Timeline for Exponent of convergence of the sequence of zeros of $e^z+z$
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
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| Aug 26 at 11:02 | comment | added | Oscar Lanzi | Please ask this as a differentcqll question. I suspect the mids will disapprove of extending the discussion here. | |
| Aug 26 at 8:54 | comment | added | Factorial_zero | Just one more query @Oscar Lanzi, I was checking the same idea for the function $e^{e^z}+z$ by the way you suggested and I got that $z=\ln\{\ln(z)+i(2n+1)\pi\}+2ki\pi$ where n, k are integers. Is the logarithm of a logarithm absolutely small compared with z itself? If so, then the exponent of convergence of zeroes of $e^{e^z}+z$ is also 1. Am I correct? | |
| Aug 23 at 6:06 | comment | added | Oscar Lanzi | Yes that is correct. | |
| Aug 23 at 4:12 | comment | added | Factorial_zero | This means that for e^z+P(z) where P(z) is any non-constant polynomial, the exponent of convergence of sequence of zeroes is always 1. Also, I know that the order of e^z+P(z) is also 1. Am I correct @Oscar Lanzi? | |
| Aug 22 at 17:49 | comment | added | Oscar Lanzi | Yes, it works out the same way. Tge logarithm of any polynomial on $z$ becomes subdominant versus $z$ itself. Oh, not officially a professor. | |
| Aug 22 at 13:11 | comment | added | Factorial_zero | Yes, nice elaboration. Thank you professor @Oscar Lanzi. One quick question about this: if I take e^z+z^2+z then does your argument work " logarithm becomes absolutely small compared with z itself " ? | |
| Aug 22 at 10:47 | history | edited | Oscar Lanzi | CC BY-SA 4.0 |
added 2 characters in body
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| Aug 22 at 10:40 | history | answered | Oscar Lanzi | CC BY-SA 4.0 |