eix = cos x + isin x
In a previous blog, I showed how it can be derived using the Taylor Series. In today's blog, I will show how it can be derived in an even simpler way using concepts from calculus.
Theorem: Euler's Formula
eix = cos x + isin x
Proof:
(1) For some number x:
let y = cos x + isin x
(2) Taking the first derivative of both sides gives us:
dy/dx = -sin x + icos x
[For details if needed:
(a) dy/dx = d(cos x + isin x)/dx
(b) d(cos x + isin x)/dx = d(cos x)/dx + d(isin x)/dx (see Lemma 3, here)
(c) d(cos x)/dx = -sin x (see Theorem 2, here)
(d) d(isin x)/dx = icos x (see Theorem 1, here) ]
(3) Since i2 = -1 by definition, we have
-sin x + icos x = i(isin x + cos x)
(4) Combining step #1, #2, and #3, we get:
dy/dx = iy
[Since:
(a) y = isin x + cos x [from step #1 above]
(b) dy/dx = -sin x + icos x [from step #2 above]
(c) dy/dx = i(isin x + cos x) [from step #3 above]
(d) dy/dx = i(y)
]
(5) If we multiply (dx/y) to each side, we get:
(1/y)dy = (i)dx
(6) Now, if we take the integral of each side we get:
ln y = ix
[For details if needed:
(a) d(ln x)/dy = 1/x [See Lemma 1, here for proof]
(b) So using the Fundamental Theorem of Calculus (see Theorem 2, here), we know that:
∫ (1/x)dx = ln x + C
(c) So, ∫ (1/y)dy = ln y + C
(d) ∫ i(dx) = ix + C [Since d(ix + C)/dx = i, see Lemma 2, here]
(e) So putting this together gives us:
ln y = ix
]
(7) Now putting e the power of both sides, we get:
y = eix [Since eln y = y]
(8) Now combining step #1 with step #7 we get:
eix = cos x + isin x
QED
References
- "Proof of Euler's Formula", August 15, 2007, Those Who Can Teach Blog.
- "Proof of Euler's Formula (II)", August 15, 2007, Those Who Can Teach Blog.
20 comments :
Dear Larry,
Thanks for linking this to my post. I'm happy that you enjoyed the proof. My student read it and pointed out that it could be made even simpler and I blogged about it here (Proof of euler's formula (part2).)
Hi Teck,
Very cool. I updated the blog with the simpler proof.
Cheers,
-Larry
Thank you for proof. Otherwise I will be crazy that if I didn't get the proof.
I like this proof much better than using Taylor's series. Thanks for showing.
Math is so beautiful. It's a shame more people don't see that. Thanks for the proof, it was very well done and easy to follow. You might consider highlighting the "details" sections in another color so those who follow can just skip those sections easier. Great work though!!
such a beautiful proof, Merci. It gonna help me in better understanding of Fourier series.
This is my most favorite proof ever. But I believe you're forgetting the necessary argument to eliminate the constant of integration, or are you leaving that implicitly? Anyway, it's easy enough:
ln(y) + C = i*x + C
reduces:
ln(y) = C + i*x
Let x = 0:
ln(y) = C
ln(cos(0) + i*sin(0)=1) = C
ln(1) = C = 0
Can you show how to derive
the identity cos jwt = 1/2 * (e^jwt + e^-jwt) ?
jwt = j times omega times time
thanks.
Where is the step 9 i can't see the step 9
Where is the step 9 i can't see the step 9
Hi Ashok,
Sorry about that. That was typo. It should have read step #7. I corrected the text.
Thanks very much for letting me know.
Cheers,
-Larry
Simply superb.. This proof is very natural and comes out of intuition that the exponential 'e' repeats itself when differentiated and so does the trig functions sin(x) and cos(x)...
Nice..
Très très intéressant, mais malheureusement en anglais.
Existe-t-il une version française ?
Merci.
Diasoluka
http://diassites.0pi.com/diasmath/fctchiffres/fctchiffres.html
Désolé, il n'y a pas de version française, mais je ne inclure un lien Babel Fish sur la gauche qui peut traduire le contenu en français.
Voici ce que la traduction de Babel Fish ressemble à ceci:
http://babelfish.yahoo.com/translate_url?doit=done&url=http% 3A% 2F% 2F2007% 2Fmathrefresher.blogspot.com% 2F10% 2Fshorter-proof-of-Eulers-formula.html & lp = en_fr
Cette réponse a été traduit en utilisant des outils Google Langue:
http://www.google.com/language_tools?hl=en
Cheers,
-Larry
thank you very much....
it helped me alot...
thanks again...
Your argument is invalid. You can't play with ln(y) that way for complex numbers. ln(y) is not a single valued function in C.
Hi can you just explain why in step 5 you multiplied dx/y into it? Thanks
I find this equally beautiful: sin(x) + i cos(x) = e^{-ix}
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