Notices
Results 1 to 9 of 9

Thread: Problem with the expansion of integration by parts

  1. #1 Problem with the expansion of integration by parts 
    Cool Dude ostkef's Avatar
    Join Date
    Mar 2013
    Posts
    154
    I've come across this funny problem while messing around with integration by parts. Probably made a mistake somewhere.




    In the integration of parts expression, it's possible to expand it further.






    Plugging the second expression into the first, we get



    I don't think this is standard notation, but it's better than writing out all the "violin holes".




    Expanding ad infinitum we get




    The problem is that it seems like we will get an infinite string of polynomials on the right hand side, but not on the left.


    "For the rest of my life I will reflect on what light is." - Albert E.
    Reply With Quote  
     

  2.  
     

  3. #2  
    Forum Professor river_rat's Avatar
    Join Date
    Jun 2006
    Location
    South Africa
    Posts
    1,517
    None of your equations are visible?


    As is often the case with technical subjects we are presented with an unfortunate choice: an explanation that is accurate but incomprehensible, or comprehensible but wrong.
    Reply With Quote  
     

  4. #3  
    Cool Dude ostkef's Avatar
    Join Date
    Mar 2013
    Posts
    154
    Quote Originally Posted by river_rat View Post
    None of your equations are visible?
    Are you using a smartphone?

    I see them perfectly fine. The equations are images.
    Last edited by ostkef; May 15th, 2013 at 07:16 AM.
    "For the rest of my life I will reflect on what light is." - Albert E.
    Reply With Quote  
     

  5. #4  
    Forum Freshman fred91's Avatar
    Join Date
    May 2013
    Location
    Cagliari (Italy)
    Posts
    19
    Don't see any mistake.. but I don't think it is a real problem, I mean, if you try to solve int(x*exp(x)) choosing exp(x) to be derived, you obtain an infinite series of integrals like int(x^n*exp(x)). Simply, when you do those integrals they produce the same "polynomials" with opposite sign and they cancel out. Am I wrong?
    Davis was questioning the increasing length of John Coltrane solos, and Trane answered "I don't know how to stop."

    Try taking the fucking horn out of your mouth.
    Reply With Quote  
     

  6. #5  
    Suspended
    Join Date
    Apr 2012
    Posts
    690
    there is plenty of equations like this take.the pythagoran theorem for example.hyp^2=x^2+y^2
    now this can be expanded infinitely by writing x+y as the hyp of another triangle and expanding out thus
    hyp^2=t^2+k^2+r^2+j^2
    now we can take constants out of this
    k^2=constant w
    j^2=constant k
    relate your r term to your t term
    r^2=n*t
    thus hyp^2=t^2+nt+c
    a quadratic equation
    one of the important things to remember about the pythagoran theorem is that while it expresses a linear distance hyp it does not express a slope therefore a spherical distance must be assumed. also a rate of change of the x and y terms does not equal a rate of change of the hyp only a change in spherical positioning. A change in the hyp however equals a rate of change in the expansion or contraction of the sphere relative to something. Addition of hyp does not make a curve only a sphere of larger area. This of course makes perfect sense if you look at a cad application. Now what we havent talked about is our two vectors (x,y,z,t)note that you can anchor either point and rotate the other on the spherical axis this means that one vector has no effect on the other,vector ie they are independant of each other and do not co-vary. also if I move the entire object through a distance then all points move through that,distance
    Last edited by fiveworlds; May 16th, 2013 at 12:04 PM.
    Reply With Quote  
     

