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Root of n degree of the number n

  1. Jun 3, 2007 #1
    Root of "n" degree of the number "n"

    Hello!
    Frankly speaking I am not sure whether this is proper forum, because I am not sure if this exercise involves calculus or not...

    The task is:
    Consider the numbers "root of "n" of n degree", n >= 2. How many of them are equal? Justify your answer.

    I wanted to use complex numbers:
    [itex]
    n=re^{\mathit{i*A}}
    [/itex]

    [itex]
    w_{k}=\sqrt[{n}]{r} * e^{(i * \frac{(2kpi + A)}{n})}
    [/itex]

    [itex]
    w_{k+1}=w_{k}\ast e^{(i\ast \frac{(2pi)}{n})}
    [/itex]

    But frankly speaking... I do not know that to do with that #-/. I know that for any rational "n", the numbers "after some time" will be the same (the angle A will be over 360 and hence repeat)... Is that good way to follow?

    Thank you for suggestions,
    Theriel
     
    Last edited: Jun 3, 2007
  2. jcsd
  3. Jun 3, 2007 #2

    cepheid

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    You seem to be on the right track. So based on this information, how many roots are distinct?
     
  4. Jun 3, 2007 #3

    HallsofIvy

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    As Cepheid pointed out, the question is "How many distinct nth roots does a number have?" It makes no sense to ask how many roots are equal.
     
  5. Jun 4, 2007 #4
    I was just thinking about the task and maybe the problem is different... Like, how many numbers ARE EQUAL. Just plot the curve for "root of "x" of x degree", n >= 2. First, it is increasing TILL THE NUMBER x=e (is it a coincidence? I do not think so), then it is descending and tends to... 1. Anyone might help with that?
     
    Last edited: Jun 4, 2007
  6. Jun 4, 2007 #5

    NateTG

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    It would probably help you if you were a bit more careful with your grouping symbols...

    As a warm-up exercise, why don't you write out the 3rd, 4th, and 5th roots of 1 and of -1. That might clarify a few things for you.
     
  7. Jun 4, 2007 #6
    NateTG:

    What do you mean by the third root of 1 and -1 (sooo many ambiguities today...) The root of degree 3?

    For 1:
    3rd: 1; -1/2 + i*sqrt(3); -1/2-i*sqrt(3),
    4th: 1, i, -1, -i,

    For -1:
    3rd: -1; 1/2- i*sqrt(3); 1/2+i*sqrt(3),
    4th: [sqrt(2)+sqrt(2)i]/2; [-sqrt(2)+sqrt(2)i]/2; [-sqrt(2)-sqrt(2)i]/2; [sqrt(2)-sqrt(2)i]/2

    Sorry, I do not get it? Graphically, the roots are just around the point 0,0, rotated.

    However, it does not satisfy the task in any of its interpretations (root of "n" of degree "n", I suppose... so these both numbers -> degree and the number under root should probably be the same...)
    ------------------

    Coming back to my second interpretation (because I think that THIS IS IT). There is no point in looking into complex numbers. The task is to say how many of roots are equal, not how many solutions they have....

    So, because I have got to know that n must be natural, and because of the fact that I mentioned, only two roots are the same:
    -root of 4 of degree 4 = root of 2 of degree 2

    Now I will probably have to prove it algebraically... So derivatives and monotonicity of the function I suppose....
     
    Last edited: Jun 4, 2007
  8. Jun 4, 2007 #7

    NateTG

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    Oh, you're trying to solve:
    [tex]\sqrt[n]{n}=\sqrt[m]{m}[/tex]
    ?
     
  9. Jun 4, 2007 #8

    Dick

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    Do you want to solve the equation x^x=c for some c and count the number of roots 'x'?
     
  10. Jun 4, 2007 #9
    Noo... Sorry guys for making problem. The task is:
    Consider the numbers n^(1/n), n>=2. How many of them are equal? Justify your answer.

    I found the second approach. The function above is decreasing from e to infinity and tends to 0. Hence, the values between 2 and 'e' repeat. Hence, there might be a value for natural number (because n is natural) which repeats. And there is. 2^(1/2) = 4 (1/4).
     
    Last edited: Jun 4, 2007
  11. Jun 4, 2007 #10

    Dick

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    Wow. You solved it before I even figured out what the question was!
     
  12. Jun 4, 2007 #11
    That's the magic of this forum ;-]. Some special vibrations helped me ^^.
    Yeah, now the point is to justify it algebraically ;-].

    So... First we check the monotonicity of the function x^(1/x). To do that, we need some bad-looking derivative of this function:
    f' = (1/x)*x^((1/x)-1)-(x^(1/x))*lnx*(1/(x^2)).
    The function changes monotonicity at f' = 0. Great!
    (1/x)*x^((1/x)-1)=(x^(1/x))*lnx*(1/(x^2))

    So, maybe:
    [tex] x = e^{(\frac{(\frac{1}{x} * x^{(\frac{1}{x}-1)})}{(x^{(\frac{1}{x})} * (\frac{1}{x^{2}}))})} [/tex]

    We derive easily that the exponent is 1, hence x = e.... So... thank you very much for your help. At least for a second ;-D. Once again, thanks for all your answers! I feel much better knowing that there is somebody who may help me ;-].
     
    Last edited: Jun 4, 2007
  13. Jun 4, 2007 #12

    Dick

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    You differentiation isn't going so well. x^(1/x-1) has no rule to back it up. Use x^(1/x)=e^(ln(x)/x) and use the chain rule.
     
  14. Jun 4, 2007 #13
    OK, I checked it. Using your way the result is the same (though more 'normally' looking at first glance).
     
    Last edited: Jun 4, 2007
  15. Jun 4, 2007 #14

    Dick

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    I guess it is. Confused me though.
     
  16. Jun 4, 2007 #15
    OK, I did what I said I would and... still, there is a problem. I got to the point, where:

    We know that there is somewhere an argument of this function which must have the same value as x=2. How to find it? I know I have to solve:
    sqrt(2) = x ^ (1/x) but... how? We may square sides and transform the exponent:
    2 = (x^2)/(x^x) but still it does not change anything (at least for me)

    Thank you for help.
     
  17. Jun 4, 2007 #16

    Dick

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    I would give up pretty quickly on trying to find a solution to x^(1/x)=c algebraically. It doesn't look like it can be expressed in terms of elementary functions. It think you would just have to do it numerically or graphically and notice that the answer was pretty close to 4 and then say, "Ah ha, 4 exactly works!".
     
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