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This is more of a maths question than a science question, but I'll ask here anyway, since I doubt it'll be covered at a High School level:

Why, precisely, is e(pi * i) + 1 = 0 ? (I'm aware that e(pi*x) = sin x + cos x, but I'm not sure why that is the case, either.)

Answer 1:

The simple answer is that your second formula is just a bit wrong: it's actually e(i*x) = cos(x) + i*sin(x). If x = pi, then cos (pi) = -1, sin(pi) = 0, and the equation simplifies to e(i*pi) = -1, which can be rearranged to e(i*pi) - 1 = 0.

The more complicated question is why e (i*x) = cos(x) + i*sin(x). This requires some calculus.

First, we have to know the definition of the exponential function exp(x) or ex ( 'exp' is a common way of writing the exponential function 'e^'). The definition of exp(x) is the function that is the derivative of itself. In other words, the value of 'e' is such that the slope of exp (x) is always exp(x). So in calculus notation, if F = exp(x), dF/dx = exp(x) = F. If F = exp (i*x), then using the chain rule, dF/dx = i*exp (x) = i*F.

OK, now suppose we have a function F = cos(x) + i*sin(x). The derivative of cos(x) is -sin (x), and the derivative of sin(x) is cos(x), so dF/dx = -sin(x) + i*cos(x). Now doing some algebra we get: dF/dx = i*( cos(x) - sin(x)/i ) = i*( cos(x) + i*sin(x) ); (to do the last step, remember that i = sqrt(-1), and multiply sin(x)/i by i/i).

So dF/dx = i*( cos(x) + i*sin(x) ) = i*F. This means that the derivatives of both cos(x) + i*sin(x) and exp(i*x) are i times the original function. So the two functions are equivalent, and e(i*x) = cos(x) + i*sin(x).

That is kind of a simple proof of the concept, if you can believe it. More formal proofs are much more complicated (and harder to write down in an email!).

This is a really useful equation for many reasons, mostly because exponential functions are much easier to use mathematically than trigonometric functions. Also, any complex number can be written as the sum of its Real and Imaginary parts, z = Re + Im = x + i*y. This in turn can be written as z = |z|*(cos(p) + i*sin (p)), where p is an angle in the complex number plane. So complex numbers can also be written using simple exponential functions, which is pretty useful, too. Wikipedia has a pretty good article on this, in

complex_number

and

Euler

if you'd like to do some more reading.


Answer 2:

Sine and cosine can be written in mathematical expansion series as the following:

ex = (sum from a = 0 to infinity) (xa)/a! = 1 + x + x2/2 + x3/6 + x4/24 + x5/120 + ...

If you substitute i*x for x, you get alternating positive and negative real and imaginary terms. If you add up all of the real terms, the result is a cosine function in which x is measured in units called radians, of which there are 2*pi in a complete cycle, and likewise the imaginary terms add up to a corresponding sine function, also in radians.

Measured in radians, cos(pi) = -1, because it's half way (i.e. 180 degrees) around the cycle, and similarly sin(pi) = 0. Therefore... e(pi*i) = cos(pi) + i*sin(pi) = -1 + i*0 = -1
-1 + 1 = 0

This function is actually EXTREMELY useful to physics, because light waves manifest as alternating electric and magnetic fields, of which the electric field follows a cosine function and the magnetic field follows a sine function at the same value of theta or x, and in which magnetism is essentially imaginary electricity.


Answer 3:

The best way to show this is with a Taylor Series expansion. Any continuous function can be expanded as an infinite sum of polynomials,

i.e. f(x) = Sum(dn/dxn(f (x))/n!|x=a (x-a)n) over all integer n. By finding the Taylor Series expansions of ex, sin(x) and cos(x), one can show

e(i*x) = cos(x) + i*sin(x). e(i*pi) = cos(pi) + i*sin(pi) = -1 + 0.

Basically, you need some calculus taking derivatives of sin, cos and ex to prove the Taylor Series expansions. Once you know e(i*x) = cos(x) + i*sin(x), though, you can go from there!


Answer 4:

You are right that your question goes beyond the high school. A full explanation could involve calculus, differential equations, and/or complex analysis. I don’t know what level of math you have had, but in my explanation, I will assume that you are familiar with trig functions (sine and cosine) and imaginary numbers.

Euler’s formula says that

e(/i/*x) = cosx + /i/*sinx,

(where x is radians). If we simply accept Euler’s formula, it is easy to see that

e(/i/*pi) = -1, because cos(pi)=-1 and sin(pi)=0.

Now, if you want to understand Euler’s formula, you need to at least know some calculus. For this part of the explanation, I will assume that you are familiar with derivatives.

Any function (meeting certain conditions, which I won’t go into here) can be represented as what is known as an infinite series known as a Taylor series:

f(x) = f(a) + f’(a)*(x-a)/1! + f’’(a)*(x-a) 2/2! +f’’’(a)*(x-a)3/3! + … + f(n) (a)*(x-a)n/n! + …

For the case of a=0, this is known as a MacLaurin series.

Recall that d/dx(e(ax))=a*e (ax) .

So, e(/i/x)=e(0) + [/i/*e(0)*x]/1!+ [(/i/) 2*e(0)*x2]/2!+ [(/i/)3*e^(0)*x3]/3!
+ [(/i/)4*e(0)*x4]/4! + [(/i/)5*e(0)*x5]/5! +[(/i/) 6*e(0)*x6]/6! + …

Simplifying (using the fact that e(0) =1,

/i/1=/i/,

/i/2=-1,

/i/3=-/i/,

/i/4=1,

/i/5=/i/,

/i/6=-1, etc)

we get:

e(/i/x) = 1 +/i/x – x2/2!
- /i/x3/3! + x4/4!
+ /i/x5/5! – x6/6! + …

Rearranging the terms:

e(/i/x) = (1– x2/2! +
x4/4! – x6/6! + …)
+ /i/(x- x3/3! + x5/5! + …)

You can verify that the MacLaurin series for

cosx= 1– x2/2! + x4/4!
– x6/6! + …),
and that the MacLaurin series for sinx= x- x3/3! + x5/5! + …

Therefore e(/i/x)=cosx + /i*/sinx, and we have proven Euler's formula.



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