This is an attempt to get the system of equations for the calculation of the coefficients aᵢ of the Taylor polynomial developed around 0.

We want the coefficients aᵢ of this Taylor expansion:

They should match this equation:

Let's start with the left side. We compose the Taylor expansion (on the first equation) with (x+1):

We use the binomial theorem

Each term on the summation is

In resume:

It makes a super nice linear system, with determinant=1.

I bet that it is well studied and known.

His inverse matrix is the same matrix, but each element has alternate sign.

Here is the inverse of the coefficients matrix, generated with PariGP

The inverse can be directly generated with m=matrix(Size, Size,n,i,(-1)^(n+i)*binomial(i-1,n-1)); if you want to test it with Sheldonison kneser.gp, use Size=60, and get the coefficients with ai=matrix(60,1,f,c,polcoeff(xsexp,f-1)).

Now, let's deal with the right side of the first equation:

This is the Taylor expansion for a^x:

Let's compose with the Taylor polynomial at the start of the post:

(We should apply the multinomial theorem)

I made an expansion of a^x with 12 terms, replaced each x with a polynomial with 12 terms:

...and threw it to PariGp to expand. I got this 24Mb text file

http://s000.tinyupload.com/index.php?fil...4013545746

Since the series are truncated, I made some guess about what each term means:

EDIT: more terms:

This last expression seems to be related to the Taylor expansion of around x=1.

So, each summatory of coefficients, for each power of x, is in some loose sense, "the derivative" of the summatory of the precedent power of x (taking da/dx=lna.a.a₁, and daᵢ/dx=aᵢ₊₁)

I wrote "in some loose sense", because there are multiplicative constants missing.

For example, the "derivative" of lna.a.a₁ "should be" (lna.a.a₂+lna².a.a₁²), but instead is (lna.a.a₂+lna².a.a₁²/n!), where n is the power exponent of a₁ in that term.

again the "derivative" of (lna.a.a₂+lna².a.a₁²/2!) "should be" (lna.a.a₃ + 2 . lna².a.a₁.a₂ +lna³.a.a₁³/2), but instead is (lna.a.a₃ + lna².a.a₁.a₂ +lna³.a.a₁³/3!)

¿Can you find the right rule for the coefficients of each power of x?

Here is an excel worksheet, which takes the coefficients calculated with Sheldonison Kneser.gp program (for base a=Pi), and compares the blue and red equations:

http://s000.tinyupload.com/index.php?fil...7813385010

Unfortunately, this website does not allow to upload most format files, so I uploaded it o Tinyupload.com, but there is no guarantee of permanence, so it may be deleted in the future.

We want the coefficients aᵢ of this Taylor expansion:

They should match this equation:

Let's start with the left side. We compose the Taylor expansion (on the first equation) with (x+1):

We use the binomial theorem

Each term on the summation is

In resume:

It makes a super nice linear system, with determinant=1.

I bet that it is well studied and known.

His inverse matrix is the same matrix, but each element has alternate sign.

Here is the inverse of the coefficients matrix, generated with PariGP

Code:

`gp > m=matrix(Size, Size,n,i,binomial(i-1,n-1))`

(20:22) gp > m^-1

[1 -1 1 -1 1 -1 1 -1 1 -1 1 -1 1 -1 1 -1 1 -1 1 -1

[0 1 -2 3 -4 5 -6 7 -8 9 -10 11 -12 13 -14 15 -16 17 -18 19

[0 0 1 -3 6 -10 15 -21 28 -36 45 -55 66 -78 91 -105 120 -136 153 -171

[0 0 0 1 -4 10 -20 35 -56 84 -120 165 -220 286 -364 455 -560 680 -816 969

[0 0 0 0 1 -5 15 -35 70 -126 210 -330 495 -715 1001 -1365 1820 -2380 3060 -3876

[0 0 0 0 0 1 -6 21 -56 126 -252 462 -792 1287 -2002 3003 -4368 6188 -8568 11628

[0 0 0 0 0 0 1 -7 28 -84 210 -462 924 -1716 3003 -5005 8008 -12376 18564 -27132

[0 0 0 0 0 0 0 1 -8 36 -120 330 -792 1716 -3432 6435 -11440 19448 -31824 50388

[0 0 0 0 0 0 0 0 1 -9 45 -165 495 -1287 3003 -6435 12870 -24310 43758 -75582

[0 0 0 0 0 0 0 0 0 1 -10 55 -220 715 -2002 5005 -11440 24310 -48620 92378

[0 0 0 0 0 0 0 0 0 0 1 -11 66 -286 1001 -3003 8008 -19448 43758 -92378

[0 0 0 0 0 0 0 0 0 0 0 1 -12 78 -364 1365 -4368 12376 -31824 75582

Now, let's deal with the right side of the first equation:

