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Anonymous Sun 15 Nov 23:04:59 2020 No.12346448 View Reply Original Report

Quoted By: >>12346498 >>12346986 >>12347939 >>12348123 >>12348730 >>12349379 >>12349527 >>12349640 >>12349973 >>12350858 >>12351338

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

Anonymous

Anonymous Sun 15 Nov 2020 23:11:14 No.12346487 Report

Quoted By: >>12346530 >>12346710 >>12348887

.9956+.0093i
It comes from eulers identity, if you're working in the reals just say it doesn't exist

Anonymous

Anonymous Sun 15 Nov 2020 23:13:27 No.12346498 Report

Quoted By: >>12346530 >>12349247

>>12346448
(-1)^(1.7)= (-1)^1 x (-1)^(.5) x (-1)^(.2) = i x (-1)

=-i

Anonymous

Anonymous Sun 15 Nov 2020 23:17:49 No.12346530 Report

Quoted By: >>12346550

>>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

Anonymous Sun 15 Nov 2020 23:20:13 No.12346550 Report

Quoted By: >>12346757

>>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

Anonymous Sun 15 Nov 2020 23:38:08 No.12346710 Report

Quoted By: >>12346799 >>12349519

>>12346487
>.9956+.0093i
That's wrong though. 1.7 should put it a bit more imaginary to 1/sqrt(2) (1-i)

Anonymous

Anonymous Sun 15 Nov 2020 23:44:50 No.12346757 Report

Quoted By:

>>12346550
oh that makes sense

Anonymous

Anonymous Sun 15 Nov 2020 23:49:33 No.12346799 Report

Quoted By:

>>12346710
bro don't argue with math

Anonymous

Anonymous Mon 16 Nov 2020 00:10:02 No.12346986 Report

Quoted By: >>12347116 >>12347622

>>12346448
brainlet here

what the actual fuck is going on

Anonymous

Anonymous Mon 16 Nov 2020 00:20:40 No.12347116 Report

Quoted By: >>12347421

>>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

Anonymous Mon 16 Nov 2020 00:37:32 No.12347421 Report

Quoted By:

>>12347116
thank you

Anonymous

Anonymous Mon 16 Nov 2020 00:49:11 No.12347622 Report

Quoted By:

>>12346986
https://www.google.com/search?q=(%2D1)%5E(1.7)

Anonymous

Anonymous Mon 16 Nov 2020 01:24:34 No.12347939 Report

Quoted By: >>12347949 >>12347958 >>12347966 >>12347994 >>12348257 >>12348335 >>12349540

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

Anonymous

Anonymous Mon 16 Nov 2020 01:25:59 No.12347949 Report

Quoted By:

>>12347939
You fucked up

Anonymous

Anonymous Mon 16 Nov 2020 01:28:58 No.12347958 Report

Quoted By: >>12347966

>>12347939
Exponents do not work that way

Anonymous

Anonymous Mon 16 Nov 2020 01:30:14 No.12347966 Report

Quoted By: >>12347973

>>12347939
>>12347958

-1^1.7 = root10((-1)^17)

Anonymous

Anonymous Mon 16 Nov 2020 01:32:30 No.12347973 Report

Quoted By: >>12347983

>>12347966
Except that $-1^{1.7} \neq -1^{0.1}$ .

Anonymous

Anonymous Mon 16 Nov 2020 01:34:36 No.12347983 Report

Quoted By:

>>12347973
Shut up

Anonymous

Anonymous Mon 16 Nov 2020 01:38:53 No.12347994 Report

Quoted By: >>12348052

>>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

Anonymous Mon 16 Nov 2020 01:55:00 No.12348052 Report

Quoted By: >>12348065 >>12349540

>>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

Anonymous Mon 16 Nov 2020 01:59:05 No.12348065 Report

Quoted By: >>12348106

>>12348052
fields dont have fractional powers you retard

Anonymous

Anonymous Mon 16 Nov 2020 02:10:09 No.12348106 Report

Quoted By: >>12348407 >>12349540

>>12348065
real numbers are a field. x^(a/b) is an element y such that y^b = x^a. git gotted

Anonymous

Anonymous Mon 16 Nov 2020 02:15:40 No.12348123 Report

Quoted By:

>>12346448
((-1)^17)^1/10
(-1)^1/10
i^1/5
maybe?

