# power series – A real variable solution to a problem posted on Terry Tao's blog

Here is my purely real variable solution.
We have this $$e ^ {x}$$ and $$f (x)$$ are both analytical, so their product $$phi (x) = e ^ {x} cdot f (x)$$ is therefore $$phi (x) = overset { infty} { subset {n = 0} { sum}} c_n x ^ n$$, or $$c_n = a_0 cdot frac {1} {(n-1)!} + … + frac {a_i} {i!} cdot frac {1} {(ni)!} + … + frac { a_n} {n!}$$.
Since $$subset {x to infty} { lim} phi (x) = subset {x to infty} { lim} dfrac {f (x)} {e ^ {- x}} = C$$, for each $$epsilon> 0$$ there is $$R> 0$$ and $$N in mathbb {N}$$ such as:
$$Bigg | overset {k} { subset {n = 0} { sum}} c_n x ^ n – C Bigg | leq Bigg | overset {k} { subset {n = 0} { sum}} c_n x ^ n – overset { infty} { subset {n = 0} { sum}} c_n x ^ n Bigg | + Bigg | overset { infty} { subset {n = 0} { sum}} c_n x ^ n – C Bigg | leq dfrac { epsilon} {2} + dfrac { epsilon} {2} = epsilon.$$

for everyone $$k> N$$ and all $$x> R$$. So,

$$C- epsilon < overset{k}{underset{n=0}{sum}}c_n x^n N, x > R$$
That implies that,

$$0 = subset {x to infty} { lim} dfrac {C- epsilon} {x ^ k}
We deduce that $$phi (x) = overset {N} { subset {n = 0} { sum}} c_n x ^ n$$ and because $$underet {x to infty} { lim} phi (x) = C$$, $$phi (x)$$ must be constant $$C.$$ Therefore, $$f (x) = Ce ^ {- x}$$, where the equations $$a_n = C (-1) ^ n$$ follows.