- #1
iknowsigularity
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Can anyone provide me with a simple explanation for why e^i(x) acts like this.
Can some explain why e^i(x) = cos(x) + isin(x)?
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- Thread starter iknowsigularity
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- Euler formula Explain
In summary, the exponential function e^ix, when graphed on a unit circle in the complex plane, maps the angle theta to the complex x and y coordinates, with x = cos(theta) and y = sin(theta). This can also be seen through the Taylor expansion of e^ix, which results in real and imaginary terms that correspond to cosine and sine. Additionally, the relation e^(i*pi) = -1 can be obtained by setting x = pi. Another remarkable relation is that i^i = e^(-pi/2). Another explanation for why e^ix acts like this is that its derivative is always i times the function itself, causing it to move at a right angle to its radius vector from 0
- #2
DuckAmuck
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If you draw a unit circle in the complex plane, the exponential maps the angle theta to the complex x and y coordinates. Keep in mind that by definition, x = cos(theta) and y = sin(theta).
e^(i theta) = x + iy
You can also see this when you taylor expand the exponential function. You get real and imaginary terms, and these are your cosine and sine respectively.
This picture demonstrates the concept better than words imo.
https://upload.wikimedia.org/wikipe...'s_formula.svg/2000px-Euler's_formula.svg.png
e^(i theta) = x + iy
You can also see this when you taylor expand the exponential function. You get real and imaginary terms, and these are your cosine and sine respectively.
This picture demonstrates the concept better than words imo.
https://upload.wikimedia.org/wikipe...'s_formula.svg/2000px-Euler's_formula.svg.png
- #3
Ssnow
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If you try using the Taylor expansion of ##e^{ix}## you will find that this is equal to the Taylor expansion of ##\cos{x}## plus ##i## times the Taylor expansion of ##\sin{x}## ...
You can find the proof on the major part of Analysis books. This formula is useful in the representation of a complex number ##z##. The remarkable relation ##e^{i\pi}=-1## can be obtained setting ##x=\pi##.
You can find the proof on the major part of Analysis books. This formula is useful in the representation of a complex number ##z##. The remarkable relation ##e^{i\pi}=-1## can be obtained setting ##x=\pi##.
- #4
DuckAmuck
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another remarkable relation is that i^i ~ 0.2. 
- #5
The Bill
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DuckAmuck said:another remarkable relation is that i^i ~ 0.2.
There's no need to be approximate: [tex]i^i=e^{\text{-}\frac{\pi}{2}}[/tex]
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FactChecker
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Another intuitive explanation: Remember that d/dx( eix ) = i*eix. That means that as real values of x increase from 0, eix starts at ei*0 = 1 and it always moves at an exact right angle to it's current radius vector from 0. So it goes around the unit circle in the complex plane. Looking at its real and imaginary part, you see that they match the cos(x) and i*sin(x), respectively.iknowsigularity said:
- #7
Julia Coggins
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So e^x=1+x+x^2/2!+x^3/3!...continues to infinity
Replacing x with ix 1+ix-x^2/2! - ix^3/3!...continues to infinity
separating the real terms from the imaginary:
1-x^2/2!+x^4/4! - x^6/6!
factor out i in the imaginary terms
i(x-x^3/3!+x^5/5!-x^7/7!)
The real terms match up exactly with the macclaurin expansion of cos(x) whereas the imaginary terms match up with the macclaurin expansion of sin(x) so
e^ix= cos(x) + i(sin(x))
Replacing x with ix 1+ix-x^2/2! - ix^3/3!...continues to infinity
separating the real terms from the imaginary:
1-x^2/2!+x^4/4! - x^6/6!
factor out i in the imaginary terms
i(x-x^3/3!+x^5/5!-x^7/7!)
The real terms match up exactly with the macclaurin expansion of cos(x) whereas the imaginary terms match up with the macclaurin expansion of sin(x) so
e^ix= cos(x) + i(sin(x))
FAQ: Can some explain why e^i(x) = cos(x) + isin(x)?
1. Why is e^i(x) equal to cos(x) + isin(x)?
The equation e^i(x) = cos(x) + isin(x) is known as Euler's formula, and it is a fundamental relationship in complex analysis. It states that the exponential function e^x can be represented as a combination of cosine and sine functions.
2. What is the significance of e^i(x)?
The exponential function e^x is a mathematical constant that appears in many areas of mathematics and science. When the exponent is imaginary, as in e^i(x), it represents a rotation in the complex plane. This is why e^i(x) is equal to cos(x) + isin(x), as it combines the real and imaginary components of the rotation.
3. Can you provide an example of how e^i(x) = cos(x) + isin(x) is used in real-life applications?
Euler's formula has many practical applications, including in electrical engineering, quantum mechanics, and signal processing. One example is in the analysis of AC circuits, where the use of complex numbers, including e^i(x), simplifies calculations and allows for more efficient solutions.
4. Is there a proof for why e^i(x) = cos(x) + isin(x)?
Yes, there are several proofs for Euler's formula, including using Taylor series, differential equations, and geometric interpretations. However, the most common and elegant proof is using the properties of complex numbers, specifically their relationship to the polar form and trigonometric functions.
5. How can I remember the relationship between e^i(x) and cos(x) + isin(x)?
One way to remember Euler's formula is by visualizing the unit circle in the complex plane. The cosine and sine terms represent the x and y coordinates of a point on the circle, and the exponential term represents the angle of rotation. Another way is to remember the mnemonic "Euler's identity: e to the i π is minus 1."