This is a numerical example for base a=e
I had shown how to get 2 sides of a system of equations for solving the coefficients. One side is blue (the linear one), and the other is red (nonlinear, and I still don't know the pattern to generate the complete system of equations. I only got up to x^7, and I can get more, but of course, what I need is a more general expression.
For base e, you get a system of equations.
you get this blue side for 7 coefficients (it is independent of the base):
and this is the red side:
\( {\color{Red} {
e^{^xe}=
e \,+\,
e.a_1 \,.\, x \,+\,
(e.a_2 + \frac{e}{2}.a_1^2) \,.\, x^2 \,+\,
(e.a_3 + e.a_1.a_2 +\frac{e}{6}.a_1^3) \,.\, x^3 \,+\,
(e.a_4 + \frac{e}{2}.a_2^2 +e.a_3.a_1+\frac{e}{2}.a_2.a_1^2+\frac{e}{24}.a_1^4) \,.\,x^4 \\
+(e.a_5 + {e}.a_3.a_2 + e.a_1.a_4+\frac{e}{2}.a_1.a_2^2+\frac{e}{2}.a_3.a_1^2+\frac{e}{6}.a_2.a_1^3+ \frac{e}{120} .a_1^5) \,.\,x^5 \\
+(e.a_6 + \frac{e}{2}.a_3^2 + e.a_2.a_4+\frac{e}{6}.a_2^3+ {e}.a_1.a_5+ {e}.a_1.a_3.a_2+\frac{e}{2}.a_4.a_1^2+\frac{e}{4}.a_2^2 . a_1^2+\frac{e}{6}.a_3.a_1^3+\frac{e}{24}.a_2.a_1^4+\frac{e}{ 720}.a_1^6) \,.\,x^6 \\
} \)
\( {\color{Red} {
+e.(a_7 + 1^2. a_4.a_3 + a_5.a_2 + \frac{1}{2}a_3.a_2^2 + a_6.a_1 + \frac{1}{2}a_3^2.a_1 + a_4.a_2.a_1 + \frac{1}{6}a_2^3.a_1 + \frac{1}{2}a_5.a_1^2 + \frac{1}{2} a_3.a_2.a_1^2 + \frac{1}{6}a_4.a_1^3 + \frac{1}{12} a_2^2.a_1^3 + \frac{1}{24}a_3.a_1^4 + \frac{1}{120}a_2.a_1^5 + \frac{1}{5040}.a_1^7 )\,.\,x^7
} \)
So you get this systems of equations (blue to the left, and red to the right):
It is a non linear system of equations, and the solution for this particular case is:
a₀= 1,00000000000000000
a₁= 1,09975111049169000
a₂= 0,24752638354178700
a₃= 0,15046151104294100
a₄= 0,12170896032120000
a₅= 0,16084324512292400
a₆= -0,02254254634348470
a₇= -0,10318144159688800
a₈= 0,06371479195361670
Compare with Sheldonison's kneser.gp solution:
This is a comparison of the coefficients obtained from both methods. Of course I only calculated (7+1dummy) coefficients, with excel precision, and Sheldonison code did 60 coefficients with 134 bits of precision.
(I estimate that I need at least between 20 and 30 coefficients, and have no clue on how to solve it).
![[Image: fHtu7JL.jpg?1]](http://i.imgur.com/fHtu7JL.jpg?1)
![[Image: CwYobxC.jpg?1]](http://i.imgur.com/CwYobxC.jpg?1)
The advantage of this method is that it does not depend of the base, so it will be able to solve negative bases.
I had shown how to get 2 sides of a system of equations for solving the coefficients. One side is blue (the linear one), and the other is red (nonlinear, and I still don't know the pattern to generate the complete system of equations. I only got up to x^7, and I can get more, but of course, what I need is a more general expression.
