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(A) Given the equation: $f(x) = f'(x) + f''(x) + f'''(x) + \dots \infty$  Taking the derivative of both sides with respect to $x$:  $f'(x) = f''(x) +

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$e^{x/2}$

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(A) Given the equation: $f(x) = f'(x) + f''(x) + f'''(x) + \dots \infty$  Taking the derivative of both sides with respect to $x$:  $f'(x) = f''(x) + f'''(x) + f''''(x) + \dots \infty$  Adding $f'(x)$ to both sides of the original equation:  $f(x) + f'(x) = f'(x) + f''(x) + f'''(x) + \dots + f'(x)$  This simplifies to:  $f(x) + f'(x) = 2f'(x)$  $f(x) = f'(x)$  This is a first-order linear differential equation:  $\frac{f'(x)}{f(x)} = 1$  Integrating both sides with respect to $x$:  $\ln|f(x)| = x + C$  Given $f(0) = 1$:  $\ln(1) = 0 + C \Rightarrow C = 0$  Thus,$\ln|f(x)| = x$,which implies $f(x) = e^x$.  Wait,re-evaluating the series:  $f(x) = f'(x) + f''(x) + f'''(x) + \dots$  $f'(x) = f''(x) + f'''(x) + f''''(x) + \dots$  Subtracting the second from the first:  $f(x) - f'(x) = f'(x)$  $f(x) = 2f'(x)$  $\frac{f'(x)}{f(x)} = \frac{1}{2}$  Integrating: $\ln f(x) = \frac{x}{2} + C$  Since $f(0) = 1$,$C = 0$.  Therefore,$f(x) = e^{x/2}$.

$A$ function $f(x)$ satisfies the condition $f(x) = f'(x) + f''(x) + f'''(x) + \dots \infty$,where $f(x)$ is an indefinitely differentiable function and the dash denotes the order of the derivative. If $f(0) = 1$,then $f(x)$ is:

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