Results 1 to 9 of 9

Math Help - no idea on this induction

  1. #1
    Junior Member
    Joined
    Mar 2007
    From
    Missouri
    Posts
    54

    no idea on this induction

    Let f(x)=xe^(-x). Prove that f^(n) (x)=(-1)^n * e^(-x) * (x-n) for every positive integer n.

    I have no idea at all on how to do this.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,605
    Thanks
    1574
    Awards
    1
    Do you know how to differentiate that function?
    Find the fourth derivative. Observe how the factors work.
    Then you can do the induction.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Mar 2007
    From
    Missouri
    Posts
    54
    I assume you mean to differentiate f(x)=xe^-x. Just curious, why the 4th derivative?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,605
    Thanks
    1574
    Awards
    1
    Quote Originally Posted by Possible actuary View Post
    I assume you mean to differentiate f(x)=xe^-x. Just curious, why the 4th derivative?
    Well when you have done it four times, you should have some idea how the factors work. You should understand how the formula you trying to prove works.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Mar 2007
    From
    Missouri
    Posts
    54
    I can see that the derivatives keep bouncing from xe^-x and -xe^-x + e^-x. So then I understand (-1^n) and (e^-x) but not (x-n).
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Senior Member ecMathGeek's Avatar
    Joined
    Mar 2007
    Posts
    436
    Quote Originally Posted by Possible actuary View Post
    Let f(x)=xe^(-x). Prove that f^(n) (x)=(-1)^n * e^(-x) * (x-n) for every positive integer n.

    I have no idea at all on how to do this.
    n = 0: f(x) = xe^(-x)
    n = 1: f'(x) = x(-e^(-x)) + e^-x = -e^(-x)(x - 1)
    n = 2: f''(x) = -e^(-x) + e^(-x)(x - 1) = e^(-x)(-1 + x - 1) = e^(-x)(x - 2)
    n = 3: f'''(x) = e^(-x) - e^(-x)(x - 2) = e^(-x)(1 - x + 2) = -e^(-x)(x - 3)
    n = 4: f''''(x) = -e^(-x) + e^(-x)(x - 3) = e^(-x)(-1 + x - 1) = e^(-x)(x - 4)
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Senior Member ecMathGeek's Avatar
    Joined
    Mar 2007
    Posts
    436
    Here's how you show it inductively:

    First, show that n = 1 works:
    f'(x) = x(-e^(-x)) + e^-x = -e^(-x)(x - 1)

    Next, assume it works for n = k:
    f^{k}(x) = (-1)^(k)*e^(-x)(x - k)

    Now, show it works for n = k + 1:
    f^{k + 1}(x) = (-1)^(k + 1)*e^(-x)[x - (k + 1)]

    Since f^{k + 1} = d/dx f^{k}, it holds that:
    f^{k + 1}(x) = d/dx [(-1)^(k)*e^(-x)(x - k)]
    = (-1)^(k)*e^(-x) - (-1)^(k)*e^(-x)(x - k)
    = -1*(-1)^(k)*e^(-x)[-1 + x - k]
    = (-1)^(k + 1)*e^(-x)[x - (k + 1)] = f^{k + 1}(x)
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Try this with your induction skills.

    Let f and g be infinitely differenciable at x.

    Find a formula for the n-th derivative of (fg) , i.e. their product.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Senior Member ecMathGeek's Avatar
    Joined
    Mar 2007
    Posts
    436
    Quote Originally Posted by ThePerfectHacker View Post
    Try this with your induction skills.

    Let f and g be infinitely differenciable at x.

    Find a formula for the n-th derivative of (fg) , i.e. their product.
    Are you asking me to try this?

    (fg)' = fg' + f'g
    (fg)'' = (fg' + f'g)' = fg'' + f'g' + f'g' + f''g = fg'' + 2f'g' + f''g
    (fg)''' = (fg'' +2f'g' + f''g)' = fg''' + f'g'' + 2f'g'' + 2f''g' + f''g' + f'''g = fg''' + 3f'g'' + 3f''g' + f'''g
    .
    .
    .
    (fg)^{n} = C(n,0)*fg^{n} + C(n,1)*f'g^{n-1) + ... + C(n,n-1)*f^{n-1}*g' + C(n,n)*f^{n}g

    Where C(n,i) is the combination (from probability theory) for the k'th term out of n terms, which is:
    C(n,i) = n!/(i!*(n - i)!)