Anonymous

Anonymous Mon 16 Nov 2020 02:54:09 No.12348257 Report

Quoted By: >>12348292

>>12347939
>3 is the quotient of 5 and 2

Anonymous

Anonymous Mon 16 Nov 2020 03:05:57 No.12348292 Report

Quoted By:

>>12348257
>7=1.7

Anonymous

Anonymous Mon 16 Nov 2020 03:17:05 No.12348335 Report

Quoted By:

>>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

Anonymous Mon 16 Nov 2020 03:34:30 No.12348407 Report

Quoted By: >>12348761

>>12348106
what does that have to do with anything
(-1)^(1/2) isnt even a real number

Anonymous

Anonymous Mon 16 Nov 2020 05:03:33 No.12348730 Report

Quoted By: >>12348781 >>12348851

>>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

Anonymous Mon 16 Nov 2020 05:09:33 No.12348761 Report

Quoted By: >>12349540

>>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

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Anonymous Mon 16 Nov 2020 05:18:22 No.12348781 Report

Quoted By: >>12348818

>>12348730
>since 1.7 is odd

Anonymous

Anonymous Mon 16 Nov 2020 05:27:30 No.12348818 Report

Quoted By: >>12348838

>>12348781
1.7 = .35*2 + 1
1.7 is odd

Anonymous

Anonymous Mon 16 Nov 2020 05:32:59 No.12348838 Report

Quoted By:

>>12348818
kek
based retard.

Anonymous

Anonymous Mon 16 Nov 2020 05:36:12 No.12348849 Report

Quoted By: >>12348866 >>12348873 >>12348887 >>12349315

(-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

Anonymous Mon 16 Nov 2020 05:36:58 No.12348851 Report

Quoted By:

>>12348730
you're odd

Anonymous

Anonymous Mon 16 Nov 2020 05:39:11 No.12348866 Report

Quoted By:

>>12348849
GANGSTA RAP NIGGA

Anonymous

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Anonymous Mon 16 Nov 2020 05:40:15 No.12348873 Report

Quoted By:

>>12348849

Anonymous

Anonymous Mon 16 Nov 2020 05:42:57 No.12348887 Report

Quoted By:

>>12346487
wrong, see this
>>12348849

Anonymous

Anonymous Mon 16 Nov 2020 06:24:37 No.12348992 Report

Quoted By: >>12349019

The answer is 9. Can't you mathlet fucks understand pedmas.

Anonymous

Anonymous Mon 16 Nov 2020 06:32:17 No.12349019 Report

Quoted By:

I'll take memes for 500, Alex.
>>12348992
What is 6÷2(1+2)?

Anonymous

Anonymous Mon 16 Nov 2020 06:45:04 No.12349055 Report

Quoted By:

(-1)^(1.7)=10?((-1)^17)=10?(-1)=undef

Anonymous

Anonymous Mon 16 Nov 2020 08:17:43 No.12349247 Report

Quoted By: >>12349315

>>12346498
Why is this wrong? Someone please tell me

Anonymous

Anonymous Mon 16 Nov 2020 08:46:30 No.12349315 Report

Quoted By:

>>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

Anonymous Mon 16 Nov 2020 09:17:17 No.12349379 Report

Quoted By:

>>12346448
>type (-1)^(1.7) into phone calculator
>red text flashes on screen
>"not a number"
scared the fuck outta me

Anonymous

Anonymous Mon 16 Nov 2020 10:28:05 No.12349519 Report

Quoted By:

>>12346710
this, I get
0.5877852522924729-0.8090169943749476i

Anonymous

Anonymous Mon 16 Nov 2020 10:29:39 No.12349527 Report

Quoted By: >>12349563

>>12346448
did we ever see this chick's tits, or no?

