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

Joshua Gerrard

Posted 9 years ago

Given i^(2/7), I get the following result: Which doesn't look correct to me. So, I entered "7th root of -1", which gave me this: 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)": 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 Imaginary power 7th root of -1

POSTED BY: Joshua Gerrard

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Greg Hurst

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

Joshua Gerrard

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

Greg Hurst

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

Joshua Gerrard

Posted 9 years ago

Ah right, that explains it. Thanks again!

POSTED BY: Joshua Gerrard

Daniel Lichtblau

Daniel Lichtblau, Wolfram Research

Posted 9 years ago

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

POSTED BY: Daniel Lichtblau

Joshua Gerrard

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

S M Blinder

S M Blinder, WRI and Univ of Michigan

Posted 9 years ago

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]

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

Joshua Gerrard

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

Joshua Gerrard

Posted 9 years ago

Nevermind, found out how to run it.

POSTED BY: Joshua Gerrard

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