Ryan Giggs Posted July 9, 2012 Report Share Posted July 9, 2012 As a homework we've been asked to prove Moivre's theorem through induction. Can anyone help me out in this? Reply Link to post Share on other sites More sharing options...
Ryan Giggs Posted July 9, 2012 Author Report Share Posted July 9, 2012 Help? Reply Link to post Share on other sites More sharing options...
Sammie Backman Posted July 9, 2012 Report Share Posted July 9, 2012 (edited) Proposition: (cos x + isin x)n = cos nx + isin nx for any positive integer nFirst observe that (cos x + isin x)¹= cos 1x + isin 1x.Assume that the proposition is true for an arbitrary positive integer k, i.e. that(cos x + isin x)k = cos kx + isin kxWe see that (cos x + isin x)k+1 = (cos x + isin x)(cos x + isin x)k = (cos x + isin x)(cos kx + isin kx) using our assumption.Now you should be able to simplify this expression the using compound angle formulae. After that you just invoke the principle of mathematical induction and write a fancy-pants statement which completes that proof.Good luck!EDIT: I just read through this post and I realized that it is quite a mess. It's late, so I'll blame the lacking grammar and any unclarities on that. Edited July 9, 2012 by Sammie Backman Reply Link to post Share on other sites More sharing options...
Ryan Giggs Posted July 10, 2012 Author Report Share Posted July 10, 2012 Thanks, but I don't see how to factorize the expression for k+1. What is it I must reach after the factorization? Reply Link to post Share on other sites More sharing options...
flinquinnster Posted July 10, 2012 Report Share Posted July 10, 2012 (edited) I myself am a little bit confused about the induction. However, I looked the proof up in my textbook and it makes sense - perhaps you could try finding a proof online. What we had to prove was (cos x + isin x)k+1= cos((k+1)x) + isin((k+1)x).I think the key is from (cos x + isin x)k+1 = (cos x + isin x)(cos x + isin x)k = (cosx + isinx)(coskx + isinkx) [using the assumption] = cosxcoskx + isinkxcosx + isinxcoskx + i2sinxsinkx = cosxcoskx - sinxsinkx + i(sinkxcosx + sinxcoskx).Then you have to use your compound angle formulae (which I nearly forgot myself!).cosxcoskx - sinxsinkx + i(sinkxcosx + sinxcoskx) = cos(x+kx) + i(sin(x+kx)) = cos((k+1)x) + isin((k+1)x).QED! Hopefully that helps. Sorry for the bad formatting, but it's really the best I can do right now. Edited July 10, 2012 by flinquinnster Reply Link to post Share on other sites More sharing options...
iHubble Posted July 10, 2012 Report Share Posted July 10, 2012 One of the IB past math HL exams had a similar question that required a similar proof using matrices Reply Link to post Share on other sites More sharing options...
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