2

I know the answer is $e^x$, but supposing I have no idea, what would be the steps required to get to that answer? Note that starting to test different functions is not a valid answer here as there are infinitely many.

I started by doing

$$ f(x)=\frac{d}{dx}f(x) $$

so

$$ \int f(x)dx=f(x) $$

and

$$\int f(x)dx=\frac{d}{dx}f(x)$$ $$f(x)=\frac{d^2}{dx^2}f(x)$$

That leads us to

$$\frac{d^n}{dx^n}f(x)=\frac{d^m}{dx^m}f(x)$$

But that gives no new information...

Garmekain
  • 3,204
  • 1
    $e^x$ is not the only answer $ce^x$ is as well. – fleablood Feb 19 '17 at 18:09
  • $f(x)=f'(x)$ so if $f(x)\neq0$ we get $f'(x)/f(x)=1.$ Integrate both sides now and you will get the answer. – KS_ Feb 19 '17 at 18:17
  • 1
    What you have done tells you that $f(0) = f'(0) = f''(0) = \ldots = f^{(n)}(0)$ so the Taylor series of $f$ about the origin is $f(x) = f(0)\sum_{n=0}^\infty \frac{x^n}{n!}$. This series converges everywhere and this expression does satisfy $f'(x) = f(x)$ so this is a possible way to answer it (this is the power-series representation of $e^x$). There are easier ways to approach this problem as given in the answers below. – Winther Feb 19 '17 at 18:24
  • This gives you a constructive method to find the solution to a broader problem. – Git Gud Feb 19 '17 at 18:30
  • @Winther: Just noticed your comment. – copper.hat Feb 19 '17 at 18:47
  • You may look at the following answer: http://math.stackexchange.com/a/1292586/72031 – Paramanand Singh Feb 21 '17 at 14:21

2 Answers2

4

You can solve this using differential equations. You have: $$\frac{df}{dx}=f$$ This is a separable ODE, so if $f\neq 0$: $$\int \frac{1}{f}~df=\int dx$$ Integrate both sides, and you should get the set of solutions.

projectilemotion
  • 15,914
  • 1
    Isn't differential equations begging the question. As it is based on knowing that $ce^x$ are the only solutions? – fleablood Feb 19 '17 at 19:19
3

Consider the function $\phi(x) = e^{-x} f(x)$, then show that $\phi' = 0$ and from that conclude that $f(x) = \phi(0) e^x$.

Git Gud asks a good question in the comments. Why $e^x$? Here is one way of guessing: Suppose that $f$ has a power series expansion around zero and see what that might be. Note that $f^{(k+1)} = f^{(k)}$, so we have $f'(0) = f(0), f''(0) = f'(0)$, etc. Then we have $f(x) = f(0) \sum_k {k^k \over k!} = f(0) e^x$. Now check it satisfies the ODE.

Note that we can write the above as $(D-I) f = 0$, where $D$ is the differential operator. In general (for linear, time invariant, etc. systems), we can write the operator in the form $p(D) f = 0$, where $p$ is a polynomial. Then we look for a solution of the form $f(x) = e^ {\alpha x}$ and notice $(p(D) f)(x) = p(\alpha) f (x)$ and we see that $f$ solves the system iff $p(\alpha) = 0$. In the above case, we have $p(x) = x-1$, so we try $\alpha = 1$.

copper.hat
  • 178,207
  • Supposing the OP has no idea, how would they know to consider $x\mapsto e^{-x}$ in the product? – Git Gud Feb 19 '17 at 18:28
  • @GitGud: This is a different tack, but I guess I would note that $f'(0) = f(0), f''(0) = f'(0)$, etc, and speculate that $f(x) = \sum_k {x^k \over k!} = e^x$. – copper.hat Feb 19 '17 at 18:39
  • The choice of $e^{-x}$ follows from the concept of an integrating factor which is one of the first concepts thought when starting with differential equations. – Winther Feb 19 '17 at 18:45