Sunday, October 14, 2007

A Shorter Proof of Euler's Formula

Euler's Formula is the very famous equation:

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


(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


(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




Teck said...

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).)

Larry Freeman said...

Hi Teck,

Very cool. I updated the blog with the simpler proof.



tornado said...
This comment has been removed by the author.
Tornado said...

Thank you for proof. Otherwise I will be crazy that if I didn't get the proof.

Dean M. said...

I like this proof much better than using Taylor's series. Thanks for showing.

Matthew said...

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!!

Elf said...

such a beautiful proof, Merci. It gonna help me in better understanding of Fourier series.

Kevin said...

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
ln(y) = C + i*x

Let x = 0:
ln(y) = C
ln(cos(0) + i*sin(0)=1) = C
ln(1) = C = 0

Chuck said...

Can you show how to derive
the identity cos jwt = 1/2 * (e^jwt + e^-jwt) ?

jwt = j times omega times time


Ashok said...

Where is the step 9 i can't see the step 9

Ashok said...

Where is the step 9 i can't see the step 9

Larry Freeman said...

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.



Sanju said...

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)...

Jean said...

Très très intéressant, mais malheureusement en anglais.

Existe-t-il une version française ?



Larry Freeman said...

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: 3A% 2F% 2F2007% 2F10% 2Fshorter-proof-of-Eulers-formula.html & lp = en_fr

Cette réponse a été traduit en utilisant des outils Google Langue:



kuldeep dhaka said...

thank you very much....
it helped me alot...
thanks again...

rfermat said...

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.

This comment has been removed by the author.

Hi can you just explain why in step 5 you multiplied dx/y into it? Thanks

Unknown said...

I find this equally beautiful: sin(x) + i cos(x) = e^{-ix}