How can I find the smallest positive solution of an equation taking four digits of the fractional part Find the smallest positive solution of 
$$\sum_{n=0}^{\infty}\frac{(-x)^n}{(n!)^2}=0$$
taking four digits of the fractional part
 A: Rewrite f(x) as
$$
f(x) = \sum\limits_{n\, = \,0}^\infty  {\frac{{\left( { - 1} \right)^{\,n} }}
{{\left( {n!} \right)^{\,2} }}x^{\,n} }  = f(x,q) + R(x,q) = \sum\limits_{n\, = \,0}^q {\frac{{\left( { - 1} \right)^{\,n} }}
{{\left( {n!} \right)^{\,2} }}x^{\,n} }  + \sum\limits_{n\, = \,q + 1}^\infty  {\frac{{\left( { - 1} \right)^{\,n} }}
{{\left( {n!} \right)^{\,2} }}x^{\,n} } 
$$
Consider that the ratio of the terms, excluding $(-1)^n$, is
$$
r = \frac{x}
{{\left( {n + 1} \right)^{\,2} }}\quad  \Rightarrow \quad r < 1\quad \left| {\;0 \leqslant x < } \right.\left( {n + 1} \right)^{\,2} 
$$
so that, for $x$ positive and less than $(n+1)^2$, the terms are decreasing in absolute value, and being the sign alternated we have that
$$
f(x,2n - 1) < f(x,2n + 1)<\dots < f(x)<\dots < f(x,2n+2) < f(x,2n )
$$
Then we can start with
$$
f(x,2) = 1 - x + \frac{{x^{\,2} }}{4}
$$
and find that its (double) root is
$$
x_{\,0,2}  = 2 < \left( {2 + 1} \right)^{\,2} 
$$
which is within the range for $r<1$.
Therefore we can feed this value to $f(x,3)$, to find a negative value, wherefrom 
we can just use the secant method ($f'(x,3)$ is negative) to estimate $x_{\,0,3}$, in case iterating it some times. But of course you can also use Newton's method: the 1st and 2nd derivatives are practically.. "already there" .
Then feed it $f(x,4)$ , and continue till the difference of the roots is less than the given threshold (i.e. <0.00001, to assure that the result is accurate to the fourth decimal digit).
And a sketch is always of great help:  

