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#2 2012-10-15 18:08:42
- bobbym
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Re: Equation
HI Leroy;
You want to solve it for x? Solve it for y? What?
In mathematics, you don't understand things. You just get used to them.
90% of mathematicians do not understand 90% of currently published mathematics.
I am willing to wager that over 75% of the new words that appeared were nothing more than spelling errors that caught on.
#3 2012-10-15 18:13:01
- zetafunc.
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Re: Equation
Do you mean integer solutions? It may help if you consider that equation mod 11...
#5 2012-10-16 00:05:35
- bobbym
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Re: Equation
Hi Leroy;
Could you please help me solve- x^2-11y^2=1?
If this is your equation
then there are no integer solutions.
If that - sign in front of the x^2 does not belong there then you are dealing with a Pell equation. This will require continued fractions to understand.
Three solutions are (1,0),(10,3) and (199,60).
There are infinitely more.
In mathematics, you don't understand things. You just get used to them.
90% of mathematicians do not understand 90% of currently published mathematics.
I am willing to wager that over 75% of the new words that appeared were nothing more than spelling errors that caught on.
#6 2012-10-18 21:14:16
- zetafunc.
- Guest
Re: Equation
How did you find those solutions? I have seen a continued fraction before (such as one for the golden ratio) but how do you use them to find solutions to a Pell equation like Leroy's?
I know x = 1 and y = 0 will solve any Pell equation, but how do I go about finding more integer solutions by hand?
#7 2012-10-18 21:26:14
- bobbym
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Re: Equation
Hi zetafunc.;
The process of getting the first non trivial solution ( fundamental solution )does involve continued fractions. But I was able to get the next solution called the fundamental solution by trial and error. After that it is a simple means to get as many solutions as you like.
In mathematics, you don't understand things. You just get used to them.
90% of mathematicians do not understand 90% of currently published mathematics.
I am willing to wager that over 75% of the new words that appeared were nothing more than spelling errors that caught on.
#9 2012-10-18 21:36:52
- bobbym
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Re: Equation
Hi;
Please hold on it is lengthy. I will post it right here.
In mathematics, you don't understand things. You just get used to them.
90% of mathematicians do not understand 90% of currently published mathematics.
I am willing to wager that over 75% of the new words that appeared were nothing more than spelling errors that caught on.
#10 2012-10-18 21:38:32
- zetafunc.
- Guest
Re: Equation
Thank you. Sorry if it is very long, I am very interested though.
#11 2012-10-18 21:44:33
- bobbym
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Re: Equation
First compute the continued fraction of
The sequence is periodic with length 2 (6,3...) The nice part is that a theorem by
Lagrange assures us that every square root like this will always have a repeating
form.
Then you get the convergents:
You pick the 2nd one ( because the period is 2 ) in the sequence 10/3 and check it in the equation with x = 10 and y = 3
so that is the fundamental solution. x = 10 and y = 3. From there you use two recurrences to get as many as you need.
I know you have many questions. This is the prettiest part of number theory. Computational Number Theory!
You might want to look at other problems I worked on:
http://www.mathisfunforum.com/viewtopic … 12#p115912
post#3.
In mathematics, you don't understand things. You just get used to them.
90% of mathematicians do not understand 90% of currently published mathematics.
I am willing to wager that over 75% of the new words that appeared were nothing more than spelling errors that caught on.
#12 2012-10-18 22:02:35
- zetafunc.
- Guest
Re: Equation
Thanks for the post.
How do you know that the continued fraction of the square root of 11 = {3, 3, 6, 3, 6, 3 ...}? How do we know there are 3s and 6s involved?
I understand where you got the convergents from (from the above sequence), but I just need help understanding why you know you have the set {3, 3, 6, 3, 6, 3, ...}.
This looks very neat... so, you are generating all the convergences, and then taking every nth convergent (where n is the cycle length). Is the cycle length always 2 for square roots?
#13 2012-10-18 22:05:02
- bobbym
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Re: Equation
See this link and you will know more.
http://www.mathisfunforum.com/viewtopic … 12#p115912
Post #3.
This is one of my favorite posts!
In mathematics, you don't understand things. You just get used to them.
90% of mathematicians do not understand 90% of currently published mathematics.
I am willing to wager that over 75% of the new words that appeared were nothing more than spelling errors that caught on.
#14 2012-10-18 22:08:32
- zetafunc.
- Guest
Re: Equation
I read that post, but I am not understanding how you got the sets of repeating forms for √11, √14, √61, etc... sorry if I am missing something obvious.
Why is the repeating form for √11 = {3, 3, 6, 3, 6, 3, 6, 3, ...}?
#15 2012-10-18 22:10:09
- bobbym
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Re: Equation
Hi;
Each part of the solution can be easily computed. Hold on I will provide the answer.
The most naive algorithm is also the simplest.
Start with:
Repeat steps 2 and 3.
In mathematics, you don't understand things. You just get used to them.
90% of mathematicians do not understand 90% of currently published mathematics.
I am willing to wager that over 75% of the new words that appeared were nothing more than spelling errors that caught on.
#18 2012-10-18 22:43:56
- zetafunc.
- Guest
Re: Equation
But, how can I convince myself that my answer is correct for
? I can see this will get very complicated with further iterations...#19 2012-10-18 22:52:46
- bobbym
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Re: Equation
c = floor(√11) = 3 ( number you hold )
Now repeat.
In mathematics, you don't understand things. You just get used to them.
90% of mathematicians do not understand 90% of currently published mathematics.
I am willing to wager that over 75% of the new words that appeared were nothing more than spelling errors that caught on.
#21 2012-10-18 23:09:01
- bobbym
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Re: Equation
Hi;
You are using a calculator of course. Better if you are using a CAS.
In mathematics, you don't understand things. You just get used to them.
90% of mathematicians do not understand 90% of currently published mathematics.
I am willing to wager that over 75% of the new words that appeared were nothing more than spelling errors that caught on.
#23 2012-10-18 23:16:53
- bobbym
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Re: Equation
For the simple one like √11 maybe. But working with symbolic radicals is even harder than working with
floating point arithmetic. What is the period is 100 or so?
In mathematics, you don't understand things. You just get used to them.
90% of mathematicians do not understand 90% of currently published mathematics.
I am willing to wager that over 75% of the new words that appeared were nothing more than spelling errors that caught on.
#25 2012-10-18 23:20:45
- bobbym
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Re: Equation
For √14 there is an easy theorem to tell you what the fundamental solution is.
For √11 it does not apply. Generally the bigger the number the longer the period.
In mathematics, you don't understand things. You just get used to them.
90% of mathematicians do not understand 90% of currently published mathematics.
I am willing to wager that over 75% of the new words that appeared were nothing more than spelling errors that caught on.