  7. #6  
    Forum Freshman
    Join Date
    Aug 2011
    Location
    East Anglia - England - Moving
    Posts
    80
    Quote Originally Posted by fiveworlds View Post
    there is plenty of equations like this take.the pythagoran theorem for example.hyp^2=x^2+y^2
    now this can be expanded infinitely by writing x+y as the hyp of another triangle and expanding out thus
    hyp^2=t^2+k^2+r^2+j^2
    now we can take constants out of this
    k^2=constant w
    j^2=constant k
    relate your r term to your t term
    r^2=n*t
    thus hyp^2=t^2+nt+c
    a quadratic equation
    one of the important things to remember about the pythagoran theorem is that while it expresses a linear distance hyp it does not express a slope therefore a spherical distance must be assumed. also a rate of change of the x and y terms does not equal a rate of change of the hyp only a change in spherical positioning. A change in the hyp however equals a rate of change in the expansion or contraction of the sphere relative to something. Addition of hyp does not make a curve only a sphere of larger area. This of course makes perfect sense if you look at a cad application. Now what we havent talked about is our two vectors (x,y,z,t)note that you can anchor either point and rotate the other on the spherical axis this means that one vector has no effect on the other,vector ie they are independant of each other and do not co-vary. also if I move the entire object through a distance then all points move through that,distance
    So Very True -
    I only ever looked at the implications of the Hypotenuse lying on the circumference of a Cricle, thus in 2D.

    The Y value ( vertical ) can always be derived from the X value ( horizontal ), both giving Coordinates of any point on the Circumference. The Centre would be the Zero Point.
    Of course, you have to take account of the Signs, if you are looking at points on the Circumference which are not in the top--left Quadrant ( where X and Y are both Positive ).
    Top Right - X is Negative, Y is Positive.
    Bottom Right - X is Negative, Y is Negative.
    Bottom Left - X is Positive, Y is Negative.

    Now - the further consideration is the "Order" of the curve.
    As can be seen we are talking about Squares and Square Roots - in other words a Circle is a curve of the "order " of TWO ( 2 ).

    In 1971 / 1972 I started some investigations of my own into what happnes if you use different Values for the "order" than two.
    I was at the Polytechnic at the time, doing my Degree in Electrical and Electronic Engineering - although I have been a Design Engineer mostly in Mechanical Engineering for some 40 years now.

    So - I wrote a Computer Program to run on their Mainframe Computer - to calculate and draw at least a thousand points on the circumference of a Circle, from the Centre Point, X = 0, Y = 0. This Program used a "normalised" Radius for simplicity.
    The big difference was that the "Order" of the Curve was selectable as an Input.

    I ran this Program thousans of times, in my free time - and got several Interesting Results.
    I had to modify the Program to convert to "absolute" values, in order to eliminate problems with Imaginary Numbers, and to eliminate problems in trying to take the Root of Minus Numbers.
    This fortunately removed the peculiare ebhaviour of the Cubes and Cube Roots of Numbers - which give reversing curve values.

    Conclusions -
    For Values of "Order" less than 2 - the four Quadrants become more flattened.
    For a Value of "Order" at 1 - we get a diamond shape - predictably.
    For Values of "Order" lower than 1 - we get more "concave" Quadrants - which come together more sharply at the junctions of the Quadrants.
    For Values of "Order" greater than 2 - we get more "convex" Quadrants, Quadrants curved more sharply in their middle - but with "flatter" areas at the junctions of the Quadrants.

    Okay - believe it or not - I took this VERY MUCH FURTHER.
    For a Value of "Order" which was 1 / 10^40 - I got a Shape which was almost like a Cross, a Crux. However - there was a very tiny amount of thickness in the middle, and the junctions of the Quadrants looked like there was an extremely tiny Radius there.
    For a Value of "Order" which was 1 x 10^40 - I got a Shape which was almost like a Square. However - there seemed to be some very slight curvature in the sides, and an extremely tiny Radius at the "corners".

    I showed all of my Results and the Printouts of these Shapes to the Professor of The Mathematics Department at the Polytechnic - this was in 1974 the following year when I was doing a Masters Degree in Computer Applications for Science and Technology ( I never finished that Course ).
    He seemed fairly interested - but at no point did he mention that these ALREADY EXIST - They are called Lame Curves.

