- #1

HeilPhysicsPhysics

- 16

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How to prove e^ix=cos x + i sin x

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- Thread starter HeilPhysicsPhysics
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- #1

HeilPhysicsPhysics

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How to prove e^ix=cos x + i sin x

- #2

d_leet

- 1,077

- 1

HeilPhysicsPhysics said:How to prove e^ix=cos x + i sin x

One way is to start with the taylor series for e

i

- #3

Data

- 998

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The proof depends on your definitions (for example, if you define [itex]\cos{x}[/itex] as [itex](e^{ix}+e^{-ix})/2[/itex] and [itex]\sin{x}[/itex] as [itex](e^{ix} - e^{-ix})/(2i)[/itex] then it's pretty easy!).

It's also clear if you start from the Taylor series of the three functions about the origin as d_leet suggested.

You can also start from the differential equation for e^x,

[tex]y^\prime = y, y(0)=1.[/tex]

Write

[itex]e^{ix} = f(x) + ig(x),[/itex]

(where f, g are real-valued)

so that [itex]f(0) = 1[/itex] and [itex]g(0) = 0[/itex]

then you must have (by the chain rule)

[tex]ie^{ix} = f^\prime(x) + ig^\prime(x),[/tex]

or, rearranging,

[tex]e^{ix} = \frac{1}{i}f^\prime(x) + g^\prime(x) = g^\prime(x) - if^\prime(x).[/tex]

Since [itex]e^0 = 1[/itex] that means [itex]g^\prime(0) = 1[/itex] and [itex]f^\prime(0) = 0[/itex]. Now differentiate again and you get

[tex]ie^{ix} = g^{\prime \prime}(x) - if^{\prime \prime}(x)[/tex]

or

[tex]e^{ix} = -f^{\prime \prime}(x) - ig^{\prime \prime}(x).[/tex]

Comparing the real and imaginary parts with our original [itex]e^{ix} = f(x) + ig(x)[/itex] you get two differential equations with initial conditions:

[tex]f^{\prime \prime}(x) = -f(x), f^{\prime}(0) = 0, f(0) = 1[/tex]

and

[tex]g^{\prime \prime} (x) = -g(x), g^{\prime}(0) = 1, g(0) = 0.[/tex]

It's easy to check that f(x) = cos(x) and g(x)= sin(x) satisfy these DEs, and by a theorem in the theory of DEs the solutions are unique.

It's also clear if you start from the Taylor series of the three functions about the origin as d_leet suggested.

You can also start from the differential equation for e^x,

[tex]y^\prime = y, y(0)=1.[/tex]

Write

[itex]e^{ix} = f(x) + ig(x),[/itex]

(where f, g are real-valued)

so that [itex]f(0) = 1[/itex] and [itex]g(0) = 0[/itex]

then you must have (by the chain rule)

[tex]ie^{ix} = f^\prime(x) + ig^\prime(x),[/tex]

or, rearranging,

[tex]e^{ix} = \frac{1}{i}f^\prime(x) + g^\prime(x) = g^\prime(x) - if^\prime(x).[/tex]

Since [itex]e^0 = 1[/itex] that means [itex]g^\prime(0) = 1[/itex] and [itex]f^\prime(0) = 0[/itex]. Now differentiate again and you get

[tex]ie^{ix} = g^{\prime \prime}(x) - if^{\prime \prime}(x)[/tex]

or

[tex]e^{ix} = -f^{\prime \prime}(x) - ig^{\prime \prime}(x).[/tex]

Comparing the real and imaginary parts with our original [itex]e^{ix} = f(x) + ig(x)[/itex] you get two differential equations with initial conditions:

[tex]f^{\prime \prime}(x) = -f(x), f^{\prime}(0) = 0, f(0) = 1[/tex]

and

[tex]g^{\prime \prime} (x) = -g(x), g^{\prime}(0) = 1, g(0) = 0.[/tex]

It's easy to check that f(x) = cos(x) and g(x)= sin(x) satisfy these DEs, and by a theorem in the theory of DEs the solutions are unique.

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- #4

Edgardo

- 705

- 15

Let [tex] z = \rm{cos}\theta + i \cdot \rm{sin}\theta [/tex]

Then derive z:

[tex]\frac{\rm{d}z}{\rm{d}\theta} = - \rm{sin}\theta + i \cdot \rm{cos}\theta = -1 (\rm{sin}\theta - i \cdot \rm{cos}\theta) = i^2 (\rm{sin}\theta - i \cdot\rm{cos}\theta ) = i (i \cdot \rm{sin}\theta - i^2 \cdot \rm{cos}\theta) = i(\rm{cos}\theta + i \cdot \rm{sin}\theta) = i \cdot z[/tex]

This is a differential equation:

[tex]\frac{\rm{d}z}{\rm{d}\theta} = i z [/tex]

whose solution is

[tex] z = e^{i \theta} [/tex]

Thus, [tex] \rm{cos}\theta + i \rm{sin}\theta = e^{i \theta} [/tex]

Then derive z:

[tex]\frac{\rm{d}z}{\rm{d}\theta} = - \rm{sin}\theta + i \cdot \rm{cos}\theta = -1 (\rm{sin}\theta - i \cdot \rm{cos}\theta) = i^2 (\rm{sin}\theta - i \cdot\rm{cos}\theta ) = i (i \cdot \rm{sin}\theta - i^2 \cdot \rm{cos}\theta) = i(\rm{cos}\theta + i \cdot \rm{sin}\theta) = i \cdot z[/tex]

This is a differential equation:

[tex]\frac{\rm{d}z}{\rm{d}\theta} = i z [/tex]

whose solution is

[tex] z = e^{i \theta} [/tex]

Thus, [tex] \rm{cos}\theta + i \rm{sin}\theta = e^{i \theta} [/tex]

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- #5

eljose

- 492

- 0

Umm.. in fact i think that the "historical" development fo the formula came with euler when studying the "Harmonic (classical) oscilator"..

[tex] y'' + \omega ^2 y =0 [/tex] trying y=exp(ax) then you get..

[tex] a^2 +\omega ^2 =0 [/tex] so [tex] + \sqrt (-1) \omega =a [/tex] (and the same

with a minus sign)

-On the other hand the long-known solution for the Harmonic movement was [tex] y= Asin (\omega t) +B Cos(\omega t) [/tex] then if you set the initial condition [tex] y(0)=0 [/tex] you get the identity for sin. and using the well-known formula [tex] cos^2 (x) + sen^2 (x) =1 [/tex] oh..if today math were so easier....i would have became famous years ago....

[tex] y'' + \omega ^2 y =0 [/tex] trying y=exp(ax) then you get..

[tex] a^2 +\omega ^2 =0 [/tex] so [tex] + \sqrt (-1) \omega =a [/tex] (and the same

with a minus sign)

-On the other hand the long-known solution for the Harmonic movement was [tex] y= Asin (\omega t) +B Cos(\omega t) [/tex] then if you set the initial condition [tex] y(0)=0 [/tex] you get the identity for sin. and using the well-known formula [tex] cos^2 (x) + sen^2 (x) =1 [/tex] oh..if today math were so easier....i would have became famous years ago....

Last edited:

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