Page 486 of Boyd's Convex Optimization book defines Newton's decrement as
$$\lambda(x)=(\nabla f(x)' \nabla^2 f(x)^{-1}\nabla f(x))^{\frac{1}{2}}$$ and then says that squaring this quantity and dividing by 2 gives estimate of how far off from the minimum you are using quadratic approximation.
$$\frac{1}{2}\lambda^2(x)=f(x)-\inf_y \hat{f}(y)$$
However, this seems to be off by factor of 2 for the case of 1 dimension and $f(x)=x^2$. IE, $\nabla f(x)=2x$, so $\lambda(x)=2x$ so
$$\frac{1}{2}\lambda^2(x)=2x^2$$
But the true answer should be $x^2$ not $2x^2$, what am I missing here?
For the case of 1 dimension and $f(x)=x^2$, $\nabla f(x)=2x$, and $\nabla^2 f(x)^{-1} = \frac{1}{2}$, so $\lambda(x)=\sqrt{2}x$.