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Topic: Euler's formula
Replies: 13   Last Post: Mar 1, 2001 7:53 PM

 Messages: [ Previous | Next ]
 Jerry Uhl Posts: 1,267 Registered: 12/3/04
Re: Euler's formula
Posted: Mar 1, 2001 7:53 PM

There is another way that I have found to be convincing to students
but not to all profs.. Here it is:

Set f[x] = e^(a x) (cos[b x] + i Sin[b x])

Calculate using standard formulas

f'[x] = (a + i b) f[x]

So f'[x] = k f[x] just like the real exponential.

Another place to see Euler's formula come to life in a very
fulfilling way is in the Feynman Lectures .(Addison-Wesley)

-Jerry Uhl

At 7:53 PM -0500 3/1/01, Bill wrote:
> > To derive Euler's formula,
> >
> > e^(ix) = cos x + i sin x,
> >
> > I used: "If f '(x) = k f(x), then f(x) = e^(kx)." Is this right? If so,
> > how may I prove it? The converse is trivial, but I don't find the above
> > statement equally trivial.
> >
> >

>
>I'm sorry, but I don't under how your derivation obtains Euler's formula
>as a result. The proof of your statement procedes along standard
>lines for solving seperable first-order differential equations. More
>trivially,
>split the equation with one side containing the dependent variable and
>the first derivative of said variable and the other side containing
>everything
>else. Take the definite integral of both sides and play around with the
>appropiate constants and you're done.
>
>The traditional proof involves taking the Taylor series of the sin and cos
>functions,
>comparing these with the Taylor series of the exponential function,
>extending the
>domain of the Taylor series to complex numbers, and, depending on the
>context
>in which Euler's formula is taught, either asserting the existence of the
>equality by
>doing some handwaving or using the techniques of analysis to prove
>equivalence.
>
>I believe that this proof is the most satisfactory and elegant; I'm sure
>that one could
>find a convulted geometric proof of the identity or a more sophistocated
>one, but
>the standard proof so integrally ties in with the idea that generally
>speaking,
>analytic functions ARE their Taylor series expansions.
>
>In the end, it's an incredible formula regardless of how one proves it.
>
>-Bill
>
>
>__________________________________________
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Jerry Uhl juhl@ncsa.uiuc.edu
Professor of Mathematics, University of Illinois at Urbana-Champaign
Member, Mathematical Sciences Education Board of National Research
Council
Calculus&Mathematica, Vector Calculus&Mathematica,
DiffEq&Mathematica, Matrices,Geometry&Mathematica, NetMath

http://www-cm.math.uiuc.edu , http://netmath.math.uiuc.edu, and
http://matheverywhere.com

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