehlberg 4-5 for higher dimension systems Thanks - I missed that - the $k_i$ should be computed for all components before moving onto $k_{i+1}$ because to compute the $k_{i+1}$ for all components we need $k_i$ for all components. Nov 18 comment Basic Residue problem The contour, $|z-1|=1$, is the unit circle in the complex plane centred at $z=1$. Are there any poles in that region? Find the poles, if any, compute the residues, and then use the residue theorem (sum of residues) multiplied by a constant... Nov 18 comment Using Runge-Kutta-Fehlberg 4-5 for higher dimension systems I've updated my question - but I'm not clear if it relates to your answer. Does my update agree with your answer? Perhaps your $k_{ijk}$ are my $a,b,c,d$, e.g. $k_{1jn}=a_j^{(n)}$ Nov 17 comment Using Runge-Kutta-Fehlberg 4-5 for higher dimension systems Thanks. Do you know of a reference which contains RK45 in 2 and 3 dimension spaces? 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 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 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 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 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 ? 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. Oct 26 comment Sketch $\{z^2|\text{Re}(z)>0\}$. Having troubles with finding what points are or aren't in the graph. You are attempting to plot 4-dimensional data since $z=x+iy$ (two dimensions), and $z^2=u+iv$ (a further two dimensions). One way to plot this is using a vector field, but that's probably not what you're after. Another way would be to plot $|z^2|$ in a three-dimensional space. Also, you say your expressions are too hard - maybe you will need to numerically evaluate your expressions. Oct 20 comment An easy way to remember PEMDAS Maybe you find BEDMAS easier to remember? Brackets, Exponentiation, Division, Multiplication, Addition, Subtraction. (I changed that from BODMAS [Brackets Off ...] to contain your inclusion of Exponentiation) Oct 20 comment Complex integration using singularities Yes, you can only use the theorem if its conditions are met, so any restrictions are perfectly acceptable. Later on you will be able to state the correct value of the integral for your special case ($|w|=1$) once you know how to compute it using the residue theorem. I suspect your lecturer will not expect you to consider the special case, but if you do know how to use the residue theorem then just evaluate that special case as well and put that in your answer ! Make sure to follow the instructions of the question carefully though - maybe they don't want you to consider that special case... Oct 20 comment Complex integration using singularities Sorry, I read your question incorrectly by forgetting your initial sentence after reading the last, which lead my thoughts down the wrong path !