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

You are not logged in.

#1 2013-11-07 13:22:14

George,Y
Member
Registered: 2006-03-12
Posts: 1,306

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

Last edited by George,Y (2013-11-07 13:22:56)


X'(y-Xβ)=0

Offline

#2 2013-11-07 14:17:18

bobbym
Administrator
From: Bumpkinland
Registered: 2009-04-12
Posts: 86,689

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.
Of course that result can be rigorously obtained, but who cares?
Combinatorics is Algebra and Algebra is Combinatorics.

Online

#3 2013-11-08 00:17:41

zetafunc.
Guest

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

Yes, that constant is called the nested radical constant, usually denoted C[sub]NR[/sub]. No closed-form expression for that constant currently exists.

#4 2013-11-08 02:32:47

bobbym
Administrator
From: Bumpkinland
Registered: 2009-04-12
Posts: 86,689

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.
Of course that result can be rigorously obtained, but who cares?
Combinatorics is Algebra and Algebra is Combinatorics.

Online

#5 2013-11-20 22:10:31

George,Y
Member
Registered: 2006-03-12
Posts: 1,306

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

Alright guys, I think I should give up.


X'(y-Xβ)=0

Offline

#6 2013-11-20 22:14:19

bobbym
Administrator
From: Bumpkinland
Registered: 2009-04-12
Posts: 86,689

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.
Of course that result can be rigorously obtained, but who cares?
Combinatorics is Algebra and Algebra is Combinatorics.

Online

#7 2013-11-22 08:26:53

Yusuke00
Member
Registered: 2013-11-19
Posts: 43

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

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

Offline

#8 2013-11-22 12:28:35

bobbym
Administrator
From: Bumpkinland
Registered: 2009-04-12
Posts: 86,689

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.
Of course that result can be rigorously obtained, but who cares?
Combinatorics is Algebra and Algebra is Combinatorics.

Online

#9 2013-11-22 12:48:41

anonimnystefy
Real Member
From: The Foundation
Registered: 2011-05-23
Posts: 15,525

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

How do you know which digits are correct?


“Here lies the reader who will never open this book. He is forever dead.

“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment

Offline

#10 2013-11-22 12:52:32

bobbym
Administrator
From: Bumpkinland
Registered: 2009-04-12
Posts: 86,689

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.
Of course that result can be rigorously obtained, but who cares?
Combinatorics is Algebra and Algebra is Combinatorics.

Online

#11 2013-11-22 12:53:16

anonimnystefy
Real Member
From: The Foundation
Registered: 2011-05-23
Posts: 15,525

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

And you entered what exactly?


“Here lies the reader who will never open this book. He is forever dead.

“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment

Offline

#12 2013-11-22 12:55:01

bobbym
Administrator
From: Bumpkinland
Registered: 2009-04-12
Posts: 86,689

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.
Of course that result can be rigorously obtained, but who cares?
Combinatorics is Algebra and Algebra is Combinatorics.

Online

#13 2013-11-22 12:58:03

anonimnystefy
Real Member
From: The Foundation
Registered: 2011-05-23
Posts: 15,525

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

Okay, but then, how do you know to how many digits the result is accurate?


“Here lies the reader who will never open this book. He is forever dead.

“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment

Offline

#14 2013-11-22 12:59:59

bobbym
Administrator
From: Bumpkinland
Registered: 2009-04-12
Posts: 86,689

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.
Of course that result can be rigorously obtained, but who cares?
Combinatorics is Algebra and Algebra is Combinatorics.

Online

#15 2013-11-22 13:02:39

anonimnystefy
Real Member
From: The Foundation
Registered: 2011-05-23
Posts: 15,525

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

Remind me what that is again.


“Here lies the reader who will never open this book. He is forever dead.

“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment

Offline

#16 2013-11-22 13:06:23

bobbym
Administrator
From: Bumpkinland
Registered: 2009-04-12
Posts: 86,689

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.
Of course that result can be rigorously obtained, but who cares?
Combinatorics is Algebra and Algebra is Combinatorics.

Online

#17 2013-11-22 13:07:34

anonimnystefy
Real Member
From: The Foundation
Registered: 2011-05-23
Posts: 15,525

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

When was that?

The next will answer.


“Here lies the reader who will never open this book. He is forever dead.

“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment

Offline

#18 2013-11-22 13:11:37

bobbym
Administrator
From: Bumpkinland
Registered: 2009-04-12
Posts: 86,689

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.
Of course that result can be rigorously obtained, but who cares?
Combinatorics is Algebra and Algebra is Combinatorics.

Online

#19 2013-11-22 13:12:39

anonimnystefy
Real Member
From: The Foundation
Registered: 2011-05-23
Posts: 15,525

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

You mean, they were specifically looking for a case when it doesn't work?


“Here lies the reader who will never open this book. He is forever dead.

“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment

Offline

#20 2013-11-22 13:13:34

bobbym
Administrator
From: Bumpkinland
Registered: 2009-04-12
Posts: 86,689

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.
Of course that result can be rigorously obtained, but who cares?
Combinatorics is Algebra and Algebra is Combinatorics.

Online

#21 2013-11-22 13:21:22

anonimnystefy
Real Member
From: The Foundation
Registered: 2011-05-23
Posts: 15,525

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.


“Here lies the reader who will never open this book. He is forever dead.

“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment

Offline

#22 2013-11-22 13:24:25

bobbym
Administrator
From: Bumpkinland
Registered: 2009-04-12
Posts: 86,689

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.
Of course that result can be rigorously obtained, but who cares?
Combinatorics is Algebra and Algebra is Combinatorics.

Online

#23 2013-11-22 13:25:51

anonimnystefy
Real Member
From: The Foundation
Registered: 2011-05-23
Posts: 15,525

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?


“Here lies the reader who will never open this book. He is forever dead.

“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment

Offline

#24 2013-11-22 13:27:26

bobbym
Administrator
From: Bumpkinland
Registered: 2009-04-12
Posts: 86,689

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.
Of course that result can be rigorously obtained, but who cares?
Combinatorics is Algebra and Algebra is Combinatorics.

Online

#25 2013-11-22 13:35:23

anonimnystefy
Real Member
From: The Foundation
Registered: 2011-05-23
Posts: 15,525

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

Hm?


“Here lies the reader who will never open this book. He is forever dead.

“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment

Offline

Board footer

Powered by FluxBB