The idea about Pythagoras stairs?

[memomath]

Hello everybody

I have looked everywhere for some good guides, so the question is:


You probably know the Fibonacci sequence (1,1,2,3,5,8…) but there is a lesser-known sequence called (Pythagoras stairs) generated by a similar recursion formula.

This is a sequence of pairs (Xn,Yn) usually arranged as follows:
1 2
2 3
5 7
12 17

And so on……. It begins with (x1,y1)=(1,1) and the recursion formula is

X{n+1} = X{n} + Y{n}
Y{n+1} = X{n}+ X{n+1}

Prove by induction that always

Y^2 = 2 X^2 ± 1

Pythagoras used this equation to generate rational approximations to (2)^1/2


Thanks in advance for usefully discuss to solve for this interesting question

 

[memomath]

Hi again

Hello everybody

I have looked everywhere for some good guides, so the question is:

You probably know the Fibonacci sequence (1,1,2,3,5,8…) but there is a lesser-known sequence called (Pythagoras stairs) generated by a similar recursion formula.

This is a sequence of pairs (Xn,Yn) usually arranged as follows:
1 2
2 3
5 7
12 17

And so on……. It begins with (x1,y1)=(1,1) and the recursion formula is

$$X\underline{n+1}$$=$$X\underline{n}$$+$$Y\underline{n}$$

$$Y\underline{n+1}$$= $$X\underline{n}$$+$$X\underline{n+1}$$


Prove by induction that always

$$Y\acute{2}$$=2$$X\acute{2}$$ $$\pm1$$

Pythagoras used this equation to generate rational approximations to $$\sqrt{2}$$


Moreover, if you would like please open the enclosed to find the corrections/clarifications of the question.

Thanks

 

[memomath]

No comments or solution until now!

hi

please open the enclosed to find the corrections/clarifications of the question.

Thanks

 

[HallsofIvy]

A lot of people will not open a Word file because they are notorious for carrying viruses.

 

[memomath]

Dear HallsofIvy

Thanks for your reply, in fact I already wrote the problem in webpage and at the same time wrote in the word file so there two options

A lot of people will not open a Word file because they are notorious for carrying viruses.

 

[memomath]

Hi again

Hello everybody

I have looked everywhere for some good guides, so the question is:

You probably know the Fibonacci sequence (1,1,2,3,5,8…) but there is a lesser-known sequence called (Pythagoras stairs) generated by a similar recursion formula.

This is a sequence of pairs (Xn,Yn) usually arranged as follows:
1 2
2 3
5 7
12 17

And so on……. It begins with (x1,y1)=(1,1) and the recursion formula is

$$X\underline{n+1}$$=$$X\underline{n}$$+$$Y\underline{n}$$

$$Y\underline{n+1}$$= $$X\underline{n}$$+$$X\underline{n+1}$$


Prove by induction that always

$$Y\acute{2}$$=2$$X\acute{2}$$ $$\pm1$$

Pythagoras used this equation to generate rational approximations to $$\sqrt{2}$$


Moreover, if you would like please open the enclosed to find the question as pdf form.

Thanks

 

[natives]

Okay here is how you do it...The method of mathematical induction for proofs follows the induction theorem which requires that for any relation dependent on variable N ,then if N=1,N=2 are TRUE and assuming N= k is TRUE then N=k+1 should also be TRUE...So if you test the Pythagoras' stairs for N=1 and N=2 then you find that it is TRUE...then assume that N=k is true such that y^2(k+1)=2x^2(k+1)+- 1...so using the above you should be able to prove that y^2(k+2)=2x^2(k+2)+-1...you can do this by taking the left hand side..You should be knowing that y(k+2)=2x(k+1)+y(k+1)..square this to have y^2(k+2) =4x^2(k+1)+4x(k+1)y(k+1)+y^2(k+1)...but from previously assuming N=k is true we had y^2(k+1)=2x^2(k+1)+-1..substitute in the previous equation...also keep in mind x(k+2)=x(k+1)+y(k+1)...use this [square it] in the above equations to get the right hand side of y^2(k+2)=2x^2(k+2)+-1 then you have proven using the induction theorem.

 

[memomath]

some wrongs here in the question

YES I AGREE WITH YOU there are mistake in the question

 

[girlmatrix]

The above answer is not correct and there are some mistakes in the question, its easy to find the mistakes in the question.

However the hint for the right question should use these formulae

Y(n + 1) X(n + 1) = (2X(n+1) + Y(n+3) (X(n+4) + Y(n+5))

YES I AGREE WITH YOU there are mistake in the question