Solving x5 in Terms of x Using Magma

[chhitiz]

if x is an integer,
x₁=6x²+8x+2
x₂=6x²₁+8x₁+2
x₃=6x²₂+8x₂+2
x₄=6x²₃+8x₃+2
x₅=6x²₄+8x₄+2
if i want to see x₅ in terms of x, can it it done using magma?
if so can somebody please give me the code?
thanks.

 

[matt grime]

Yes. Define a poly ring, define f to be 6x^2 + 8x+2, then evaluate f(f(f(f(f(x)))))

 

[chhitiz]

i am totally ignorant of the code,and have absolutely no idea how to define a polyring. please, can somebody tell me the code?

 

[HallsofIvy]

What, you want to solve a problem by pouring molten rock on it? How do you turn in the solution to your teacher?

 

[matt grime]

If you don't know how to use magma, then why do you want to use magma? Start with the magma documentation if you want magma help.

The answer can be worked out by hand easily (if laboriously).

If you want to learn magma this seems like a strange way to go about it.

 

[squidsoft]

if x is an integer,
x₁=6x²+8x+2
x₂=6x²₁+8x₁+2
x₃=6x²₂+8x₂+2
x₄=6x²₃+8x₃+2
x₅=6x²₄+8x₄+2
if i want to see x₅ in terms of x, can it it done using magma?
if so can somebody please give me the code?
thanks.

This is Mathematica code (and output). Hopefully you can convert it to the hot stuff.

In[10]:= Expand[Nest[6*#1^2 + 8*#1 + 2 & , x, 5]]

Out[10]= 3074457345618258602 + 
   147573952589676412928*x + 
   3431094397709976600576*x^2 + 
   51466415965649649008640*x^3 + 
   559697273626439932968960*x^4 + 
   4701457098462095436939264*x^5 + 
   31734835414619144199340032*x^6 + 
   176808368738592374824894464*x^7 + 
   828789228462151756991692800*x^8 + 
   3315156913848607027966771200*x^9 + 
   11437291352777694246485360640*x^10 + 
   34311874058333082739456081920*x^11 + 
   90068669403124342191072215040*x^12 + 
   207850775545671558902474342400*x^13 + 
   423124793075117102051465625600*x^14 + 
   761624627535210783692638126080*x^15 + 
   1213839250134242186510142013440*x^16 + 
   1713655411954224263308435783680*x^17 + 
   2142069264942780329135544729600*x^18 + 
   2367550240199915100623496806400*x^19 + 
   2308361484194917223107909386240*x^20 + 
   1978595557881357619806779473920*x^21 + 
   1483946668411018214855084605440*x^22 + 
   967791305485446661862011699200*x^23 + 
   544382609335563747297381580800*x^24 + 
   261303652481070598702743158784*x^25 + 
   105526475040432357168415506432*x^26 + 
   35175491680144119056138502144*x^27 + 
   9422006700038603318608527360*x^28 + 
   1949380696559711031436247040*x^29 + 
   292407104483956654715437056*x^30 + 
   28297461724253869811171328*x^31 + 
   1326443518324400147398656*x^32

 

[chhitiz]

Yes. Define a poly ring, define f to be 6x^2 + 8x+2, then evaluate f(f(f(f(f(x)))))

ok, i got how to declare a polyring and defined f<x> :=6*x^2+8*x+2;
but how to evaluate f(f(f(f(x))));