Drorbn - User contributions [en]

0708-1300/Just for Fun10

2008-03-08T17:45:35Z

Oantolin:

First label the edges of the tree with ones. Then, for any edge not belonging to the tree, not already labeled and such that a contiguous square has the rest of the edges labeled, label it with the sum of those edges. Observe that in this process the labeling of all the edges will be well defined and every edge will be labeled since we started from a tree that was maximal. The non selected edges form like passages on a labyrinth through which you can move along. Observe that there are at least one edge in the boundary of the square that is not part of the tree. Consider the numbers on those edges (the ones in the boundary of the square and not on the tree). Let's assume that all of these edges has label strictly less than <math>n</math>. Observe that when you move one step along the labyrinth then the number in the edge you are on can be reduced by at least two units. Then none of the passages starting outside te square reach the center of it. Remove from the figure all those passages and the edges that are common to two passages. Then you get a non-empty (contains some squares) part of the square that is completely surrounded by the tree. But this is a contradiction because this would be a cycle in the tree.

If this solution is correct and you already read it it doesn't means that the fun is over because this problem is certainly here because it can be solved with the aid of homology theory. So, we still need to find that proof.

Note: this proof is incorrect. It does show one of the numbers assigned to the edges is at least n, so to complete the proof we need to show that those numbers are the distances along the tree or at least bounded above by them. Unfortunately that is not true, those numbers are generally larger than the distance along the tree (see the red edge in the following figure).

[[Image:Maze.png]]

File:Maze.png

2008-03-08T17:43:59Z

Oantolin:

0708-1300/Class Photo

2007-10-06T00:38:53Z

Oantolin: /* Who We Are... */

{{0708-1300/Navigation}}

Our class on September 27, 2007:

[[Image:0708-1300-ClassPhoto.jpg|thumb|centre|600px|Class Photo: click to enlarge]]

Please identify yourself in this photo! There are two ways to do that:

* [[Special:Userlogin|Log in]] to this Wiki and edit this page. Put your name, userid, email address and location in the picture in the alphabetical list below.
* Send [[User:Drorbn|Dror]] an email message with this information.

The first option is more fun but less private.

===Who We Are...===

{| align=center border=1
|-
!First name
!Last name
!UserID
!Email
!In the photo
!Comments
{{Photo Entry|last=Bar-Natan|first=Dror|userid=Drorbn|email=drorbn @ math.toronto.edu|location=facing everybody, as the photographer|comments=Take this entry as a model and leave it first. Otherwise alphabetize by last name. Feel free to leave some fields blank}}
{{Photo Entry|last=Bazett|first=Trefor|userid=Trefor|email=trefor.bazett @ toronto.ca|location=tallest person a little right of center in a beige shirt|comments=}}
{{Photo Entry|last=Bjorndahl|first=Adam|userid=ABjorndahl|email=adam.bjorndahl @ utoronto.ca|location=back row, fifth from the left, under the "f(tp)dt"|comments=Looking forward to a great year!}}
{{Photo Entry|last=Choi|first=Brian|userid=Brianchoi|email=brianymc.choi@utoronto.ca|location=In the middle of the front row, the weird looking (!) guy with brown shirt over blue and white|comments=}}
{{Photo Entry|last=Chow|first=Aaron|userid=aaron.chow|email=aaron @ utoronto.ca|location=Third from right, in a black shirt.|comments=Hope we have a good year together!}}
{{Photo Entry|last=Fisher|first=Jonathan|userid=jonathan.fisher|email=jonathan.fisher @ utoronto.ca|location=6th from the right, brown shirt, eyes closed|comments=}}
{{Photo Entry|last=Isgur|first=Abraham|userid=Abisgu|email=abraham.isgur@ math.toronto.edu|location=2nd person in the back row, from the right, the one with the beard and long hair|comments=}}
{{Photo Entry|last=Mann|first=Katie|userid=katiemann|email=katie.mann@ utoronto.ca|location=middle, wearing "Eulers" shirt|comments=}}
{{Photo Entry|last=Mourtada|first=Mariam|userid=Mourtada|email=mariam.mourtada@ utoronto.ca|location=I am the girl in the front middle, wearing a blue shirt and catching my hands|comments=I am not wearing glasses! }}
{{Photo Entry|last=Pym|first=Brent|userid=Bpym|email=bpym @ math.toronto.edu|location=10th from the right (cumulatively), under the <math>T_p(M)\!</math>|comments=Adding this entry was my first-ever edit of a Wiki!}}
{{Photo Entry|last=Snow|first=Megan|userid=megan|email=megansnow @ gmail.com|location=back row, slightly right of centre, wearing a blue shirt over a black one|comments=}}
{{Photo Entry|last=Vera Pacheco|first=Franklin|userid=Franklin|email=franklin.vp @ gmail.com|location=Xth from left to right|comments=To find me you must first go to [[http://www.deathball.net/notpron/]] solve the first 4 pages. Once this done you will know how to find me. Once this done go back to NOTPRON an solve the rest of the puzzle}}
{{Photo Entry|last=Watts|first=Jordan|userid=Jwatts|email=jwatts @ math.toronto.edu|location=in the back, 2nd or 3rd from the left, depending on your convention|comments=My glasses become invisible in pictures.}}
{{Photo Entry|last=Wong|first=Silian|userid=kuramay|email=kurama_y @ hotmail.com|location=One of the Asian-looking girls...with sparkling teeth(??)|comments=I'll write up some comments after their existences}}
{{Photo Entry|last=Kinzebulatov|first=Damir|userid=Dkinz|email=dkinz @ math.toronto.edu|location=In the middle in red shirt|comments=}}
{{Photo Entry|last=Liu|first=Xiao|userid=Ninetiger|email=ninetiger.liu@utoronto.ca|location=In the first row. A boy in orange T-shirt|comments=}}
{{Photo Entry|last=Li|first=Zhiqiang|userid=li-zhiqiang|email=lizhiqiangfly @ gmail.com|location=2nd from the left, 1st boy in the front row.|comments=}}
{{Photo Entry|last=Antolin Camarena|first=Omar|userid=oantolin|email=oantolin @ math.utoronto.ca|location=furthest person to the right|comments=I'm the TA}}
|}

0708-1300/Class Photo

2007-10-06T00:38:11Z

Oantolin: /* Who We Are... */

The Existence of the Exponential Function

2007-01-30T03:57:35Z

Oantolin: /* Computing the Homology */

{{Paperlets Navigation}}

==Introduction==

The purpose of this [[paperlet]] is to use some homological algebra in order to prove the existence of a power series <math>e(x)</math> (with coefficients in <math>{\mathbb Q}</math>) which satisfies the non-linear equation

{{Equation|Main|<math>e(x+y)=e(x)e(y)</math>}}

as well as the initial condition

{{Equation|Init|<math>e(x)=1+x+</math>''(higher order terms)''.}}

Alternative proofs of the existence of <math>e(x)</math> are of course available, including the explicit formula <math>e(x)=\sum_{k=0}^\infty\frac{x^k}{k!}</math>. Thus the value of this paperlet is not in the result it proves but rather in the '''allegorical story''' it tells: that there is a technique to solve functional equations such as {{EqRef|Main}} using homology. There are plenty of other examples for the use of that technique, in which the equation replacing {{EqRef|Main}} isn't as easy. Thus the exponential function seems to be the easiest illustration of a general principle and as such it is worthy of documenting.

