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Power Series?

  1. Aug 7, 2009 #1
    1. The problem statement, all variables and given/known data
    Given f(x)=a0x0+a1x1+a2x2...anxn
    p.s: anxn is to represent a sub n multiply x^n
    f(1)=8 ,
    f(35)=e^e
    Find f(7)

    I dont know what are the theorem that i have to know and how can i proceed

    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Aug 7, 2009 #2
    Do you mean:

    [tex]f(x)=a_0x^0+a_1x^1+a_2x^2+...+a_nx^n[/tex]

    It is actually:

    [tex]\sum_{i=0}^{n}a_ix^i[/tex]

    but doesn't matter. The point is:

    [tex]8=a_0+a_1+a_2+...+a_n[/tex]

    and

    [tex]e^e=a_035^0+a_135^1+a_235^2+...+a_n35^n[/tex]

    [tex]y=a_07^0+a_17^1+a_27^2+...+a_n7^n[/tex]


    Can you continue now?

    P.S 35 = 7 * 5
     
    Last edited: Aug 7, 2009
  4. Aug 7, 2009 #3
    Thanks Дьявол!
    How do I proceed from here?

    [tex]
    e^e=a_05^07^0+a_15^17^1+a_25^27^2+...+a_n5^n7^n
    [/tex]
     
    Last edited: Aug 7, 2009
  5. Aug 7, 2009 #4

    Mark44

    Staff: Mentor

    Is your function in fact a finite sum? IOW, is it defined this way (finite sum)?
    [tex]f(x) = a_0 + a_1 x + a_2 x^2 + ... + a_n x^n[/tex]
    or this way (infinite sum)?
    [tex]f(x) = a_0 + a_1 x + a_2 x^2 + ... + a_n x^n + ...[/tex]

    The reason I ask is that there is an infinite sum representation for ee.
     
  6. Aug 7, 2009 #5
    Thanks! It is the finite series. [tex]f(x) = a_0 + a_1 x + a_2 x^2 + ... + a_n x^n[/tex]
     
    Last edited: Aug 8, 2009
  7. Aug 8, 2009 #6
    Where did you find this task from?

    What equations you should use to solve the task?

    What is [itex]a_0+a_1+a_2+...+a_n[/itex], is it geometric or arithmetic series?

    Can you provide more information about this task?
     
  8. Aug 8, 2009 #7
    It is not stated if it is a geometric or arithmetic series.

    The question goes like this(without a single omission):

    Let f(x)=[itex]a_0+a_1x^1+a_2x^2+...+a_nx^n[/itex] ,where [itex]a_0,a_1,a_2,...,a_n[/itex]
    are nonnegative integers. If f(1)=8 and f(35)=[itex]e^e[/itex] , find f(7).

    Answer is 512.

    Thanks in advance
     
  9. Aug 8, 2009 #8
    It is not stated if it is a geometric or arithmetic series.

    The question goes like this(without a single omission):

    Let f(x)=[itex]a_0+a_1x+a_2x^2+...+a_nx^n[/itex] ,where [itex]a_0,a_1,a_2,...,a_n[/itex]
    are nonnegative integers. If f(1)=8 and f(35)=[itex]e^e[/itex] , find f(7).

    Answer is 512.

    Thanks in advance
     
  10. Aug 8, 2009 #9

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    If the a's are nonnegative integers and x=35, then the value of your polynomial is an integer. e^e is not an integer, if e is the usual constant represented by e=2.7182... What does that e^e mean?
     
  11. Aug 8, 2009 #10
    Hi Dick! I am not certain but is it a infinite series?
     
  12. Aug 8, 2009 #11

    VietDao29

    User Avatar
    Homework Helper

    Please type in exactly what the problem tells you. Otherwise, we cannot help you much.. As a matter of fact, we cannot guess what you problem might be.. :( If you are not even certain about the problem, how can we suppose to know it? :frown:
     
  13. Aug 8, 2009 #12
    [tex] e^x = \sum_{n=0}^\infty \frac{x^n}{n!} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots,[/tex]

    is infinite.

    Could you possibly provide relevant equations or something?
     
  14. Aug 8, 2009 #13
    The question goes like this(without a single omission):

    Let f(x)=[itex]a_0+a_1x+a_2x^2+...+a_nx^n[/itex] ,where [itex]a_0,a_1,a_2,...,a_n[/itex]
    are nonnegative integers. If f(1)=8 and f(35)=[itex]e^e[/itex] , find f(7).

    Answer is 512.

    Thanks in advance
     
    Last edited: Aug 8, 2009
  15. Aug 8, 2009 #14

    VietDao29

    User Avatar
    Homework Helper

    As Dick has pointed out,

    Since a0, a1, ..., an are all non-negative integers. it must follow that: f(35) is an integer.

    I'm positively sure that you have copied the problem down incorrectly. I may suggest you ask some of your friends to get the correct version of it, then re-post it here; or you can contact your professor. :)
     
  16. Aug 8, 2009 #15
    Even if he have the correct value for f(35), I am quite sure that the task can't be solved. I tried various techniques but still no result.

    Regards.
     
  17. Aug 9, 2009 #16
    Thanks Guys! I verified with my professor and he said he made a mistake.
    f(35) ought to be [tex]6^6[/tex] and not [tex]e^e[/tex]
     
  18. Aug 9, 2009 #17

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Now we are getting somewhere. Since the a's are nonnegative integers and f(1)=0 you can see n is at most 7 and all a's sum to 8. There's only a finite (but large) number of ways to do that. But you can also notice 35^4>6^6. What does that tell you? After that I'd just start guessing by adding combinations of 35^0, 35^1, 35^2, 35^3 to try and find the a's. In fact, since 8<35 you can actually think of the polynomial as expressing the number 6^6 in base 35. (Base as in decimal, binary, etc).
     
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