Results 1 to 5 of 5

Math Help - Calculus Class Bonus Question

  1. #1
    Newbie
    Joined
    Apr 2009
    Posts
    5

    Calculus Class Bonus Question

    My teacher gave us this bonus question (said getting help was fine as long as we understand the solution), but I think I'm stumped.

    "Let f(x) = 1 / sqrt(x)
    Determine a function of n for the nth derivative of this function."

    I wrote out the first 5 derivatives to find a pattern. Here's the function I've come up with so far:
    f(x) to the nth derivative = (-1)^n * ( ) * ( 1/ sqrt(x)^ (2n+1))

    I know the (1)^n takes care of the alternating signs, and the ( 1/ sqrt(x)^ (2n+1)) takes care of the power of x. I just can't think of a function to describe the middle term which should be the multiplication of the previous coefficient with the coefficient(?) of the derivative.

    Some help would be greatly appreciated Thanks!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Oct 2005
    From
    Earth
    Posts
    1,599
    Welcome to MHF. You have a good start.

    I wouldn't think of the general derivative in terms of \sqrt{x}. You start out with x^{-\frac{1}{2}} and each time you perform another derivative the initial numerator is increased by 2 and the denominator stays the same. For example the first derivative has a power of (-3/2) and a coefficient of (-1/2). The second derivative has a power of (-5/2) and a coefficient of (-1/2)*(-3/2). So how does the power of the derivative relate to the last coefficient? The coefficients follow a pattern and I think the best way to express them generally is using factorials.

    Are you familiar with factorials and can you do some more now or are you still stuck?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Apr 2009
    Posts
    5
    Hmmm, Ok. So seeing as the coefficients for the first 6 derivatives are 1, 1/2, 3/4, 15/8, 105/6, 945/32 I would think the easiest way to express the denominator would just be 1/(2^n). As far as factorials are concerned, we haven't done much beyond having them in infinite series problems... I was thinking a factorial was the only way to gain larger and larger increases in the numerator, but I have no idea as to how to express that.

    Hopefully I'm thinking in the right direction?

    Thanks!
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    Oct 2005
    From
    Earth
    Posts
    1,599
    Quote Originally Posted by The0wn4g3 View Post
    Hmmm, Ok. So seeing as the coefficients for the first 6 derivatives are 1, 1/2, 3/4, 15/8, 105/6, 945/32 I would think the easiest way to express the denominator would just be 1/(2^n). As far as factorials are concerned, we haven't done much beyond having them in infinite series problems... I was thinking a factorial was the only way to gain larger and larger increases in the numerator, but I have no idea as to how to express that.

    Hopefully I'm thinking in the right direction?

    Thanks!
    Don't simplify the coefficients. Think of them in terms of multiplication. So for n=1, meaning the first derivative, we get f' = -\frac{1}{2} x^{-\frac{3}{2}}. For the second we get f'' = \left( -\frac{3}{2}*-\frac{1}{2} \right) x^{-\frac{5}{2}}. So you should see the pattern being that the last coefficient is going to be \frac{2n-1}{2}, and the next term will be \frac{(2n-1)-2}{2} and so on until you reach 1/2. The sign of the total multiplication result will alternate like you said.

    Look at the expression \left( \frac{(2n-1)}{2} \right) ! This is going to look like \frac{2n-1}{2} * \frac{2(n-1)-1}{2} * \frac{2(n-2)-1}{2}.... Take note that when you distribute the 2 to the parentheses in the numerators you get the pattern \frac{2n-1}{2}*\frac{2n-3}{2}*\frac{2n-5}{2}... The factorial is defined for how you'll use it to only work with positive numbers. So after the numerator goes below 0, the process will stop.

    Make sense so far?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Apr 2009
    Posts
    5
    Oh wow, that's cool. I don't think I ever would have thought of it that way. I was trying to calculate the numerator and the denominator separately, but that makes a lot more sense!

    Thanks a ton!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: December 13th 2011, 09:11 PM
  2. Bonus Question
    Posted in the Pre-Calculus Forum
    Replies: 16
    Last Post: August 26th 2010, 06:31 PM
  3. Very Hard Bonus Question
    Posted in the Math Topics Forum
    Replies: 1
    Last Post: February 24th 2010, 10:24 PM
  4. Need help with e for my Calculus class.
    Posted in the Calculus Forum
    Replies: 1
    Last Post: January 27th 2010, 08:57 PM
  5. Bonus Logarithem Question
    Posted in the Algebra Forum
    Replies: 3
    Last Post: December 7th 2006, 06:58 PM

Search Tags


/mathhelpforum @mathhelpforum