11/05/2007, 07:28 AM

Gottfried,

I really need to thank you for opening my eyes up to a different way of viewing this problem! I was on a flight across the U.S. (from California to Pennsylvania), and then a 2-hour drive to my final destination, and I was thinking over these matrices the whole time. And then it finally made sense!

For me, it has been like magic that Andrew's solution works. I mean, I understood the derivation of the matrix from complicated derivatives and equations, etc., but it just seemed too much like a coincidence. I couldn't understand why it worked, only that it did seem to work.

But now I can "see" it!

It's simply a generic matrix solver for the Abel function. Assume we have a function , with an associated power series . Write the coefficients of the power series as a column vector, .

Now, to compose two power series, we can use basic matrix math. Start with the "inside" function, and write all the integer powers of the series as columns. By powers, I'm talking about good old fashioned polynomial multiplication.

Now, each column represents the function to a given power, and we can use each column in place of a variable. Multiply the matrix by the column vector for the "outside" series, and you have the composition.

For e^x, let's call the power series E(x), with column vector form E~.

Then, the matrix for the powers of E(x) is:

[E~^0, E~^1, E~^2, ...]

For example truncating the power series for e^x at the 6th term, we have E=[1, 1, 1/2, 1/6, 1/24, 1/120]~, and the matrix is then given by:

This matches the matrix B you have. The easy way to see why this works is to consider that (e^x)^2 is e^(2x), or in general, (e^x)^k is e^(kx). Since the terms of the power series for exp are (x^n)/n!, the terms of the powers of the power series would be (kx)^n/n!, with coefficients in the matrix being simply (k^n)/n!, Here, k is the column and n is the row (both zero-based).

Now comes the interesting part. Since F(e^x) = F(x)+1, we can solve for F(e^x)-F(x) = 1

Given P_F as the power series for F, we can find F(e^x) by multiplying , and then subtract , where I is the identity matrix. Set this equal to a column vector of [1, 0, 0, ..]~, and then solve for :

This is exactly what Andrew's matrix does (with the exception of explicitly removing the first column; see below). And now I see exactly why it should work, assuming A) that the infinite system has a unique solution, and B) that the partial solutions converge on this unique solution as we increase the matrix size.

It took me a while to see why it works. I was in the car at this point, so I was having to do this in my head.

Essentially, we subtract 1 from the diagonal of B. This simulates finding F(e^x)-F(x). We set the matrix times an unknown column vector equal to "1", which is a column vector with a 1 in the first row.

Now, the top-left entry of the matrix will become 0, since we subtracted 1 from 1. This makes sense, because if F(e^x)-F(x)=1, then we don't know what the constant term is. So we just chop off that first column (associated with the constant), and in our P_F column vector, we make note that the first entry will now correspond to the first power of x.

Then, to see that this equals Andrew's solution, we pre-multiply by the diagonal factorial matrix. This gives us the Vandermonde matrix, ZV as Gottfried writes it, with factorials subtracted from the subdiagonal. It's the subdiagonal, because it was originally the diagonal, but we chopped off the first column.

We even get the factor of the log of the base to successively higher factors. Going back to the power series for exp, if we used base 2 instead, for example, then each row would have be divided by a higher power of ln(2). We subtract 1's from the diagonal, then multiply by the diagonal factorial matrix, and multiply by a diagonal matrix of powers of ln(2), to get back to the ZV matrix, with factorials times powers of ln(2) subtracted from the subdiagonal.

I'm quite excited, because assuming this is correct, it gives us a template for solving the continuous iteration of tetration as well. Instead of column vectors as powers of the exponential, we could use column vectors as the powers of sexp, and by doing so, find a "pentalog" such that S(sexp(x))-S(x)=1. I haven't tried, so I'm just making a conjecture here.

In fact, I'm wondering if this is a generic method for solving Abel functions, given a well-defined power series of the iterating function.

Although, now that I think about it, I wonder if someone has covered this before in this forum, and I didn't understand it at the time, so I had to derive it on my own to understand it...

I really need to thank you for opening my eyes up to a different way of viewing this problem! I was on a flight across the U.S. (from California to Pennsylvania), and then a 2-hour drive to my final destination, and I was thinking over these matrices the whole time. And then it finally made sense!

