Overflow Error Complex Exponentiation
here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site About Us Learn more about Stack Overflow the company Business Learn more about hiring developers or posting ads with us Stack Overflow Questions Jobs Documentation Tags Users Badges Ask Question x Dismiss Join the Stack Overflow Community Stack Overflow is a community of 6.2 million programmers, just like you, helping each other. Join them; it only takes a minute: Sign up is OverflowError actually raised? up vote 5 down vote favorite According to the python documentation exception OverflowError Raised when the result of an arithmetic operation is too large to be represented. This cannot occur for long integers (which would rather raise MemoryError than give up) and for most operations with plain integers, which return a long integer instead. Because of the lack of standardization of floating point exception handling in C, most floating point operations also aren’t checked. Indeed, this error made sense when overflowing integers were not converted to long automatically. Similarly, floats overflow to inf. I don't really see any situation where the standard interpreter may still raise OverflowError. Is there such a case somewhere ? Just a curiosity. python share|improve this question asked Aug 21 '11 at 12:52 Stefano Borini 55.2k46210329 add a comment| 1 Answer 1 active oldest votes up vote 7 down vote accepted Python 2.7.2 (v2.7.2:8527427914a2, Jun 11 2011, 15:22:34) [GCC 4.2.1 (Apple Inc. build 5666) (dot 3)] on darwin Type "help", "copyright", "credits" or "license" for more information. >>> float(10**1000) Traceback (most recent call last): File "
07:15 PMHello Ubuntu Forums community, I am making a python program that requires the math.pow function. I need it to go up to 2^31 (2 to the 31st power) and preferably even farther. When I run the program in the terminal, it says "OverflowError: math range error" Any way to fix this? Thanks in advance. BachstelzeSeptember 25th, 2010, 07:20 PMWhat are you doing, exactly? math.pow(2, 32) doesn't raise an exception for me. Queue29September 25th, 2010, 07:32 PMPython supports arbitrary precision numbers out of the box. What version http://stackoverflow.com/questions/7138387/is-overflowerror-actually-raised are you using? And can you post the code that seems to be troublesome? ssamSeptember 25th, 2010, 07:57 PMPython supports arbitrary precision numbers out of the box. What version are you using? And can you post the code that seems to be troublesome? thats not right. its just long ints that have infinite range. http://docs.python.org/library/stdtypes.html#numeric-types-int-float-long-complex >>> math.sin(pow(2,1000)) -0.15920170308624243 >>> math.sin(pow(2,10000)) Traceback https://ubuntuforums.org/archive/index.php/t-1581805.html (most recent call last): File "
here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site About Us Learn more about Stack Overflow the http://math.stackexchange.com/questions/169623/how-complex-exponential-converges-and-sum-of-exponents-rule-holds company Business Learn more about hiring developers or posting ads with us Mathematics Questions Tags Users Badges Unanswered Ask Question _ Mathematics Stack Exchange is a question and answer site for people studying math at https://books.google.com/books?id=YJ_0O9Z-knoC&pg=PA445&lpg=PA445&dq=overflow+error+complex+exponentiation&source=bl&ots=qhBZXgAE--&sig=pUxmDD4AfhhqxPSOhpzwB_-ON_M&hl=en&sa=X&ved=0ahUKEwibyd_U1OXPAhVE5xoKHSrlDNgQ6AEIVTA any level and professionals in related fields. Join them; it only takes a minute: Sign up Here's how it works: Anybody can ask a question Anybody can answer The best answers are voted up and rise overflow error to the top How complex exponential converges and “sum of exponents” rule holds up vote 1 down vote favorite How is it the complex exponential converges for any value of $z$ in the complex plane? $$e^{z} = 1 + \frac{z}{1!} + \frac{z^2}{2!} \cdots\cdots$$ How is it the "sum of exponents" rule holds for complex exponential, that is $e^{w}e^{z} = e^{w+z} $? ...using only the definition of $e^z$? complex-numbers exponentiation share|cite|improve this question overflow error complex edited Jul 11 '12 at 20:03 anon 63.8k395179 asked Jul 11 '12 at 20:00 Aftnix 1617 Ratio test, binomial theorem. –anon Jul 11 '12 at 20:01 Hint (for the second part): If you multiply the two power series together and collect up terms, and use the binomial theorem, you should be able to prove this. It can be justified by absolute convergence. –Old John Jul 11 '12 at 20:02 i know 2nd expression can be proved by binomial theorem, but failed to do so. and for the first one, if only the postulates of "complex arithmetic" is given, how it converges? –Aftnix Jul 11 '12 at 20:03 Latex output is not correct in my browser, sometimes its ok, sometimes it not. –Aftnix Jul 11 '12 at 20:05 @Aftnix If you are familiar with the proof of the ratio test in the real case, you should see that it carries over easily for complex series. –user17794 Jul 11 '12 at 20:17 add a comment| 3 Answers 3 active oldest votes up vote 2 down vote accepted $$e^z=\sum_{n=0}^\infty\frac{z^n}{n!}$$ Putting $$a_n:=\frac{1}{n!}\Longrightarrow R:=\frac{1}{\lim_{n\to\infty}\sqrt[n]{|a_n|}}=\lim_{n\to\infty}\sqrt[n]{n!}=\infty$$ So the series has infinite convergence radius and is thus absolutely convergent for any $\,z\in\Bbb C\,$ , and from here it follows that $$e^{w+z}=\sum_{n=0}^\infty\frac{(w+z)^n}{n!}=\sum_{n=0}^\infty\sum_{k=0
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