big o - Using big-O to prove N^2 is O(2^N) -


I can see clearly in comparison to N ^ 2, surrounded by c2 ^ N, but I have given it a formal definition How can I prove the use of big-o I just got it M.I. I can prove by

Here's my attempt ... by definition, there is no n & gt; For n0, a stable c is present which is f (n) & lt; = Cg (n) where f (n) = n ^ 2 and g (n) = 2 ^ n

Should I log in both ways and solve for C?

And another question about the Fibonacci sequence, I want to solve the repetition relation. 

  int fib (int n) {if (n & lt; = 1) return n; And reboot fib (N-1) + fib (N-2); The equation is  
  T (n) = t (n -1) + t (n -2) + c // where c < / (N) + 2 T (N3) + T (N-4) + 3C  / Code>  
 and another  
  T (n) = t (n3) + 3 t (n4) + 3 t (n-5) + T (N-6) + 6C   
 Then I'm lost to make the normal equation. Is the pattern like some sort of Pascal triangle?  
  T (n) = t (ni) + at (ni-1) + bt (ni2) + ... + kt (nii) + c   f (x) Ïμ O (g (x))>  
 
 
 < / Em> You need to find ...  
     
  •  ... something  c & gt; 0  and  
  •  ... some  x  0     
     such  f ( X) & lt; C for all (x)   x & gt; X  0  .  
     In this case, you can select  c = 1  and  x  0  = 2  you need to prove it Is that  
      x  2  & lt; 2  x   for all  x & gt; 2   
     At this point you can log both sides (because log ( x )> log ( y ) , Then  x> Y .) Assume that you are using base-2 log, you get the following  
      log (x  2 ) & lt; By logging (2  x )   
     and the standard laws of logarithm, you  
      2 · log (x) & lt ;  log (2) = 1  can be written as  
      2 · Log (x) log (2)   
     X) & lt; X   
     If we set  x  = 2, then  
      2 · Log (2)  = 2 gets faster than   << p> and  x   log (x)  we know that  2 · Log (x) & lt; X  all  x & gt; 2 .

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