Define a sequence of functions by f0x-1- f1x-x- - fnx -2-1-fn-1xfn-1x- for n - 1- Prove that each fnx is a polynomial with integer coefficients-

We have, f_n^2(x) f_n 1(x)f_n+1(x)=1=f_n 1^2(x) f_n 2(x)f_n(x) . ∴ f_n(x)[ f_n(x)+f_n 2(x) ]=f_n 1[ f_n 1(x)+f_n+1(x) ] .We write this as f_n 1(x)+f ...

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Byju's Answer

Standard XII

Mathematics

Definition of Functions

Define a sequ...

Question

Define a sequence $⟨ f_{0} (x), f_{1} (x), f_{2} (x), \dots ⟩$ of functions by $f_{0} (x) = 1, f_{1} (x) = x, {(\begin{matrix} f_{n} (x) \end{matrix})}^{2} - 1 = f_{n + 1} (x) f_{n - 1} (x)$ , for $n \geq 1$ .
Prove that each $f_{n} (x)$ is a polynomial with integer coefficients.

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Solution

We have,
$f_{n}^{2} (x) - f_{n - 1} (x) f_{n + 1} (x) = 1 = f_{n - 1}^{2} (x) - f_{n - 2} (x) f_{n} (x)$ .
$∴ f_{n} (x) (\begin{matrix} f_{n} (x) + f_{n - 2} (x) \end{matrix}) = f_{n - 1} (\begin{matrix} f_{n - 1} (x) + f_{n + 1} (x) \end{matrix})$ .
We write this as
$\frac{f_{n - 1} (x) + f_{n + 1} (x)}{f_{n} (x)} = \frac{f_{n - 2} (x) + f_{n} (x)}{f_{n - 1} (x)}$ .
Using induction, we get
$\frac{f_{n - 1} (x) + f_{n + 1} (x)}{f_{n} (x)} = \frac{f_{0} (x) + f_{2} (x)}{f_{1} (x)}$ .
Observe that
$f_{2} (x) = \frac{f_{1}^{2} (x) - 1}{f_{0} (x)} = x^{2} - 1$ .
Hence
$\frac{f_{n - 1} (x) + f_{n + 1} (x)}{f_{n} (x)} = \frac{1 + (x^{2} - 1)}{x} = x$ .
Thus we obtain
$f_{n + 1} (x) = x f_{n} (x) - f_{n - 1} (x)$ .
Since $f_{0} (x), f_{1} (x) a n d f_{2} (x)$ are polynomials with integer coefficients, induction again shows that $f_{n} (x)$ is a polynomial with integer coefficients.
Note: We can get $f_{n} (x)$ explicitly:
$f_{n} (x) = x^{n} - (\begin{matrix} n - 1 1 \end{matrix}) x^{n - 2} + (\begin{matrix} n - 2 2 \end{matrix}) x^{n - 4} - (\begin{matrix} n - 3 3 \end{matrix}) x^{n - 6} + \dots$

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Define a sequence ⟨f0(x),f1(x),f2(x),…⟩ of functions by f0(x)=1,f1(x)=x, (fn(x))2−1=fn+1(x)fn−1(x), for n≥1. Prove that each fn(x) is a polynomial with integer coefficients.

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Prove that each $f_{n} (x)$ is a polynomial with integer coefficients.