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2013-08-04T02:59:18.863
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2017-09-06T21:42:24.247
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2017-09-06T21:42:24.247
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Here's my attempt: A function, <code>f(n)</code> is <code>O(n)</code>, if and only if there exists a constant, <code>c</code>, such that <code>f(n) <= c*g(n)</code>. Using this definition, could we say that the function <code>f(2^(n+1))</code> is <code>O(2^n)</code>? <ol> <li>In other words, does a constant <code>'c'</code> exist such that <code>2^(n+1) <= c*(2^n)</code>? Note the second function (<code>2^n</code>) is the function after the Big O in the above problem. This confused me at first.</li> <li>So, then use your basic algebra skills to simplify that equation. <code>2^(n+1)</code> breaks down to <code>2 * 2^n</code>. Doing so, we're left with: <code>2 * 2^n <= c(2^n)</code></li> <li>Now its easy, the equation holds for any value of <code>c</code> where <code>c >= 2</code>. So, yes, we can say that <code>f(2^(n+1))</code> is <code>O(2^n)</code>.</li> </ol> Big Omega works the same way, except it evaluates <code>f(n)</code> >= <code>c*g(n)</code> for some constant <code>'c'</code>. So, simplifying the above functions the same way, we're left with (note the >= now): <code>2 * 2^n >= c(2^n)</code> So, the equation works for the range <code>0 <= c <= 2</code>. So, we can say that <code>f(2^(n+1))</code> is Big Omega of <code>(2^n)</code>. Now, since BOTH of those hold, we can say the function is Big Theta (<code>2^n</code>). If one of them wouldn't work for a constant of <code>'c'</code>, then its not Big Theta. The above example was taken from the Algorithm Design Manual by Skiena, which is a fantastic book. Hope that helps. This really is a hard concept to simplify. Don't get hung up so much on what <code>'c'</code> is, just break it down into simpler terms and use your basic algebra skills. 
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1. PODifference between Big-Theta and Big O notation in simple language
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