Question

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?

Answer 1

Almost. If `k` is any constant then `f(x) = k e^x` satisfies `f'(x) = f(x)`.

Uniqueness of `e^x` is given by requiring `f(x)` to be defined everywhere, and `f(0) = 1`.

Explanation:

Any function of the form `f(x) = ke^x` satisfies `f'(x) = f(x)`

In order for `e^x` to be the only solution we need an extra condition such as `f(0) = 1`

Suppose `f(x)` is a well behaved function from `RR` to `RR` with `f(0) = 1`.

We can write the Maclaurin series for `f(x)` as:

`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 `f'(x) = f(x)`, we have:

`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 `e^x`

Then if `k` is any constant, we find:

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

So:

`f(x) = k e^x`

is also a solution of `f(x) = f'(x)`

Footnote

There are some non well behaved functions that also satisfy `f'(x) = f(x)`

For example:

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