Thus below we will pretend not to know the exponential function and/or its relationship with the differential equation <math>e'=e</math>.

==The Scheme==

We aim to construct <math>e(x)</math> and solve {{EqRef|Main}} inductively, degree by degree. Equation {{EqRef|Init}} gives <math>e(x)</math> in degrees 0 and 1, and the given formula for <math>e(x)</math> indeed solves {{EqRef|Main}} in degrees 0 and 1. So booting the induction is no problem. Now assume we've found a degree 7 polynomial <math>e_7(x)</math> which solves {{EqRef|Main}} up to and including degree 7, but at this stage of the construction, it may well fail to solve {{EqRef|Main}} in degree 8. Thus modulo degrees 9 and up, we have

{{Equation|M|<math>e_7(x+y)-e_7(x)e_7(y)=M(x,y)</math>,}}

where <math>M(x,y)</math> is the "mistake for <math>e_7</math>", a certain homogeneous polynomial of degree 8 in the variables <math>x</math> and <math>y</math>.

Our hope is to "fix" the mistake <math>M</math> by replacing <math>e_7(x)</math> with <math>e_8(x)=e_7(x)+\epsilon(x)</math>, where <math>\epsilon_8(x)</math> is a degree 8 "correction", a homogeneous polynomial of degree 8 in <math>x</math> (well, in this simple case, just a multiple of <math>x^8</math>).

{{Begin Side Note|35%}}*1 The terms containing no <math>\epsilon</math>'s make a copy of the left hand side of {{EqRef|M}}. The terms linear in <math>\epsilon</math> are <math>\epsilon(x+y)</math>, <math>-e_7(x)\epsilon(y)</math> and <math>-\epsilon(x)e_7(y)</math>. Note that since the constant term of <math>e_7</math> is 1 and since we only care about degree 8, the last two terms can be replaced by <math>-\epsilon(y)</math> and <math>-\epsilon(x)</math>, respectively. Finally, we don't even need to look at terms higher than linear in <math>\epsilon</math>, for these have degree 16 or more, high in the stratosphere.
{{End Side Note}}
So we substitute <math>e_8(x)=e_7(x)+\epsilon(x)</math> into <math>e(x+y)-e(x)e(y)</math> (a version of {{EqRef|Main}}), expand, and consider only the low degree terms - those below and including degree 8:*1

{{Equation*|<math>e_8(x+y)-e_8(x)e_8(y)=M(x,y)-\epsilon(y)+\epsilon(x+y)-\epsilon(x)</math>.}}

We define a "differential" <math>d:{\mathbb Q}[x]\to{\mathbb Q}[x,y]</math> by <math>(df)(x,y)=f(y)-f(x+y)+f(x)</math>, and the above equation becomes

{{Equation*|<math>e_8(x+y)-e_8(x)e_8(y)=M(x,y)-(d\epsilon)(x,y)</math>.}}

{{Begin Side Note|35%}}*2 It is worth noting that in some a priori sense the existence of an exponential function, a solution of <math>e(x+y)=e(x)e(y)</math>, is quite unlikely. For <math>e</math> must be an element of the relatively small space <math> {\mathbb Q}[[x]] </math> of power series in one variable, but the equation it is required to satisfy lives in the much bigger space <math> {\mathbb Q}[[x,y]] </math>. Thus in some sense we have more equations than unknowns and a solution is unlikely. How fortunate we are!
{{End Side Note}}
To continue with our inductive construction we need to have that <math>e_8(x+y)-e_8(x)e_8(y)=0</math>. Hence the existence of the exponential function hinges upon our ability to find an <math>\epsilon</math> for which <math>M=d\epsilon</math>. In other words, we must show that <math>M</math> is in the image of <math>d</math>. This appears hopeless unless we learn more about <math>M</math>, for the domain space of <math>d</math> is much smaller than its target space and thus <math>d</math> cannot be surjective, and if <math>M</math> was in any sense "random", we simply wouldn't be able to find our correction term <math>\epsilon</math>.*2

As we shall see momentarily by "finding syzygies", <math>\epsilon</math> and <math>M</math> fit within the 0th and 1st chain groups of a rather short complex

{{Equation*|<math>\left(\epsilon\in C_1={\mathbb Q}[[x]]\right)\longrightarrow\left(M\in C_2={\mathbb Q}[[x,y]]\right)\longrightarrow\left(C_3={\mathbb Q}[[x,y,z]]\right)</math>,}}

whose first differential was already written and whose second differential is given by <math>(d^2m)(x,y,z)=m(y,z)-m(x+y,z)+m(x,y+z)-m(x,y)</math> for any <math>m\in{\mathbb Q}[[x,y]]</math>. We shall further see that for "our" <math>M</math>, we have <math>d^2M=0</math>. Therefore in order to show that <math>M</math> is in the image of <math>d^1</math>, it suffices to show that the kernel of <math>d^2</math> is equal to the image of <math>d^1</math>, or simply that <math>H^2=0</math>.

==Finding a Syzygy==

So what kind of relations can we get for <math>M</math>? Well, it measures how close <math>e_7</math> is to turning sums into products, so we can look for preservation of properties that both addition and multiplication have. For example, they're both commutative, so we should have <math>M(x,y)=M(y,x)</math>, and indeed this is obvious from the definition. Now let's try associativity, that is, let's compute <math>e_7(x+y+z)</math> associating first as <math>(x+y)+z</math> and then as <math>x+(y+z)</math>. In the first way we get

<math>e_7(x+y+z)=M(x+y,z)+e_7(x+y)e_7(z)=M(x+y,z)+\left(M(x,y)+e_7(x)e_7(y)\right)e_7(z).</math>

In the second we get

<math>e_7(x+y+z)=M(x,y+z)+e_7(x)e_7(y+z)=M(x+y,z)+e_7(x)\left(M(y,z)+e_7(y)e_7(z)\right)</math>.

Comparing these two we get an interesting relation for <math>M</math>: <math>M(x+y,z)+M(x,y)e_7(z) = M(x,y+z) + e_7(x)M(y,z) </math>. Since we'll only use <math>M</math> to find the next highest term, we can be sloppy about all but the first term of <math>M</math>. This means that in the relation we just found we can replace <math>e_7</math> by its constant term, namely 1. Upon rearranging, we get the relation promised for <math>M</math>: <math> d^2M = M(y,z)-M(x+y,z)+M(x,y+z)-M(x,y) = 0</math>.

==Computing the Homology==

Now let's prove that <math>H^2=0</math> for our (piece of) chain complex. That is, letting <math>M(x,y) \in \mathbb{Q}[[x,y]]</math> be such that <math>d^2M=0</math>, we'll prove that for some <math>E(x) \in \mathbb{Q}[[x]]</math> we have <math> d^1E = M </math>.

