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/sci/ - Science & Math

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Anonymous Sun Nov 15 18:04:59 2020 No.12346448 [Reply] [Original]

What is (-1)^(1.7) and why?

>>	Anonymous Sun Nov 15 18:11:14 2020 No.12346487 .9956+.0093i It comes from eulers identity, if you're working in the reals just say it doesn't exist

>>	Anonymous Sun Nov 15 18:13:27 2020 No.12346498 >>12346448 (-1)^(1.7)= (-1)^1 x (-1)^(.5) x (-1)^(.2) = i x (-1) =-i

>>	Anonymous Sun Nov 15 18:17:49 2020 No.12346530 >>12346487 but .9956 and .0093 dont add to 1 so how is it possible? >>12346498 whoa this makes no sense but let's go with it!

>>	Anonymous Sun Nov 15 18:20:13 2020 No.12346550 >>12346530 Why would they have to add to one? If you square them they should add up to 1 cause that's how the Pythagorean theorem works but I rounded so it's gonna be slightly less

>>	Anonymous Sun Nov 15 18:38:08 2020 No.12346710 >>12346487 >.9956+.0093i That's wrong though. 1.7 should put it a bit more imaginary to 1/sqrt(2) (1-i)

>>	Anonymous Sun Nov 15 18:44:50 2020 No.12346757 >>12346550 oh that makes sense

>>	Anonymous Sun Nov 15 18:49:33 2020 No.12346799 >>12346710 bro don't argue with math

>>	Anonymous Sun Nov 15 19:10:02 2020 No.12346986 >>12346448 brainlet here what the actual fuck is going on

Anonymous Sun Nov 15 19:20:40 2020 No.12347116

>>12346986
Doing exponential "^" with a non full number (1, 2, 3... 2 = square, 3 = cubic...) and a negative number at base "-1" can't be done with normal numbers and calculation. It's just not defined. Electronic calculator will deliver "Error".

You need imaginary number called "i" to do it.

Sorry for bad language, me from Austria.

>>	Anonymous Sun Nov 15 19:37:32 2020 No.12347421 >>12347116 thank you

>>	Anonymous Sun Nov 15 19:49:11 2020 No.12347622 >>12346986 https://www.google.com/search?q=(%2D1)%5E(1.7)

Anonymous Sun Nov 15 20:24:34 2020 No.12347939

>>12346448
Consider the fact that [math]2^{3}=\frac {2^{5}} {2^{2}}[/math]. It's obvious that you can exchange any single exponent into a pair of exponents for which the quotient is the original exponent.
Therefore:
[math]-1^{1.7} = \frac {-1^{17}} {-1^{10}}[/math].

>>	Anonymous Sun Nov 15 20:25:59 2020 No.12347949 >>12347939 You fucked up

>>	Anonymous Sun Nov 15 20:28:58 2020 No.12347958 >>12347939 Exponents do not work that way

>>	Anonymous Sun Nov 15 20:30:14 2020 No.12347966 >>12347939 >>12347958 -1^1.7 = root10((-1)^17)

>>	Anonymous Sun Nov 15 20:32:30 2020 No.12347973 >>12347966 Except that [math]-1^{1.7} \neq -1^{0.1}[/math].

>>	Anonymous Sun Nov 15 20:34:36 2020 No.12347983 >>12347973 Shut up

>>	Anonymous Sun Nov 15 20:38:53 2020 No.12347994 >>12347939 >It's obvious that you can exchange any single exponent into a pair of exponents for which the quotient is the original exponent. no it isnt, because that isnt even true

>>	Anonymous Sun Nov 15 20:55:00 2020 No.12348052 >>12347994 Yes it's completely true. I won't get into orders of elements in a field, because your tiny brain couldn't handle it.

>>	Anonymous Sun Nov 15 20:59:05 2020 No.12348065 >>12348052 fields dont have fractional powers you retard

>>	Anonymous Sun Nov 15 21:10:09 2020 No.12348106 >>12348065 real numbers are a field. x^(a/b) is an element y such that y^b = x^a. git gotted

>>	Anonymous Sun Nov 15 21:15:40 2020 No.12348123 >>12346448 ((-1)^17)^1/10 (-1)^1/10 i^1/5 maybe?

>>	Anonymous Sun Nov 15 21:54:09 2020 No.12348257 >>12347939 >3 is the quotient of 5 and 2

>>	Anonymous Sun Nov 15 22:05:57 2020 No.12348292 >>12348257 >7=1.7

>>	Anonymous Sun Nov 15 22:17:05 2020 No.12348335 >>12347939 you fool, it'd be (-1)^(17/10), if you're into that shit. the real question is does the 17 come first, or the 1/10

>>	Anonymous Sun Nov 15 22:34:30 2020 No.12348407 >>12348106 what does that have to do with anything (-1)^(1/2) isnt even a real number

>>	Anonymous Mon Nov 16 00:03:33 2020 No.12348730 >>12346448 Holy shit you bunch of brainlets y=(-1)^n If n is even y=1 If n is odd y=-1 If n=1.7 , since 1.7 is odd, y=(-1)^1.7=-1 QED

>>	Anonymous Mon Nov 16 00:09:33 2020 No.12348761 >>12348407 >fields don't have fractional powers debunked. they can, and R+ and C provably ALWAYS have fractional powers >what does that have to do with anything C is a field as well.

