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I am given this question: Let a function $ T_n(x) $ be defined by: $$ T_{n+1}(x) = \frac{1}{x} T_{n-1}(x) - \frac{2}{7} T_n(x), $$ where $ T_0(x) = 1, T_1(x) = x $.

I need to construct a recursive function for $ T_n(x) $, one way by using If and Which and another way using function overloading. 

t[n_, x_] := Which[n = 0, 1, n = 1, x, n > 1, 1/x*(t[n - 2]) - 2/7*(t[n - 1])]

First, I have made this function, however, I'm not sure how to use it such that it will store the values for $ T_1 $ and $ T_0 $, at the moment it doesn't store these and therefore won't calculate the value of $ T_n $.

Thanks!

Αλέξανδρος Ζεγγ
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4 Answers4

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Clear["Global`*"]

eqns = {t[n + 1] == 1/x t[n - 1] - 2/7 t[n], t[0] == 1, t[1] == x};

For comparison, use RSolve to obtain the general solution

sol = RSolve[eqns, t, n][[1]];

t[n] /. sol

(* (1/(2 Sqrt[
 49 + x]))7^-n (-Sqrt[x] (-1 - Sqrt[49 + x]/Sqrt[x])^n - 
   7 x^(3/2) (-1 - Sqrt[49 + x]/Sqrt[x])^n + 
   Sqrt[49 + x] (-1 - Sqrt[49 + x]/Sqrt[x])^n + 
   Sqrt[x] (-1 + Sqrt[49 + x]/Sqrt[x])^n + 
   7 x^(3/2) (-1 + Sqrt[49 + x]/Sqrt[x])^n + 
   Sqrt[49 + x] (-1 + Sqrt[49 + x]/Sqrt[x])^n) *)

Verifying the result

eqns /. sol // Simplify

(* {True, True, True} *)

Defining with Which

Clear[t]

t[n_Integer?NonNegative] := t[n] =
  Which[
   n == 0, 1,
   n == 1, x,
   n > 1, 1/x*(t[n - 2]) - 2/7*(t[n - 1])]

m = 8;

seq = t /@ Range[m] // Simplify

(* {x, 1/x - (2 x)/7, 
 1 - 2/(7 x) + (4 x)/49, -(4/7) + 1/x^2 + 4/(49 x) - (8 x)/343, (-1372 + 
  2345 x + 588 x^2 + 16 x^3)/(
 2401 x^2), -(32/343) + 1/x^3 + 12/(49 x^2) - 2042/(2401 x) - (32 x)/
  16807, (-100842 + 106673 x + 57400 x^2 + 3920 x^3 + 64 x^4)/(
 117649 x^3), -((-823543 - 403368 x + 913752 x^2 + 191632 x^3 + 9408 x^4 + 
   128 x^5)/(823543 x^4))} *)

Comparing the results with the general solution from RSolve

seq == ((t /. sol) /@ Range[m]) // Simplify

(* True *)

Using FindSequenceFunction to generalize from the sequence

sol2 = FindSequenceFunction[seq, n]

(* (1/(2 Sqrt[
 49 + x]))7^-n (-Sqrt[x] (-1 - Sqrt[49 + x]/Sqrt[x])^n - 
   7 x^(3/2) (-1 - Sqrt[49 + x]/Sqrt[x])^n + 
   Sqrt[49 + x] (-1 - Sqrt[49 + x]/Sqrt[x])^n + 
   Sqrt[x] (-1 + Sqrt[49 + x]/Sqrt[x])^n + 
   7 x^(3/2) (-1 + Sqrt[49 + x]/Sqrt[x])^n + 
   Sqrt[49 + x] (-1 + Sqrt[49 + x]/Sqrt[x])^n) *)

Comparing the RSolve result with the FindSequenceFunction result

(t /. sol)[n] == sol2

(* True *)
Bob Hanlon
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4

You can find the exact solution directly:

RSolve[{t[n + 1] == t[n - 1]/x - 2/7 t[n], t[0] == 1, t[1] == x}, t[n], n] // FullSimplify

(*    {{t[n] -> (7^-n (Sqrt[49 + x] (-1 - 1/Sqrt[x/(49 + x)])^n - 
                 Sqrt[x] (1 + 7 x) (-1 - 1/Sqrt[x/(49 + x)])^n + 
                 Sqrt[49 + x] (-1 + 1/Sqrt[x/(49 + x)])^n + 
                Sqrt[x] (1 + 7 x) (-1 + 1/Sqrt[x/(49 + x)])^n))/(2 Sqrt[49 + x])}}    *)
Roman
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2

RSolve might be the best worker for this job and also do not forget RecurrenceTable.

But here I offer a solution via the method of transfer matrix. Denoting $ v_n = (T_{n-1}\quad T_n)^\top $ so that $ v_1 = (1\quad x)^\top $, and the transfer matrix $ M $, $$ M = \begin{pmatrix} 0 & 1 \\ 1/x & -2/7 \end{pmatrix}, $$ the recursive relation can be rewritten as $$ v_{n+1} = M\cdot v_n = M^2\cdot v_{n-1} = \cdots = M^n\cdot v_1. $$

So now programming gets involved, the final expression of $ T_n $ can be directly obtained by MatrixPower:

enter image description here

Αλέξανδρος Ζεγγ
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0
T[0, x] = 1;
T[1, x] = x;
T[n_, x_] := T[n, x] = 1/x T[n - 2, x] - 2/7 T[n - 1, x]

T[4, x] // Simplify
Cesareo
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