For base e, you get a system of equations.
you get this blue side for 7 coefficients (it is independent of the base):
Code:
(21:23) gp > Size=9
%3 = 9
(21:24) gp > m=matrix(Size, Size,n,i,binomial(i-1,n-1))
%4 =
[1 1 1 1 1 1 1 1 1]
[0 1 2 3 4 5 6 7 8]
[0 0 1 3 6 10 15 21 28]
[0 0 0 1 4 10 20 35 56]
[0 0 0 0 1 5 15 35 70]
[0 0 0 0 0 1 6 21 56]
[0 0 0 0 0 0 1 7 28]
[0 0 0 0 0 0 0 1 8]
[0 0 0 0 0 0 0 0 1]and this is the red side:
\( {\color{Red} {
e^{^xe}=
e \,+\,
e.a_1 \,.\, x \,+\,
(e.a_2 + \frac{e}{2}.a_1^2) \,.\, x^2 \,+\,
(e.a_3 + e.a_1.a_2 +\frac{e}{6}.a_1^3) \,.\, x^3 \,+\,
(e.a_4 + \frac{e}{2}.a_2^2 +e.a_3.a_1+\frac{e}{2}.a_2.a_1^2+\frac{e}{24}.a_1^4) \,.\,x^4 \\
+(e.a_5 + {e}.a_3.a_2 + e.a_1.a_4+\frac{e}{2}.a_1.a_2^2+\frac{e}{2}.a_3.a_1^2+\frac{e}{6}.a_2.a_1^3+ \frac{e}{120} .a_1^5) \,.\,x^5 \\
+(e.a_6 + \frac{e}{2}.a_3^2 + e.a_2.a_4+\frac{e}{6}.a_2^3+ {e}.a_1.a_5+ {e}.a_1.a_3.a_2+\frac{e}{2}.a_4.a_1^2+\frac{e}{4}.a_2^2 . a_1^2+\frac{e}{6}.a_3.a_1^3+\frac{e}{24}.a_2.a_1^4+\frac{e}{ 720}.a_1^6) \,.\,x^6 \\
} \)
\( {\color{Red} {
+e.(a_7 + 1^2. a_4.a_3 + a_5.a_2 + \frac{1}{2}a_3.a_2^2 + a_6.a_1 + \frac{1}{2}a_3^2.a_1 + a_4.a_2.a_1 + \frac{1}{6}a_2^3.a_1 + \frac{1}{2}a_5.a_1^2 + \frac{1}{2} a_3.a_2.a_1^2 + \frac{1}{6}a_4.a_1^3 + \frac{1}{12} a_2^2.a_1^3 + \frac{1}{24}a_3.a_1^4 + \frac{1}{120}a_2.a_1^5 + \frac{1}{5040}.a_1^7 )\,.\,x^7
} \)
So you get this systems of equations (blue to the left, and red to the right):
Code:
[1 1 1 1 1 1 1 1 1] [ 1] [e]
[0 1 2 3 4 5 6 7 8] [a₁] [e.a₁]
[0 0 1 3 6 10 15 21 28] [a₂] [e.a₂+e/2.a₁²]
[0 0 0 1 4 10 20 35 56] [a₃] [e.a₃+e.a₁.a₂+e/6.a₁³]
[0 0 0 0 1 5 15 35 70] * [a₄] = [e.a₄+e/2.a₂²+e.a₃.a₁+e/2.a₂.a1²+e/24.a₁⁴]
[0 0 0 0 0 1 6 21 56] [a₅] [...]
[0 0 0 0 0 0 1 7 28] [a₆] [...]
[0 0 0 0 0 0 0 1 8] [a₇] [...]
[0 0 0 0 0 0 0 0 1] [a₈] [...]It is a non linear system of equations, and the solution for this particular case is:
a₀= 1,00000000000000000
a₁= 1,09975111049169000
a₂= 0,24752638354178700
a₃= 0,15046151104294100
a₄= 0,12170896032120000
a₅= 0,16084324512292400
a₆= -0,02254254634348470
a₇= -0,10318144159688800
a₈= 0,06371479195361670
Compare with Sheldonison's kneser.gp solution:
Code:
a0= 1.000000000000000000000000000000000000000000000000000000000000000
a1= 1.091767351258320991801384550027151644384731177193748032852736750
a2= 0.2714832129016945953317066836235490061739872161990598817820694029
a3= 0.2124532481762562843089676377409482685666305767555868534429427250
a4= 0.06954037613998737372867423270746944766611954537384154515961000453
a5= 0.04429195209047330440644034438551461331201912452396122603445409319
a6= 0.01473674209638939115209628691553407087036283284802524310071901416
a7= 0.008668781817225260366380392529639960019193562974233192587504340111
a8= 0.002796479398385459694825991301149597269227881323260081302040577606
a9= 0.001610631290584272072162645164026110049792896861436056957822790182
a10= 0.0004899272314843773346986672258324787189179236966986597429129832332
a11= 0.0002881810711540458113452640412964722721968311050442105340877181595
a12= 8.009461253854333344427358300999263618841918537261023130790163984 E-5
a13= 5.029114179380540369459011462420351199253676250298573251465345775 E-5
a14= 1.218379034490009161619171109859335192564698424745403501454238589 E-5
a15= 8.665533667381574685245804554105291695042724874461047071879364312 E-6
a16= 1.687782319317538991789009317583785512852144820496380916395175482 E-6
a17= 1.493253248573492581066504431732751485456118359869377142782801855 E-6
a18= 1.987607642049274553198189794968196417348167228423221823766208771 E-7
a19= 2.608673560043263731645821608532933364457829738518570518951024854 E-7
