I saw such an interesting proposition:
Any given real number sequence $\left\{ a_{n}\right\} $, can construct a smooth function so that the coefficient of its Taylor series is $\left\{ a_{n}\right\} $.
Someone prompts me to construct a bulge function $\phi(x) $:
$\phi(x) :\mathbb{R} \rightarrow \mathbb{R} $ is a smooth function, $0\leqslant\phi(x) \leqslant 1$, and $$\phi \left( x\right) =1, \;x\in \left[ -\tfrac{1}{2} ,\tfrac{1}{2} \right];\;\phi \left( x\right) =0,\,\left\vert x\right\vert \geqslant 1. $$
Set $\ a_{n}\in \mathbb{R}\left( n\geqslant 0\right) $, let $\xi_{n} =\ n\ +\ \sum^{n}_{i=0} \left\vert a_{i}\right\vert$. Define a function$$f(x) =\sum^{\infty }_{n=0} \frac{a_{n}}{n!} \phi \left( \xi_{n} x\right) x^{n}$$
When $\ n\geqslant 1\ $, $\left\vert x\right\vert \leqslant \xi^{-1}_{n} $,
$$\left\vert \frac{a_{n}}{n!} \phi \left( \xi_{n} x\right) x^{n}\right\vert \leqslant \frac{\left\vert a_{n}\right\vert }{n!} |x|^{n}\leqslant \frac{\left\vert a_{n}\right\vert }{n!} \frac{1}{\left( \xi_{n} \right)^{n} } \leqslant \frac{1}{n!} $$ when $\left\vert x\right\vert >\xi^{-1}_{n}$, the upper inequality also holds.
So the series defining $f$ is uniformly convergent, $f(x)$ is continuous, and $f\left( 0\right) =a_{0}$.
If it can be proven that $f$ is infinitely differentiable, and $f^{(n)}(0) =a_{n}$, ${\forall}n\geqslant 1$, the conclusion is proved. But I encountered difficulties here!
I tried a proof method, but I'm not sure if there's a problem.
It only needs to prove that $\sum^{\infty }_{i=1} f^{\left( n\right) }_{i}\left( x\right) $ is uniformly convergent ,${\forall}n\geqslant 1$. So I tried to estimate$|\ f^{\left( n\right) }_{i}\left( x\right) |$:
First, from the definition of $\phi$, it can be seen that $\exists G>0, \left\vert \phi^{\left( n-k\right) } \left( \xi_{i} x\right) \right\vert \leqslant G,{\forall}k\leqslant n$
We have $C^{k}_{n}\leqslant 2^{n}$ and set $\xi_{i}>i>2$, by Newton Leibniz formula,
when i is sufficiently large, $\frac{1}{(i-k)!} < 1$,$\left\vert \xi_{i} x\right\vert < \frac{1}{2},$ $\left\vert a_{i}x\right\vert <\frac{\left\vert a_{i}\right\vert }{2\xi_{i} } <2$ $$|\ f^{\left( n\right) }_{i}\left( x\right) |=\left\vert \sum^{n}_{k=0} C^{k}_{n}\frac{a_{i}}{i!} \phi^{\left( n-k\right) } \left( \xi_{i} x\right) \left( \xi_{i} \right)^{n-k} x^{i-k}\frac{i!}{(i-k)!} \right\vert $$ $$=\left\vert \sum^{n}_{k=0} C^{k}_{n}\frac{a_{i}}{(i-k)!} \phi^{\left( n-k\right) } \left( \xi_{i} x\right)^{} \left( \xi_{i} x\right)^{n-k} x^{i-n}\right\vert $$ $$\leqslant \left\vert \sum^{n}_{k=0} a_{i}x\cdot 2^{n}\phi^{\left( n-k\right) } \left( \xi_{i} x\right) \left( \frac{1}{\xi_{i} } \right)^{i-n-1} \right\vert $$ $$\leqslant \left\vert \sum^{n}_{k=0} 2^{n+1}\phi^{\left( n-k\right) } \left( \xi_{i} x\right) \left( \frac{1}{\xi_{i} } \right)^{i-n-1} \right\vert $$ $$\leqslant \left\vert \sum^{n}_{k=0} 2^{n+1}G \cdot\left( \frac{1}{2 } \right)^{i-n-1} \right\vert $$ $$=n\cdot 2^{2n+2}G\left( \frac{1}{2} \right)^{i} $$it indicates that $\sum^{\infty }_{i=1} f^{\left( n\right) }_{i}\left( x\right) $ is uniformly convergent.
May I ask if my proof is correct? Thank you in advance.