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Math Help - how to find the function

  1. #1
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    how to find the function

    if f(1)=1 and f(n+1)=2f(n)+1, if n>=1 then f(n)=
    if i put some values of n then i got the answer but is there is any independent way of finding the answer.
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  2. #2
    Senior Member abhishekkgp's Avatar
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    Quote Originally Posted by ayushdadhwal View Post
    if f(1)=1 and f(n+1)=2f(n)+1, if n>=1 then f(n)=
    if i put some values of n then i got the answer but is there is any independent way of finding the answer.
    f(2)=2f(1)+1
    f(3)=2f(2)+1
    .
    .
    .
    f(n+1)=2f(n)+1.
    multiply the second last equation by 2, the third last equation by 4 , the fourth last equation by 8 and so on... then add all the equations.
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  3. #3
    Behold, the power of SARDINES!
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    I am not sure how else to solve this problem but to use the guess that the solution is of the form

    f(n)=Ar^n

    If you plug this into the homogenous equation (get rid of the 1) we get

    r^n+1=2r^n \iff r^{n}(r-2)=0 \implies r=2

    Since the nonhomogenous term is a constant we guess that the particular solution is also a constant so we have that

    f(n)=A2^n+B if we plug this in we get

    A2^{n+1}+B=2(Ar^n+B)+1 \iff A2^{n+1}+B=A2^{n+1}+2B+1 \iff B=-1

    So we get

    f(n)=A2^{n}-1

    Finally if we use the initial condition we get that

    f(1)=2A-1=1 \implies A=1


    f(n)=2^{n}-1
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