I'm looking for a function $f$, whose third derivative is $f$ itself, while the first derivative isn't.

Is there any such function? Which one(s)? If not, how can we prove that there is none?


  • $x\longmapsto c\cdot e^x, c \in R$ are the functions whose derivative is itself.

  • $x\longmapsto \cosh(x)={e^x+e^{-x}\over 2}$ and $x\longmapsto \sinh(x)={e^x-e^{-x}\over 2}$ have their second derivatives equal to themselves.

  • $x\longmapsto f(x)$, has its third derivative equal to itself.

  • $x\longmapsto \cos(x)$ and $x\longmapsto \sin(x)$ have their fourth derivatives equal to themselves.

  • 3
    $\begingroup$ See this. $\endgroup$ – David Mitra Dec 24 '14 at 18:56
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    $\begingroup$ You came tantalisingly close to answering that question by yourself. If you had instead listed the examples as $$e^{1x}$$ $$e^{-1x}$$$$e^{?x}$$$$e^{\pm ix}$$, a pattern becomes apparent. $n$-th derivatives equal to self admit solutions $e^{\alpha x}$ such that $\alpha^n = 1$. As you may see, 1 is a first root of 1, -1 is a square root of 1, and $i$ and $-i$ are fourth roots of 1. You're asking here for the case $n=3$. What is the cube root of 1 then? $\endgroup$ – Iwillnotexist Idonotexist Dec 25 '14 at 7:24
  • $\begingroup$ Now how about that question: Find a function whose fourth derivative is itself. $\endgroup$ – user203255 Dec 26 '14 at 6:32
  • $\begingroup$ @user203255 make it a seperate question, it is interesting enough for that $\endgroup$ – Willemien Dec 27 '14 at 19:28
  • $\begingroup$ I will happily upvote any valid answer that shows some graphs. $\endgroup$ – joeytwiddle Dec 28 '14 at 6:42

Try this: $f(x) = \sum_{m=0}^\infty \frac{x^{3m}}{(3m)!}$. This function is like the exponential function, which can be defined as $e^x=\sum_{m=0}^\infty \frac{x^m}{m!}$, but only taking every third term of the sum. Differentiating term-by-term verifies the desired property.

Generally, the functions $\Lambda_m^k(x) := \sum_{n=0}^\infty \frac{x^{nm+k}}{(nm+k)!}$ for $0\leq k < m$ define the set of $m$ linearly independent solutions to $\frac{d^mf}{df^m} = f$ with $\frac{d^jf}{df^j} \neq f$ for $j < m$. Again, it is easy to see this using term-by-term differentiation, but actually computing the functions in this form is unfeasible. The theory of ODE's ensures that all solutions are linear combinations of these lambda-functions, but not all linear combinations satisfy our second condition.

Note that all the functions you mention can be realized as linear combinations of these lambda-functions:

$e^x = \Lambda_1^0(x)$; $\cosh(x) = \Lambda_2^0(x)$; $\sinh(x) = \Lambda_2^1(x)$;

$\sin(x) = \Lambda_4^1(x) - \Lambda_4^3(x); \cos(x) = \Lambda_4^0(x) - \Lambda_4^2(x)$

PS: If there are an concerns about convergence of these sums, just note that they are essentially the sum for the exponential function but with terms missing. Since the exponential sum is absolutely convergent for all $x$, the sub-sums are also absolutely convergent and may be differentiated term-by-term.

  • 2
    $\begingroup$ Is this an "if and only if" relation? Will all answers be linear combinations of your Lambdas? $\endgroup$ – Richard Dec 25 '14 at 3:58
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    $\begingroup$ They will. It's well known that an ordinary equation of order n has n linearly independent solutions (under some continuity conditions that I don't recall the exact details of but are certainly satisfied here). I have shown n solutions that are linearly independent, so they must generate all solutions. $\endgroup$ – Zach Effman Dec 25 '14 at 4:16
  • $\begingroup$ In the previous comment, should that have been ”ordinary +differential+ equation”? Though it is well known, it is not an area I am familiar with. Is there a reference for where I can learn more? $\endgroup$ – Richard Dec 25 '14 at 5:26
  • $\begingroup$ Sorry, right after I typed that I realized it sounded condescending. I think it's pretty fundamental ODE theory, but I couldn't find a proof with a quick google search (hence well known meaning "I know this, but I can't find a reference"). Wolfram Math World states it without proof here: mathworld.wolfram.com/OrdinaryDifferentialEquation.html $\endgroup$ – Zach Effman Dec 25 '14 at 5:30
  • $\begingroup$ And yes, thank you, that should be ordinary differential equation $\endgroup$ – Zach Effman Dec 25 '14 at 5:30

