e^pi=-1

par/ par/ par/ Hence:par/
Where is the error?


Thanks,
Yatir

David

Dear Yatir,

I think the problem arises when squaring each member of an equation. Doing so introduces solutions which are not valid for the unsquared equation. For instance, let consider
x-1=0, whose unique solution is x=1, and
x^2-1=0, which is satisfied also by x=-1.
This may be a hint, though I'm not sure it could help in your case.

kind regards, Patrizio

David, Please explain yourself, I'm not familiar with the fact that "the logarithm function isn't single-valued", and why is that?


Thanks,
Yatir

I'm sorry but I didn't understand what you meant.
Please explain detail more detail.

Thanks,
Yatir

Sorry.

Suppose we define a function f: P -> C where P is the set of all the people alive in the world and C is the set of all countries in the world. Let f(p) be the country of birth of person p. (Leaving aside people born on frontiers in aeroplanes or at sea it is pretty clear that f is well defined). The problem of constructing an inverse function to f is analagous to the problem of constructing an inverse to exp. What is bothering is that most (with out much doubt all) countries have provided a place of birth for more than one living person. It doesn't make sense to say 'Given the country Nigeria find me the person born there' in the same way it makes no sense to say 'Given that exp(y)=1 find the value of y.'

to add to what william has said....

real number set is a subset of complex number set..now let us take
a=r×e^iq
now if we take log then,
loga=logr+iq
but,
r×e^iq=r×e^i(2np+q)... where n is any real number (you can check this result fairly easily)
therefore,
loga=logr+i(2np+q)
so as you see log function is multi valued...

love arun

I'm sorry if I didn't explain myself right. I know what a multivalued function is, what I meant is why is e^ix is this kind of function and e^x is not.
Why is e^iq=e^i(2np+q)?
Thanks,
Yatir

When you specify a function, you must specify its domain. e^x , viewed as a function from the reals R to R, is one-to-one and is invertible on its range, which is the set of positive reals. Thus it posesses an inverse, the log function, which is single-valued if you take the domain of the exponential function to be R.

However, e^z viewed as a function from the complex numbers C to C is not one-to-one (the technical term is that it is non-injective). Thus it doesn't have a proper, single-valued inverse, and the log function doesn't have a unique value in this case.

If you want, you could compare this with the function f(x) = x² . If we regard f as a function with domain the positive reals R⁺ , f : R⁺ -> R⁺ is injective (and invertible). So its inverse (the square root function) is well-defined, i.e. single-valued.

However, if we extend the domain of f to the whole real line, f : R -> R⁺ , f is not invertible, as there are two distinct values x such that x^2 = 1 (for example). Note that the codomain is still R⁺ in this example; the problem is not that -1 doesn't have a square root - this is not important as -1 isn't in the codomain. The problem is that +1 has two square roots, so the inverse is not uniquely defined.
Hence in this case the square-root function doesn't have a unique value.

So, how is it proven, without cosidering the fact that:
e^ix =cosx+isinx
the e^ix is periodic and that cycles every 2pi

Thanks,
Yatir

Well, the definition of e^x more or less immediately gives the result e^ix = cos x + i sin x, so this is more or less the only way to prove the periodicity.

David

If you want a heuristic argument, using the fact that d/dx(e^{m x} ) = m e^x, you can think about a particle travelling around a circle in the Argand diagram. Specifically, if a particle has position z = A e^{i t} where A > 0 is real, then dz/dt = i z so the particle is travelling in a direction perpendicular to its position vector from the origin therefore it is travelling in a circle. So it's distance from the origin stays fixed at A, and its speed |dz/dt| = |i z| = A. It has a distance of 2pA to travel once round the circle and it is going at speed A so the time it takes to loop once round the circle is 2p. Therefore the function A e^{i t} is periodic in t with period 2p.

cosz=(e^{i z}+e^{-i z})/2, sinz=(e^{i z}-e^{-i z})/(2i)

and proceed to prove that they are periodic with period 2p, where p is the minimal positive root of sinz = 0, from their absolutely convergent power series. Finally, we conclude that e^{i z} must be periodic (since it is the sum of two periodic functions having the same period). This is the only way I know of which does not take the function cos and sin to be periodic as their starting point.

Kerwin

Thanks Guys,

Yatir