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PostPosted: Sat, 24 Apr 2010 00:11:20 UTC 
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Let T map the space of 2nd degree polynomials into itself by the rule:

T(y)= a + (2*a+b)x + (2a+b+c)x^2 where y=ax^2+bx+c.

Informally, T maps y to the sum of itself, its first derivative, and its second derivative.
Find an explicit formula for T^n(y), i.e. the composition of T n times applied to y.

Spoiler:
Answer: T^n(y)=ax^2+(2na+b)x+((n+1)!a + nb + c)x^2

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PostPosted: Sat, 24 Apr 2010 03:25:49 UTC 
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So?

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PostPosted: Sat, 24 Apr 2010 12:17:23 UTC 
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srjcstud wrote:
T(y)= a + (2*a+b)x + (2a+b+c)x^2 where y=ax^2+bx+c.

Informally, T maps y to the sum of itself, its first derivative, and its second derivative.

Did you mean T(y)=ax^2 + (2*a+b)x + (2a+b+c)?


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PostPosted: Sat, 24 Apr 2010 19:28:31 UTC 
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I guess you are talking about the map
T\colon ax^2+bx+c \mapsto ax^2 + (2a+b)x + (2a+b+c)
since this is what you wrote informally.

This map can be described as linear map (a,b,c)\mapsto (a,2a+b,2a+b+c), or
in the matrix form
(a,b,c)\mapsto(a,b,c)\begin{pmatrix}
1 & 2 & 2\\
0 & 1 & 1\\
0 & 0 & 1
\end{pmatrix}=(a,b,c)A.

So your question is equivalent to computing A^n. I tried this, however, my result is different from yours...?

Spoiler:
A=I+B where
B=\begin{pmatrix}
0 & 2 & 2\\
0 & 0 & 1\\
0 & 0 & 0
\end{pmatrix}. We have
B^2=\begin{pmatrix}
0 & 0 & 6\\
0 & 0 & 0\\
0 & 0 & 0
\end{pmatrix}
and B^3=0.
From binomial theorem (which can be used, since every matrix commutes with I) we get
A^n=(I+B)^n=I+nB+\binom{n}2B^2=
\begin{pmatrix}
  1 & 2n & 3n(n-1) \\
  0 & 1 & n \\
  0 & 0 & 1 
\end{pmatrix}
then T^n(y)=ax^2+(2n+b)x+[3n(n-1)a+nb+c].


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PostPosted: Sun, 25 Apr 2010 18:28:00 UTC 
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I get a different matrix A for the transformation.

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