Question
(a) Show that the function $ f(x) = \sum_{n = 0}^{\infty} \frac {x^n}{n!} $is a solution of the differential equation $ f'(x) = f(x) $(b) Show that $ f(x) = e^x. $
Step 1
To do this, we take the derivative of $ f(x) $: \begin{align*} f'(x) &= \sum_{n = 1}^{\infty} \frac {nx^{n-1}}{(n-1)!} \\ &= \sum_{n = 1}^{\infty} \frac {x^{n-1}}{(n-1)!} \\ &= \sum_{n = 0}^{\infty} \frac {x^n}{n!} \\ &= f(x) \end{align*} So, we have shown that $ Show more…
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