Intuitive understanding of Euler's identity?

[HuskyLab]

I'm trying to get a more intuitive understanding of Euler's identity, more specifically, what raising e to the power of i means and why additionally raising by an angle in radians rotates the real value into the imaginary plane. I understand you can derive Euler's formula from the cosx, sinx and ex Taylor series with the addition of i to form the identity. I understand the algebra but is this property only exclusive to e^x ? Does raising let's say 2^{i*pi} mean anything at all? The only thing we are changing is the base, after all e is just a constant. I had a quick look at the Taylor series for 2^x but by a quick comparison, there were some nasty constants, no direct relation seemed apparent. If I said anything that's wrong just me know. Thanks.

 

[andrewkirk]

Does raising let's say 2^{i*pi} mean anything at all?

One way to get an intuitive sense is as follows:

$e^x$ is defined to be
$$\lim_{n\to\infty}\left(1+\frac xn\right)^n$$
So to give a value for $e^{ix}$ we look at the sequence $(s_n)_{n\in\mathbb N}$ such that $s_n=\left(1+\frac {ix}n\right)^n$. We can write $s_n = a_n+b_ni$ in real and imaginary parts and then prove, by expanding each sequence element using the binomial expansion, that for real $x$, the sequences $(a_n)$ and $(b_n)$ converge, say to values $a$ and $b$. Then we conclude that
$$e^{ix} =
\lim_{n\to\infty}\left(1+\frac {ix}n\right)^n
=\lim_{n\to\infty}\left(a_n+ib_n\right)
=\lim_{n\to\infty}a_n+i\lim_{n\to\infty}b_n
=a+ib$$

Having done that, we can then define $2^{ix}=\left(e^{\log 2}\right)^{ix}=e^{i(x\log 2)}$ and apply the definition of $e^{ix}$ that we just made, to get a value for this.

Note that we did not need to use either Taylor series or trig functions anywhere in this.

 

[HuskyLab]

One way to get an intuitive sense is as follows:

$e^x$ is defined to be
$$\lim_{n\to\infty}\left(1+\frac xn\right)^n$$
So to give a value for $e^{ix}$ we look at the sequence $(s_n)_{n\in\mathbb N}$ such that $s_n=\left(1+\frac {ix}n\right)^n$. We can write $s_n = a_n+b_ni$ in real and imaginary parts and then prove, by expanding each sequence element using the binomial expansion, that for real $x$, the sequences $(a_n)$ and $(b_n)$ converge, say to values $a$ and $b$. Then we conclude that
$$e^{ix} =
\lim_{n\to\infty}\left(1+\frac {ix}n\right)^n
=\lim_{n\to\infty}\left(a_n+ib_n\right)
=\lim_{n\to\infty}a_n+i\lim_{n\to\infty}b_n
=a+ib$$

Having done that, we can then define $2^{ix}=\left(e^{\log 2}\right)^{ix}=e^{i(x\log 2)}$ and apply the definition of $e^{ix}$ that we just made, to get a value for this.

Note that we did not need to use either Taylor series or trig functions anywhere in this.

Thanks, is that a log(base 10) in $e^{i(x\log 2)}$? I understand the algebra you mention although I'm trying to gain an understanding of how best to understand imaginary space, particularly its application to wavefunctions expressed in exponential form. One thing I find very strange is that if you include a log constant in the first expression for $e^{ix}$ you get $({e^{log10}})^{ix}$ it doesn't seem to follow any relation in regards to $n^{ix}=\left(e^{\log n}\right)^{ix}=e^{i(x\log n)}$ for n=2,3,4,etc If the log was a natural log on the other hand I do see a pattern.

 

[Mark44]

Thanks, is that a log(base 10) in $e^{i(x\log 2)}$? I understand the algebra you mention although I'm trying to gain an understanding of how best to understand imaginary space, particularly its application to wavefunctions expressed in exponential form. One thing I find very strange is that if you include a log constant in the first expression for $e^{ix}$ you get $({e^{log10}})^{ix}$ it doesn't seem to follow any relation in regards to $n^{ix}=\left(e^{\log n}\right)^{ix}=e^{i(x\log n)}$ for n=2,3,4,etc If the log was a natural log on the other hand I do see a pattern.

I'm reasonably certain that andrewkirk's intention is that log 2 means the natural log of 2. My preference is that if you mean "natural log," you should write ln.

Also, $e^{ix} \ne (e^{\log 10})^{ix}$ unless "log" here means "log₁₀", which is not what andrewkirk was intending.

 

[HuskyLab]

I'm reasonably certain that andrewkirk's intention is that log 2 means the natural log of 2. My preference is that if you mean "natural log," you should write ln.

Also, $e^{ix} \ne (e^{\log 10})^{ix}$ unless "log" here means "log₁₀", which is not what andrewkirk was intending.

Thanks for the reply. Ok, yeah, it's just that it wasn't written as ln in the equation, so one would assume it's $log_{10}$. I have always known log with no subscript to mean $log_{10}$ (because it's the most common), not sure about other people.

 

[Mark44]

Thanks for the reply. Ok, yeah, it's just that it wasn't written as ln in the equation, so one would assume it's $log_{10}$. I have always known log with no subscript to mean $log_{10}$ (because it's the most common), not sure about other people.

It really depends on the context. In more advanced math books, $\log$ can mean $\ln$, the natural logarithm. In some computer science textbooks, $\log$ can mean $\log_2$. When I first learned about logarithms, many years ago, $\log$ meant the common log, $\log_{10}$. I'm told that at least in some cases, this isn't true any longer.

 

[Mark44]

I'm trying to get a more intuitive understanding of Euler's identity, more specifically, what raising e to the power of i means and why additionally raising by an angle in radians rotates the real value into the imaginary plane.

It's really the difference between the exponential function of a real argument versus the exponential function extended to the complex numbers.

As you are aware, the Maclaurin series of $e^x$ is $1 + x + \frac {x^2}{2!} + \dots$. The series for $e^{x + iy}$ can be obtained simply by substitution into the series I showed, and can be manipulated to obtain $e^{x + iy} = \cos(x) + i\sin(y)$. If we let x = 0 and y = $\pi$, it's easy to see that $e^{i\pi} = \cos(0) + i\sin(\pi) = 1$, a very elegant equation involving some of the most fundamental constants in mathematics.

It's also easy to see that $e^{i\theta}$ represents a point on the unit circle identified by letting a ray from the origin to (1, 0) sweep around CCW through an angle of $\theta$ radians.

BTW, the usual description is "complex plane," not "imaginary plane."