Help with function question $f(2x + 1) = 2f(x) + 1$ [closed]

$\begingroup$

I've been working on this question for a few hours but I can't figure out the answer.

$f(2x + 1) = 2f(x) + 1$ for all $x$ and $f(0) = 2$ what does $f(3)$ equal?

$\endgroup$ 3

3 Answers

$\begingroup$

Well, $f(3)=f(2\cdot 1+1)=2f(1)+1$. Now, $f(1)=f(2\cdot 0+1)=2f(0)+1=5$. Thus $f(3)=11$.

$\endgroup$ 2 $\begingroup$

You have $$ f(2x+1) = 2f(x)+1$$

Let $$x=0$$ to get $$f(1)= 2f(0)+1 = 2(2)+1 =5$$

Let $x=1$ to get $$f(3)= 2f(1)+1 = 2(5)+1 =11$$

$\endgroup$ $\begingroup$

iteration speed of outside function is the same as inside function just they don't start with the same initial values F"Inside" starts with initial value of 0 and f"outside" starts with the initial value of 2 so you just solve for first order non homogenous recurrence relation of the form F(c+1)=2F(c)+1, with the seeds FI(0)=0, FO(0)=2. FI(t)=2^t-1, FO(t)=3*2^t-1, putting it in the form of parametric curve we get [FI(t),FO(t)], and solving for it as form of y(x). FI inverse of tI(x) = log base 2(x+1). 2^log base 2(x+1)=x+1, 3(x+1)-1=3x+2. so your final answer is 3x+2. checking 3(2x+1)+2=6x+3+2=6x+5. 2(3x+2)+1=6x+4+1=6x+5.

$\endgroup$ 1