<< Chapter < Page Chapter >> Page >

Lemma 3.1 [link] Formula 1.1.2 Given a continued fraction [ r 1 , r 2 , r 3 , . . . ] , h v and k v satisfy the following recursions:

h v = r v h v - 1 + h v - 2 , h - 1 = 1 , h - 2 = 0
k v = r v k v - 1 + k v - 2 , k - 1 = 0 , k - 2 = 1 .

Lemma 3.2 [link] Formula 1.2.1 Given a continued fraction as above, A v satisfies the following relations:

A δ A v v o d d A δ A v v e v e n .

Moreover, this means that A v A v + 1 for odd v.

The following is a claim about the n 's from the Euclidean Algorithm.

Lemma 3.3 In the above notation, n v = ( - 1 ) v ( p h v - k v q ) .

This will be done by induction with two base cases. If v = 1 , then

n 1 = q - p r 1 = - ( p r 1 - q ) = - ( p h 1 - k 1 q ) .

If v = 2 , then

n 2 = p - r 2 n 1 = p - r 2 ( q - r 1 p ) = p ( r 1 r 2 + 1 ) - r 2 q = ( - 1 ) 2 ( p h 2 - k 2 q ) .

Now we will assume the claim is true for v and v + 1 and we will prove that it is true for v + 2 . From the Euclidean algorithm, we have that

n v + 2 = n v - r v + 2 n v + 1
= ( - 1 ) v ( p h v - k v q ) - ( - 1 ) v + 1 r v + 2 ( p h v + 1 - k v + 1 q ) .

For v even, using Lemma [link] we obtain

n v + 2 = p ( h v + r v + 2 h v + 1 ) - ( k v + r v + 2 k v + 1 ) q = p h v + 2 - k v + 2 q .

If v is odd, using Lemma [link] we have

n v + 2 = - p ( h v + r v + 2 h v + 1 ) + ( k v + r v + 2 k v + 1 ) q = - ( p h v + 2 - k v + 2 q ) .

Since v and v + 2 will be either both even or both odd, this proves the claim. The following lemmas relate recursions that occur in the proof of Theorem [link] to continued fraction expressions.

Lemma 3.4 Suppose { C i } is a sequence of integers satisfying the following recurrence relation

C i = ( C i - 1 + n i - 1 ) r i + C i - 2 , C 1 = r 1 p , C 2 = r 2 q .

Then

C v = p h v v o d d q k v v e v e n .

This is done by induction again. For v = 1 and v = 2 , the initial conditions of the recursion of C give C 1 = p h 1 and C 2 = q k 2 . Now we assume that the claim is true for v and v + 1 and we will prove that it is true for v + 2 . First, we have

C v + 2 = ( C v + 1 + n v + 1 ) r v + 2 - C v = [ C v + 1 + ( - 1 ) v + 1 ( p h v + 1 - k v + 1 q ) ] r v + 2 - C v

by the original recursion of C i and Lemma [link] . There are two cases. First we will consider the case where v is odd. Then

C v + 2 = ( q k v + 1 + p h v + 1 - k v + 1 q ) r v + 2 + p h v

by the inductive hypothesis. It follows from Lemma [link] that

C v + 2 = p ( h v + 1 r v + 2 + h v ) = p h v + 2 .

Now consider the case where v is even. Then

C v + 2 = ( p h v + 1 - p h v + 1 + k v + 1 q ) r v + 2 + q k v

by the inductive hypothesis. It follows from Lemma [link] that

C v + 2 = q ( k v + 1 r v + 2 + k v ) = q k v + 2 .

This proves the claim.

Lemma 3.5 Suppose that { K i } is a sequence of integers satisfying the following recurrence relation

K i = ( K i - 1 + 1 ) r i + K i - 2 , K 1 = r 1 , K 2 = r 2 ( r 1 + 1 ) .

Then K v + 1 = h v + k v .

We again use induction with two base cases. First, if v = 1 , then

K 1 + 1 = r 1 + 1 = h 1 + k 1 .

Second, if v = 2 , then

K 2 + 1 = r 2 ( r 1 + 1 ) + 1 = r 2 r 1 + r 2 + 1 = h 2 + k 2 .

Now we will assume that the claim is true for v and v + 1 and we will prove that it is true for v + 2 . By the original recursion of K i ,

K v + 2 + 1 = ( K v + 1 + 1 ) r v + 2 + K v + 1 .

Then by the inductive hypothesis,

K v + 2 + 1 = ( h v + 1 + k v + 1 ) r v + 2 + h v + k v = ( h v + 1 r v + 2 + h v ) + ( k v + 1 r v + 2 + k v ) = h v + 2 + k v + 2

by the recursion of k v and h v from Lemma [link] . This proves the claim.

