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Taylor polynomial. System of equations for the coefficients.
#1
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
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
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:
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.
Reply


Messages In This Thread
Taylor polynomial. System of equations for the coefficients. - by marraco - 04/30/2015, 03:24 AM

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