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Math Help - function equation functions

  1. #1
    rcs
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    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.
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  2. #2
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    Quote Originally Posted by rcs View Post
    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) = ....
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    Quote Originally Posted by rcs View Post
    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?
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  4. #4
    rcs
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    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
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    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}<br />
f(1) &=& 5 \\<br />
f(2) &=&2(5)+1 &=& 11\\<br />
f(3) &=& 2(11)+1 &=& 23 \\<br />
f(4) &=& 2(23)+1 &=& 47 \\<br />
f(5) &=& 2(47)+1 &=& 95 \\<br />
f(6) &=& 2(95)+1 &=& 191 \\<br />
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

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  7. #7
    rcs
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    thanks you so so much Soroban!!! ilove you all here! i can now have a good night sleep
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