Write the two power series as <math> E(x) = \sum{\frac{e_i}{i!}x^i} </math> and <math> M(x,y) = \sum{\frac{m_{ij}}{i!j!}x^i y^j} </math>, where the <math>e_i</math> are the unknowns we wish to solve for.

The coefficient of <math>x^i y^j z^k</math> in <math> M(y,z)-M(x+y,z)+M(x,y+z)-M(x,y)</math> is

<math> \frac{\delta_{i0} m_{jk}}{j!k!} - {{i+j} \choose i} \frac{m_{i+j,k}}{(i+j)!k!} +
{{j+k} \choose j} \frac{m_{i,j+k}}{i!(j+k)!} - \frac{\delta_{k0} m_{ij}}{i!j!}. </math>

Here, <math> \delta_{i0} </math> is a Kronecker delta: 1 if <math> i=0 </math> and 0 otherwise. Since <math> d^2 M = 0 </math>, this coefficient should be zero. Multiplying by <math> i!j!k! </math> (and noting that, for example, the first term doesn't need an <math> i! </math> since the delta is only nonzero when <math> i=0 </math>) we get:

<math> \delta_{i0} m_{jk} - m_{i+j,k} + m_{i,j+k} + \delta_{k0} m_{ij}=0</math>.

An entirely analogous procedure tells us that the equations we must solve boil down to <math> \delta_{i0} e_j - e_{i+j} + \delta_{j0} e_i = m_{ij} </math>.

By setting <math> i=j=0 </math> in this last equation we see that <math> e_0 = m_{00} </math>. Now let <math>i</math> and <math>j</math> be arbitrary ''positive'' integers. This solves for most of the coefficients: <math> e_{i+j} = -m_{ij} </math>. Any integer at least two can be written as <math> i+j </math>, so this determines all of the <math> e_m </math> for <math> m \ge 2 </math>. We just need to prove that <math> e_m </math> is well defined, that is, that <math> m_{ij} </math> doesn't depend on <math> i </math> and <math> j </math> but only on their sum.

But when <math> i </math> and <math> k </math> are strictly positive, the relation for the <math> m_{ij} </math> reads <math> m_{i+j,k} = m_{i,j+k} </math>, which show that we can "transfer" <math> j </math> from one index to the other, which is what we wanted.

It only remains to find <math> e_1 </math> but it's easy to see this is impossible: if <math> E </math> satisfies <math> d^1E = M </math>, then so does <math> E(x)+kx </math> for any <math> k </math>, so <math> e_1 </math> is abritrary. How do our coefficient equations tell us this?

Well, we can't find a single equation for <math> e_1 </math>! We've already tried taking both <math> i </math> and <math> j </math> to be zero, and also taking them both positive. We only have taking one zero and one positive left. Doing so gives two neccessary conditions for the existence of the <math> e_m </math>: <math> m_{0r} = m_{r0} = 0 </math> for <math> r>0 </math>. So no <math> e_1 </math> comes up, but we're still not done. Fortunately setting one of <math> i </math> and <math> k </math> to be zero and one positive in the realtion for the <math> m_{ij} </math> does the trick.

The Existence of the Exponential Function

2007-01-30T03:57:06Z

Oantolin: /* Computing the Homology */

The Existence of the Exponential Function

2007-01-30T03:54:25Z

Oantolin: /* Computing the Homology */

{{Paperlets Navigation}}

==Introduction==

The purpose of this [[paperlet]] is to use some homological algebra in order to prove the existence of a power series <math>e(x)</math> (with coefficients in <math>{\mathbb Q}</math>) which satisfies the non-linear equation

{{Equation|Main|<math>e(x+y)=e(x)e(y)</math>}}

as well as the initial condition

{{Equation|Init|<math>e(x)=1+x+</math>''(higher order terms)''.}}

Alternative proofs of the existence of <math>e(x)</math> are of course available, including the explicit formula <math>e(x)=\sum_{k=0}^\infty\frac{x^k}{k!}</math>. Thus the value of this paperlet is not in the result it proves but rather in the '''allegorical story''' it tells: that there is a technique to solve functional equations such as {{EqRef|Main}} using homology. There are plenty of other examples for the use of that technique, in which the equation replacing {{EqRef|Main}} isn't as easy. Thus the exponential function seems to be the easiest illustration of a general principle and as such it is worthy of documenting.

Thus below we will pretend not to know the exponential function and/or its relationship with the differential equation <math>e'=e</math>.

==The Scheme==

We aim to construct <math>e(x)</math> and solve {{EqRef|Main}} inductively, degree by degree. Equation {{EqRef|Init}} gives <math>e(x)</math> in degrees 0 and 1, and the given formula for <math>e(x)</math> indeed solves {{EqRef|Main}} in degrees 0 and 1. So booting the induction is no problem. Now assume we've found a degree 7 polynomial <math>e_7(x)</math> which solves {{EqRef|Main}} up to and including degree 7, but at this stage of the construction, it may well fail to solve {{EqRef|Main}} in degree 8. Thus modulo degrees 9 and up, we have

{{Equation|M|<math>e_7(x+y)-e_7(x)e_7(y)=M(x,y)</math>,}}

where <math>M(x,y)</math> is the "mistake for <math>e_7</math>", a certain homogeneous polynomial of degree 8 in the variables <math>x</math> and <math>y</math>.

Our hope is to "fix" the mistake <math>M</math> by replacing <math>e_7(x)</math> with <math>e_8(x)=e_7(x)+\epsilon(x)</math>, where <math>\epsilon_8(x)</math> is a degree 8 "correction", a homogeneous polynomial of degree 8 in <math>x</math> (well, in this simple case, just a multiple of <math>x^8</math>).

{{Begin Side Note|35%}}*1 The terms containing no <math>\epsilon</math>'s make a copy of the left hand side of {{EqRef|M}}. The terms linear in <math>\epsilon</math> are <math>\epsilon(x+y)</math>, <math>-e_7(x)\epsilon(y)</math> and <math>-\epsilon(x)e_7(y)</math>. Note that since the constant term of <math>e_7</math> is 1 and since we only care about degree 8, the last two terms can be replaced by <math>-\epsilon(y)</math> and <math>-\epsilon(x)</math>, respectively. Finally, we don't even need to look at terms higher than linear in <math>\epsilon</math>, for these have degree 16 or more, high in the stratosphere.
{{End Side Note}}
So we substitute <math>e_8(x)=e_7(x)+\epsilon(x)</math> into <math>e(x+y)-e(x)e(y)</math> (a version of {{EqRef|Main}}), expand, and consider only the low degree terms - those below and including degree 8:*1

{{Equation*|<math>e_8(x+y)-e_8(x)e_8(y)=M(x,y)-\epsilon(y)+\epsilon(x+y)-\epsilon(x)</math>.}}

We define a "differential" <math>d:{\mathbb Q}[x]\to{\mathbb Q}[x,y]</math> by <math>(df)(x,y)=f(y)-f(x+y)+f(x)</math>, and the above equation becomes

{{Equation*|<math>e_8(x+y)-e_8(x)e_8(y)=M(x,y)-(d\epsilon)(x,y)</math>.}}

{{Begin Side Note|35%}}*2 It is worth noting that in some a priori sense the existence of an exponential function, a solution of <math>e(x+y)=e(x)e(y)</math>, is quite unlikely. For <math>e</math> must be an element of the relatively small space <math> {\mathbb Q}[[x]] </math> of power series in one variable, but the equation it is required to satisfy lives in the much bigger space <math> {\mathbb Q}[[x,y]] </math>. Thus in some sense we have more equations than unknowns and a solution is unlikely. How fortunate we are!
{{End Side Note}}
To continue with our inductive construction we need to have that <math>e_8(x+y)-e_8(x)e_8(y)=0</math>. Hence the existence of the exponential function hinges upon our ability to find an <math>\epsilon</math> for which <math>M=d\epsilon</math>. In other words, we must show that <math>M</math> is in the image of <math>d</math>. This appears hopeless unless we learn more about <math>M</math>, for the domain space of <math>d</math> is much smaller than its target space and thus <math>d</math> cannot be surjective, and if <math>M</math> was in any sense "random", we simply wouldn't be able to find our correction term <math>\epsilon</math>.*2

As we shall see momentarily by "finding syzygies", <math>\epsilon</math> and <math>M</math> fit within the 0th and 1st chain groups of a rather short complex

{{Equation*|<math>\left(\epsilon\in C_1={\mathbb Q}[[x]]\right)\longrightarrow\left(M\in C_2={\mathbb Q}[[x,y]]\right)\longrightarrow\left(C_3={\mathbb Q}[[x,y,z]]\right)</math>,}}

whose first differential was already written and whose second differential is given by <math>(d^2m)(x,y,z)=m(y,z)-m(x+y,z)+m(x,y+z)-m(x,y)</math> for any <math>m\in{\mathbb Q}[[x,y]]</math>. We shall further see that for "our" <math>M</math>, we have <math>d^2M=0</math>. Therefore in order to show that <math>M</math> is in the image of <math>d^1</math>, it suffices to show that the kernel of <math>d^2</math> is equal to the image of <math>d^1</math>, or simply that <math>H^2=0</math>.

==Finding a Syzygy==

So what kind of relations can we get for <math>M</math>? Well, it measures how close <math>e_7</math> is to turning sums into products, so we can look for preservation of properties that both addition and multiplication have. For example, they're both commutative, so we should have <math>M(x,y)=M(y,x)</math>, and indeed this is obvious from the definition. Now let's try associativity, that is, let's compute <math>e_7(x+y+z)</math> associating first as <math>(x+y)+z</math> and then as <math>x+(y+z)</math>. In the first way we get

<math>e_7(x+y+z)=M(x+y,z)+e_7(x+y)e_7(z)=M(x+y,z)+\left(M(x,y)+e_7(x)e_7(y)\right)e_7(z).</math>

In the second we get

<math>e_7(x+y+z)=M(x,y+z)+e_7(x)e_7(y+z)=M(x+y,z)+e_7(x)\left(M(y,z)+e_7(y)e_7(z)\right)</math>.

Comparing these two we get an interesting relation for <math>M</math>: <math>M(x+y,z)+M(x,y)e_7(z) = M(x,y+z) + e_7(x)M(y,z) </math>. Since we'll only use <math>M</math> to find the next highest term, we can be sloppy about all but the first term of <math>M</math>. This means that in the relation we just found we can replace <math>e_7</math> by its constant term, namely 1. Upon rearranging, we get the relation promised for <math>M</math>: <math> d^2M = M(y,z)-M(x+y,z)+M(x,y+z)-M(x,y) = 0</math>.

==Computing the Homology==

Now let's prove that <math>H^2=0</math> for our (piece of) chain complex. That is, letting <math>M(x,y) \in \mathbb{Q}[[x,y]]</math> be such that <math>d^2M=0</math>, we'll prove that for some <math>E(x) \in \mathbb{Q}[[x]]</math> we have <math> d^1E = M </math>.

Write the two power series as <math> E(x) = \sum{\frac{e_i}{i!}x^i} </math> and <math> M(x,y) = \sum{\frac{m_{ij}}{i!j!}x^i y^j} </math>, where the <math>e_i</math> are the unknowns we wish to solve for.

The coefficient of <math>x^i y^j z^k</math> in <math> M(y,z)-M(x+y,z)+M(x,y+z)-M(x,y)</math> is

<math> \frac{\delta_{i0} m_{jk}}{j!k!} - {{i+j} \choose i} \frac{m_{i+j,k}}{(i+j)!k!} +
{{j+k} \choose j} \frac{m_{i,j+k}}{i!(j+k)!} - \frac{\delta_{k0} m_{ij}}{i!j!}. </math>

Here, <math> \delta_{i0} </math> is a Kronecker delta: 1 if <math> i=0 </math> and 0 otherwise. Since <math> d^2 M = 0 </math>, this coefficient should be zero. Multiplying by <math> i!j!k! </math> (and noting that, for example, the first term doesn't need an <math> i! </math> since the delta is only nonzero when <math> i=0 </math>) we get:

<math> \delta_{i0} m_{jk} - m_{i+j,k} + m_{i,j+k} + \delta_{k0} m_{ij}=0</math>.

An entirely analogous procedure tells us that the equations we must solve boil down to <math> \delta_{i0} e_j - e_{i+j} + \delta_{j0} e_i = m_{ij} </math>.

By setting <math> i=j=0 </math> in this last equation we see that <math> e_0 = m_{00} </math>. Now let <math>i</math> and <math>j</math> be arbitrary ''positive'' integers. This solves for most of the coefficients: <math> e_{i+j} = -m_{ij} </math>. Any integer at least tw0 can be written as <math> i+j </math>, so this determines all of the <math> e_m </math> for <math> m \ge 2 </math>. We just need to prove that <math> e_m </math> is well defined, that is, that <math> m_{ij} </math> doesn't depend on <math> i </math> and <math> j </math> but only on their sum <math> m </math>.

But when <math> i </math> and <math> k </math> are strictly positive, the relation for the <math> m_{ij} </math> reads <math> m_{i+j,k} = m_{i,j+k} </math>, which show that we can "transfer" <math> j </math> from one index to the other, i.e., that <math> m_{rs} </math> depends only on <math> r+s </math> when <math> r </math> and <math> s </math> are positive.

It only remains to find <math> e_1 </math> but it's easy to see this is impossible: if <math> E </math> satisfies <math> d^1E = M </math>, then so does <math> E(x)+kx </math> for any <math> k </math>, so <math> e_1 </math> is abritrary. How do our coefficient equations tell us this?

Well, we can't find a single equation for <math> e_1 </math>! We've already tried taking both <math> i </math> and <math> j </math> to be zero, and also taking them both positive. We only have taking one zero and one positive left. Doing so gives two neccessary conditions for the existence of the <math> e_m </math>: <math> m_{0r} = m_{r0} = 0 </math> for <math> r>0 </math>. So no <math> e_1 </math> comes up, but we're still not done. But setting one of <math> i </math> and <math> k </math> to be zero and one positive in the realtion for the <math> m_rs </math> does the trick.

