If f0x - x2 and fn -1 x - f0fnx for n - 0- 1- 2- -Then- find the value of log 2 f 5 2-

f_0=x^2 f_1=f_0(f_0) =x^4 =x^2^2 f_2=f_0(f_0(f_0))) =(x^4)^2=x^8 =x^2^3 Hence f_n(x)=x^2^n+1 Now nbsp; f_5(x)=x^2^6 ...

You visited us 2 times! Enjoying our articles? Unlock Full Access!

Definition of Functions

If f0x = x2...

Question

If $f_{0} (x) = x^{2}$ and $f_{n + 1} (x) = f_{0} (f_{n} (x))$ for $n = 0, 1, 2, \dots$
Then, find the value of ${log}_{2} f_{5} (2)$ .

Open in App

Solution

Suggest Corrections

Similar questions

x \in R, x \neq 0, x \neq 1

f_{0} (x) = \frac{1}{1 - x}

f_{n + 1} (x) = f_{0} (f_{n} (x)), n = 0, 1, 2, . . . .

f_{100} (3) + f_{1} (\frac{2}{3}) + f_{2} (\frac{3}{2})

f_{0} (x) = \frac{x}{(x + 1)}

f_{n + 1} = f_{0} \circ f_{n} (x)

n = 0, 1, 2, \dots

f_{n} (x)

f_{n - 1} (x) = ln (f_{n} (x)) \forall n \in N

f_{0} (x) = x - 1

\frac{d}{d x} (f_{n} (x))

f_{n - 1} (x) = ln (f_{n} (x)) \forall n \in N

f_{0} (x) = x - 1

\frac{d}{d x} (f_{n} (x))

⟨ f_{0} (x), f_{1} (x), f_{2} (x), \dots ⟩

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)

n \geq 1

f_{n} (x)

Join BYJU'S Learning Program