# Question : Why 0! = 1 ?

1. Jul 29, 2004

### sunny86

ln(ln(ln(ln x+4 ))) = e

Last edited: Jul 30, 2004
2. Jul 29, 2004

3. Jul 29, 2004

### sunny86

Thanks a lot...

4. Jul 30, 2004

### Saint Medici

I'm slightly confused...are you asking why 0!=1, or are you asking what the solution for "x" is in the problem $$ln(ln(ln(ln [x+4])))=e$$? 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." $$ln(ln(ln(ln [x+4])))=e$$ turns into e^$$[ln(ln(ln(ln [x+4])))]=e^e$$, which is really $$ln(ln(ln [x+4])))=e^e$$ 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?

5. Jul 31, 2004

6. Jul 31, 2004

### Zurtex

$$x = e^{e^{e^{e^e}}} - 4$$

$$x \approx 5.14843556 \cdot 10^{23}$$

7. Aug 2, 2004

### sunny86

Ok..thank you..

8. Aug 4, 2004

### Petrushka

That value of x certainly doesn't work in my calculator, or on Mathematica.

9. Aug 4, 2004

### Zurtex

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.

10. Aug 4, 2004

### Zurtex

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:

$$x \approx e^{5 * 10^{1000000}}$$

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:

$$\ln \ln \ln \ln x \approx e$$

Therefore if my answer is right:

$$\ln \ln \ln \ln e^{5 * 10^{1000000}} \approx e$$

$$\ln \ln \ln 5 * 10^{1000000} \approx e$$

$$\ln \ln [1000000 \ln 50] \approx e$$

I put this into my calc and got:

$$\ln \ln [1000000 \ln 50] = 2.71995 \ldots$$

Last edited: Aug 4, 2004
11. Aug 4, 2004

### Gokul43201

Staff Emeritus
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.

12. Aug 4, 2004

### dannyboy

Hiya.

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 ($$\Gamma(n)$$) come in! Basically, it can be shown that

$$\\\Gamma(n-1)=\int^{\infty}_{0}x^{n}\exp(-x)dx=n!$$
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
$$\Gamma(-1)=\int^{\infty}_{0}\exp(-x)dx=1=0!$$;
i.e. 0!=1. (QED)
In general,
$$\Gamma(n-1)=n!$$;
it can also be shown that
$$\Gamma(n+1)=n\Gamma(n)$$, or $$\Gamma(n)=(1/n)\Gamma(n+1)$$;
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
$$\Gamma(1/2)=\sqrt{\pi}$$.

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

"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

Last edited: Aug 5, 2004