The Existence of the Exponential Function

2007-01-30T03:37:52Z

Oantolin: /* Computing the Homology */

{{Paperlets Navigation}}

==Introduction==

The purpose of this [[paperlet]] is to use some homological algebra in order to prove the existence of a power series <math>e(x)</math> (with coefficients in <math>{\mathbb Q}</math>) which satisfies the non-linear equation

{{Equation|Main|<math>e(x+y)=e(x)e(y)</math>}}

as well as the initial condition

{{Equation|Init|<math>e(x)=1+x+</math>''(higher order terms)''.}}

Alternative proofs of the existence of <math>e(x)</math> are of course available, including the explicit formula <math>e(x)=\sum_{k=0}^\infty\frac{x^k}{k!}</math>. Thus the value of this paperlet is not in the result it proves but rather in the '''allegorical story''' it tells: that there is a technique to solve functional equations such as {{EqRef|Main}} using homology. There are plenty of other examples for the use of that technique, in which the equation replacing {{EqRef|Main}} isn't as easy. Thus the exponential function seems to be the easiest illustration of a general principle and as such it is worthy of documenting.

Thus below we will pretend not to know the exponential function and/or its relationship with the differential equation <math>e'=e</math>.

==The Scheme==

We aim to construct <math>e(x)</math> and solve {{EqRef|Main}} inductively, degree by degree. Equation {{EqRef|Init}} gives <math>e(x)</math> in degrees 0 and 1, and the given formula for <math>e(x)</math> indeed solves {{EqRef|Main}} in degrees 0 and 1. So booting the induction is no problem. Now assume we've found a degree 7 polynomial <math>e_7(x)</math> which solves {{EqRef|Main}} up to and including degree 7, but at this stage of the construction, it may well fail to solve {{EqRef|Main}} in degree 8. Thus modulo degrees 9 and up, we have

{{Equation|M|<math>e_7(x+y)-e_7(x)e_7(y)=M(x,y)</math>,}}

where <math>M(x,y)</math> is the "mistake for <math>e_7</math>", a certain homogeneous polynomial of degree 8 in the variables <math>x</math> and <math>y</math>.

Our hope is to "fix" the mistake <math>M</math> by replacing <math>e_7(x)</math> with <math>e_8(x)=e_7(x)+\epsilon(x)</math>, where <math>\epsilon_8(x)</math> is a degree 8 "correction", a homogeneous polynomial of degree 8 in <math>x</math> (well, in this simple case, just a multiple of <math>x^8</math>).

{{Begin Side Note|35%}}*1 The terms containing no <math>\epsilon</math>'s make a copy of the left hand side of {{EqRef|M}}. The terms linear in <math>\epsilon</math> are <math>\epsilon(x+y)</math>, <math>-e_7(x)\epsilon(y)</math> and <math>-\epsilon(x)e_7(y)</math>. Note that since the constant term of <math>e_7</math> is 1 and since we only care about degree 8, the last two terms can be replaced by <math>-\epsilon(y)</math> and <math>-\epsilon(x)</math>, respectively. Finally, we don't even need to look at terms higher than linear in <math>\epsilon</math>, for these have degree 16 or more, high in the stratosphere.
{{End Side Note}}
So we substitute <math>e_8(x)=e_7(x)+\epsilon(x)</math> into <math>e(x+y)-e(x)e(y)</math> (a version of {{EqRef|Main}}), expand, and consider only the low degree terms - those below and including degree 8:*1

{{Equation*|<math>e_8(x+y)-e_8(x)e_8(y)=M(x,y)-\epsilon(y)+\epsilon(x+y)-\epsilon(x)</math>.}}

We define a "differential" <math>d:{\mathbb Q}[x]\to{\mathbb Q}[x,y]</math> by <math>(df)(x,y)=f(y)-f(x+y)+f(x)</math>, and the above equation becomes

{{Equation*|<math>e_8(x+y)-e_8(x)e_8(y)=M(x,y)-(d\epsilon)(x,y)</math>.}}

{{Begin Side Note|35%}}*2 It is worth noting that in some a priori sense the existence of an exponential function, a solution of <math>e(x+y)=e(x)e(y)</math>, is quite unlikely. For <math>e</math> must be an element of the relatively small space <math> {\mathbb Q}[[x]] </math> of power series in one variable, but the equation it is required to satisfy lives in the much bigger space <math> {\mathbb Q}[[x,y]] </math>. Thus in some sense we have more equations than unknowns and a solution is unlikely. How fortunate we are!
{{End Side Note}}
To continue with our inductive construction we need to have that <math>e_8(x+y)-e_8(x)e_8(y)=0</math>. Hence the existence of the exponential function hinges upon our ability to find an <math>\epsilon</math> for which <math>M=d\epsilon</math>. In other words, we must show that <math>M</math> is in the image of <math>d</math>. This appears hopeless unless we learn more about <math>M</math>, for the domain space of <math>d</math> is much smaller than its target space and thus <math>d</math> cannot be surjective, and if <math>M</math> was in any sense "random", we simply wouldn't be able to find our correction term <math>\epsilon</math>.*2

As we shall see momentarily by "finding syzygies", <math>\epsilon</math> and <math>M</math> fit within the 0th and 1st chain groups of a rather short complex

{{Equation*|<math>\left(\epsilon\in C_1={\mathbb Q}[[x]]\right)\longrightarrow\left(M\in C_2={\mathbb Q}[[x,y]]\right)\longrightarrow\left(C_3={\mathbb Q}[[x,y,z]]\right)</math>,}}

whose first differential was already written and whose second differential is given by <math>(d^2m)(x,y,z)=m(y,z)-m(x+y,z)+m(x,y+z)-m(x,y)</math> for any <math>m\in{\mathbb Q}[[x,y]]</math>. We shall further see that for "our" <math>M</math>, we have <math>d^2M=0</math>. Therefore in order to show that <math>M</math> is in the image of <math>d^1</math>, it suffices to show that the kernel of <math>d^2</math> is equal to the image of <math>d^1</math>, or simply that <math>H^2=0</math>.

==Finding a Syzygy==

So what kind of relations can we get for <math>M</math>? Well, it measures how close <math>e_7</math> is to turning sums into products, so we can look for preservation of properties that both addition and multiplication have. For example, they're both commutative, so we should have <math>M(x,y)=M(y,x)</math>, and indeed this is obvious from the definition. Now let's try associativity, that is, let's compute <math>e_7(x+y+z)</math> associating first as <math>(x+y)+z</math> and then as <math>x+(y+z)</math>. In the first way we get

<math>e_7(x+y+z)=M(x+y,z)+e_7(x+y)e_7(z)=M(x+y,z)+\left(M(x,y)+e_7(x)e_7(y)\right)e_7(z).</math>

In the second we get

<math>e_7(x+y+z)=M(x,y+z)+e_7(x)e_7(y+z)=M(x+y,z)+e_7(x)\left(M(y,z)+e_7(y)e_7(z)\right)</math>.