This is the Taylor expansion for a^x:

Let's compose with the Taylor polynomial at the start of the post:

(We should apply the multinomial theorem)

I made an expansion of a^x with 12 terms, replaced each x with a polynomial with 12 terms:

Code:

`(01:40) gp > exp(lna*x)`

%1 = 1 + lna*x + 1/2*lna^2*x^2 + 1/6*lna^3*x^3 + 1/24*lna^4*x^4 + 1/120*lna^5*x^5 + 1/720*lna^6*x^6 + 1/5040*lna^7*x^7 + 1/40320*lna^8*x^8 + 1/362880*lna^9*x^9 + 1/3628800*lna^10*x^10 + 1/39916800*lna^11*x^11

Code:

`1 + lna*(1+a1*x+a2*x^2+a3*x^3+a4*x^4+a5*x^5+a6*x^6+a7*x^7+a8*x^8+a9*x^9+a10*x^10+a11*x^11) `

+1/2*lna^2*(1+a1*x+a2*x^2+a3*x^3+a4*x^4+a5*x^5+a6*x^6+a7*x^7+a8*x^8+a9*x^9+a10*x^10+a11*x^11)^2

+1/6*lna^3*(1+a1*x+a2*x^2+a3*x^3+a4*x^4+a5*x^5+a6*x^6+a7*x^7+a8*x^8+a9*x^9+a10*x^10+a11*x^11)^3

+1/24*lna^4*(1+a1*x+a2*x^2+a3*x^3+a4*x^4+a5*x^5+a6*x^6+a7*x^7+a8*x^8+a9*x^9+a10*x^10+a11*x^11)^4

+1/120*lna^5*(1+a1*x+a2*x^2+a3*x^3+a4*x^4+a5*x^5+a6*x^6+a7*x^7+a8*x^8+a9*x^9+a10*x^10+a11*x^11)^5

+1/720*lna^6*(1+a1*x+a2*x^2+a3*x^3+a4*x^4+a5*x^5+a6*x^6+a7*x^7+a8*x^8+a9*x^9+a10*x^10+a11*x^11)^6

+1/5040*lna^7*(1+a1*x+a2*x^2+a3*x^3+a4*x^4+a5*x^5+a6*x^6+a7*x^7+a8*x^8+a9*x^9+a10*x^10+a11*x^11)^7

+1/40320*lna^8*(1+a1*x+a2*x^2+a3*x^3+a4*x^4+a5*x^5+a6*x^6+a7*x^7+a8*x^8+a9*x^9+a10*x^10+a11*x^11)^8

+1/362880*lna^9*(1+a1*x+a2*x^2+a3*x^3+a4*x^4+a5*x^5+a6*x^6+a7*x^7+a8*x^8+a9*x^9+a10*x^10+a11*x^11)^9

+1/3628800*lna^10*(1+a1*x+a2*x^2+a3*x^3+a4*x^4+a5*x^5+a6*x^6+a7*x^7+a8*x^8+a9*x^9+a10*x^10+a11*x^11)^10

+1/39916800*lna^11*(1+a1*x+a2*x^2+a3*x^3+a4*x^4+a5*x^5+a6*x^6+a7*x^7+a8*x^8+a9*x^9+a10*x^10+a11*x^11)^11

...and threw it to PariGp to expand. I got this 24Mb text file

http://s000.tinyupload.com/index.php?fil...4013545746

Since the series are truncated, I made some guess about what each term means:

EDIT: more terms:

This last expression seems to be related to the Taylor expansion of around x=1.

So, each summatory of coefficients, for each power of x, is in some loose sense, "the derivative" of the summatory of the precedent power of x (taking da/dx=lna.a.a₁, and daᵢ/dx=aᵢ₊₁)

I wrote "in some loose sense", because there are multiplicative constants missing.

For example, the "derivative" of lna.a.a₁ "should be" (lna.a.a₂+lna².a.a₁²), but instead is (lna.a.a₂+lna².a.a₁²/n!), where n is the power exponent of a₁ in that term.

again the "derivative" of (lna.a.a₂+lna².a.a₁²/2!) "should be" (lna.a.a₃ + 2 . lna².a.a₁.a₂ +lna³.a.a₁³/2), but instead is (lna.a.a₃ + lna².a.a₁.a₂ +lna³.a.a₁³/3!)

¿Can you find the right rule for the coefficients of each power of x?

Here is an excel worksheet, which takes the coefficients calculated with Sheldonison Kneser.gp program (for base a=Pi), and compares the blue and red equations:

http://s000.tinyupload.com/index.php?fil...7813385010

Unfortunately, this website does not allow to upload most format files, so I uploaded it o Tinyupload.com, but there is no guarantee of permanence, so it may be deleted in the future.

I have the result, but I do not yet know how to get it.