# exponential function periodic or not

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#### 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 .

#### hamidr_karami

##### Banned 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

#### v_c 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,
$v_c$

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#### neils_arm_strong

##### Full Member level 5 x is complex,not completely imagianry.

#### whynot910

##### Junior Member level 2 if x=a+jb,j=sqrt(-1), e^x is periodical so long as b is not equal 0

#### electronics_kumar 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...

#### jayc

##### Member level 3 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.

##### Member level 5 Physics aside, according to mathematics the exponential function can be periodic in complex domain essentially a circle.

#### jraks

##### Newbie level 6 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

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

#### newmin

##### Newbie level 4 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.

#### claudiocamera

##### Full Member level 4 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.

#### eecs4ever

##### Full Member level 3 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.

#### claudiocamera

##### Full Member level 4 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...

#### newmin

##### Newbie level 4 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.

#### eecs4ever

##### Full Member level 3 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.

#### newmin

##### Newbie level 4 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.

#### claudiocamera

##### Full Member level 4 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

#### LouisSheffield

##### Member level 5 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).

### neils_arm_strong

points: 2

#### neils_arm_strong

##### Full Member level 5 Yes.That is what .Thanks for the help

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