Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ • π ƒ ¹ ² ³ °
 

You are not logged in. #2 20131108 13:17:18
Re: √1+(√2+(√3+(√4+... = ? Harder than it seemsHi; In mathematics, you don't understand things. You just get used to them. I have the result, but I do not yet know how to get it. All physicists, and a good many quite respectable mathematicians are contemptuous about proof. #3 20131108 23:17:41
Re: √1+(√2+(√3+(√4+... = ? Harder than it seemsYes, that constant is called the nested radical constant, usually denoted C_{NR}. No closedform expression for that constant currently exists. #4 20131109 01:32:47
Re: √1+(√2+(√3+(√4+... = ? Harder than it seemsHi; and the Fibonacci Nested Radical In mathematics, you don't understand things. You just get used to them. I have the result, but I do not yet know how to get it. All physicists, and a good many quite respectable mathematicians are contemptuous about proof. #6 20131121 21:14:19
Re: √1+(√2+(√3+(√4+... = ? Harder than it seemsHi; In mathematics, you don't understand things. You just get used to them. I have the result, but I do not yet know how to get it. All physicists, and a good many quite respectable mathematicians are contemptuous about proof. #8 20131123 11:28:35
Re: √1+(√2+(√3+(√4+... = ? Harder than it seemsGoogle and M. In mathematics, you don't understand things. You just get used to them. I have the result, but I do not yet know how to get it. All physicists, and a good many quite respectable mathematicians are contemptuous about proof. #9 20131123 11:48:41
Re: √1+(√2+(√3+(√4+... = ? Harder than it seemsHow do you know which digits are correct? The limit operator is just an excuse for doing something you know you can't. “It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman “Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment #10 20131123 11:52:32
Re: √1+(√2+(√3+(√4+... = ? Harder than it seemsI am not following you? The calculation was done by M. In mathematics, you don't understand things. You just get used to them. I have the result, but I do not yet know how to get it. All physicists, and a good many quite respectable mathematicians are contemptuous about proof. #11 20131123 11:53:16
Re: √1+(√2+(√3+(√4+... = ? Harder than it seemsAnd you entered what exactly? The limit operator is just an excuse for doing something you know you can't. “It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman “Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment #12 20131123 11:55:01
Re: √1+(√2+(√3+(√4+... = ? Harder than it seemsJust what you see in post #2 but to many more iterations. In mathematics, you don't understand things. You just get used to them. I have the result, but I do not yet know how to get it. All physicists, and a good many quite respectable mathematicians are contemptuous about proof. #13 20131123 11:58:03
Re: √1+(√2+(√3+(√4+... = ? Harder than it seemsOkay, but then, how do you know to how many digits the result is accurate? The limit operator is just an excuse for doing something you know you can't. “It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman “Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment #14 20131123 11:59:59
Re: √1+(√2+(√3+(√4+... = ? Harder than it seemsBy using the double rule. In mathematics, you don't understand things. You just get used to them. I have the result, but I do not yet know how to get it. All physicists, and a good many quite respectable mathematicians are contemptuous about proof. #15 20131123 12:02:39
Re: √1+(√2+(√3+(√4+... = ? Harder than it seemsRemind me what that is again. The limit operator is just an excuse for doing something you know you can't. “It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman “Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment #16 20131123 12:06:23
Re: √1+(√2+(√3+(√4+... = ? Harder than it seemsYou run the calculation to n iterations and then you run it to 2n. You compare digits. I have only ever seen it fail one time. In mathematics, you don't understand things. You just get used to them. I have the result, but I do not yet know how to get it. All physicists, and a good many quite respectable mathematicians are contemptuous about proof. #17 20131123 12:07:34
Re: √1+(√2+(√3+(√4+... = ? Harder than it seemsWhen was that? The limit operator is just an excuse for doing something you know you can't. “It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman “Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment #18 20131123 12:11:37
Re: √1+(√2+(√3+(√4+... = ? Harder than it seemsI can not remember, it was a pathological case where someone was making a stupid math point and totally ignoring the thing called practicality. In mathematics, you don't understand things. You just get used to them. I have the result, but I do not yet know how to get it. All physicists, and a good many quite respectable mathematicians are contemptuous about proof. #19 20131123 12:12:39
Re: √1+(√2+(√3+(√4+... = ? Harder than it seemsYou mean, they were specifically looking for a case when it doesn't work? The limit operator is just an excuse for doing something you know you can't. “It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman “Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment #20 20131123 12:13:34
Re: √1+(√2+(√3+(√4+... = ? Harder than it seemsSurely you have seen that idiotic example the Borweins came up with? In mathematics, you don't understand things. You just get used to them. I have the result, but I do not yet know how to get it. All physicists, and a good many quite respectable mathematicians are contemptuous about proof. #21 20131123 12:21:22
Re: √1+(√2+(√3+(√4+... = ? Harder than it seemsNo. What was it? The limit operator is just an excuse for doing something you know you can't. “It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman “Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment #22 20131123 12:24:25
Re: √1+(√2+(√3+(√4+... = ? Harder than it seemsSo did I. In mathematics, you don't understand things. You just get used to them. I have the result, but I do not yet know how to get it. All physicists, and a good many quite respectable mathematicians are contemptuous about proof. #23 20131123 12:25:51
Re: √1+(√2+(√3+(√4+... = ? Harder than it seemsOkay, just checking. The limit operator is just an excuse for doing something you know you can't. “It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman “Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment #24 20131123 12:27:26
Re: √1+(√2+(√3+(√4+... = ? Harder than it seemsYes, he shows a sum that agrees with pi to over 1 billion places and then veers away. In mathematics, you don't understand things. You just get used to them. I have the result, but I do not yet know how to get it. All physicists, and a good many quite respectable mathematicians are contemptuous about proof. #25 20131123 12:35:23
Re: √1+(√2+(√3+(√4+... = ? Harder than it seemsHm? The limit operator is just an excuse for doing something you know you can't. “It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman “Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment 