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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.

    cookiemonster
     
  7. Jul 31, 2004 #6

    Zurtex

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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

    Zurtex

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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

    Zurtex

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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

    Gokul43201

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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
    Hiya.

    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

    [tex]\\\Gamma(n-1)=\int^{\infty}_{0}x^{n}\exp(-x)dx=n![/tex]
    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
    [tex]\Gamma(-1)=\int^{\infty}_{0}\exp(-x)dx=1=0![/tex];
    i.e. 0!=1. (QED)
    In general,
    [tex]\Gamma(n-1)=n![/tex];
    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
    [tex]\Gamma(1/2)=\sqrt{\pi}[/tex].

    Gamma functions are usually tabulated between 0 and 1; using the recursive formulas above, you can then find, say,
    [tex](-3/2)!=\Gamma(-1/2)=2\Gamma(1/2)=2\sqrt{\pi}[/tex].

    "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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