Results 1 to 6 of 6
Like Tree2Thanks
  • 1 Post By hollywood
  • 1 Post By kspkido

Math Help - Approximation Error and Tough Integral

  1. #1
    Junior Member EliteAndoy's Avatar
    Joined
    Feb 2013
    From
    United States
    Posts
    56
    Thanks
    3

    Approximation Error and Tough Integral

    Hello Everyone! Today We are asked to find \int_{1}^{2}  e^\frac{1}{x}dx. We are given two option. First is by evaluating the definite integral itself or we can do numerical approximation using both Trapezoidal and Mid point Rules with sub interval n valued at 10. Since I've figured out that integrating such integral would be a , I just gone on and used the Numerical Approximation way. Everything is well until I am asked to find the error for both T_{10} and M_{10}. I know the formula and the value of for the second derivative of the function e^\frac{1}{x} to use it as a rule to find constant K from the formula |E_T|\leq\frac{K(b-a)^3}{12n^2} and |E_M|\leq\frac{K(b-a)^3}{24n^2}. Now I am stuck at how to find K when K\geq|\frac{e^\frac{1}{x}(2x+1)}{x^4}| at interval [1,2]. Anyone got any idea to find K? Thanks y'all in advance!

    Edit: I have an idea on how to solve this one. As we can see, x is bounded in [1,2], and that 1/x would only max out at x=1 since anything more than 1 would make x less so we can conclude that \frac{1}{x}\leq 1. Since K is the max value of the second derivative, then max value is acquired when 1/x is evaluated at x=1. So that K\geq|(e^\frac{1}{x}(\frac{2}{\frac{1}{x}}+1)) (\frac{1}{x})^4| so if we substitute 1 to all 1/x, we get: K=3e. Is that the right way to do it?Hehe it is late in the morning and I am still pumped to do this problem!
    Last edited by EliteAndoy; February 19th 2013 at 10:51 PM. Reason: Other solution doesn't work
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Mar 2010
    Posts
    993
    Thanks
    244

    Re: Approximation Error and Tough Integral

    I think you need to find an upper bound for K, not a lower bound.

    But you can't get a bound of any kind by setting x to the same value when you have the product of a decreasing and an increasing function. For example, if you have y=x(1-x) on [0,1], if you set x=0 across the board, you get a bound of zero, but y goes above zero. The correct procedure in this simplistic case is to recognize that x is always less than or equal to 1, and 1-x is always less than or equal to 1. So x(1-x) is always less than or equal to 1 times 1, which is 1. In this case we know the function has a maximum of \frac{1}{4}, so it is indeed always less than or equal to 1.

    - Hollywood
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member EliteAndoy's Avatar
    Joined
    Feb 2013
    From
    United States
    Posts
    56
    Thanks
    3

    Re: Approximation Error and Tough Integral

    Hi Hollywood! Well I think from the definitions of Error bounds, K is just a constant in which the second derivative of integrand, f(x), maxes out. And if we look at the second derivative of f(x), we can see that we have product of 3 kinds of functions-exponential(decreasing), linear(increasing), and rational(decreasing). If we take the limit of f"(x) as x goes to infinity, the value would go to zero. That means that for any increase in x in the bounds [1,2], the second derivative of f(x) only decreases, meaning, in the bounds of the integral [n-k,n], the second derivative maxes out on the smallest value of n-k which is 1. So to find the max value for K, we have to plug in 1 for any x in the second derivative. If we do that correctly, we should have K=3e. And since the Error is only bounded by the formula that involves max K, then we can just plug in K to all variable K in the formula. Is that reasoning good? Not really sure.
    Last edited by EliteAndoy; February 20th 2013 at 06:46 AM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member
    Joined
    Mar 2010
    Posts
    993
    Thanks
    244

    Re: Approximation Error and Tough Integral

    Showing the limit of f''(x) as x goes to infinity is not good enough to do what you're doing. You would have to show that it is a decreasing function (i.e. that f'''(x)<0). Since it is a decreasing function, your answer is correct.

    Another way to look at it is to write |f''(x)| as:

    \left|e^\frac{1}{x}\left( \frac{2}{x^3} + \frac{1}{x^4} \right)\right|

    and all the functions are decreasing, so you can just plug in 1, the smallest value for x.

    And yes, I also get 3e for the numerical answer.

    - Hollywood
    Thanks from EliteAndoy
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Aug 2012
    From
    Manila, Philippines
    Posts
    24
    Thanks
    1

    Re: Approximation Error and Tough Integral

    Hey pal. As far as Im concern, K is the abs(f''(x))<=K
    Thanks from EliteAndoy
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Junior Member EliteAndoy's Avatar
    Joined
    Feb 2013
    From
    United States
    Posts
    56
    Thanks
    3

    Re: Approximation Error and Tough Integral

    Thanks both of you for your answers. From what I understand, since K is defined from the inequality \mid f''(x) \mid \leq K then K must be the highest value of the second derivative of the function.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Approximation and error
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: October 9th 2011, 04:14 AM
  2. Approximation and error formula
    Posted in the Calculus Forum
    Replies: 2
    Last Post: March 28th 2011, 03:58 AM
  3. Error approximation (hermite)
    Posted in the Calculus Forum
    Replies: 0
    Last Post: July 29th 2009, 02:42 AM
  4. Replies: 0
    Last Post: April 1st 2009, 04:22 AM
  5. Help! Finding error of approximation
    Posted in the Calculus Forum
    Replies: 6
    Last Post: March 29th 2009, 09:00 PM

Search Tags


/mathhelpforum @mathhelpforum