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Math Help - Sequence and bounds (2)

  1. #1
    Member akhayoon's Avatar
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    Sequence and bounds (2)

    can some one prove to me why

    (1.2)^{-n} is a decreasing sequence?

    I know it's obvious but in my textbook they prove it by finding the derivative

    which should be \frac{-(1.2)^{-n}}{ln1.2} which is increasing though right?
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  2. #2
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    Surely the derivative is just (-n)(1.2)^{-n-1}? Which is negative for n>0, hence decreasing? Unless I'm misreading the question as I've no clue how logs appeared...

    But your derivative is also always negative; either way it's decreasing.
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  3. #3
    Member akhayoon's Avatar
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    I don't think u could do that...or I'm pretty sure...because the variable is a power not a base
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    Oh right that was stupid of me, sorry! I redid it and got the same as you. log(1.2) is positive and (1.2)^n is also positive, so you definitely have a negative answer which shows decreasing. I don't see the problem.
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  5. #5
    Member akhayoon's Avatar
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    no but (1.2)^{-n} gets thrown to the denominator right? and theres already a negative sign in front of it
    doesn't that make it increase to 0?
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  6. #6
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    Yes as n tends to infinity the derivative tends to 0, but it is never actually 0, it just gets closer and closer. Any value of n you choose will give a negative answer.
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  7. #7
    Grand Panjandrum
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    Quote Originally Posted by akhayoon View Post
    can some one prove to me why

    (1.2)^{-n} is a decreasing sequence?

    I know it's obvious but in my textbook they prove it by finding the derivative

    which should be \frac{-(1.2)^{-n}}{ln1.2} which is increasing though right?
    Your sequence is obtained by sampling f(x)=1.2^{-x} this function is decreasing because its derivative is negative. Because the function is decreasing the sequence obtained by sampling it at 1, 2, .. is also decreasing.

    (you cannot differentiate a sequence because the index is a discrete variable and you need a continuous variable to differentiate)

    RonL
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