If \(f(x)=f^{\prime}(x)+f^{\prime \prime}(x)+f^{\prime \prime \prime}(x)+\ldots\) and \(f(0)=1\), then \(f(x)\) is equal to

Given, \(\quad f(x)=f^{\prime}(x)+f^{\prime \prime}(x)+f^{\prime \prime \prime}(x)+\ldots\)
If \(\quad f(x)=e^{x / 2}\)
Then, \(f^{\prime}(x)=\frac{1}{2} e^{x / 2}, f^{\prime \prime}(x)=\frac{1}{2^{2}} \cdot e^{x / 2}\),
\(\begin{aligned}
f^{\prime \prime \prime}(x) &=\frac{1}{2^{3}} e^{x / 2} \ldots \text { so on } \\
\therefore \quad f(x) &=f^{\prime}(x)+f^{\prime \prime}(x)+f^{\prime \prime \prime}(x)+\ldots \\
&=\frac{1}{2} e^{x / 2}+\frac{1}{2^{2}} \cdot e^{x / 2}+\frac{1}{2^{3}} \cdot e^{x / 2}
\end{aligned}\)
\(=e^{x / 2}\left(\frac{1}{2}+\frac{1}{2^{2}}+\frac{1}{2^{3}}+\ldots\right)\)
\(=e^{x / 2}\left\{\frac{\frac{1}{2}}{1-\frac{1}{2}}\right\}=e^{x / 2} \cdot\left(\frac{\frac{1}{2}}{\frac{1}{2}}\right)=e^{x / 2} \cdot 1\)
\(\therefore \quad f(x)=e^{x / 2}\)
and \(f(0)=e^{0}=1\)