Comparing these two we get an interesting relation for <math>M</math>: <math>M(x+y,z)+M(x,y)e_7(z) = M(x,y+z) + e_7(x)M(y,z) </math>. Since we'll only use <math>M</math> to find the next highest term, we can be sloppy about all but the first term of <math>M</math>. This means that in the relation we just found we can replace <math>e_7</math> by its constant term, namely 1. Upon rearranging, we get the relation promised for <math>M</math>: <math> d^2M = M(y,z)-M(x+y,z)+M(x,y+z)-M(x,y) = 0</math>.

==Computing the Homology==

Now let's prove that <math>H^2=0</math> for our (piece of) chain complex. That is, letting <math>M(x,y) \in \mathbb{Q}[[x,y]]</math> be such that <math>d^2M=0</math>, we'll prove that for some <math>E(x) \in \mathbb{Q}[[x]]</math> we have <math> d^1E = M </math>.

Write the two power series as <math> E(x) = \sum{\frac{e_i}{i!}x^i} </math> and <math> M(x,y) = \sum{\frac{m_{ij}}{i!j!}x^i y^j} </math>, where the <math>e_i</math> are the unknowns we wish to solve for.

The coefficient of <math>x^i y^j z^k</math> in <math> M(y,z)-M(x+y,z)+M(x,y+z)-M(x,y)</math> is

<math> \frac{\delta_{i0} m_{jk}}{j!k!} - {{i+j} \choose i} \frac{m_{i+j,k}}{(i+j)!k!} +
{{j+k} \choose j} \frac{m_{i,j+k}}{i!(j+k)!} - \frac{\delta_{k0} m_{ij}}{i!j!}. </math>

Here, <math> \delta_{i0} </math> is a Kronecker delta: 1 if <math> i=0 </math> and 0 otherwise. Since <math> d^2 M = 0 </math>, this coefficient should be zero. Multiplying by <math> i!j!k! </math> (and noting that, for example, the first term doesn't need an <math> i! </math> since the delta is only nonzero when <math> i=0 </math>) we get <math> \delta_{i0} m_{jk} - m_{i+j,k} + m_{i,j+k} + \delta_{k0} m_{ij}</math>.

An entirely analogous procedure tells us that the equations we must solve boil down to <math> \delta_{i0} e_j - e_{i+j} + \delta_{j0} e_i = m_{ij} </math>.

The Existence of the Exponential Function

2007-01-30T03:37:15Z

Oantolin: /* Computing the Homology */

{{Paperlets Navigation}}

==Introduction==

The purpose of this [[paperlet]] is to use some homological algebra in order to prove the existence of a power series <math>e(x)</math> (with coefficients in <math>{\mathbb Q}</math>) which satisfies the non-linear equation

{{Equation|Main|<math>e(x+y)=e(x)e(y)</math>}}

as well as the initial condition

{{Equation|Init|<math>e(x)=1+x+</math>''(higher order terms)''.}}

Alternative proofs of the existence of <math>e(x)</math> are of course available, including the explicit formula <math>e(x)=\sum_{k=0}^\infty\frac{x^k}{k!}</math>. Thus the value of this paperlet is not in the result it proves but rather in the '''allegorical story''' it tells: that there is a technique to solve functional equations such as {{EqRef|Main}} using homology. There are plenty of other examples for the use of that technique, in which the equation replacing {{EqRef|Main}} isn't as easy. Thus the exponential function seems to be the easiest illustration of a general principle and as such it is worthy of documenting.

Thus below we will pretend not to know the exponential function and/or its relationship with the differential equation <math>e'=e</math>.

==The Scheme==

We aim to construct <math>e(x)</math> and solve {{EqRef|Main}} inductively, degree by degree. Equation {{EqRef|Init}} gives <math>e(x)</math> in degrees 0 and 1, and the given formula for <math>e(x)</math> indeed solves {{EqRef|Main}} in degrees 0 and 1. So booting the induction is no problem. Now assume we've found a degree 7 polynomial <math>e_7(x)</math> which solves {{EqRef|Main}} up to and including degree 7, but at this stage of the construction, it may well fail to solve {{EqRef|Main}} in degree 8. Thus modulo degrees 9 and up, we have

{{Equation|M|<math>e_7(x+y)-e_7(x)e_7(y)=M(x,y)</math>,}}

where <math>M(x,y)</math> is the "mistake for <math>e_7</math>", a certain homogeneous polynomial of degree 8 in the variables <math>x</math> and <math>y</math>.

Our hope is to "fix" the mistake <math>M</math> by replacing <math>e_7(x)</math> with <math>e_8(x)=e_7(x)+\epsilon(x)</math>, where <math>\epsilon_8(x)</math> is a degree 8 "correction", a homogeneous polynomial of degree 8 in <math>x</math> (well, in this simple case, just a multiple of <math>x^8</math>).

{{Begin Side Note|35%}}*1 The terms containing no <math>\epsilon</math>'s make a copy of the left hand side of {{EqRef|M}}. The terms linear in <math>\epsilon</math> are <math>\epsilon(x+y)</math>, <math>-e_7(x)\epsilon(y)</math> and <math>-\epsilon(x)e_7(y)</math>. Note that since the constant term of <math>e_7</math> is 1 and since we only care about degree 8, the last two terms can be replaced by <math>-\epsilon(y)</math> and <math>-\epsilon(x)</math>, respectively. Finally, we don't even need to look at terms higher than linear in <math>\epsilon</math>, for these have degree 16 or more, high in the stratosphere.
{{End Side Note}}
So we substitute <math>e_8(x)=e_7(x)+\epsilon(x)</math> into <math>e(x+y)-e(x)e(y)</math> (a version of {{EqRef|Main}}), expand, and consider only the low degree terms - those below and including degree 8:*1

{{Equation*|<math>e_8(x+y)-e_8(x)e_8(y)=M(x,y)-\epsilon(y)+\epsilon(x+y)-\epsilon(x)</math>.}}

We define a "differential" <math>d:{\mathbb Q}[x]\to{\mathbb Q}[x,y]</math> by <math>(df)(x,y)=f(y)-f(x+y)+f(x)</math>, and the above equation becomes

{{Equation*|<math>e_8(x+y)-e_8(x)e_8(y)=M(x,y)-(d\epsilon)(x,y)</math>.}}

{{Begin Side Note|35%}}*2 It is worth noting that in some a priori sense the existence of an exponential function, a solution of <math>e(x+y)=e(x)e(y)</math>, is quite unlikely. For <math>e</math> must be an element of the relatively small space <math> {\mathbb Q}[[x]] </math> of power series in one variable, but the equation it is required to satisfy lives in the much bigger space <math> {\mathbb Q}[[x,y]] </math>. Thus in some sense we have more equations than unknowns and a solution is unlikely. How fortunate we are!
{{End Side Note}}
To continue with our inductive construction we need to have that <math>e_8(x+y)-e_8(x)e_8(y)=0</math>. Hence the existence of the exponential function hinges upon our ability to find an <math>\epsilon</math> for which <math>M=d\epsilon</math>. In other words, we must show that <math>M</math> is in the image of <math>d</math>. This appears hopeless unless we learn more about <math>M</math>, for the domain space of <math>d</math> is much smaller than its target space and thus <math>d</math> cannot be surjective, and if <math>M</math> was in any sense "random", we simply wouldn't be able to find our correction term <math>\epsilon</math>.*2

