Math2.org Math Tables: Complexity |
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Consider the function on the right hand side (RHS)
f(x) = cos( x ) + i sin( x )
Differentiate this function
f ' (x) = -sin( x ) + i cos( x) = i f(x)
So, this function has the property that its derivative is i times the original function.
What other type of function has this property?
A function g(x) will have this property if
dg / dx = i g
This is a differential equation that can be solved with seperation of variables
(1/g) dg = i dx
(1/g) dg = i dx
ln| g | = i x + C
| g | = e^{i x + C} = e^{C} e^{i x}
| g | = C_{2} e^{i x}
g = C_{3} e^{i x}
So we need to determine what value (if any) of the constant C_{3} makes g(x) = f(x).
If we set x=0 and evaluate f(x) and g(x), we get
f(x) = cos( 0 ) + i sin( 0 ) = 1
g(x) = C_{3} e^{i 0} = C_{3}
These functions are equal when C_{3} = 1.
Therefore,
cos( x ) + i sin( x ) = e^{i x}
(This is the usual justification given in textbooks.)
By use of Taylors Theorem, we can show the following to be true for all real numbers:sin x = x - x^{3}/3! + x^{5}/5! - x^{7}/7! + x^{9}/9! - x^{11}/11! + ...Knowing that, we have a mechanism to determine the value of e^{i}, because we can express it in terms of the above series:
cos x = 1 - x^{2}/2! + x^{4}/4! - x^{6}/6! + x^{8}/8! - x^{10}/10! + ...
e^{x} = 1 + x + x^{2}/2! + x^{3}/3! + x^{4}/4! + x^{5}/5! + x^{6}/6! + x^{7}/7! + x^{8}/8! + x^{9}/9! + x^{10}/10! + x^{11}/11! + ...e^(i) = 1 + (i) + (i)^{2}/2! + (i)^{3}/3! + (i)^{4}/4! + (i)^{5}/5! + (i)^{6}/6! + (i)^{7}/7! + (i)^{8}/8! + (i)^{9}/9! + (i)^{10}/10! + (i)^{11}/11! + ...We know how to evaluate an imaginary number raised to an integer power, which is done as such:i^{1} = iWe can see that it repeats every four terms. Knowing this, we can simpliy the above expansion:
i^{2} = -1 terms repeat every four
i^{3} = -i
i^{4} = 1
i^{5} = i
i^{6} = -1
etc...e^(i) = 1 + i - ^{2}/2! - i^{3}/3! + ^{4}/4! + i^{5}/5! - ^{6}/6! - i^{7}/7! + ^{8}/8! + i^{9}/9! - ^{10}/10! - i^{11}/11! + ...It just so happens that this power series can be broken up into two very convenient series:Now, look at the series expansions for sine and cosine. The above above equation happens to include those two series. The above equation can therefore be simplified to
e^(i) =
[1 - ^{2}/2! + ^{4}/4! - ^{6}/6! + ^{8}/8! - ^{10}/10! + ...]
+
[i - i^{3}/3! + i^{5}/5! - i^{7}/7! + i^{9}/9! - i^{11}/11! + ...]e^(i) = cos() + i sin()An interesting case is when we set = , since the above equation becomese^( i) = -1 + 0i = -1.which can be rewritten ase^( i) + 1 = 0. special casewhich remarkably links five very fundamental constants of mathematics into one small equation.
Again, this is not necessarily a proof since we have not shown that the sin(x), cos(x), and e^{x} series converge as indicated for imaginary numbers.