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Is This possible?

  1. Dec 14, 2006 #1
    Is This possible???

    Hello,

    Is it possible to diffrentiate or intigrate the following equation?

    f \left( x \right) =\sum _{r=1}^{x-1}{\frac { \left( x-1 \right) !\,f

    \left( x-r \right) }{ \left( x-1-r \right) !\,r!}}-2\,\sum _{r=1}^{1/

    2\,x-3/4-1/4\,\cos \left( {\frac {22}{7}}\,x \right) }{\frac { \left(

    x-1 \right) !\,f \left( x-2\,r \right) }{ \left( x-2\,r-1 \right) !\,

    \left( 2\,r \right) !}}



    I'll be thankfull for your answer.

    edit: I've attached the equation as jpeg.
     

    Attached Files:

    Last edited: Dec 14, 2006
  2. jcsd
  3. Dec 14, 2006 #2
    it's not even possible to read it :biggrin:
     
  4. Dec 14, 2006 #3
    here is the same eq. in string:

    f(x) = sum((x-1)!/(x-1-r)!/r!*f(x-r),r = 1 .. x-1)-2*sum((x-1)!/(x-2*r-1)!/(2*r)!*f(x-2*r),r = 1 .. 1/2*x-3/4-1/4*cos(22/7*x))
     
  5. Dec 14, 2006 #4
    [itex]f \left( x \right) =\sum _{r=1}^{x-1}{\frac { \left( x-1 \right) !\,f

    \left( x-r \right) }{ \left( x-1-r \right) !\,r!}}-2\,\sum _{r=1}^{1/

    2\,x-3/4-1/4\,\cos \left( {\frac {22}{7}}\,x \right) }{\frac { \left(

    x-1 \right) !\,f \left( x-2\,r \right) }{ \left( x-2\,r-1 \right) !\,

    \left( 2\,r \right) !}}[/itex]

    you need to put itex and /itex around it, both in square brackets
     
  6. Dec 14, 2006 #5

    CRGreathouse

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    I wouldn't do it inline -- here's how it looks in full form ([ tex ] ... [ /tex ]):

    [tex]f \left( x \right) =\sum _{r=1}^{x-1}{\frac { \left( x-1 \right) !\,f \left( x-r \right) }{ \left( x-1-r \right) !\,r!}}-2\,\sum _{r=1}^{1/2\,x-3/4-1/4\,\cos \left( {\frac {22}{7}}\,x \right) }{\frac { \left(x-1 \right) !\,f \left( x-2\,r \right) }{ \left( x-2\,r-1 \right) !\, \left( 2\,r \right) !}}[/tex]
     
  7. Dec 15, 2006 #6
    thankyou. i'm still getting my head around the tex tags on this forum myself :biggrin:

    some notes though, it's not an equation, it's a function definition. also all those factorials make it defined on some subset of the integers so you couldn't really differentiate or integrate without some jiggery-pokery
     
  8. Dec 15, 2006 #7
    Thanks for your help, Now Im able to write it correctly;

    So is it possible to be diffrentiated?
     
  9. Dec 15, 2006 #8
    as it is no. the factorial has no derivative as it's not defined on any interval of R. i suggest you post info on motivation and context, rather than a raw expression.
     
  10. Dec 15, 2006 #9
    How do you expect to differentiate a discreet function?
     
  11. Dec 16, 2006 #10

    uart

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    Also note the second summation upper limit is a bit suspect, it should be an integer so you should be using floor() or ceil() or round() or something there.

    Additionally the use of [tex]1/2x[/tex] (again in that second sum upper limit) is a bit ambiguous. I think that strictly speaking it should mean [tex]x/2[/tex] but many people write that expression (format) when they really mean [tex]1/(2x)[/tex].

    One last thing, that fraction 22/7 (in the cosine argument of the upper limit of the second sum) isn't supposed to be an "exact" rational reresentation of [tex]\pi[/tex] is it? I certainly hope not anyway.


    BTW. How about telling us where your goofy function comes from. That is, what does it represent?
     
    Last edited: Dec 16, 2006
  12. Dec 16, 2006 #11
    Well sure Uart, But there is a long story behind this function...

    I'm a collage student & doing my O'levels write now, In my math book there were several patterns and that we were to find the next ones...

    that was when I was able to start my way which somehow results to this function;

    It's for polynomial sequences only and gives you the nth term using n-1 terms.

    I'd be glad to give you much more information on how it is derived if you are interested.

    By the way you are abseloutly correct about x/2 and 22/7 :)
    you'll have to round it off to the nearest whole number for the second sum.

    & thankyou for the information
     
  13. Dec 16, 2006 #12

    uart

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    Ok mubashirmansoor that second summation is making more sense now that you've explained those things.

    Note that cos(Pi x) only takes the values or +/- 1 for integer x. So if x is even then x/2 is integer and -3/4 - 1/4 * cos(Pi x) is also integer. Additionally when x is odd then x/2 is half integer and -3/4 - 1/4 * cos(Pi x) is also half integer. So in all cases the upper limit is indeed integer and no rounding flooring of ceiling are required.

    Now this function is looking more interesting.
     
  14. Dec 17, 2006 #13
    Yes.... You are correct, I had not looked at it from this point of view... I had checked some values in a calculator and it gave me some errors... But you are abseloutly correct.

    How many of such function definitions are discovered? And do these look intresting to the scientific comunity?

    An example of such sequences is say x^2:

    1 4 9 ..... whats the next term????

    By taking x = 4 in the function given above we have:

    3(9) - 3(4) + 1 which is equal to 16 .
     
  15. Dec 21, 2006 #14

    ssd

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    I shall give you a problem to find your answer. Let, x be a variable and x>0.
    then

    x+x+x+......+ upto x terms =x^2
    Differentiating both sides with respect to x
    1+1+1+......+ upto x terms =2x
    or, x=2x
    or, 1=2, x can be cancelled out since x>0.
    Find the fallacy.
     
  16. Dec 23, 2006 #15
    Possible falacy:

    the x on the left handside & the right handside are different from each other...

    hence better to say : y=2x

    Correct?
     
  17. Dec 26, 2006 #16

    ssd

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    Not correct...... x's on both sides are same. Example: 5+5+...upto 5 terms=5^2.
    Answer: The left hand side the expression uses the fact that the sum is upto x terms, that is, x is an integer.... hence x is not differentiable.
     
  18. Dec 27, 2006 #17

    CRGreathouse

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    I would have said "x is fixed", not "x is an integer".
     
  19. Dec 27, 2006 #18

    ssd

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    I am sorry but I dont follow you......why x should be fixed? x is a variable and can take values 1,2,3,..., ad inf (since given that x>0). How can we disagree on the fact that x is an integer since the sum runs for x terms!

    Particular example: Let x denote the number of points turning up (on the top face) from the throw of a six faced unbiased die. Now, x is an integer valued variable with possible values of 1,2,...,6. In this case also we have, x+x+....+ x terms = x^2.
     
    Last edited: Dec 28, 2006
  20. Dec 28, 2006 #19

    CRGreathouse

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    I would have said that the fact that x is not fixed is the reason you can't differentiate it, instead of that it is an integer.
     
  21. Dec 28, 2006 #20
    What do you mean by "x is not fixed"? What do you mean by fixed?
     
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