As we shall see momentarily by "finding syzygies", <math>\epsilon</math> and <math>M</math> fit within the 0th and 1st chain groups of a rather short complex

{{Equation*|<math>\left(\epsilon\in C_1={\mathbb Q}[[x]]\right)\longrightarrow\left(M\in C_2={\mathbb Q}[[x,y]]\right)\longrightarrow\left(C_3={\mathbb Q}[[x,y,z]]\right)</math>,}}

whose first differential was already written and whose second differential is given by <math>(d^2m)(x,y,z)=m(y,z)-m(x+y,z)+m(x,y+z)-m(x,y)</math> for any <math>m\in{\mathbb Q}[[x,y]]</math>. We shall further see that for "our" <math>M</math>, we have <math>d^2M=0</math>. Therefore in order to show that <math>M</math> is in the image of <math>d^1</math>, it suffices to show that the kernel of <math>d^2</math> is equal to the image of <math>d^1</math>, or simply that <math>H^2=0</math>.

==Finding a Syzygy==

So what kind of relations can we get for <math>M</math>? Well, it measures how close <math>e_7</math> is to turning sums into products, so we can look for preservation of properties that both addition and multiplication have. For example, they're both commutative, so we should have <math>M(x,y)=M(y,x)</math>, and indeed this is obvious from the definition. Now let's try associativity, that is, let's compute <math>e_7(x+y+z)</math> associating first as <math>(x+y)+z</math> and then as <math>x+(y+z)</math>. In the first way we get

<math>e_7(x+y+z)=M(x+y,z)+e_7(x+y)e_7(z)=M(x+y,z)+\left(M(x,y)+e_7(x)e_7(y)\right)e_7(z).</math>

In the second we get

<math>e_7(x+y+z)=M(x,y+z)+e_7(x)e_7(y+z)=M(x+y,z)+e_7(x)\left(M(y,z)+e_7(y)e_7(z)\right)</math>.

Comparing these two we get an interesting relation for <math>M</math>: <math>M(x+y,z)+M(x,y)e_7(z) = M(x,y+z) + e_7(x)M(y,z) </math>. Since we'll only use <math>M</math> to find the next highest term, we can be sloppy about all but the first term of <math>M</math>. This means that in the relation we just found we can replace <math>e_7</math> by its constant term, namely 1. Upon rearranging, we get the relation promised for <math>M</math>: <math> d^2M = M(y,z)-M(x+y,z)+M(x,y+z)-M(x,y) = 0</math>.

==Computing the Homology==

Now let's prove that <math>H^2=0</math> for our (piece of) chain complex. That is, letting <math>M(x,y) \in \mathbb{Q}[[x,y]]</math> be such that <math>d^2M=0</math>, we'll prove that for some <math>E(x) \in \mathbb{Q}[[x]]</math> we have <math> d^1E = M </math>.

Write the two power series as <math> E(x) = \sum{\frac{e_i}{i!}x^i} </math> and <math> M(x,y) = \sum{\frac{m_{ij}}{i!j!}x^i y^j} </math>, where the <math>e_i</math> are the unknowns we wish to solve for.

The coefficient of <math>x^i y^j z^k</math> in <math> M(y,z)-M(x+y,z)+M(x,y+z)-M(x,y)</math> is

<math> \frac{\delta_{i0} m_{jk}}{j!k!} - {{i+j} \choose i} \frac{m_{i+j,k}}{(i+j)!k!} +
\binom{j+k}{j} \frac{m_{i,j+k}}{i!(j+k)!} - \frac{\delta_{k0} m_{ij}}{i!j!}. </math>

Here, <math> \delta_{i0} </math> is a Kronecker delta: 1 if <math> i=0 </math> and 0 otherwise. Since <math> d^2 M = 0 </math>, this coefficient should be zero. Multiplying by <math> i!j!k! </math> (and noting that, for example, the first term doesn't need an <math> i! </math> since the delta is only nonzero when <math> i=0 </math>) we get <math> \delta_{i0} m_{jk} - m_{i+j,k} + m_{i,j+k} + \delta_{k0} m_{ij}</math>.

An entirely analogous procedure tells us that the equations we must solve boil down to <math> \delta_{i0} e_j - e_{i+j} + \delta_{j0} e_i = m_{ij} </math>.

The Existence of the Exponential Function

2007-01-30T03:36:13Z

Oantolin: /* Computing the Homology */

{{Paperlets Navigation}}

==Introduction==

The purpose of this [[paperlet]] is to use some homological algebra in order to prove the existence of a power series <math>e(x)</math> (with coefficients in <math>{\mathbb Q}</math>) which satisfies the non-linear equation

{{Equation|Main|<math>e(x+y)=e(x)e(y)</math>}}

as well as the initial condition

{{Equation|Init|<math>e(x)=1+x+</math>''(higher order terms)''.}}

Alternative proofs of the existence of <math>e(x)</math> are of course available, including the explicit formula <math>e(x)=\sum_{k=0}^\infty\frac{x^k}{k!}</math>. Thus the value of this paperlet is not in the result it proves but rather in the '''allegorical story''' it tells: that there is a technique to solve functional equations such as {{EqRef|Main}} using homology. There are plenty of other examples for the use of that technique, in which the equation replacing {{EqRef|Main}} isn't as easy. Thus the exponential function seems to be the easiest illustration of a general principle and as such it is worthy of documenting.

Thus below we will pretend not to know the exponential function and/or its relationship with the differential equation <math>e'=e</math>.

==The Scheme==

We aim to construct <math>e(x)</math> and solve {{EqRef|Main}} inductively, degree by degree. Equation {{EqRef|Init}} gives <math>e(x)</math> in degrees 0 and 1, and the given formula for <math>e(x)</math> indeed solves {{EqRef|Main}} in degrees 0 and 1. So booting the induction is no problem. Now assume we've found a degree 7 polynomial <math>e_7(x)</math> which solves {{EqRef|Main}} up to and including degree 7, but at this stage of the construction, it may well fail to solve {{EqRef|Main}} in degree 8. Thus modulo degrees 9 and up, we have

{{Equation|M|<math>e_7(x+y)-e_7(x)e_7(y)=M(x,y)</math>,}}

where <math>M(x,y)</math> is the "mistake for <math>e_7</math>", a certain homogeneous polynomial of degree 8 in the variables <math>x</math> and <math>y</math>.

Our hope is to "fix" the mistake <math>M</math> by replacing <math>e_7(x)</math> with <math>e_8(x)=e_7(x)+\epsilon(x)</math>, where <math>\epsilon_8(x)</math> is a degree 8 "correction", a homogeneous polynomial of degree 8 in <math>x</math> (well, in this simple case, just a multiple of <math>x^8</math>).

