Continue to Site

# exponential function periodic or not

Status
Not open for further replies.

#### neils_arm_strong

##### Full Member level 5
My friend argued with me that e^x is periodic with a period of 2πj.I do not agree with him.
Can we define such periodicity where the time period is imaginary? e^x is not even periodic .

hi
if x is Complex then x=a+bi
where, i=sqrt(-1)
hence, exp(x+2ni)=exp(a+bi+2ni)=exp(a+bi)*exp(2ni)=exp(a+bi)
exp(it)=cos(t)+i sin(t)
hence, the function is periodic.
but if x is real the function is not periodic.
hamid karami

### neils_arm_strong

Points: 2
If you plot $e^{j t}$ on the real (x) vs. imaginary plane, you will see that it will be a circle with radius 1 centered at the origin. When time $t=0$ you will be at $x=1,y=0$; at $t=\pi/2$ you will be at $x=0,y=1$; at $t=\pi$ you will be at $x=-1,y=0$; etc ... etc

You see that the point in the complex $x-y$ plane will take $2 \pi$ seconds to get back to where it started at $t=0$. So yes, the function is periodic.

Incidentally, as the point is rotating counter-clockwise in the $x-y$ plane, the instantaneous value of $x$ represents a cosine wave and the instantaneous value of $y$ represents a sine wave -- both with period of $2 \pi$.

Best regards,
$v_c$

Last edited by a moderator:

x is complex,not completely imagianry.

if x=a+jb,j=sqrt(-1), e^x is periodical so long as b is not equal 0

neils_arm_strong said:
My friend argued with me that e^x is periodic with a period of 2πj.I do not agree with him.
Can we define such periodicity where the time period is imaginary? e^x is not even periodic .
i think that exponential function is a nonperiodic...since it keeping on increasing or decreasing even at infinity...

Strictly speaking, exp(x) cannot be considered periodic. By definition, if exp(x) is periodic, then there must be some period, T, for which exp(x + T) = exp(x).

Equivalently, exp(x) = exp(x) * exp(T) => exp(T) = 1. This is only true if T = j*n*2*pi, where n is any integer. However, it does not make much sense to define an imaginary period.

Physics aside, according to mathematics the exponential function can be periodic in complex domain essentially a circle.

common neils we argued this at college and you still dont get it ,e^x is periodic in mathematical and physical science sense but in reality it is not, if you want more of this lets see you in college

i dont know how to visualise this physically, can anyone explain me??

I don't see how e^x can be periodic for any x = a +jb. e^x is periodic if and only if x is a pure imaginary number that is when a= 0.

For a continuous function f(x) to be continuous, f(x) must be equal to f(x+T) where T is a positive real number.

whynot910 said:
if x=a+jb,j=sqrt(-1), e^x is periodical so long as b is not equal 0
If if x=a+jb, e^x is periodic only if a=0 and b≠0. If a <0 the signal is called dumped sine signal, wich means a signal signal whose amplitude decreasy exponetialy. On the other hand, If a>0 the periodic signal e^jb will have its amplitude increasing exponentially until saturation or theoretically until infinite.

a doesnt have to equal 0 for e^x to be periodic...

e^x , where x is a element of C (complex numbers)

this function is periodic long the j (imaginary axis) .

but its not periodic along the real axis.

for example,
e^1000 is going ot be much bigger than e^5, and as you increase
along the real axis , e^x will only get bigger.

But along the imaginary axis, e^x is periodic.

for example, e^(7) = e^(7+j*2*pi) = e^(7+j*2*pi*k) where k is any real integer.

if you restrict x to be a real number, then e^x is NOT periodic.
but if x is complex, then its periodic along the imaginary axis.

I mantain that "a" must be zero, the point here is if the function is periodic or not and not its behavior in a particular axis.

the definition of periodic function is f(t)=f(t + KT) where T is the period and K is an integer number.

Now, tell me that this definition stands for e^(a+jb)t if "a" is not zero.

e^(a+jb)t can be written as (e^at)*(e^jbt) which is the product of a non periodic funtion with a periodic function, now the ultimate question: can the product of a non periodic function with a periodic function results in a periodic function? I don't think so...

in complete agreement with camera.

And for those who think e^x is periodic function for all x=a+jb , with b not equal to zero, please review the definition of a periodic function. It can be found in any signals and systems book.

claudiocamera said:
I mantain that "a" must be zero, the point here is if the function is periodic or not and not its behavior in a particular axis.

the definition of periodic function is f(t)=f(t + KT) where T is the period and K is an integer number.

Now, tell me that this definition stands for e^(a+jb)t if "a" is not zero.

I agree that f(t) = e^(a+jb)t is periodic iff a == 0.

this is different from f(x) = e^(a+jb)x .

if f(x) = e^x then f(x+i*2*pi) = f(x) since e^(x+i*2pi) = e^x.
so f(x) is periodic long the imaginary axis, with a period of 2*pi.

If you look at your quotation about definition of periodic function, it is obvious that i*2pi is not equal to kT where k is an integer number and T is a real positive number.

Yes, there is no such thing as a period of i*2pi... period is always a real positive number. And you cannot say it is periodic along imaginary axis, unless it is a new definition of periodicity introduced by you.....

eecs4ever said:
claudiocamera said:
I mantain that "a" must be zero, the point here is if the function is periodic or not and not its behavior in a particular axis.

the definition of periodic function is f(t)=f(t + KT) where T is the period and K is an integer number.

Now, tell me that this definition stands for e^(a+jb)t if "a" is not zero.

I agree that f(t) = e^(a+jb)t is periodic iff a == 0.

this is different from f(x) = e^(a+jb)x .

if f(x) = e^x then f(x+i*2*pi) = f(x) since e^(x+i*2pi) = e^x.
so f(x) is periodic long the imaginary axis, with a period of 2*pi.

Lets see again the definition of a periodic function:

the definition of periodic function is f(t)=f(t + KT) where T is the period and K is an integer number.

if f(x) = e^x then f(x+i*2*pi) = f(x) since e^(x+i*2pi) = e^x.

I agree but the period woud be i*2*pi and not 2*pi so we would think about having a imaginary period !!! So im my opinion there is a concept bias here. You can analyse e^x considering :

x as a real variable
x as a complex variable

I've noticed that two seperate problems are being evaluated:
In the change of variables x=a+jb, first consider the trivial case - x is constant.
For this case the exponential is also constant, and at best its period is infinite.
Allowing x to vary linearly for simplicity, the two competing views seem to be:
1) a is constant, but b varies linearly, and
2) both a and b vary linearly (and let's allow separate slopes)

For the first case, exp(a) is of constant amplitude, and exp(jb) is truly periodic.
( remembering that exp(a+jb) = exp(a)exp(jb) )

For the second, we have the case of damped or exponentially growing oscillation.
It is not periodic in the sense that f(a+jb) is never of the same amplitude and phase for some subsequent a and b.

The concept of being "periodic on the imaginary axis" is flawed.
A phasor diagram of the second case would show a spiral in amplitude for all phases - real as well as imaginary.

The problem could certainly be reformulated, though, where true periodicity is achieved. Yet, this is answering a different question (and problem) than the one posed (for example, demodulating the amplitude change by a function of the original function's derivative).

Points: 2