1. ## function equation functions

Let f be a function de ned on the set of integers such that f(1) = 5
and f(x + 1) = 2f(x) + 1 for all integers x. What is the value of
f(7) - f(0)?

i bet the answer is 381 but i couldn't make how to the solution is done.

2. Originally Posted by rcs
Let f be a function de ned on the set of integers such that f(1) = 5
and f(x + 1) = 2f(x) + 1 for all integers x. What is the value of
f(7) - f(0)?

i bet the answer is 381 but i couldn't make how to the solution is done.
Since you know f(1) you can calculate f(2). Knowing f(2) you can calculate f(3) etc. So you can get f(7). And obviously f(1) = 2f(0) + 1 => f(0) = ....

3. Originally Posted by rcs
Let f be a function dened on the set of integers such that f(1) = 5
and f(x + 1) = 2f(x) + 1 for all integers x. What is the value of
f(7) - f(0)?

i bet the answer is 381 but i couldn't make how to the solution is done.
How did you get 381? A guess?

4. a friend of mine told me that it is 381,,, but she couldnt show to me how she solved it,,, that is why i just guessed it. hope i could get the right flow of the solution. thanks

5. Hello, rcs!

If you can read the problem, you can baby-talk your way through it.

$\text{Let }f(x)\text{ be a function de{f}ined on the set of integers}$
$\text{such that: }\:f(1) = 5\,\text{ and }\,f(x\!+\!1) \:=\: 2\!\cdot\!f(x) + 1\,\text{ for all integers }x.$

$\text{What is the value of: }\:f(7) - f(0)\,?$

The first term is 5.
Each subsequent term is twice the preceding term, plus 1.

We have:

. . $\begin{array}{ccccc}
f(1) &=& 5 \\
f(2) &=&2(5)+1 &=& 11\\
f(3) &=& 2(11)+1 &=& 23 \\
f(4) &=& 2(23)+1 &=& 47 \\
f(5) &=& 2(47)+1 &=& 95 \\
f(6) &=& 2(95)+1 &=& 191 \\
f(7) &=& 2(192)+1 &=& 383 \end{array}$

What is $f(0)$ ?

Since $f(1) \:=\:2\!\cdot\!f(0) + 1 \:=\:5$, we have: . $f(0) = 2$

Therefore: . $f(7) - f(0) \;=\;383 -2 \;=\;381$

6. thanks you so so much Soroban!!! ilove you all here! i can now have a good night sleep