Things that are equal to 1

bahamagreen

You mean ##\sqrt{\pi}/2## ? Wolframalpha shows it as that... but it shows Gamma for (-1/3)! as ##-\gamma (4/3)##

Mute

Nope, notice there's a minus sign: "(-1/2)!", if interpreted as ##\Gamma(1-1/2)##, is equal to ##\sqrt{\pi}##. "(+1/2)!", if interpreted as ##\Gamma(1+1/2)##, is equal to ##\sqrt{\pi}/2##.

bahamagreen

Hmm, I guess I don't understand Gamma...

The results of -(1/2)! and (-1/2)! are different as you indicate.

Gamma takes precedence in the order of operations?
I'll take a look at Gamma.

Mute

'Gamma' is a function, so "##\Gamma(x)##" is the same sort of notation as "f(x)", except that f(x) is a general notation for a function while ##\Gamma(x)## generally refers to a specific function defined in terms of an integral (and the analytic continuation if we consider complex number inputs to the function).

It can be shown that for x = n, where n is an integer, ##\Gamma(n+1) = n!##. The trick with the non-integer factorials comes from abusing this notation in the case where x is not an integer, i.e., writing ##\Gamma(x+1) = x!##. From this it may be easier to see why (-x)! is different from -(x!).

Edit: to keep this post somewhat on the actual topic, one of the forms of 1 that I use often enough is introducing ##1 = z^\ast/z^\ast## when I want to rewrite a complex number ##1/z## in a more convenient form with the imaginary and real parts readily obvious:

$$\frac{1}{z} = \frac{1}{z}\times 1 = \frac{1}{z} \frac{z^\ast}{z^\ast} = \frac{z^\ast}{|z|^2}.$$

dkotschessaa

Cool discussion, and thanks for humoring me. :)

-Dave K

mfb

$$0.\bar{9}$$
$$\lim_{n \to \infty} \sqrt[n]{n}$$
More general, for every real a:
$$\lim_{n \to \infty} \sqrt[n]{n^a}

JNeutron2186

-e^(i*pi*2k) where k is an integer

micromass

Are you sure about that - ??

JNeutron2186

Pretty sure its a more general euler's equation

micromass

What happens if k=0?

Gackhammer

Ʃ(1/2^k) from k = 1 to infinity

Whovian

I'll agree, they made a minor fail, it should be ##e^{2\cdot k\cdot\pi\cdot i}##, not the negative of that.