I Intuitive understanding of Euler's identity?

AI Thread Summary
Euler's identity connects the exponential function with trigonometric functions, illustrating how raising e to the power of iθ results in a rotation in the complex plane. The discussion clarifies that while Euler's formula is derived from Taylor series, it can also be understood through limits and binomial expansion without relying on trigonometric functions. The conversation explores whether raising other bases, like 2, to an imaginary power holds significance, concluding that it can be expressed in terms of e using logarithms. The distinction between natural logarithm and common logarithm is also highlighted, emphasizing the importance of context in mathematical notation. Understanding these concepts is crucial for grasping the applications of complex exponentials in fields like quantum mechanics.
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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.
 
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HuskyLab said:
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.
 
andrewkirk said:
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.
 
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HuskyLab said:
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.
 
Mark44 said:
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.
 
HuskyLab said:
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.
 
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HuskyLab said:
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."
 
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