>>	Anonymous Mon Nov 16 00:18:22 2020 No.12348781 File: 22 KB, 512x512, brainlet3.jpg [View same] [iqdb] [saucenao] [google] >>12348730 >since 1.7 is odd

>>	Anonymous Mon Nov 16 00:27:30 2020 No.12348818 >>12348781 1.7 = .35*2 + 1 1.7 is odd

>>	Anonymous Mon Nov 16 00:32:59 2020 No.12348838 >>12348818 kek based retard.

Anonymous Mon Nov 16 00:36:12 2020 No.12348849

(-1)^17/10
do you see that ^1/10, nigger?
this is like (-1)^17(1/10)
so (-1)^17 will always be an odd an negative
it’s like saying [(-1)^17]^(1/10)
now you have (-1)^(1/10)
now we have this last part and
we derive this with de moivre’s to find all its roots
electrical engineering nigger here
that’s why

>>	Anonymous Mon Nov 16 00:36:58 2020 No.12348851 >>12348730 you're odd

>>	Anonymous Mon Nov 16 00:39:11 2020 No.12348866 >>12348849 GANGSTA RAP NIGGA

>>	Anonymous Mon Nov 16 00:40:15 2020 No.12348873 File: 561 KB, 1136x640, B968DA2D-34AA-4251-AE42-C4BA7B05BFBA.png [View same] [iqdb] [saucenao] [google] >>12348849

>>	Anonymous Mon Nov 16 00:42:57 2020 No.12348887 >>12346487 wrong, see this >>12348849

>>	Anonymous Mon Nov 16 01:24:37 2020 No.12348992 The answer is 9. Can't you mathlet fucks understand pedmas.

>>	Anonymous Mon Nov 16 01:32:17 2020 No.12349019 I'll take memes for 500, Alex. >>12348992 What is 6÷2(1+2)?

>>	Anonymous Mon Nov 16 01:45:04 2020 No.12349055 (-1)^(1.7)=10√((-1)^17)=10√(-1)=undef

>>	Anonymous Mon Nov 16 03:17:43 2020 No.12349247 >>12346498 Why is this wrong? Someone please tell me

Anonymous Mon Nov 16 03:46:30 2020 No.12349315

>>12349247
It’s wrong because he’s making the assumption that (-1)^(1/10) has only one root, which is not true.

see >>12348849
de moivre’s theorem on roots of fractional powers explains the answer OP is getting. This is taught in the second year of engineering

>>	Anonymous Mon Nov 16 04:17:17 2020 No.12349379 >>12346448 >type (-1)^(1.7) into phone calculator >red text flashes on screen >"not a number" scared the fuck outta me

>>	Anonymous Mon Nov 16 05:28:05 2020 No.12349519 >>12346710 this, I get 0.5877852522924729-0.8090169943749476i

>>	Anonymous Mon Nov 16 05:29:39 2020 No.12349527 >>12346448 did we ever see this chick's tits, or no?

>>	Anonymous Mon Nov 16 05:33:02 2020 No.12349538 For some reason Python seems to think (-1)**0.5 is -1.0, but you get the sensible answer if you use (-1+0j).

>>	Anonymous Mon Nov 16 05:33:15 2020 No.12349540 >>12347939 >>12348052 >>12348106 >>12348761 kill yourself pathetic undergrad

Anonymous Mon Nov 16 05:36:39 2020 No.12349544

(-1)^yx.n
y can be any sequence of numbers, doesn't change a thing. x is vertical in the table, n is horizontal.
>b-but (-1)^1.7 is undefined!!!
Seethe, I defined it as -Ɨ. I chose a capital I with strike to differentiate it from i but indicate it has sth to do with it. For yx being negative, the symbols are reversed (+ to -, - to +)
0 1 2 3 4 5 6 7 8 9
0 1 -Ɨ -1 Ɨ 1 -Ɨ -1 Ɨ 1 -Ɨ
1-1 Ɨ 1 -Ɨ -1 Ɨ 1 -Ɨ -1 Ɨ
2 1 -Ɨ -1 Ɨ 1 -Ɨ -1 Ɨ 1 -Ɨ
3-1 Ɨ 1 -Ɨ -1 Ɨ 1 -Ɨ -1 Ɨ
4 1 -Ɨ -1 Ɨ 1 -Ɨ -1 Ɨ 1 -Ɨ
5-1 Ɨ 1 -Ɨ -1 Ɨ 1 -Ɨ -1 Ɨ
6 1 -Ɨ -1 Ɨ 1 -Ɨ -1 Ɨ 1 -Ɨ
7-1 Ɨ 1 -Ɨ -1 Ɨ 1 -Ɨ -1 Ɨ
8 1 -Ɨ -1 Ɨ 1 -Ɨ -1 Ɨ 1 -Ɨ

