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  •  » √1+(√2+(√3+(√4+... = ? Harder than it seems

#1 2013-11-08 12:22:14

George,Y
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√1+(√2+(√3+(√4+... = ? Harder than it seems

Last edited by George,Y (2013-11-08 12:22:56)


X'(y-Xβ)=0

#2 2013-11-08 13:17:18

bobbym
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Re: √1+(√2+(√3+(√4+... = ? Harder than it seems

Hi;

I think you will have to settle for a numerical result. I am not finding a simple form in terms of known constants.


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 2013-11-08 23:17:41

zetafunc.
Guest

Re: √1+(√2+(√3+(√4+... = ? Harder than it seems

Yes, that constant is called the nested radical constant, usually denoted CNR. No closed-form expression for that constant currently exists.

#4 2013-11-09 01:32:47

bobbym
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Re: √1+(√2+(√3+(√4+... = ? Harder than it seems

Hi;

Looking at the literature there does not even seem to be a good upper and lower bound.

Also interesting is the Prime Nested Radical



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.

#5 2013-11-21 21:10:31

George,Y
Super Member

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Re: √1+(√2+(√3+(√4+... = ? Harder than it seems

Alright guys, I think I should give up.


X'(y-Xβ)=0

#6 2013-11-21 21:14:19

bobbym
Administrator

Online

Re: √1+(√2+(√3+(√4+... = ? Harder than it seems

Hi;

Give up when you have the answer? Why?


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.

#7 2013-11-23 07:26:53

Yusuke00
Member

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Re: √1+(√2+(√3+(√4+... = ? Harder than it seems

Guess because he can't understand how u got that results? xD

#8 2013-11-23 11:28:35

bobbym
Administrator

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Re: √1+(√2+(√3+(√4+... = ? Harder than it seems

Google 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 2013-11-23 11:48:41

anonimnystefy
Real Member

Online

Re: √1+(√2+(√3+(√4+... = ? Harder than it seems

How 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 2013-11-23 11:52:32

bobbym
Administrator

Online

Re: √1+(√2+(√3+(√4+... = ? Harder than it seems

I 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 2013-11-23 11:53:16

anonimnystefy
Real Member

Online

Re: √1+(√2+(√3+(√4+... = ? Harder than it seems

And 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 2013-11-23 11:55:01

bobbym
Administrator

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Re: √1+(√2+(√3+(√4+... = ? Harder than it seems

Just 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 2013-11-23 11:58:03

anonimnystefy
Real Member

Online

Re: √1+(√2+(√3+(√4+... = ? Harder than it seems

Okay, 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 2013-11-23 11:59:59

bobbym
Administrator

Online

Re: √1+(√2+(√3+(√4+... = ? Harder than it seems

By 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 2013-11-23 12:02:39

anonimnystefy
Real Member

Online

Re: √1+(√2+(√3+(√4+... = ? Harder than it seems

Remind 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 2013-11-23 12:06:23

bobbym
Administrator

Online

Re: √1+(√2+(√3+(√4+... = ? Harder than it seems

You 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 2013-11-23 12:07:34

anonimnystefy
Real Member

Online

Re: √1+(√2+(√3+(√4+... = ? Harder than it seems

When was that?

The next will answer.


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 2013-11-23 12:11:37

bobbym
Administrator

Online

Re: √1+(√2+(√3+(√4+... = ? Harder than it seems

I 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 2013-11-23 12:12:39

anonimnystefy
Real Member

Online

Re: √1+(√2+(√3+(√4+... = ? Harder than it seems

You 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 2013-11-23 12:13:34

bobbym
Administrator

Online

Re: √1+(√2+(√3+(√4+... = ? Harder than it seems

Surely 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 2013-11-23 12:21:22

anonimnystefy
Real Member

Online

Re: √1+(√2+(√3+(√4+... = ? Harder than it seems

No. What was it?

By the way, I am getting

for the first 41 decimal digits.


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 2013-11-23 12:24:25

bobbym
Administrator

Online

Re: √1+(√2+(√3+(√4+... = ? Harder than it seems

So did I.

You did not read about the billion decimal places of pi?


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 2013-11-23 12:25:51

anonimnystefy
Real Member

Online

Re: √1+(√2+(√3+(√4+... = ? Harder than it seems

Okay, just checking.

I remember reading about BBP type formulas. Has that anything to do with what you are saying?


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 2013-11-23 12:27:26

bobbym
Administrator

Online

Re: √1+(√2+(√3+(√4+... = ? Harder than it seems

Yes, 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 2013-11-23 12:35:23

anonimnystefy
Real Member

Online

Re: √1+(√2+(√3+(√4+... = ? Harder than it seems

Hm?


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
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