Derivatives on HP 42S

08262018, 04:54 AM
Post: #19




RE: Derivatives on HP 42S
(08252018 09:20 PM)Albert Chan Wrote: Can you explain the word analytic ? Consider the complex valued function \(w=z^2\). You can calculate both the real and imaginary part of \(w=u+iv\): \(\begin{aligned} u &= x^2y^2 \\ v &= 2xy \end{aligned}\) But these functions \(u(x, y)\) and \(v(x, y)\) are not independent. Instead the Cauchyâ€“Riemann equations hold true: \(\begin{aligned} \frac {\partial u}{\partial x}&=\frac{\partial v}{\partial y} \\ \frac {\partial u}{\partial y}&=\frac{\partial v}{\partial x} \end{aligned}\) And indeed: \(\begin{aligned} u_x &=2x=v_y \\ u_y &=2y=v_x \end{aligned}\) Thus this function is analytic. However this function isn't: \(\begin{aligned} u &= x^2+xy^2 \\ v &= 2xy \end{aligned}\) Because \(u_x=2x+1\) but still \(v_y=2x\). For the other function that I mentioned, e.g. \(\Re[z]\) we have: \(\begin{aligned} u &= x \\ v &= 0 \end{aligned}\) This isn't analytic since \(u_x=1\) but \(v_y=0\). In short: If you define the function in terms of \(z\) the function is most probably analytic. However if you try to stitch together a complex function based on \(x\) and \(y\) chances are high that it's not analytic. Quote:Is f(x) = x^(1/3) an analytic function ? Yes. Its derivative is: \(\frac{d}{dz}\left(\sqrt[3]{z}\right)=\frac{1}{3z^{\frac{2}{3}}}\) Quote:If x is complex, is it true that f(x) same as f(x) ? No. Consider \(f(z)=z^2\). Here we have \(f(z)=f(z)\). Cheers Thomas 

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