$$f(x)=e^{\omega x}$$ where $\omega$ is a primitive third root of unity. We have $$f'(x)=\omega e^{\omega x}, ~~ f''(x)=\omega^2 e^{\omega x}, ~~ f'''(x)=\omega^3e^{\omega x}=e^{\omega x}$$

  • 22
    $\begingroup$ real functions $$ e^{-x/2} \; \cos \left( \frac{x \sqrt 3}{2} \right) $$ and $$ e^{-x/2} \; \sin \left( \frac{x \sqrt 3}{2} \right) $$ along with the unwanted $e^x$ $\endgroup$ – Will Jagy Dec 24 '14 at 19:21
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    $\begingroup$ just to clarify how @WillJagy found his answers: one complex third root of unity is $\omega=-\frac12(-1+i\sqrt3)$ so $e^{\omega x}=e^{-x/2}(\cos(x\sqrt{3}/2)+i\sin(x\sqrt{3}/2))$ and since $f$ satisfies $f'''=f$ it follows that $\Re\{f\},\Im\{f\}$ satisfy it as well $\endgroup$ – oldrinb Dec 25 '14 at 4:37

The statement means $f''' = f$. We can solve this differential equation using its characteristic equation:

$$f''' - f = 0 \implies r^3-1 = 0 \implies r^3 = 1$$

for $r \in \mathbb{C}$. So $f(x) = Ce^{rx}$. Noting that $f' = Cre^{rx}$, $f'' = Cr^2 e^{rx}$, and $f'''(x) = Cr^3 e^{rx} = Ce^{rx}$, minding that $r^3 = 1$.

  • $\begingroup$ $C$ can be arbitrary. $\endgroup$ – Hans Lundmark Dec 24 '14 at 18:57
  • $\begingroup$ @NajibIdrissi A typo -- good catch! $\endgroup$ – Emily Dec 24 '14 at 18:58
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    $\begingroup$ explaining what have the r's to do with the f's would help $\endgroup$ – Rolazaro Azeveires Dec 25 '14 at 15:05

Let $F=(f,f',f'')^T$ then we have $F'=(f',f'',f)^T$ so we get $F'=AF$ where

$$A=\begin{pmatrix}0&1&0\\0&0&1\\1&0&0\end{pmatrix}$$ hence we get






$m=1$ or

$m=-0.5\pm \frac{\sqrt{3}}{2}i$

if we take the second term, the solution become

$y=e^{-0.5x}( C_{1}\cos \frac{\sqrt{3}}{2}x+C_{2}\sin \frac{\sqrt{3}}{2}x)$


We want to solve $f''' = f$ with $f' \neq f$. Using standard methods for solving ODE with constant coefficients (some of which are reported here) one obtains

$$ f(x) = \frac{1}{3} e^{-t/2} \left(e^{3 t/2} \left(f''(0)+f'(0)+f(0)\right)+\sqrt{3} \sin \left(\frac{\sqrt{3} t}{2}\right) \left(f'(0)-f''(0)\right)+\cos \left(\frac{\sqrt{3} t}{2}\right) \left(-f''(0)-f'(0)+2 f(0)\right)\right) $$

Now we compute $f'(x)$ :

$$ f'(x) = \frac{1}{3} e^{-t/2} \left(\sqrt{3} \left(f''(0)-f(0)\right) \sin \left(\frac{\sqrt{3} t}{2}\right)+e^{3 t/2} \left(f''(0)+f'(0)+f(0)\right)-\cos \left(\frac{\sqrt{3} t}{2}\right) \left(f''(0)-2 f'(0)+f(0)\right)\right) $$

