೧ ಯಾವುದಿದ್ಯಾವುದು ಯಾವ ಶಕ್ತಿ ಇದಾವುದು? ಇದ್ದರೇಕಿಂದಿಷ್ಟು ಕುರುಡಾಗಿರುವುದು? ಇರದಿದ್ದರೇಕಿದಿಷ್ಟು ಬೇಕೆನಿಸುತಿಹುದು? ಇದೇಕಿದಿಷ್ಟು ಮೂಖಾಗಿಹುದು? ಇರಳೀಕೆ-ಇರನಿವನು-ಇರದಿಹುದು ಎನ್ನಲೇನಿಹುದು? ೨ ಸೆರಗ ಕೈ ಚೆಲ್ಲಿ ಕೃಷ್ಣಾ ಎಂದ ಮಾತ್ರವೇ ಕೃಪೆಯೇ? ಭಕ್ತಿ ಕ್ಷೀರ ಸಾಗರದಿ ಮಿಂದ ಮಾತ್ರವೇ ಮಡಿಯೇ? ಬೇಡದೆ ಕೈ ಕಡಿವ ಕಾಳಿಯಾಗಬೇಕೆಂಬುದು ಅಪನಂಬಿಕೆಯೇ? ಇದ್ದರೇನಿರಲಿಲ್ಲದಿದ್ದರೇನೆಂಬ ಬದುಕು ನಾಸ್ತಿಕತೆಯೇ? ಇರುವಳೀಕೆ-ಇರುವನಿವನು-ಇದ್ದೀತಿದು ಎನ್ನಲೇನಾದರಿದಿಯೇ? ~ ಮಂಜುನಾಥ ಎಂ ಆರ್ English explanation (Not translated version of the poem): 1 What is this? Which power is it? Why is it so blind? If it exists Why does it feel like a needed thing? If it does not exist Why is it so silent? What proof is there? To say he doesn't exist - she doesn't exist - it doesn't exist. 2 Is it necessary to shout for Krishna and drop hands on the saree to get grace? ( Context: Draupadi vastraharana ) Is deep diving into the milky ocean of devotion necessary for a pious life? Is it disbelief if she became Kali and cut the hands of the adversary? ( Imagine if Draupadi thought abou...
Solution to Question 1:
In Question 1, I asked whether it is possible to build a Taylor series expansion of $$ f(z) = \sqrt{(\pi + e z) + i(\pi - e z)} $$ around the point \( z_0 = 1 + i \), and if so, how far it converges.
The function involves a square root, which is multi-valued, so its analyticity depends on the branch chosen. We shall consider the principal branch of the square root, which places a branch cut along the set of complex numbers where the argument of the root lies on the non-positive real axis.
Let \( z = x + i y \). Then, $$ (\pi + e z) + i(\pi - e z) = (\pi + e x + e y) + i(\pi - e x + e y), $$ which we rewrite as \( A + i B \), where $$ A = \pi + e x + e y, \quad B = \pi - e x + e y. $$ The branch cut occurs where \( B = 0 \) and \( A \leq 0 \). Solving, $$ \pi - e x + e y = 0 \Rightarrow y = x - \frac{\pi}{e}, $$ and substituting into \( A \leq 0 \) gives: $$ \pi + e x + e\left(x - \frac{\pi}{e}\right) = 2 e x \leq 0 \Rightarrow x \leq 0. $$ So the branch cut lies along the line $$ y = x - \frac{\pi}{e}, \quad x \leq 0. $$ To determine the radius of convergence of the Taylor series around \( z_0 = 1 + i \), we compute the Euclidean distance from \( z_0 \) to its nearest point on this branch cut. Since the function is analytic in the open disk centered at \( z_0 \) that avoids the cut, this distance gives the radius of convergence.
Therefore, the Taylor series exists and converges in a neighborhood of \(1 + i\), and its radius of convergence is equal to the distance from \(1 + i\) to the nearest point on the branch cut, which is \(-i \frac{\pi}{e}\) [1, p.167]. Thus, $$ R = \left| (1 + i) - \left( -i \frac{\pi}{e} \right) \right| = \sqrt{1 + \left(1 + \frac{\pi}{e}\right)^2}. $$ Reference:
[1] S. Ponnusamy, Foundations of Complex Analysis (2nd edition), Narosa Publishing House, New Delhi, 2005.
In Question 1, I asked whether it is possible to build a Taylor series expansion of $$ f(z) = \sqrt{(\pi + e z) + i(\pi - e z)} $$ around the point \( z_0 = 1 + i \), and if so, how far it converges.
The function involves a square root, which is multi-valued, so its analyticity depends on the branch chosen. We shall consider the principal branch of the square root, which places a branch cut along the set of complex numbers where the argument of the root lies on the non-positive real axis.
Let \( z = x + i y \). Then, $$ (\pi + e z) + i(\pi - e z) = (\pi + e x + e y) + i(\pi - e x + e y), $$ which we rewrite as \( A + i B \), where $$ A = \pi + e x + e y, \quad B = \pi - e x + e y. $$ The branch cut occurs where \( B = 0 \) and \( A \leq 0 \). Solving, $$ \pi - e x + e y = 0 \Rightarrow y = x - \frac{\pi}{e}, $$ and substituting into \( A \leq 0 \) gives: $$ \pi + e x + e\left(x - \frac{\pi}{e}\right) = 2 e x \leq 0 \Rightarrow x \leq 0. $$ So the branch cut lies along the line $$ y = x - \frac{\pi}{e}, \quad x \leq 0. $$ To determine the radius of convergence of the Taylor series around \( z_0 = 1 + i \), we compute the Euclidean distance from \( z_0 \) to its nearest point on this branch cut. Since the function is analytic in the open disk centered at \( z_0 \) that avoids the cut, this distance gives the radius of convergence.
Therefore, the Taylor series exists and converges in a neighborhood of \(1 + i\), and its radius of convergence is equal to the distance from \(1 + i\) to the nearest point on the branch cut, which is \(-i \frac{\pi}{e}\) [1, p.167]. Thus, $$ R = \left| (1 + i) - \left( -i \frac{\pi}{e} \right) \right| = \sqrt{1 + \left(1 + \frac{\pi}{e}\right)^2}. $$ Reference:
[1] S. Ponnusamy, Foundations of Complex Analysis (2nd edition), Narosa Publishing House, New Delhi, 2005.
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