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where 
be a continuous function such that for every 
,![\[
f(x) + f\left(x + \frac{1}{3}\right) + f\left(x + \frac{2}{3}\right) = 0.
\]](http://latex.artofproblemsolving.com/b/0/7/b070aac1d51df83c88940284e7fd6a71a44703af.png)
Show that 
is periodic and find the least positive period.
such that for every polynomial 
with integer coefficients and for every integer 
the integer![\[p^{(j)}(k)=\left. \frac{d^j}{dx^j}p(x) \right|_{x=k}\]](http://latex.artofproblemsolving.com/a/2/7/a2793d7f46345bcba4ccaa3082cc47afcd35074b.png)
(the -th derivative of at 
) is divisible by 
of continuously differentiable functions by the following recurrence: 
exists for every 
and determine the limit function.
, induction: 
and 
, indeed 
integrating we get the result.
exists for every . Then by 1, 3 and Lebesgue dominated convergence theorem
, from here 
is differentiable and 
, since also 
, we can conclude that 
. Then by induction, we find for every fixed 
: 
is increasing sequence and 
. Hence limit function exists.
lemma we will have for every fixed 
from math induction. Hence 
!
lemma we will have
![\[e^{x+\frac{x^2}{2}+\dotsc+\frac{x^n}{n}}=e^{x+\frac{x^2}{2}+\dotsc}+\mathcal{O}(x^{n+1})=e^{-\log(1-x)}+\mathcal{O}(x^{n+1})=\frac{1}{1-x}+\mathcal{O}(x^{n+1})=1+x+x^2+\dotsc+x^n+\mathcal{O}(x^{n+1})\]](http://latex.artofproblemsolving.com/6/e/b/6eb48c135c0fea4e9cad4a03881449c7413915a4.png)
So the difference of the LHS and the RHS is a power series containing only those terms with exponent 
or higher from the LHS. But all coefficients of the LHS (as a power series in 
) are non-negative. Hence the result.
So the difference of the LHS and the RHS is a power series containing only those terms with exponent or higher from the LHS. But all coefficients of the LHS (as a power series in ) are non-negative. Hence the result.
.
, hence 
is always positive and then always increasing. Furthermore we have 
, so is convex.
for all , then
so 
as well.
for all ; note this is true for 
. This claim is equivalent to 
, which will follow from proving that 
for all 
and integrating. But this rearranges to 
so we are done by induction.
exists for every (it equals 
); let it equal 
and note that it's nondecreasing. I claim that we actually have 
uniformly converging to on any ![$[0,t] \subset [0,1)$](http://latex.artofproblemsolving.com/5/1/b/51b354611731e2dba4ce689c78c24625387d26af.png)
. Indeed fix some 
. Partition ![$[0,t]$](http://latex.artofproblemsolving.com/9/3/9/9398cf678fee12444e20e80c8ec38bb2803749fb.png)
into finitely many intervals ![$[t_i,t_{i+1}]$](http://latex.artofproblemsolving.com/9/a/b/9aba5be221b8c6726708a61dabf49e970e6978f1.png)
such that 
, which is possible. Then pick some 
such that for all 
and 
, 
, which is also possible since there are finitely many 
. Since is increasing it follows that 
for all ![$x \in [0,t]$](http://latex.artofproblemsolving.com/d/2/1/d21152ef5ecea31cc8238e7f9e79fc5b3effa371.png)
, and this quantity is positive. Pick arbitrarily and any 
and observe that
By sending 
we conclude that 
. This should hold for all 
by picking an appropriate choice of , so by taking logarithms and differentiating we find that , and since it follows that . 
for 
and 
for 
gives ![\[f_n^{(2)}(x) = f_n^{(1)}(x)\cdot f_{n-1}(x) + f_n(x)\cdot f_{n-1}^{(1)}(x) = \left(f_n(x)\cdot f_{n-1}(x)\right)\cdot f_{n-1}(x) + f_n(x)\cdot \left(f_{n-1}(x) \cdot f_{n-2}(x)\right).\]](http://latex.artofproblemsolving.com/8/b/4/8b466940aacf413686aeda8d5af94aad02745ac5.png)
We got 2 summands, each of which is a product of 3 terms. In general, we would get 
summands, each of which is a product of 
terms of form 
(some of which may be zero, as 
![\[\dfrac{1-x^n}{1-x} = 1 + x + x^2 + \dots + x^{n-1} < f_n(x) < 1+x+x^2+\dots=\dfrac{1}{1-x}.\]](http://latex.artofproblemsolving.com/1/5/7/157ee34091356430634c21c38cd39331ee1c94f8.png)
gives 
