Newton’s method

Taylor series

A Taylor series is a representation of a function as an infinite sum of terms that are calculated from the values of the function’s derivatives at a single point

$$f(a) \approx \sum\limits_{n=0}^{\infty}{\frac{f^{(n)}(a)}{n!}(x-a)^n}$$

where $f^{n}(a)$ donates the $n^{th}$ derivative of $f$ evaluated at the point $a$.

And here’s a very intuitive example:

Taylor series
The exponential function $e^x$ (in blue), and the sum of the first $n + 1$ terms of its Taylor series at $0$ (in red).

Newton’s method


In calculus, Newton’s method is an iterative method for finding the roots of a differentiable function $f$.

In machine learning, we apply Newton’s method to the derivative $f’$ of the cost function $f$.

One-dimension version

In the one-dimensional problem, Newton’s method attempts to construct a sequence ${x_n}$ from an initial guess $x_0$ that converges towards some value $\hat{x}$ satisfying $f'(\hat{x})=0$. In another word, we are trying to find a stationary point of $f$.

Consider the second order Taylor expansion $f_T(x)$ of $f$ around $x_n$ is

$$f_T(x)=f_T(x_n+\Delta x) \approx f(x_n) + f'(x_n) \Delta x + \frac{1}{2}f”(x_n) \Delta x^2$$

So now, we are trying to find a $\Delta x$ that sets the derivative of this last expression with respect to $\Delta x$ equal to zero, which means

$$\frac{\rm{d}}{\rm{d} \Delta x}(f(x_n) + f'(x_n) \Delta x + \frac{1}{2}f”(x_n) \Delta x^2) = f'(x_n)+f”(x_n)\Delta x = 0$$

Apparently, $\Delta x = -\frac{f'(x_n)}{f”(x_n)}$ is the solution. So it can be hoped that $x_{n+1} = x_n+\Delta x = x_n – \frac{f'(x_n)}{f”(x_n)}$ will be closer to a stationary point $\hat{x}$.

High dimensional version

Still consider the second order Taylor expansion $f_T(x)$ of $f$ around $x_n$ is

$$f_T(x)=f_T(x_n+\Delta x) \approx f(x_n) + \Delta x^{\mathsf{T}} \nabla f(x_n) + \frac{1}{2} {\rm H} f(x_n) \Delta x (\Delta x)^{\mathsf{T}}$$

where ${\rm H} f(x)$ donates the Hessian matrix of $f(x)$ and $\nabla f(x)$ donates the gradient. (See more about Taylor expansion at

So $\Delta x = – [{\rm H}f(x_n)]^{-1}\nabla f(x_n)$ should be a good choice.


As is known to us all, the time complexity to get $A^{-1}$ given $A$ is $O(n^3)$ when $A$ is a $n \times n$ matrix. So when the data set is of too many dimensions, the algorithm will work quite slow.


The reason steepest descent goes wrong is that the gradient for one weight gets messed up by simultaneous changes to all the other weights.

And the Hessian matrix determines the sizes of these interactions so that Newton’s method minimize these interactions as much as possible.


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