Approximated second derivative of $y(0)''$ function $y(x)$ at x = 0 difference quotient, using the values of $y(x)$ at the sites of the three-point template $ {x}_{1} = \frac {-4h} { 5}, {x} _ {2} = 0, {x} _ {3} = \frac {h} {5} $.
Calculate the difference quotient value (approximate value of $y '' (0)$) for $ y (x) = {x} ^ {3} $ when $h = 0.05$. The result is rounded to hundredths.
Decision:
$y''= \frac{{y} _ {i +1} -2 {y} _ {i} + {y} _ {i-1}} {{h} ^ { 2}}$
$y _ {i + 1} = {\frac {-4 (0.05)} {5}} ^ {3}$
$y _ {i} = 0$
$y _ {i-1} = {\frac {(0.05)} {5}} ^ {3}$
=> $y'' = - 0.03$
The obvious answer is $-0.03$.
But the authors insist that the answer is $-0.06$.
Please help me understand where they got the multiplication by $2$ ? I can not understand. Thanks in advance.