For the few rare souls who actually have visited this blog more than once. Why on earth did you come back? The content of this blog is perhaps what you might aptly call eclectic, yet another apt title might be completely disordered badly written ruminations of a crazed and unstable mind. Because of this I would be willing to bet that the thing that gives cogency to the group that is my readers is none other than personal acquaintance which lends interest to the blog even though. The content is terrible. Is this the case? Please tell me in the comments.

P.S. The title of this post is intended to imply that I am acting as a sort of content tyrant. I cannot stop being the content dictator but if the content is bad then that makes me the content tyrant since my rule is unjust. So give me some feedback and hopefully I can become the benevolent dictator instead of the unjust tyrant.

## Tuesday, October 28, 2008

### Do You Know Me Personally?

The poll with that as the question has finally come to a close after a year of polling.

The poll got a total of 22 votes which I hope is less than the number of people that actually visited the blog in a year. Still 22 people isn't bad and I am particularly pleased that only 7 (thats 7/22 or approximately 100/pi % thats 31% ish) of them claimed to know me while the majority 15 (68% ish) of them claimed to not know me at all. This is heartening, since it means my blog is not entirely invisible on the web. Furthermore it means that the blog has more pull than merely the pull afforded by personal acquaintance with this nut.

The poll got a total of 22 votes which I hope is less than the number of people that actually visited the blog in a year. Still 22 people isn't bad and I am particularly pleased that only 7 (thats 7/22 or approximately 100/pi % thats 31% ish) of them claimed to know me while the majority 15 (68% ish) of them claimed to not know me at all. This is heartening, since it means my blog is not entirely invisible on the web. Furthermore it means that the blog has more pull than merely the pull afforded by personal acquaintance with this nut.

## Tuesday, October 7, 2008

### Fractional Dimensional Spaces and Fourier's Trick

One doesn't usually think of the existence of the standard basis for euclidean vector spaces as related to fourier series. It is easy to show that there is a rather direct correspondence. If we take a slightly non standard method of providing a coordinate system to the points in R

Now consider what happens when we take our above functions mapped to points. We can obtain the dot product of the two points as the integral of the product of the functions associated to the points over the unit sphere of vectors. If the space under consideration is integer dimensional space then relatively simple exercises in linear algebra show that any point can be expressed as the linear combination of d linearly independent vectors (d being the dimension of the space). So (as is kind of obvious) if a (finite) standard basis exists then the dimension of the space is integer.

This is all based on the fact that the space has a vector space structure set up on it. I posit that vector space structure does not in itself imply the dimension of the space to be of integer dimension. If this is true then it means that any non-integer dimensional space would have to have an infinite basis

^{n}we let every point p be labeled with a the function defined using the usual dot product as f_{p}(x) = p * x where * is the vector dot product and x is any other point in R^{n}Now to obtain the dot product of two points we express each point as a linear combination of the standard basis vectors and then take the sum of the products of the corresponding coefficients in the standard basis expansion. More to the point if you take the dot product of one of the basis vectors and any point you obtain the coordinate of that point with respect to that basis vector. Thus the dot product of two points is the sum over the standard basis of the products of the dot products of the two points under consideration and the i'th vector in the standard basis.Now consider what happens when we take our above functions mapped to points. We can obtain the dot product of the two points as the integral of the product of the functions associated to the points over the unit sphere of vectors. If the space under consideration is integer dimensional space then relatively simple exercises in linear algebra show that any point can be expressed as the linear combination of d linearly independent vectors (d being the dimension of the space). So (as is kind of obvious) if a (finite) standard basis exists then the dimension of the space is integer.

This is all based on the fact that the space has a vector space structure set up on it. I posit that vector space structure does not in itself imply the dimension of the space to be of integer dimension. If this is true then it means that any non-integer dimensional space would have to have an infinite basis

## Thursday, October 2, 2008

### Problem Solving

At least once in a while I should post something with some personal interest. It is strange how reading something which is happening in someones life that you don't even know can be entertaining. I suppose what is on my mind at the moment is problem solving.

There is an undergraduate problem solving contest here at the U where a new problem is posed once a month during the school year. People submit solutions at the end of the month and are assigned points based on their solutions to the problems 3 points for a correct one and +e points if you win. I have found that I can solve a significant fraction of the problems but that also sometimes I spend dozens of hours working on a problem only to be unable to solve it. This last problem really made me angry when I found out how it admits a really simple solution and even one that I sort of thought about before. you have a 7x7 checkerboard with a black square in the corner. What single tiles can you remove in order to make the board tileable by 2x1 dominoes? The answer is you can remove any black square and not any red squares. I worked on the problem for hours and made it more and more complicated... but I couldn't figure out how to prove that removing a red tile made the board untileable. The amazingly simple thing to realize is that there is one more black square to begin with on the board and every dominoe covers a red and black square thus removing a red square leaves the board untilable since it leaves the number of black and red squares unequal. An even more elegant solution deals with finding even paths on the board which gives you both the black and the red tile cases in one go. The really infuriating thing is that I considered both the fact that every dominoe covers a red and a black tile and the fact that the existence of eulerian paths on the dual graph of the board is related to the tiling.... and yet I didn't solve the damn thing. I guess I am really just not capable of doing personal happenings rants. I just get caught up in the details of whatever little abstract thing I mention. I'll do better next time.

There is an undergraduate problem solving contest here at the U where a new problem is posed once a month during the school year. People submit solutions at the end of the month and are assigned points based on their solutions to the problems 3 points for a correct one and +e points if you win. I have found that I can solve a significant fraction of the problems but that also sometimes I spend dozens of hours working on a problem only to be unable to solve it. This last problem really made me angry when I found out how it admits a really simple solution and even one that I sort of thought about before. you have a 7x7 checkerboard with a black square in the corner. What single tiles can you remove in order to make the board tileable by 2x1 dominoes? The answer is you can remove any black square and not any red squares. I worked on the problem for hours and made it more and more complicated... but I couldn't figure out how to prove that removing a red tile made the board untileable. The amazingly simple thing to realize is that there is one more black square to begin with on the board and every dominoe covers a red and black square thus removing a red square leaves the board untilable since it leaves the number of black and red squares unequal. An even more elegant solution deals with finding even paths on the board which gives you both the black and the red tile cases in one go. The really infuriating thing is that I considered both the fact that every dominoe covers a red and a black tile and the fact that the existence of eulerian paths on the dual graph of the board is related to the tiling.... and yet I didn't solve the damn thing. I guess I am really just not capable of doing personal happenings rants. I just get caught up in the details of whatever little abstract thing I mention. I'll do better next time.

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