Proof of theorem

Without loss of generality, assume that q p . For this proof, we will use a “reduced” notion of blow-ups in which consecutive blow-ups that use the same variable will be reduced to a single blow-up. For example, if we have three typical blow-up steps with y = s x , s = t x , and t = a x , we will combine these to be y = a x 3 . Note that the multiplicities of these blow-ups are the same. We will soon justify the use of these blow up “chains”.

We will use the equations from the Euclidean algorithm as in the previous section. The number of r i 's used will be the number of blow-ups needed, and the value of r i will be the degree of the variable at that particular blow-up (e.g. r i = 3 in the previous example). Note that the r i values can also be expressed as the continued fraction of q p as [ r 1 , r 2 , . . . , r δ ] . Let C i be the multiplicity of the i t h exceptional divisor and let K i be the log canonical multiplicity of the differential form of the i t h blow-up. We claim that the multiplicities K i and C i satisfy the following recursions:

K i = ( K i - 1 + 1 ) r i + K i - 2 , K 1 = r 1 , K 2 = r 2 ( r 1 + 1 ) ,
C i = ( C i - 1 + n i - 1 ) r i + C i - 2 , C 1 = r 1 p , C 2 = r 2 q .

First, it is easy to check that K 1 = r 1 and K 2 = r 2 ( r 1 + 1 ) . Then, after i - 1 blow-ups, our differential form ihas the form

t i - 2 K i - 2 t i - 1 K i - 1 d t i - 1 d t i - 2 .

By definition of r i , the next iterated blow up includes the substitution t i - 1 = t i t i - 2 r i . We obtain the differential form

t i - 2 K i - 2 ( t i t i - 2 r i ) K i - 1 t i - 2 r i d t i - 2 d t i = t i K i - 1 t i - 2 ( K i - 1 + 1 ) r i + K i - 2 d t i - 2 d t i .

Thus, K i = ( K i - 1 + 1 ) r i + K i - 2 .

It is also easy to check that C 1 = p r 1 and C 2 = q r 2 . After i - 1 blow-ups the equation is

t i - 2 C i - 2 t i - 1 C i - 1 ( t i - 2 n i - 2 - t i - 1 n i - 1 ) .

By definition of r i , the next iterated blow up includes the substitution of t i - 1 = t i t i - 2 r i . We obtain the new equation

t i C i - 1 t i - 2 C i - 2 + r i C i - 1 ( t i - 2 n i - 2 - t i n i - 1 t i - 2 r i n i - 1 )
= t i C i - 1 t i - 2 ( C i - 1 + n i - 1 ) r i + C i - 2 ( t i - 2 n i - t i n i - 1 ) .

Thus, C i = ( C i - 1 + n i - 1 ) r i + C i - 2 .

The recursion relations for C i and K i allow us to apply Lemma [link] to C i and Lemma [link] to K i . Now we can justify the notion of blow-ups that groups together consecutive substitutions of the same variable into “chains.” It is not entirely clear that substitutions within a given chain do not yield smaller contributions to the LCT. To show that the smallest contribution to the LCT from a particular chain comes from its final substitution, we need to show that each successive substitution within such a chain decreases the contribution of the chain to the LCT. Let j be the number of blow-ups within the i t h chain. To prove the above statement, it would suffice to show that

( K i - 1 + 1 ) j + K i - 2 + 1 ( C i - 1 + n i - 1 ) j + C i - 2 ( K i - 1 + 1 ) ( j + 1 ) + K i - 2 + 1 ( C i - 1 + n i - 1 ) ( j + 1 ) + C i - 2 ,

which is true if and only if

( C i - 1 + n i - 1 ) ( K i - 2 + 1 ) C i - 2 ( K i - 1 + 1 ) .

We will show that this is true using two cases. First, assume that i is even. Then the above equation becomes

( q k i - 1 ) ( h i - 2 + k i - 2 ) ( q k i - 2 ) ( h i - 1 + k i - 1 )

A i - 2 + 1 A i - 1 + 1 ,

which is true for even i by Lemma [link] . Now we will assume i is odd. Then the previous expression becomes

( p h i - 1 ) ( h i - 2 + k i - 2 ) ( p h i - 2 ) ( h i - 1 + k i - 1 )

1 A i - 2 + 1 1 A i - 1 + 1

A i - 2 A i - 1 ,

which is true for odd i by Lemma [link] . This shows that the LCT will be determined by these chains.

