Physicist: This equation (
First, you’ll find (by plugging them into a graphing calculator and graphing) that:
1)
2)
3)
Where 
These are called “Taylor expansions” of “Sine”, “Cosine”, and “e to the x”. If you were to continue the patterns above forever, then you would find that the equality is exact. There is some very exciting math to back me up on this, but for now just trust.
Know that 
Check it!:
and grouping terms that have “
Holy crap! 
This is the “Euler Equation”. Or one of them at least. Just plug in “
So the trick is Euler’s equation, which is (surprisingly) true.
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whats wrong with this proof:
e^ipi=-1
-e^ipi=1
(-1)*e^ipi=e^0
e^ipi*e^ipi=e^0
e^ipi+ipi=e^0
ipi+ipi=0 (this is false)
@patty. Where did (e to the ipi) squared come from?
look at line 3 to line 4, since (-1) equals (e^ipi), i can rewrite as such, does that help?
Logarithms are either multi-valued or use the ‘principle value’ of the argument which is restricted to -π<ϕ≤π. By this I mean that you have Ln(re^(iθ))=ln(r)+i(θ+2πn) with an arbitrary integer n, or ln(re^(iθ))=ln(r)+iθ where θ satisfies -π<θ≤π.
Because of that, the last line must be either i(π+π+2πn)=i(0+2πm), where n and m are integers, or 0=0. These are both true (for the appropriate values of m and n).
that makes sense, but what if m and n were really high negative #’s( like negative infinity) that we are left with 0^i=1, could this this true?
i^23 = -i, then I can redefine it as (i^4)*^23/4 , so that its value doesn’t change. But since i^4 = 1, then (i^4)*^23/4 will become 1^23/4 which is 1. thus, as we see, we got two answers. How is it possible?
Whenever you see a “1/N” in an exponent, you’re doing an Nth root, which drops N different roots. In this case the solutions you get are 1^23/4 = {1, i, -1, -i}.
The difference comes about because while taking a power produces one answer, but taking a root produces several.
The exponential function over the complex field has period 2pi.
So e^ipi*e^ipi = e^(2pi)i = e^0 = 1
More generally e^(2npi)i = 1 for n = 1, 2, 3, ….
I had always questioned the equation but never actually knew it had a good proof!Thank you so much for the post!
Can a mathematician help a brother out?
How to plot this function?
∫eˣ = f (u ⁿ)
Patty, your proof really vexed me, too. What is wrong with it is step 3, where you changed 1 into e^0. This only occasionally results in a valid solution.
To give you a more general example, suppose you had an equation x^2 = 1. Obviously, the solution is ±1, but would it be valid to say that x^2 = x^0 and then 2 = 0? Of course not!