For me, it has been like magic that Andrew's solution works. I mean, I understood the derivation of the matrix from complicated derivatives and equations, etc., but it just seemed too much like a coincidence. I couldn't understand why it worked, only that it did seem to work.

But now I can "see" it!

It's simply a generic matrix solver for the Abel function. Assume we have a function , with an associated power series . Write the coefficients of the power series as a column vector, .

Now, to compose two power series, we can use basic matrix math. Start with the "inside" function, and write all the integer powers of the series as columns. By powers, I'm talking about good old fashioned polynomial multiplication.

Now, each column represents the function to a given power, and we can use each column in place of a variable. Multiply the matrix by the column vector for the "outside" series, and you have the composition.

For e^x, let's call the power series E(x), with column vector form E~.

Then, the matrix for the powers of E(x) is:

[E~^0, E~^1, E~^2, ...]

For example truncating the power series for e^x at the 6th term, we have E=[1, 1, 1/2, 1/6, 1/24, 1/120]~, and the matrix is then given by:

Code:

`1 1 1 1 1 1`

0 1 2 3 4 5

0 1/2 2 9/2 8 25/2

0 1/6 4/3 9/2 32/3 125/6

0 1/24 2/3 27/8 32/3 625/24

0 1/120 4/15 81/40 128/15 625/24

This matches the matrix B you have. The easy way to see why this works is to consider that (e^x)^2 is e^(2x), or in general, (e^x)^k is e^(kx). Since the terms of the power series for exp are (x^n)/n!, the terms of the powers of the power series would be (kx)^n/n!, with coefficients in the matrix being simply (k^n)/n!, Here, k is the column and n is the row (both zero-based).

Now comes the interesting part. Since F(e^x) = F(x)+1, we can solve for F(e^x)-F(x) = 1

Given P_F as the power series for F, we can find F(e^x) by multiplying , and then subtract , where I is the identity matrix. Set this equal to a column vector of [1, 0, 0, ..]~, and then solve for :

This is exactly what Andrew's matrix does (with the exception of explicitly removing the first column; see below). And now I see exactly why it should work, assuming A) that the infinite system has a unique solution, and B) that the partial solutions converge on this unique solution as we increase the matrix size.

It took me a while to see why it works. I was in the car at this point, so I was having to do this in my head.

Essentially, we subtract 1 from the diagonal of B. This simulates finding F(e^x)-F(x). We set the matrix times an unknown column vector equal to "1", which is a column vector with a 1 in the first row.

Now, the top-left entry of the matrix will become 0, since we subtracted 1 from 1. This makes sense, because if F(e^x)-F(x)=1, then we don't know what the constant term is. So we just chop off that first column (associated with the constant), and in our P_F column vector, we make note that the first entry will now correspond to the first power of x.

Then, to see that this equals Andrew's solution, we pre-multiply by the diagonal factorial matrix. This gives us the Vandermonde matrix, ZV as Gottfried writes it, with factorials subtracted from the subdiagonal. It's the subdiagonal, because it was originally the diagonal, but we chopped off the first column.

We even get the factor of the log of the base to successively higher factors. Going back to the power series for exp, if we used base 2 instead, for example, then each row would have be divided by a higher power of ln(2). We subtract 1's from the diagonal, then multiply by the diagonal factorial matrix, and multiply by a diagonal matrix of powers of ln(2), to get back to the ZV matrix, with factorials times powers of ln(2) subtracted from the subdiagonal.

I'm quite excited, because assuming this is correct, it gives us a template for solving the continuous iteration of tetration as well. Instead of column vectors as powers of the exponential, we could use column vectors as the powers of sexp, and by doing so, find a "pentalog" such that S(sexp(x))-S(x)=1. I haven't tried, so I'm just making a conjecture here.

In fact, I'm wondering if this is a generic method for solving Abel functions, given a well-defined power series of the iterating function.

Although, now that I think about it, I wonder if someone has covered this before in this forum, and I didn't understand it at the time, so I had to derive it on my own to understand it...

~ Jay Daniel Fox