We will now show that the final chain determines the LCT. Thus, we want to show that α δ α v for all v . First, we will introduce the notation for A v = h v k v to be the expression of the v t h convergent. We must consider two cases. First, if v is odd, then

α δ α v

h δ + k δ p h δ h v + k v p h v

1 + 1 A δ 1 + 1 A v

1 A δ 1 A v

A δ A v .

This last statement is true by Lemma [link] , so for v odd, α δ α v . If v is even, then

α δ α v

h δ + k δ q k δ h v + k v q k v

1 + A δ 1 + A v

A δ A v

A δ A v .

This last statement is also true by Lemma [link] , so for v even, α δ α v . Thus, α δ α v for all v < δ .

It only remains to show that α δ = 1 p + 1 q . We know that q p = h δ k δ , which is to say q k δ = p h δ , so by Lemma [link] C δ = q k δ for both odd and even δ . Thus,

α δ = K δ + 1 C δ = h δ + k δ q k δ = h δ k δ + 1 q = q p + 1 q = 1 p + 1 q .

This concludes the proof of the theorem.

Questions & Answers

where we get a research paper on Nano chemistry....?
Maira Reply
what are the products of Nano chemistry?
Maira Reply
There are lots of products of nano chemistry... Like nano coatings.....carbon fiber.. And lots of others..
learn
Even nanotechnology is pretty much all about chemistry... Its the chemistry on quantum or atomic level
learn
Google
da
no nanotechnology is also a part of physics and maths it requires angle formulas and some pressure regarding concepts
Bhagvanji
hey
Giriraj
Preparation and Applications of Nanomaterial for Drug Delivery
Hafiz Reply
revolt
da
Application of nanotechnology in medicine
what is variations in raman spectra for nanomaterials
Jyoti Reply
I only see partial conversation and what's the question here!
Crow Reply
what about nanotechnology for water purification
RAW Reply
please someone correct me if I'm wrong but I think one can use nanoparticles, specially silver nanoparticles for water treatment.
Damian
yes that's correct
Professor
I think
Professor
Nasa has use it in the 60's, copper as water purification in the moon travel.
Alexandre
nanocopper obvius
Alexandre
what is the stm
Brian Reply
is there industrial application of fullrenes. What is the method to prepare fullrene on large scale.?
Rafiq
industrial application...? mmm I think on the medical side as drug carrier, but you should go deeper on your research, I may be wrong
Damian
How we are making nano material?
LITNING Reply
what is a peer
LITNING Reply
What is meant by 'nano scale'?
LITNING Reply
What is STMs full form?
LITNING
scanning tunneling microscope
Sahil
how nano science is used for hydrophobicity
Santosh
Do u think that Graphene and Fullrene fiber can be used to make Air Plane body structure the lightest and strongest. Rafiq
Rafiq
what is differents between GO and RGO?
Mahi
what is simplest way to understand the applications of nano robots used to detect the cancer affected cell of human body.? How this robot is carried to required site of body cell.? what will be the carrier material and how can be detected that correct delivery of drug is done Rafiq
Rafiq
if virus is killing to make ARTIFICIAL DNA OF GRAPHENE FOR KILLED THE VIRUS .THIS IS OUR ASSUMPTION
Anam
analytical skills graphene is prepared to kill any type viruses .
Anam
Any one who tell me about Preparation and application of Nanomaterial for drug Delivery
Hafiz
what is Nano technology ?
Bob Reply
write examples of Nano molecule?
Bob
The nanotechnology is as new science, to scale nanometric
brayan
nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale
Damian
Is there any normative that regulates the use of silver nanoparticles?
Damian Reply
what king of growth are you checking .?
Renato
What fields keep nano created devices from performing or assimulating ? Magnetic fields ? Are do they assimilate ?
Stoney Reply
why we need to study biomolecules, molecular biology in nanotechnology?
Adin Reply
?
Kyle
yes I'm doing my masters in nanotechnology, we are being studying all these domains as well..
Adin
why?
Adin
what school?
Kyle
biomolecules are e building blocks of every organics and inorganic materials.
Joe
how did you get the value of 2000N.What calculations are needed to arrive at it
Smarajit Reply
Privacy Information Security Software Version 1.1a
Good
Got questions? Join the online conversation and get instant answers!
Jobilize.com Reply

Get the best Algebra and trigonometry course in your pocket!





Source:  OpenStax, The art of the pfug. OpenStax CNX. Jun 05, 2013 Download for free at http://cnx.org/content/col10523/1.34
Google Play and the Google Play logo are trademarks of Google Inc.

Notification Switch

Would you like to follow the 'The art of the pfug' conversation and receive update notifications?

Ask