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## Function

Ordinary Differerential Equations
Papapetros Vaggelis
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### Function

Find the differentiable function $\displaystyle{f:\mathbb{R}\to \mathbb{R}}$ which satisfies

$\displaystyle{f(0)=1}$ and $\displaystyle{f^\prime(x)=f^2(x)\,f(-x)\,,\forall\,x\in\mathbb{R}}$.
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### Re: Function

Let us prove first that the function $g(x)=f(x)f(-x)$ is constant. Indeed taking the derivative we have that

\begin{align*}
g'(x) &= f(x) f(-x) \\
&= f'(x) f(-x) - f(x) f'(-x)\\
&= \cancel{f^2(x) f^2(-x) - f^2(x) f^2(-x)}\\
&= 0
\end{align*}

Thus $g(x)=c$ forall $x \in \mathbb{R}$. For $x=0$ we have that $g(0)=f^2(0)=1$. Thus $g(x)=1$ forall $x \in \mathbb{R}$. Hence $f(-x) f(x)=1$. This relation tells us that $f$ cannot have any roots within its domain. Since $f(0)=1>0$ we also conclude that $f$ is positive. The initial condition gives us

$$f'(x)=f(x)$$

and this implies $f(x)=e^x , \; x \in \mathbb{R}$ a function that satisfies all conditions given.
$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^s}= \prod_{p \; \text{prime}}\frac{1}{1-p^{-s}}$
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