a20= 1.470995414254190186141218818247639464249470033298457423356618453 E-8
a21= 4.683449732741350625509370993006576850859342869408936466444526422 E-8
a22= -1.549241665546769521805465176448252112209562873661024180269440066 E-9
a23= 8.741510781350935912992558117122349583609707251224417742494147299 E-9
a24= -1.125787310103062317575134515738378022835159203164294437887008675 E-9
a25= 1.707959267270728412565608778729726960430408931538582375779659881 E-9
a26= -3.778583154922985176492143492500296036058409993500345462555638121 E-10
a27= 3.495778765110216317873145649935531461006488191616787742019992521 E-10
a28= -1.053770123445001506629425792917061970255456556066414419570018689 E-10
a29= 7.459097147607505280732283202189701957070179874063753558099592924 E-11
a30= -2.717598206577734869329877172492724299204938962536131010671806592 E-11
a31= 1.646076610661447130388508182175742339593708736967841194185926886 E-11
a32= -6.741873152405052999147453463677312005415532501572568628993007553 E-12
a33= 3.725328723319468544317086960688595471124134410671562760851241992 E-12
a34= -1.639087326793590223458207893420015494534981157084934821663605376 E-12
a35= 8.583638311358568060488665543255939724011799192373411667290837403 E-13
a36= -3.943738739105384313579489883439749145625771519735734916959207765 E-13
a37= 2.002523128021887055893526704591127887891070857821009841312866823 E-13
a38= -9.441962242924065023715111579990954292712006432497954997730218600 E-14
a39= 4.712054745849371340817414393412613311684787369946584046802094568 E-14
a40= -2.256291882035597080043272706155245325625189406689252333379604487 E-14
a41= 1.115468850616536996293093710617388942901924987321949988173687870 E-14
a42= -5.390745557016350491840931638954438791176540897408393713471988968 E-15
a43= 2.652158491516681872817207767927177972525297352287457523410538087 E-15
a44= -1.288910765544553681933994492897094530731798239089453308045916388 E-15
a45= 6.326678501956660453007840302627916831673424077807564985330246473 E-16
a46= -3.085457150492335988961833456292213681855982339566944345179362092 E-16
a47= 1.513176771782740527337006888790219839276363051364244735670911612 E-16
a48= -7.396534137094751433579658688811828771999777094874553108587585999 E-17
a49= 3.626987671054187604858900781140139207944144420079830647639126832 E-17
a50= -1.775725598676298403622157433145771025580236336491145640801940287 E-17
a51= 8.709879544396054644216187847273723532343829279009721890829235905 E-18
a52= -4.269289282339156322311235369711703332985224326874010775539646161 E-18
a53= 2.095044162575528110270707271710047732715299297962567107013275560 E-18
a54= -1.027883709282258792101512551665977860702273246084671587434733006 E-18
a55= 5.046824247438176461629463228342973276669090061957443080345413943 E-19
a56= -2.478050595821552304759308268057851431159467131543548274881868429 E-19
a57= 1.217394203039331646009323993543125208245611419932029287538356629 E-19
a58= -5.981648632303782947485566789252901875389559922820136772179247709 E-20
a59= 2.940264344513896261748193069655017265616156575476060556667439589 E-20This is a comparison of the coefficients obtained from both methods. Of course I only calculated (7+1dummy) coefficients, with excel precision, and Sheldonison code did 60 coefficients with 134 bits of precision.
(I estimate that I need at least between 20 and 30 coefficients, and have no clue on how to solve it).
![[Image: fHtu7JL.jpg?1]](http://i.imgur.com/fHtu7JL.jpg?1)
![[Image: CwYobxC.jpg?1]](http://i.imgur.com/CwYobxC.jpg?1)
The advantage of this method is that it does not depend of the base, so it will be able to solve negative bases.
I have the result, but I do not yet know how to get it.