From which

$$ f(x) - f'(x) = \frac{1}{3} e^{-t/2} \left(3 \left(f(0)-f'(0)\right) \cos \left(\frac{\sqrt{3} t}{2}\right)+\sqrt{3} \sin \left(\frac{\sqrt{3} t}{2}\right) \left(-2 f''(0)+f'(0)+f(0)\right)\right) $$

The solutions are given by the first equation with

$$ f(0)-f'(0)\neq 0 \,\, \mathrm{or }\,\, -2 f''(0)+f'(0)+f(0)\neq 0 $$


If you are looking for a real function, following is one:

$$ f(x) = e^{cos(\frac{2}{3}\pi)x} * cos(sin(\frac{2}{3}\pi)x) $$

Plot of f, f', f'', f''':

Plot of f, f', f'', f'''

You can verify it by: derivative-calculator.net

For more common problem: Function whose nth derivative is itself, following is one:

$$ f(x) = e^{cos(\frac{2\pi}{n})x} * cos(sin(\frac{2\pi}{n})x) $$

With n = 1, 2, 4:

$$ f(x)_{n=1} = e^{cos(2\pi)x} * cos(sin(2\pi)x) = e^x * cos(0) = e^x $$

$$ f(x)_{n=2} = e^{cos(\pi)x} * cos(sin(\pi)x) = e^{-x} * cos(0) = e^{-x} $$

$$ f(x)_{n=4} = e^{cos({\pi \over 2})x} * cos(sin({\pi \over 2})x) = e^0 * cos(x) = cos(x) $$

short explain:

$$ (e^{wx})' = w * e^{wx} \\ (e^{wx})^{(n)} = w^n * e^{wx} \\ $$

We known $w^n = 1$ has n roots:

$$ w = cos(\frac{2\pi*t}{n}) + isin(\frac{2\pi*t}{n}), t = 0, 1, 2...n-1 $$


$$ \begin{align} w_1 & = cos({2\pi \over n}) + isin({2\pi \over n}) \\ w_2 & = cos({2\pi \over n}) - isin({2\pi \over n}) \\ a & = cos({2\pi \over n}) \\ b & = sin({2\pi \over n}) \\ w_1 & = a + bi \\ w_2 & = a - bi \\ \end{align} $$

we got: $$ \begin{align} f(x) & = {{e^{w_1x} + e^{w_2x}} \over 2} \\ & = {{e^{(a+bi)x} + e^{(a-bi)x}} \over 2} \\ & = {{e^{ax} * e^{bix} + e^{ax} * e^{-bix}} \over 2} \\ & = {{e^{ax} * (e^{bix} + e^{-bix})} \over 2} \\ & = {e^{ax} * (cos(bx) + isin(bx) + cos(-bx) + isin(-bx)) \over 2} \\ & = e^{ax} * cos(bx) \\ & = e^{cos(\frac{2\pi}{n})x} * cos(sin(\frac{2\pi}{n})x) \end{align} $$

  • $\begingroup$ Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to. $\endgroup$ – José Carlos Santos Oct 8 '18 at 13:56
  • $\begingroup$ While this link may answer the question, it is better to include the essential parts of the answer here and provide the link for reference. Link-only answers can become invalid if the linked page changes. - From Review $\endgroup$ – Lee David Chung Lin Oct 8 '18 at 15:02
  • $\begingroup$ (The comment above is "automatically" generated by the review process. This one is my personal comment) Please realize that your answer is already completely contained in more than one existing answers. Your post provides no new information (redundant) and is ill-formatted. I understand that this question is mathematically interesting, but next time please read others' answers before you make a contribution. $\endgroup$ – Lee David Chung Lin Oct 8 '18 at 15:07
  • $\begingroup$ Hi @LeeDavidChungLin, I just update my answer $\endgroup$ – pjincz Oct 8 '18 at 15:49
  • $\begingroup$ I'm glad you made the effort to improve the post. Allow me to remind you that (1) It's hard to see the downvote go away (it wasn't by me) because people rarely go back to change their votes (2) Don't care too much about the points (3) StackExchange is a community moderated site, that is, a good proportion of "active users" can express their opinions on whether or not to close your post. This is out of my hands now. btw you probably have noticed that I'm a relatively low rep user. $\endgroup$ – Lee David Chung Lin Oct 8 '18 at 16:05

protected by Saad Oct 8 '18 at 13:40

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