Home » Ask & Discuss » Mathematics. » Algebra « Back to Discussion

Algebra

Forum Expert

Joined:	28 Feb 2007
Post:	2173

11 Feb 2008 20:01:26 IST

0 People liked this

470

Sequences - nice one

None

The sequence $\{ a_n \}^{\infty}_{n=1}, \{ b_n \}^{\infty}_{n=1}$ are defined through $a_1 = \alpha, b_1 = \beta,$ and $a_{n+1} = \alpha a_n - \beta b_n, b_{n+1} = \beta a_n + \alpha b_n$ for all $n \geq 1.$ How many pairs $(\alpha, \beta)$ of real numbers are there such that

$a_{1997} = b_1, \ \text{and} \ b_{1997} = a_1 \text{?}$

Share this article on:

Comments (6)

Abhijith

Blazing goIITian

Joined: 23 Jan 2007

Posts: 678

11 Feb 2008 21:42:05 IST

Like 2 people liked this

The sequences

and

are defined such that

and

for all $n geq 1.$ How many pairs of $(alpha, eta)$ of real numbers are there such that $a_{1997} = b_1, ext{and} b_{1997} = a_1 ext{?}$

-----------------------------------------------------------------------------------------------------------------------------------------------------------

Multiplying

with

we get,

.
Multiplying

with

we get,

Hence,

Hence the series is a G.P with common ratio

Similar procedure gives,

The question requires that

and

Thus,

Hari Shankar

Forum Expert

Joined: 28 Feb 2007

Posts: 2173

11 Feb 2008 21:45:43 IST

Like 0 people liked this

Good. But I can wait for more replies I think

Hari Shankar

Forum Expert

Joined: 28 Feb 2007

Posts: 2173

12 Feb 2008 09:06:16 IST

Like 3 people liked this

Well this is yet another instance of a sequence where trigonometric substitutions can save us a lot of sweat. For those who came in late, the reason I say "yet another" are http://www.goiit.com/posts/list/trignometry-if-a-b-are-positive-quantities-and-if-41432.htm and http://www.goiit.com/posts/list/algebra-salutes-assured-41758.htm#205608.

and

If we dispose of the trivial case

= 0, This form should remind us of the formulas of cos(A+B) and sin(A+B).

So, how to get these expressions into a trigonometric form. If we let

= r cos

and

= r sin

where r =

²+

² then we can see that a₂ = r²(cos²

- sin²

) = r²cos(2

). Similarly b₂ = r²(sin

cos

+sin

cos

) = r²sin(2

In this way, we get a_n = rⁿcosn

and b_n = rⁿsinn

Hence a₁₉₉₇ = b₁ and b₁₉₉₇ = a₁ is the same as

r¹⁹⁹⁷cos1997

= rsin

and

r¹⁹⁹⁷sin1997

= rcos

Squaring and adding gives r =

²+

² = 1

Hence the eqns are cos1997

= sin

sin1997

= cos

This gives 1997

= 2n

/2 -

or 1998

= 2n

/1998+

/3996

Since we are looking for distinct values of

= cos

and

= sin

, we need only take values of n =1,2,3,...,1998.

So, the total number of distinct ordered pairs (

) including (0,0) is 1999.

While I was gloating over my solution I looked at the solution provided by the author which I have copied in the post below. I'll look quite a fool then!

Hari Shankar

Forum Expert

Joined: 28 Feb 2007

Posts: 2173

12 Feb 2008 09:09:30 IST

Like 4 people liked this

The "official" solution:

Let $z = \alpha + \beta i$ . Then $z_1 = z$ and $z_{n+1} = z \cdot z_n \implies z_n = z^n$ . We therefore wish to solve

$z^{1997} = i \bar{z}$ .

Case: $|z| = 0$ . Then $z = 0$ .

Case: $|z| = 1$ . Then $z^{1998} = i$ , which has $1998$ complex roots for a total of $\boxed{1999}$ pairs.

There you are, the solution in 1/10th our effort!

Sairam

Blazing goIITian

Joined: 14 Sep 2007

Posts: 573

12 Feb 2008 11:52:06 IST

Like 0 people liked this

hi sir

i solved it using trigonometric substitutions much similiar to what u hav dun.

I defined a new sequence ci such that an= cn sin

and bn=cncos

= r cos

and

= r sin

proceeding on similiar lines and eliminating r,

we get the same ans.

Hari Shankar

Forum Expert

Joined: 28 Feb 2007

Posts: 2173

12 Feb 2008 12:10:04 IST

Like 1 people liked this

that's good. so keep your eyes skinned for such possibilities. I used to find it useful to just stare at the problem without putting pen to paper while the brain searched for a match with something familiar.

Keep it up!

Quick Reply

Reply

	Some HTML allowed.
	Keep your comments above the belt or risk having them deleted.
	Signup for a avatar to have your pictures show up by your comment
	If Members see a thread that violates the Posting Rules, bring it to the attention of the Moderator Team