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Functions ?

  1. Apr 1, 2009 #1
    functions ???

    1. The problem statement, all variables and given/known data

    I apologize for the title, I don't know what to call these kind of problems.

    Q3. Suppose f is function that satisfies the equation

    equation 1

    For all numbers x and y, suppose also that

    equation 2

    Find: (i). f(0), (ii). f’(0), (iii). f’(x)



    2. Relevant equations

    ?

    3. The attempt at a solution

    I have no idea. So what?
     

    Attached Files:

  2. jcsd
  3. Apr 1, 2009 #2

    lanedance

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    Re: functions ???

    hi Aryt

    pictures take a while to get viewable.... so can't see yet, can you enter the problem otherwise?
     
  4. Apr 1, 2009 #3
    Re: functions ???

    equation 1: f(x+y)=f(x)+f(y)+x^2 y+xy^2

    equation 2: lim f(x)/x =1 while x tends to zero
     
  5. Apr 1, 2009 #4

    lanedance

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    Re: functions ???

    is that all the info...? if so, then how about considering f(1+0)?

    then think about a derivative from first limits at zero...

    anyway, see what you think ;)
     
  6. Apr 1, 2009 #5
    Re: functions ???

    I think we should write sth like: f(-1+1) = f(0) = f(1)- f(-1)+(1)^2 (-1) + (1)(-1^2)
    which will be f(1)-f(-1). and we know that when x tends to 0, f(x)/x = 1
    so f(1)/1 = 1 and also f(-1)/-1 = 1
    Therefore f(0) = 1-1=0 or I dunno, maybe 1+1=2 .

    Is it a possible answer for (i)?

    ahhhhhh, what kind of question is this? :(
     
  7. Apr 1, 2009 #6
    Re: functions ???

    I think that it is rather a good question.
    Get to the defintion of f'(x).

    f'(x)= [f(x+h) - f(h)}/h . Where h tends towards zero. Now think what can you do to f(x+h). Once done with that. You wil get to a dead end. But you can get rid of that by knowing the value of f(0). The value of f(0) cab be found out by replacing x and y in the original equation by 0 and 0. Once done with that you will get the value of F(0).
     
  8. Apr 1, 2009 #7
    Re: functions ???

    well, if we go for 0 and 0, we will have: f(0) = f(0)+f(0) that is f(0)=2f(0)
    and 1=2 ! Great!
     
  9. Apr 1, 2009 #8
    Re: functions ???

    Well its obvious isnt it. You are divinding by zero. Hence you get ridiculous statements like 1=2. So f(0)=0
     
  10. Apr 1, 2009 #9
    Re: functions ???

    so f'(0) is also zero?

    and f'(x)=1?

    I think I got it. Thanks
     
  11. Apr 1, 2009 #10

    lanedance

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    Re: functions ???

    i think you're there, but here's what I meant

    [tex] f(x+y)=f(x)+f(y)+x^2 y+xy^2 [/tex]
    [tex] f(0+1)=f(0)+f(1)+0^2.1+0.1^2 [/tex]
    [tex] f(1)=f(0)+f(1) [/tex]
    [tex] 0=f(0)[/tex]
     
  12. Apr 1, 2009 #11
    Re: functions ???

    I haven't calculated the value of f'(x). Even if it is equal to 1 how can the value of f'(0) be equal to zero considering that f'(x)=1
     
  13. Apr 1, 2009 #12

    lanedance

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    Re: functions ???

    yeah sorry i misread your last post

    how did you get f'(0) = 0, is that a guess?
    calculate f'(0) first as Fedex implies (use the limit information and the first principles defintino of a derivative)
     
    Last edited: Apr 2, 2009
  14. Apr 2, 2009 #13
    Re: functions ???

    f'(x)= [f(x+h) - f(h)]/h while h tends to zero.

    so f'(0)=[f(0+0)-f(0)]/0
    f'(0)= 0/0

    ???
     
  15. Apr 2, 2009 #14

    lanedance

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    Re: functions ???

    remember its not h=0, its the limit as h goes to zero
     
  16. Apr 2, 2009 #15
    Re: functions ???

    We're under heavy fire C.O., Clarification is needed. :D

    lol

    Isn't it a bit stupid to say:
    We have: f(0) = 0 and Equation 1:

    we have lim [f(0+h)-f(h)]/h while h maps to 0.
    Thus (h tends to 0): lim f(h)/h - lim f(h)/h = 1-1 = 0
     
  17. Apr 2, 2009 #16
    Re: functions ???

    I think I've found the problem. You've cited the equation wrongly. It should be:

    [f(x+h)-f(x)]/h while h tends to 0. According to equation 1, we have:

    [f(x)+f(h)+x^2 h + xh^2 - f(x)] / h => lim f(h)/h + lim (x^2+xh) while h maps to zero.

    So f'(x)=1+x^2

    and accordign to this we have: f'(0)=1+0=1

    Right?
     
  18. Apr 2, 2009 #17
    Re: functions ???

    Buffon
     
  19. Apr 2, 2009 #18
    Re: functions ???

    Genuis

    PS Sorry for "buffon" I just wanted to bring a dramatic end.
     
  20. Apr 2, 2009 #19
    Re: functions ???

    No, It's OK. Thanks for the tips. :D

    A better dramatic end: lol

    Although they're killing my innovation here. For one assignment (C++ programming), I've used only one line to compile a rather complicated program without sth useless (i.e. End of File function: eof.). And the tutor told me very directly, “why are you trying to come up with your own ways”? I gave you a kind of lecture for this question. When you can't solve sth in my way, why you do it in your way?
    ahhhh, it's innovation, if you add only one space to the input file, your program will fail, but mine will remain ok, I said! :)
     
    Last edited: Apr 2, 2009
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