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Thread: Proving inequality for all values greater than n0

  1. #1
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    Question Proving inequality for all values greater than n0

    Hi,

    I hope someone can help.

    I'm trying to figure out how to formally prove that $$\log_2(n) - \log_{10}(n) \geq \frac{\log_2(n)}{2^n}$$

    I know intuitively that the right-hand side will become 0 for very large values of n, and the left-hand side will be increasing, but I am not sure how to prove this. Any ideas?


    Below I have the following definitions which may or not be helpful in proving:

    increasing function: $$\forall x_1, x_2 \in D, x_1 < x_2 \Rightarrow h(x_1) < h(x_2)$$
    delta-epsilon definition as x approaches infinity: $$\forall \epsilon > 0, \exists M \in \mathbb{R}, x > M \Rightarrow |f(x) - L| < \epsilon$$

    (Also, sorry if the title of this post was not that descriptive. I did not once mention an $n_0$ in my question here so I apologize!)
    Last edited by otownsend; Oct 30th 2018 at 08:19 PM.
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  2. #2
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    Re: Proving inequality for all values greater than n0

    No, the right hand side does NOT "become 0 for very large values of n". It's limit, as n goes to infinity, is 0 but it is never 0 no matter how large n is.

    My first thought is to write $\displaystyle log_{10}(n)$ as $\displaystyle \frac{log_2(n)}{log_2(10)}$ so the putative inequality is $\displaystyle log_2(n)- \frac{log_2(n)}{log_2(10)}\ge \frac{log_2(n)}{2^n}$. Now, for n= 1, all the logarithms are 0 so we have equality, 0= 0. For n larger than 1, $\displaystyle log_2(n)$ is positive so we can divide both sides by it leaving $\displaystyle 1- \frac{1}{log_2(10)}\ge \frac{1}{2^n}$. You should have no problem with that.
    Thanks from topsquark
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