    To prove this:

    1. First, show it works for n = 1:
    (fg)^{1} = C(1,0)*fg' + C(1,1)*f'g = fg' + f'g

    2. Second, assume it's true for n = k
    (fg)^{k} = C(k,0)*fg^{k} + C(k,1)*f'g^{k - 1} + ... + C(k,k-1)*f^{k-1}g' + C(k,k)*f^{k}g

    3. Third, show it's true for n = k + 1
    (fg)^{k + 1} = C(k+1,0)*fg^{k + 1} + C(k+1,1)*f'g^{k} + C(k+1,2)*f''g^{k - 1}... + C(k+1,k-1)*f^{k - 1}g'' + C(k+1,k)*f^{k}g' + C(k+1,k+1)*f^{k + 1}g

    We have that (this is going to be a lot of writing):
    (fg)^{k+1} = d/dx(fg)^{k}
    = (C(k,0)*fg^{k+1} + C(k,0)*f'g^{k}) + (C(k,1)*fg^{k} + C(k,1)*f'g^{k-1}) + ... + (C(k,k-1)*f^{k-1}g'' + C(k,k-1)*f^{k}g') + (C(k,k)*f^{k}g' + C(k,k)*f^{k+1}g)
    = [C(k,0)]*fg^{k+1} + [C(k,0) + C(k,1)]*f'g^{k} + ... + [C(k,k-1) + C(k,k)]*f^{k}g' + [C(k,k)]*f^{k+1}g

    Now, it needs to be shown that C(k,i) + C(k,i+1) = C(k+1,i+1), where 0 <= i < k. I'll do this by direct proof.
    Note that C(k,i) = k!/(i!*(k - i)!)

    a. show it works for k = 1 (note that i = 0 since i + 1 = 1)
    C(1,0) + C(1,1) = 1!/(0!*(1-1)!) + 1!/((0+1)!*(-0)!) = 1 + 1 = 2 = 2!/(1!(2-1)!) = C(2,1)

    b. show it works for any value of k = j
    C(j,i) + C(j,i+1) = j!/(i!(j - i)!) + j!/((i + 1)!(j - i - 1)!)
    = j![(i + 1 + j - i)]/((i + 1)!(j - i)!)
    = (j + 1)!/((i + 1)!((j + 1) - (i + 1))!)
    = C(j+1,i+1) Q.E.D.

    3. (continued) Therefore, the problem above becomes:
    = [C(k,0)]*fg^{k+1} + [C(k,0) + C(k,1)]*f'g^{k} + ... + [C(k,k-1) + C(k,k)]*f^{k}g' + [C(k,k)]*f^{k+1}g
    = C(k+1,0)*fg^{k+1} + C(k+1,1)*f'g^{k} + ... + C(k+1,k)f^{k}g' + C(k+1,k+1)f^{k+1}g
    = (fg)^{k+1}

    Q.E.D
    Last edited by ecMathGeek; April 18th 2007 at 04:50 PM. Reason: Simplify some of the steps.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Help with a D.E., no idea what to do with it
    Posted in the Differential Equations Forum
    Replies: 6
    Last Post: July 7th 2011, 04:28 AM
  2. any idea
    Posted in the Algebra Forum
    Replies: 1
    Last Post: June 4th 2008, 04:59 AM
  3. Please Help No Idea!!
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: March 7th 2008, 10:16 PM
  4. anybody with any idea please help!!!
    Posted in the Pre-Calculus Forum
    Replies: 4
    Last Post: November 1st 2007, 04:27 AM
  5. No idea what to do
    Posted in the Algebra Forum
    Replies: 2
    Last Post: October 21st 2006, 05:50 AM

Search Tags


/mathhelpforum @mathhelpforum