Chapter 1, continued

• Please browse the rest of the introduction.
  • Notice the theme about small data vs large data.
  • Remember any algorithm will work on sufficiently small data.
• Please skim section 1.2
  • 1.2.3, you will need Σⁿ_i=1i = n(n+1)/2
  • Let's just check the proof of this:

    By Induction.
    
    Remember: 
        prove a base case 
        assume true for n=k k>base case 
        prove for n=k+1 
    
    Prove for n=1
            1 = 1(2)/2
    
    Assume true for n=k k>1
           Σki=1i = k(k+1)/2
    Prove true for n=k+1;
       Show: Σk+1i=1i   = (k+1)(k+2)/2
    
          since  Σki=1i     = k(k+1)/2
          then   Σki=1i+k+1 = k(k+1)/2 + k+1
    
                 Σki=1i + k+1 = Σk+1i=1i  
    
    	     Σk+1i=1i  = [k(k+1) + 2(k+1) ]/2
                          = (k2+k+2k+2)/2
    		      = (k2+3k+2)/2
    		      = (k+1)(k+2)/2
    

  • This is called strong induction.
  • Sometimes we use weak induction
    • Prove for the base case (i)
    • Assume true for all j where i≥j≥k
    • Prove true for k+1
    • Example 2, (from the book)

      Prove the for any Fibonacci number Fk, Fk < (5/3)i
      
      (remember 
                1, i = 1
           Fi = 1, i = 2
                Fi-1+Fi-2, i > 2
      
      
       Base Case: i = 1;
              1 < (5/3)1
       Assume true for all values 1≥j≥k
       Prove true for k+1
           Fk+1 = Fk + Fk-1
           Fk + Fk-1 < (5/3)k + (5/3)k-1
                     < (3/5)(5/3) (5/3)k + (3/5)2(5/3)2(5/3)k-1
      	       = (3/5)(5/3)k+1+(3/5)2(5/3)k+1
      	       = (5/3)k+1(3/5 + 9/25)
      	       = (5/3)k+1(15+9)/25
      	       = (5/3)k+1(24)/25
      	       < (5/3)k+1
      

• We will be doing some proving throughout the semester.