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Question : Why 0! = 1 ?

  1. Jul 29, 2004 #1
    Question : Need your help.

    Can anyone please help me to solve this problem..please.. :yuck: :yuck:

    ln(ln(ln(ln x+4 ))) = e
    Last edited: Jul 30, 2004
  2. jcsd
  3. Jul 29, 2004 #2
  4. Jul 29, 2004 #3
    Thanks a lot...
  5. Jul 30, 2004 #4
    I'm slightly confused...are you asking why 0!=1, or are you asking what the solution for "x" is in the problem [tex]ln(ln(ln(ln [x+4])))=e[/tex]? In the case of the former, use that link. In the case of the latter, all I did was eliminate logarithms by using each side as a power of "e." [tex]ln(ln(ln(ln [x+4])))=e[/tex] turns into e^[tex][ln(ln(ln(ln [x+4])))]=e^e[/tex], which is really [tex]ln(ln(ln [x+4])))=e^e[/tex] and so on until you end up with ..... well, I can't get this to format correctly, but the way I have it pictured in my head, it's a stairstep of 5 e's {e to power e to power e to power e to power e......so, I guess that makes it e^(e^(e^(e^(e))))}. Hm...I haven't slept in awhile, can someone verify me?
  6. Jul 31, 2004 #5
    Please don't edit your posts to ask new questions. Just start a new topic.

  7. Jul 31, 2004 #6


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    Yep your right, the answer is:

    [tex]x = e^{e^{e^{e^e}}} - 4[/tex]

    [tex]x \approx 5.14843556 \cdot 10^{23}[/tex]
  8. Aug 2, 2004 #7
    Ok..thank you..
    I am sorry about the mistake i made..i am soory.
  9. Aug 4, 2004 #8
    Are you sure about that?

    That value of x certainly doesn't work in my calculator, or on Mathematica.
  10. Aug 4, 2004 #9


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    No I am most defintly not sure, but I imagine it would not be very accurate for the orriginal equation. I couldn't find my calculator so I used Google's internal one, I'll have a look around and get back to you.
  11. Aug 4, 2004 #10


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    Yeah the approximation was WAY out. Service to say my calculator couldn't work x out but after a bit of messing about with the formulae to do some trial and error, x is far closer to:

    [tex]x \approx e^{5 * 10^{1000000}}[/tex]

    So it is big lol. (Again I'm not 100% sure I am right but I am more confident than last time).

    Edit: I did a little test to see if I was right. My thought was:

    [tex]\ln \ln \ln \ln x \approx e[/tex]

    Therefore if my answer is right:

    [tex]\ln \ln \ln \ln e^{5 * 10^{1000000}} \approx e[/tex]

    [tex]\ln \ln \ln 5 * 10^{1000000} \approx e[/tex]

    [tex]\ln \ln [1000000 \ln 50] \approx e[/tex]

    I put this into my calc and got:

    [tex]\ln \ln [1000000 \ln 50] = 2.71995 \ldots[/tex]
    Last edited: Aug 4, 2004
  12. Aug 4, 2004 #11


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    I too tried Google but found it didn't go far enough. Zurtex, I think you got your brackets wrong. I say this because I got your answer when I got my brackets wrong. The answer is too big to express as a power of 10. So, I'm sure the point of the question is not to be to write the number in decimal notation.
  13. Aug 4, 2004 #12

    If you're asking why 0!=1, then here is the answer.

    Typically, at high school level, you are told that a factorial is simply the product of all integers down to 1 (e.g. 6!=6*5*4*3*2*1) and that 0! is "defined" to be equal to unity, right?

    That's all well and good for the nonnegative integers; but what if we want to find, say, (-3)! or (1/2)!? The recursive definition doesn't work!

    That's where GAMMA FUNCTIONS ([tex]\Gamma(n)[/tex]) come in! Basically, it can be shown that

    for ALL n larger than -1. (The integral is improper for -1<x<0, but can be shown to converge there.) So, punching n=0 to our new definition for n!, we get
    i.e. 0!=1. (QED)
    In general,
    it can also be shown that
    [tex]\Gamma(n+1)=n\Gamma(n)[/tex], or [tex]\Gamma(n)=(1/n)\Gamma(n+1)[/tex];
    i.e., we have a recursive definition for gamma functions. (Note that this means that the gamma function of any integer less than or equal to zero is infinite; that is, the factorial of a negative integer is infinite.) Also, it can be shown that

    Gamma functions are usually tabulated between 0 and 1; using the recursive formulas above, you can then find, say,

    "Mathematical Methods In The Physical Sciences" (2nd ed., published by Wiley) by Mary Boas explains this beautifully. (I strongly recommend it: an absolute bible to me during my university years!)

    Hope that helps!

    dannyboy :smile:
    Last edited: Aug 5, 2004
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