>>	Anonymous Mon Nov 16 05:37:49 2020 No.12349548 >>12349538 disregard that, I brainfarted and typed -1**.5

>>	Anonymous Mon Nov 16 05:43:22 2020 No.12349563 File: 29 KB, 430x510, f1d5af177bb95ce15037cd50dc84fe64.jpg [View same] [iqdb] [saucenao] [google] >>12349527 >2020 >still no nudes why is she doing this to me

>>	Anonymous Mon Nov 16 06:58:30 2020 No.12349640 >>12346448 Non-ironic subliminal messaging edit. Based or not?

>>	Anonymous Mon Nov 16 07:56:37 2020 No.12349740 >>12349563 She's a business woman. I've seen her tits though. There's one pic where you can see one nipple.

>>	Anonymous Mon Nov 16 10:14:15 2020 No.12349973 >>12346448 I want to have sex really bad. Who do I have to kill in combat in order to win a bride as tribute?

>>	Anonymous Mon Nov 16 10:59:29 2020 No.12350062 >>12349973 >Who do I have to kill in combat in order to win a bride as tribute? You already lost, fren.

>>	Anonymous Mon Nov 16 11:18:56 2020 No.12350117 >>12349563 >she

>>	Anonymous Mon Nov 16 11:45:37 2020 No.12350185 File: 27 KB, 428x406, markus_thumbs_up.jpg [View same] [iqdb] [saucenao] [google] Is this like division with zero? When the universe collapses or you cross the streams?

>>	Anonymous Mon Nov 16 13:59:51 2020 No.12350649 >>12349540 Why are you so upset bro? Just pick up a book on field theory. It should only take a seasoned grad like you 10 minutes to complete.

>>	Anonymous Mon Nov 16 14:04:38 2020 No.12350669 >>12350649 there's no need to pick up a book on field theory, because field theory has nothing to do with fractional powers

>>	Anonymous Mon Nov 16 14:07:41 2020 No.12350677 >>12350669 It does when the set over two operations is a field.

>>	Anonymous Mon Nov 16 14:46:15 2020 No.12350793 >>12349563 moar

Anonymous Mon Nov 16 15:10:13 2020 No.12350858
File: 46 KB, 1173x320, tuelogo.png [View same] [iqdb] [saucenao] [google]

>>12346448
[math](-1)^{1.7}= e^{1.7\cdot ln(-1)}[/math]
[math]ln(-1) = \pi \cdot i[/math] Use taylor expansion of e and plug in pi*i if you dont believe it.

Using a result from euler which he proved by lookin at taylor expansions we know.

[math]e^{ix} = cos(x)+i\cdot sin(x) \hspace{3em} \forall x \in \mathbb{R} \\ \therefore e^{1.7\cdot ln(-1)} = cos(1.7)+i\cdot sin(1.7) [/math]
using a finite taylor series we can approximate the result as
[math]cos(1.7)+i\cdot sin(1.7)\approx 0.587785252 - 0.809016994 i[/math]

>>	Anonymous Mon Nov 16 15:12:46 2020 No.12350872 File: 1.19 MB, 1631x1816, 803203ed0a04b1d15177fd9b54189dac.png [View same] [iqdb] [saucenao] [google] >>12350858 I have made a mistake in the TeX the content of sines and cosines should be 1.7pi instead of 1.7 I am sorry ther result is still correct

>>	Anonymous Mon Nov 16 15:13:56 2020 No.12350879 >>12350793 google image search liz vicious just don't look at any recent pictures

>>	Anonymous Mon Nov 16 17:21:02 2020 No.12351338 >>12346448 (-1)^1.7 =(-1)^17/10 =(-1)^17 / (-1)^10 =-1/1 =-1 Everything else is retarded or bait.

>>	Anonymous Mon Nov 16 18:27:39 2020 No.12351612 >>12351338 [math]-1^{\frac {17}{10}} \neq \frac {-1^{17}} {-1^{10}} [/math] What you want is [math]-1^{\frac {17}{10}} = \frac {-1^{17}} {-1^{15.3}} = \frac {-1^{11.7}} {-1^10}}[/math]

>>	Anonymous Tue Nov 17 02:15:55 2020 No.12353584 >>12350858 why the logo of Eindhoven University of Technology?

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