# Things that are equal to 1

by dkotschessaa
Tags: equal, things
HW Helper
P: 1,391
 Quote by bahamagreen You mean ##\sqrt{\pi}/2## ? Wolframalpha shows it as that...
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##.
 P: 422 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.
HW Helper
P: 1,391
 Quote by 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?a I'll take a look at Gamma.
'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:

$$\frac{1}{z} = \frac{1}{z}\times 1 = \frac{1}{z} \frac{z^\ast}{z^\ast} = \frac{z^\ast}{|z|^2}.$$
P: 513
 Quote by 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: $$\frac{1}{z} = \frac{1}{z}\times 1 = \frac{1}{z} \frac{z^\ast}{z^\ast} = \frac{z^\ast}{|z|^2}.$$
Cool discussion, and thanks for humoring me. :)

-Dave K
 Mentor P: 9,634 $$0.\bar{9}$$ $$\lim_{n \to \infty} \sqrt[n]{n}$$ More general, for every real a: \lim_{n \to \infty} \sqrt[n]{n^a}
 P: 6 -e^(i*pi*2k) where k is an integer
PF Patron
Thanks
Emeritus
P: 15,671
 Quote by JNeutron2186 -e^(i*pi*2k) where k is an integer
Are you sure about that - ??
P: 6
Pretty sure its a more general euler's equation

 Quote by micromass Are you sure about that - ??
PF Patron
Thanks
Emeritus
P: 15,671
 Quote by JNeutron2186 Pretty sure its a more general euler's equation

What happens if k=0?
 P: 13 Ʃ(1/2^k) from k = 1 to infinity
P: 634
 Quote by micromass What happens if k=0?
I'll agree, they made a minor fail, it should be ##e^{2\cdot k\cdot\pi\cdot i}##, not the negative of that.

 Related Discussions General Discussion 6 Linear & Abstract Algebra 7 Calculus 17 General Discussion 47 General Discussion 10