Anonymous

Anonymous Mon 16 Nov 2020 10:33:02 No.12349538 Report

Quoted By: >>12349548

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

Anonymous Mon 16 Nov 2020 10:33:15 No.12349540 Report

Quoted By: >>12350649

>>12347939
>>12348052
>>12348106
>>12348761
kill yourself pathetic undergrad

Anonymous

Anonymous Mon 16 Nov 2020 10:36:39 No.12349544 Report

Quoted By:

(-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 -?. 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 -? -1 ? 1 -? -1 ? 1 -?
1-1 ? 1 -? -1 ? 1 -? -1 ?
2 1 -? -1 ? 1 -? -1 ? 1 -?
3-1 ? 1 -? -1 ? 1 -? -1 ?
4 1 -? -1 ? 1 -? -1 ? 1 -?
5-1 ? 1 -? -1 ? 1 -? -1 ?
6 1 -? -1 ? 1 -? -1 ? 1 -?
7-1 ? 1 -? -1 ? 1 -? -1 ?
8 1 -? -1 ? 1 -? -1 ? 1 -?

Anonymous

Anonymous Mon 16 Nov 2020 10:37:49 No.12349548 Report

Quoted By:

>>12349538
disregard that, I brainfarted and typed -1**.5

Anonymous

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Anonymous Mon 16 Nov 2020 10:43:22 No.12349563 Report

Quoted By: >>12349740 >>12350117 >>12350793

>>12349527
>2020
>still no nudes

why is she doing this to me

Anonymous

Anonymous Mon 16 Nov 2020 11:58:30 No.12349640 Report

Quoted By:

>>12346448
Non-ironic subliminal messaging edit. Based or not?

Anonymous

Anonymous Mon 16 Nov 2020 12:56:37 No.12349740 Report

Quoted By:

>>12349563
She's a business woman.

I've seen her tits though. There's one pic where you can see one nipple.

Anonymous

Anonymous Mon 16 Nov 2020 15:14:15 No.12349973 Report

Quoted By: >>12350062

>>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

Anonymous Mon 16 Nov 2020 15:59:29 No.12350062 Report

Quoted By:

>>12349973
>Who do I have to kill in combat in order to win a bride as tribute?
You already lost, fren.

Anonymous

Anonymous Mon 16 Nov 2020 16:18:56 No.12350117 Report

Quoted By:

>>12349563
>she

Anonymous

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Anonymous Mon 16 Nov 2020 16:45:37 No.12350185 Report

Quoted By:

Is this like division with zero? When the universe collapses or you cross the streams?

Anonymous

Anonymous Mon 16 Nov 2020 18:59:51 No.12350649 Report

Quoted By: >>12350669

>>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

Anonymous Mon 16 Nov 2020 19:04:38 No.12350669 Report

Quoted By: >>12350677

>>12350649
there's no need to pick up a book on field theory, because field theory has nothing to do with fractional powers

Anonymous

Anonymous Mon 16 Nov 2020 19:07:41 No.12350677 Report

Quoted By:

>>12350669
It does when the set over two operations is a field.

Anonymous

Anonymous Mon 16 Nov 2020 19:46:15 No.12350793 Report

Quoted By: >>12350879

>>12349563
moar

Anonymous

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Anonymous Mon 16 Nov 2020 20:10:13 No.12350858 Report

Quoted By: >>12350872 >>12353584

>>12346448
$(-1)^{1.7}= e^{1.7\cdot ln(-1)}$
$ln(-1) = \pi \cdot i$ 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.

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

Anonymous

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Anonymous Mon 16 Nov 2020 20:12:46 No.12350872 Report

Quoted By:

>>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

Anonymous Mon 16 Nov 2020 20:13:56 No.12350879 Report

Quoted By:

>>12350793
google image search liz vicious

just don't look at any recent pictures

Anonymous

Anonymous Mon 16 Nov 2020 22:21:02 No.12351338 Report

Quoted By: >>12351612

>>12346448
(-1)^1.7
=(-1)^17/10
=(-1)^17 / (-1)^10
=-1/1
=-1
Everything else is retarded or bait.

Anonymous

Anonymous Mon 16 Nov 2020 23:27:39 No.12351612 Report

Quoted By:

>>12351338
$-1^{\frac {17}{10}} \neq \frac {-1^{17}} {-1^{10}}$
What you want is
$-1^{\frac {17}{10}} = \frac {-1^{17}} {-1^{15.3}} = \frac {-1^{11.7}} {-1^10}}$

Anonymous

Anonymous Tue 17 Nov 2020 07:15:55 No.12353584 Report

Quoted By:

>>12350858
why the logo of Eindhoven University of Technology?

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