{{Begin Side Note|35%}}*1 The terms containing no <math>\epsilon</math>'s make a copy of the left hand side of {{EqRef|M}}. The terms linear in <math>\epsilon</math> are <math>\epsilon(x+y)</math>, <math>-e_7(x)\epsilon(y)</math> and <math>-\epsilon(x)e_7(y)</math>. Note that since the constant term of <math>e_7</math> is 1 and since we only care about degree 8, the last two terms can be replaced by <math>-\epsilon(y)</math> and <math>-\epsilon(x)</math>, respectively. Finally, we don't even need to look at terms higher than linear in <math>\epsilon</math>, for these have degree 16 or more, high in the stratosphere.
{{End Side Note}}
So we substitute <math>e_8(x)=e_7(x)+\epsilon(x)</math> into <math>e(x+y)-e(x)e(y)</math> (a version of {{EqRef|Main}}), expand, and consider only the low degree terms - those below and including degree 8:*1

{{Equation*|<math>e_8(x+y)-e_8(x)e_8(y)=M(x,y)-\epsilon(y)+\epsilon(x+y)-\epsilon(x)</math>.}}

We define a "differential" <math>d:{\mathbb Q}[x]\to{\mathbb Q}[x,y]</math> by <math>(df)(x,y)=f(y)-f(x+y)+f(x)</math>, and the above equation becomes

{{Equation*|<math>e_8(x+y)-e_8(x)e_8(y)=M(x,y)-(d\epsilon)(x,y)</math>.}}

{{Begin Side Note|35%}}*2 It is worth noting that in some a priori sense the existence of an exponential function, a solution of <math>e(x+y)=e(x)e(y)</math>, is quite unlikely. For <math>e</math> must be an element of the relatively small space <math> {\mathbb Q}[[x]] </math> of power series in one variable, but the equation it is required to satisfy lives in the much bigger space <math> {\mathbb Q}[[x,y]] </math>. Thus in some sense we have more equations than unknowns and a solution is unlikely. How fortunate we are!
{{End Side Note}}
To continue with our inductive construction we need to have that <math>e_8(x+y)-e_8(x)e_8(y)=0</math>. Hence the existence of the exponential function hinges upon our ability to find an <math>\epsilon</math> for which <math>M=d\epsilon</math>. In other words, we must show that <math>M</math> is in the image of <math>d</math>. This appears hopeless unless we learn more about <math>M</math>, for the domain space of <math>d</math> is much smaller than its target space and thus <math>d</math> cannot be surjective, and if <math>M</math> was in any sense "random", we simply wouldn't be able to find our correction term <math>\epsilon</math>.*2

As we shall see momentarily by "finding syzygies", <math>\epsilon</math> and <math>M</math> fit within the 0th and 1st chain groups of a rather short complex

{{Equation*|<math>\left(\epsilon\in C_1={\mathbb Q}[[x]]\right)\longrightarrow\left(M\in C_2={\mathbb Q}[[x,y]]\right)\longrightarrow\left(C_3={\mathbb Q}[[x,y,z]]\right)</math>,}}

whose first differential was already written and whose second differential is given by <math>(d^2m)(x,y,z)=m(y,z)-m(x+y,z)+m(x,y+z)-m(x,y)</math> for any <math>m\in{\mathbb Q}[[x,y]]</math>. We shall further see that for "our" <math>M</math>, we have <math>d^2M=0</math>. Therefore in order to show that <math>M</math> is in the image of <math>d^1</math>, it suffices to show that the kernel of <math>d^2</math> is equal to the image of <math>d^1</math>, or simply that <math>H^2=0</math>.

==Finding a Syzygy==

So what kind of relations can we get for <math>M</math>? Well, it measures how close <math>e_7</math> is to turning sums into products, so we can look for preservation of properties that both addition and multiplication have. For example, they're both commutative, so we should have <math>M(x,y)=M(y,x)</math>, and indeed this is obvious from the definition. Now let's try associativity, that is, let's compute <math>e_7(x+y+z)</math> associating first as <math>(x+y)+z</math> and then as <math>x+(y+z)</math>. In the first way we get

<math>e_7(x+y+z)=M(x+y,z)+e_7(x+y)e_7(z)=M(x+y,z)+\left(M(x,y)+e_7(x)e_7(y)\right)e_7(z).</math>

In the second we get

<math>e_7(x+y+z)=M(x,y+z)+e_7(x)e_7(y+z)=M(x+y,z)+e_7(x)\left(M(y,z)+e_7(y)e_7(z)\right)</math>.

Comparing these two we get an interesting relation for <math>M</math>: <math>M(x+y,z)+M(x,y)e_7(z) = M(x,y+z) + e_7(x)M(y,z) </math>. Since we'll only use <math>M</math> to find the next highest term, we can be sloppy about all but the first term of <math>M</math>. This means that in the relation we just found we can replace <math>e_7</math> by its constant term, namely 1. Upon rearranging, we get the relation promised for <math>M</math>: <math> d^2M = M(y,z)-M(x+y,z)+M(x,y+z)-M(x,y) = 0</math>.

==Computing the Homology==

Now let's prove that <math>H^2=0</math> for our (piece of) chain complex. That is, letting <math>M(x,y) \in \mathbb{Q}[[x,y]]</math> be such that <math>d^2M=0</math>, we'll prove that for some <math>E(x) \in \mathbb{Q}[[x]]</math> we have <math> d^1E = M </math>.

Write the two power series as <math> E(x) = \sum{\frac{e_i}{i!}x^i} </math> and <math> M(x,y) = \sum{\frac{m_{ij}}{i!j!}x^i y^j} </math>, where the <math>e_i</math> are the unknowns we wish to solve for.

The coefficient of <math>x^i y^j z^k</math> in <math> M(y,z)-M(x+y,z)+M(x,y+z)-M(x,y)</math> is

<math> \frac{\delta_{i0} m_{jk}}{j!k!} - \binom{i+j}{i} \frac{m_{i+j,k}}{(i+j)!k!} +
\binom{j+k}{j} \frac{m_{i,j+k}}{i!(j+k)!} - \frac{\delta_{k0} m_{ij}}{i!j!}. </math>

Here, <math> \delta_{i0} </math> is a Kronecker delta: 1 if <math> i=0 </math> and 0 otherwise. Since <math> d^2 M = 0 </math>, this coefficient should be zero. Multiplying by <math> i!j!k! </math> (and noting that, for example, the first term doesn't need an <math> i! </math> since the delta is only nonzero when <math> i=0 </math>) we get <math> \delta_{i0} m_{jk} - m_{i+j,k} + m_{i,j+k} + \delta_{k0} m_{ij}</math>.

An entirely analogous procedure tells us that the equations we must solve boil down to <math> \delta_[i0} e_j - e_{i+j} + \delta_{j0} e_i = m_{ij} </math>.

The Existence of the Exponential Function

2007-01-30T03:20:34Z

Oantolin: /* Finding a Syzygy */

The Existence of the Exponential Function

2007-01-30T03:07:16Z

Oantolin: /* Finding a Syzygy */

The Existence of the Exponential Function

2007-01-30T02:57:27Z

Oantolin: /* Computing the Homology */

The Existence of the Exponential Function

2007-01-23T21:54:20Z

Oantolin: /* Computing the Homology */