Newton's Interpolation Formula: Difference between the forward and the backward formula

theoretically assuming that the calculations are done in a infinite arithmetic then the polynomials are the same so they give the same answer .

But since the computer and applied arithmetic are finite then when calculating the polynomials they do no turn out to be equal (at least not in all the cases). and note that the more arithmetic operations you do the more you loose accuracy!

so if the input x is closer to $ x_i $ (one of the data we already have); then choosing $x_i$ as $x_0$ gives a better accuracy if we are using the forward differences formula in our finite arithmetic system .(or in the case of back ward differences choosing $x_i$ as $x_n$) (or even in the case of Centered Differences choosing $x_i$ as the middle data)

since

e.g. in the case of $x_i$ as $x_0$ using the forward differences formula ; the $f(x_0)$ is a single term (with no additional arithmetic to loose accuracy like other terms) and since we are assuming that if x is close to $x_i$ then $f(x)$ is also close to $f(x_i)$. so we get the least possible error in calculating $f(x)$.(which is also the case for Centered Differences and back ward differences).

Additional notes:

if we use the newton's divided differences formula then indexing the data really does not matter . so to raise the accuracy we simply re-index the data the way its more accurate and find the polynomial.

BUT if we are using the assumption that $x_{i+1} - x_{i} = h$(where h is the step size).

then we cannot re-index the data as we wish an the data should be either increasing or decreasing .so to get more accuracy and to be able to choose the single term ($f(x_{0})$ or $f(x_{n})$ or ...) to be the closest data; we need to change the polynomial formula to meet our needs.