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Things that are equal to 1

by dkotschessaa
Tags: equal, things
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Mute
#19
Feb26-13, 11:00 AM
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P: 1,391
Quote Quote by bahamagreen View Post
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##.
bahamagreen
#20
Feb26-13, 11:04 AM
P: 529
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
#21
Feb26-13, 11:49 AM
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Quote Quote by bahamagreen View Post
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:

$$\frac{1}{z} = \frac{1}{z}\times 1 = \frac{1}{z} \frac{z^\ast}{z^\ast} = \frac{z^\ast}{|z|^2}.$$
dkotschessaa
#22
Feb27-13, 02:37 PM
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Quote Quote by Mute View Post
'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
mfb
#23
Feb27-13, 03:32 PM
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P: 11,576
$$0.\bar{9}$$
$$\lim_{n \to \infty} \sqrt[n]{n}$$
More general, for every real a:
$$\lim_{n \to \infty} \sqrt[n]{n^a}
JNeutron2186
#24
Feb27-13, 04:39 PM
P: 7
-e^(i*pi*2k) where k is an integer
micromass
#25
Feb27-13, 04:41 PM
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Quote Quote by JNeutron2186 View Post
-e^(i*pi*2k) where k is an integer
Are you sure about that - ??
JNeutron2186
#26
Feb27-13, 04:44 PM
P: 7
Pretty sure its a more general euler's equation

Quote Quote by micromass View Post
Are you sure about that - ??
micromass
#27
Feb27-13, 04:45 PM
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Quote Quote by JNeutron2186 View Post
Pretty sure its a more general euler's equation

What happens if k=0?
Gackhammer
#28
Feb27-13, 11:13 PM
P: 13
Ʃ(1/2^k) from k = 1 to infinity
Whovian
#29
Feb28-13, 10:55 AM
P: 642
Quote Quote by micromass View Post
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.


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