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An interesting limit from quadratic functions...

  • 20-12-2010 12:30AM
    #1
    Registered Users, Registered Users 2 Posts: 3,745 ✭✭✭


    Given that the rightmost root of [LATEX]\displaystyle ax^2+bx+c[/LATEX] is

    [LATEX]\displaystyle \frac{-b+\sqrt{b^2-4ac}}{2a}[/LATEX]

    and that the only root of [LATEX]\displaystyle bx+c[/LATEX] is

    [LATEX]\displaystyle \frac{-c}{b}[/LATEX]

    Then surely

    [LATEX]\displaystyle \lim_{a \to 0}\left(\frac{-b+\sqrt{b^2-4ac}}{2a}\right)=\frac{-c}{b}[/LATEX]

    for b>0! What a bizzare limit! Like, it comes out of nowhere! How does the b even go below the line?!

    I wrote a short C++ program to test it out for b=2; c=1 and it started converging (slowly).

    Wolfram Alpha plot:
    gif&s=41&w=410&h=190


Comments

  • Registered Users, Registered Users 2 Posts: 966 ✭✭✭equivariant


    Given that the rightmost root of [LATEX]\displaystyle ax^2+bx+c[/LATEX] is

    [LATEX]\displaystyle \frac{-b+\sqrt{b^2-4ac}}{2a}[/LATEX]

    and that the only root of [LATEX]\displaystyle bx+c[/LATEX] is

    [LATEX]\displaystyle \frac{-c}{b}[/LATEX]

    Then surely

    [LATEX]\displaystyle \lim_{a \to 0}\left(\frac{-b+\sqrt{b^2-4ac}}{2a}\right)=\frac{-c}{b}[/LATEX]

    for b>0! What a bizzare limit! Like, it comes out of nowhere! How does the b even go below the line?!

    I wrote a short C++ program to test it out for b=2; c=1 and it started converging (slowly).

    Wolfram Alpha plot:
    gif&s=41&w=410&h=190

    I guess that multiplying above and below by the conjugate (i.e. [latex]-b - \sqrt{b^2-4ac}[/latex]) will explain it?


  • Closed Accounts Posts: 4,204 ✭✭✭FoxT


    yes, it jumps out immediately. Interesting, counterintuitive limit though!


  • Registered Users, Registered Users 2 Posts: 3,745 ✭✭✭Eliot Rosewater


    Ah damn, now it's not a mystery any more. :(

    :pac:

    Seriously though, cheers for that - even with the proof, it still looks counter-intuitive on the face of it, as FoxT says.


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