Message Boards Message Boards

0
|
11840 Views
|
9 Replies
|
0 Total Likes
View groups...
Share
Share this post:
GROUPS:

Is this a bug? i^(2/7) yields different results to 7th root of -1

Posted 9 years ago

Given i^(2/7), I get the following result:

i^(2/7)

Which doesn't look correct to me. So, I entered "7th root of -1", which gave me this:

7th root of -1

Which is what I expected. So I thought, perhaps these aren't equivalent and I've gone wrong somewhere? I entered "(7th root of -1) = i^(2/7)":

equality

WolframAlpha appears to be contradicting itself! Looking more closely, the two seem to be rotations of one another.

Is this a bug, or am I missing something here?

Links

Equality

equality

Imaginary power

i^(2/7)

7th root of -1

7th root of -1

POSTED BY: Joshua Gerrard
9 Replies
Posted 9 years ago

Thank you for finding this. We've fixed this today, and it should make it to the live site in about a week.

POSTED BY: Greg Hurst
Posted 9 years ago

Glad to help! If you're able to tell me, what was causing the issue? Otherwise thank you for such a great tool.

POSTED BY: Joshua Gerrard
Posted 9 years ago

The numerator in the fractional exponent was being overlooked, so you were seeing the root plot for i^(1/7) instead of i^(2/7).

POSTED BY: Greg Hurst
Posted 9 years ago

Ah right, that explains it. Thanks again!

POSTED BY: Joshua Gerrard

i agree something is amiss. Reported it as needing investigation.

POSTED BY: Daniel Lichtblau
Posted 9 years ago

Thank you for your help, it did strike me as odd and my more mathematically adept peers at work agreed. Hopefully this will lead to an improvement in WolframAlpha in some way or form!

POSTED BY: Joshua Gerrard

Noting that E^(I Pi(1+2n))=-1 for {n,0,6}

Do[Print[Re[E^(I Pi (1 + 2 n)/7)], "   ", 
  Im[E^(I Pi (1 + 2 n)/7)]], {n, 0, 6}]

Cos[\[Pi]/7]   Sin[\[Pi]/7]

Sin[\[Pi]/14]   Cos[\[Pi]/14]

-Sin[(3 \[Pi])/14]   Cos[(3 \[Pi])/14]

-1   0

-Sin[(3 \[Pi])/14]   -Cos[(3 \[Pi])/14]

Sin[\[Pi]/14]   -Cos[\[Pi]/14]

Cos[\[Pi]/7]   -Sin[\[Pi]/7]
POSTED BY: S M Blinder
Posted 9 years ago

Is this some form of source code?

Apologies if I'm missing something obvious, I am not particularly knowledgeable in the world of maths (this came from me experimenting with fractional powers of complex numbers to learn about them more deeply).

POSTED BY: Joshua Gerrard
Posted 9 years ago

Nevermind, found out how to run it.

POSTED BY: Joshua Gerrard
Reply to this discussion
Community posts can be styled and formatted using the Markdown syntax.
Reply Preview
Attachments
Remove
or Discard

Group Abstract Group Abstract