# Is #e^x# the unique function of which derivative is itself? Can you prove it?

##
Is #e^x# the uniq function of which derivative is itself? Can you prove it? #f'(x)=f(x)=>f(x)=e^x# , is this true everytime? Is there any second function like this?

Is

##### 1 Answer

Almost. If

Uniqueness of

#### Explanation:

Any function of the form

In order for

Suppose

We can write the Maclaurin series for

#f(x) = sum_(n=0)^oo a_n x^n" "# for some constants#a_0, a_1,...#

Then:

#1 = f(0) = a_0#

and:

#f'(x) = sum_(n=0)^oo (d/(dx) a_n x^n)#

#color(white)(f'(x)) = sum_(n=0)^oo (n a_n x^(n-1))#

#color(white)(f'(x)) = sum_(n=0)^oo (n+1) a_(n+1) x^n#

Since we want

#sum_(n=0)^oo (n+1) a_(n+1) x^n = sum_(n=0)^oo a_n x^n#

So equating the coefficients, we find:

#{ (a_0 = 1), (a_(n+1) = a_n/(n+1)" for "n >= 0") :}#

So:

#1/a_0 = 1/1 = 0!#

#1/a_1 = 1/a_0 = 1!#

#1/a_2 = 2/a_1 = 2!#

#1/a_3 = 3/a_2 = 3!#

and:

#1/a_(n+1) = (n+1) (1/a_n) = (n+1)!#

Hence:

#f(x) = sum_(n=0)^oo x^n/(n!)#

which is one of the definitions of

Then if

#d/(dx) k e^x = k d/(dx) e^x = k e^x#

So:

#f(x) = k e^x#

is also a solution of

**Footnote**

There are some non well behaved functions that also satisfy

For example:

#f(x) = { ("undefined" " if " x=0), (abs(e^x)" if "x != 0) :}#