    I pursued my interest in these Curves - but my PC then did not really have anough "computing power", and furthermore, I could not Program it in any Computer Language with which I am familiar.

    I was able to generate some "milder" forms of these curves in Exel - and by plotting the results as a 3-D Graph, I was able to demonstate for my own interest, that they can be used as "Ship Curves" - to genearet the curves which could be used in laying-out a Ships's Hull. To do this - a Scaling Factor has to be included, to make the X Coordinates ( length ) a multiple of the calculated values.
    For Ship Curves - you really only need to find the two SETS of Curves which meet at the Bow, or the two SETS of Curves which meet at the Stern. In other words - Left Hand Quadrants or Right-Hand Quadrants.
    For the Bow Section - Upper Decks you want the Curves to be "fuller", or more "convex" - but when you get near to the Keel, you want the Curves to be more "concave" - leading into a sharp Bow.
    At the Stern, you may have to modify the Curves to insert a Radial cruve portion. However at the Stern, you want the Curves to be more gradual in any case.

    I have deduced that ALL of these Curves, even "Extreme" ones show a constantly increasing or decreasing Value for the Rate of Change of Curvature - in other words the Third Differential or "Jerk" Factor is smooth and predictable.
    If we were to use some of these Curves ( even if Scaled ) to lay-out our Roads and Railways Tracks - then they vcould be designed with smooth transitions, and reduced lateral forces ( more accurately - gradually increasing or decreasing lateral forces ).
    The present method of using intermediate "Transition Curves" for Railway Tracks seems very "rough and ready" by comparison.

    So Sorry that I have somehow "wasted" a fair bit of my time and thoughts on something which was already "Invented".
    I wish someone had TOLD ME that it had already been done.
    And yet - I found time and time again, that all who I spoke to, thought the whole idea was either pointless, of little value, or even stupid.
    If I wasn't so stupid - I might know what I was doing
    Reply With Quote  
     

  8. #7  
    Suspended
    Join Date
    Apr 2012
    Posts
    690
    I once knew somebody who thought they knew everything did really well in tests etc. I asked this person to explain the implications of fermat's last theorem and was met with a blank stare. You cant expect others to know everything just as you cant expect it of yourself. You cant blame yourself for not knowing everything nor should you pretend to know everything.
    Reply With Quote  
     

  9. #8  
    Forum Professor river_rat's Avatar
    Join Date
    Jun 2006
    Location
    South Africa
    Posts
    1,517
    With a bit of work you can actually make your expansions look like something a lot familiar:

    Ignoring all the technical details we can start with a nice function that satisfies the fundamental theorem of calculus, so

    We can do a substitution and rewrite this equation slightly as and then we start iterating the product rule.



    But we notice we are merely generating the Taylor Series of the Function in question and the integral form of the remainder
    As is often the case with technical subjects we are presented with an unfortunate choice: an explanation that is accurate but incomprehensible, or comprehensible but wrong.
    Reply With Quote  
     

  10. #9  
    Suspended
    Join Date
    Apr 2012
    Posts
    690
    we could have a game of this The fibonacci sequence f(n+1)=f(n)+f(n-1)
    Reply With Quote  
     

Similar Threads

  1. Integration problem
    By Intricacy in forum Mathematics
    Replies: 1
    Last Post: April 27th, 2013, 08:22 AM
  2. Expansion problem question
    By Ascended in forum Physics
    Replies: 11
    Last Post: October 4th, 2012, 04:52 AM
  3. Replies: 30
    Last Post: October 19th, 2011, 09:18 PM
  4. Doubt about integration by parts
    By doctor_cat in forum Mathematics
    Replies: 8
    Last Post: January 16th, 2009, 07:15 AM
  5. Integration by parts
    By EV33 in forum Mathematics
    Replies: 4
    Last Post: March 3rd, 2008, 01:36 AM
Bookmarks
Bookmarks
Posting Permissions
  • You may not post new threads
  • You may not post replies
  • You may not post attachments
  • You may not edit your posts
  •