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A divided difference table arranges divided differences by starting interpolation node $i$ and order $k$. Its base column is $D_{i,0}=f(x_i)$; every later entry uses two neighboring entries from the previous column:The denominator spans all $k+1$ interpolation nodes of that entry, even though the numerator uses adjacent table entries. The interpolation nodes must be distinct; they need not be equally spaced or sorted.^[ref]

Take the polynomial $f(x)=x^3-2x+1$ at the interpolation nodes $(-1,0,2,3)$, giving values $(2,1,5,22)$. Its divided difference table isThe first divided differences are $(1-2)/(0+1)=-1$, $(5-1)/(2-0)=2$, and $(22-5)/(3-2)=17$. The second divided differences use the wider spans: $(2+1)/(2+1)=1$ and $(17-2)/(3-0)=5$. The third divided difference is $(5-1)/(3+1)=1$. Reading the top row gives the Newton polynomial coefficients $(2,-1,1,1)$, rather than all the entries in any one column.

The resulting Newton polynomial grows through four partial interpolants:At $x=2$, $p_1(2)=-1$ misses $f(2)=5$ by $6$; the correction $(x+1)x$ supplies exactly $6$ there and vanishes at the earlier interpolation nodes $-1,0$. At $x=3$, $p_2(3)=10$ misses $f(3)=22$ by $12$; the correction $(x+1)x(x-2)$ supplies exactly $12$ there and vanishes at all three earlier interpolation nodes. Each added term therefore preserves the previous agreements while satisfying the next one.

For $n+1$ interpolation nodes, the triangle contains $n+(n-1)+\cdots +1=n(n+1)/2$ recurrence steps. Once the top row is known, nested evaluation of the Newton polynomial uses only $n$ multiplications and $n$ additions per argument. The triangle organizes the work; the factors explain why the construction works.

The Newton polynomial expresses the interpolation polynomial in products that successively vanish at all previously used interpolation nodes. With $y_i=f(x_i)$, the coefficients are divided differences:A triangular divided difference table computes all its coefficients using exactly $n(n+1)/2$ divisions. The Newton polynomial and the Lagrange polynomial have the same polynomial degree bound and the same nodal values, hence are equal by uniqueness of polynomial interpolation.

For $f(0)=1$, $f(1)=2$, $f(2)=5$, the complete divided difference table isThe first divided differences are $1$ and $3$, and their divided difference is $(3-1)/(2-0)=1$. Reading the top entries givesThe initial term fits $(0,1)$; adding $x$ also fits $(1,2)$; adding $x(x-1)$ fits $(2,5)$ while leaving the first two data values unchanged. This is the practical beauty of the Newton polynomial: each new correction respects all previous interpolation constraints.

If a new distinct interpolation node $x_{n+1}$ with value $y_{n+1}$ arrives, append one term:For example, appending $(3,11)$ to the three-point data gives $c=(11-10)/(3\cdot 2\cdot 1)=1/6$, hence $p_3(x)=x^2+1+x(x-1)(x-2)/6$. Changing the order of the interpolation nodes changes the Newton polynomial coefficients and products, but not the final interpolation polynomial in exact arithmetic.

The products also permit nested evaluation analogous to Horner's method. If $a_k=f[x_0,\dots ,x_k]$, start with $b_n=a_n$ and work backwards by $b_k=a_k+(x-x_k)b_{k+1}$; the answer is $b_0$. The three-point example becomes $1+x[1+(x-1)]$, needing two multiplications for each evaluation. This organization is useful when adding observations incrementally or evaluating a stored Newton polynomial many times; the Runge phenomenon still depends on the interpolation nodes rather than on which representation is used.

To justify the update formula, its added product vanishes at every previous interpolation node. Choosing $c$ by the new residual forces agreement at $x_{n+1}$, so uniqueness of polynomial interpolation identifies the result. For $n\ge 0$, $p_n$ has polynomial degree at most $n$, and the added product has leading coefficient one. Thus the new coefficient $c$ is exactly $f[x_0,\dots ,x_{n+1}]$. The initial state is $p_0(x)=f(x_0)$, with the empty product interpreted as one.^[ref]

Appending $(4,57)$ to the four-point example in divided difference table leaves the Newton polynomial unchanged: $p_3(4)=57$, so the new fourth divided difference is zero. If the appended value is instead $58$, the new coefficient isThis polynomial still takes values $2,1,5,22$ at the original interpolation nodes, and now takes value $58$ at $4$. A zero update is a valid completed step: adding another exact sample of the original polynomial need not increase the polynomial degree.

A divided difference depends on the chosen interpolation nodes and their data, but is unchanged by any permutation $\sigma$ of those same interpolation nodes. The explicit sum in divided difference makes this visible: a permutation merely rearranges its summands.^[ref]

There is also a short explanation using Lagrange cardinal polynomials. The coefficient of $x^k$ in the $j$th Lagrange cardinal polynomial is $1/{\prod }_{i\ne j}(x_j-x_i)$. Taking that coefficient in their weighted sum gives exactly the explicit divided difference formula. Since polynomial interpolation has a unique answer for a fixed set of distinct interpolation nodes, this coefficient cannot depend on their ordering. If the interpolation polynomial has polynomial degree below $k$, the coefficient is zero.

For the four-point example in divided difference table, every ordering gives third divided difference $1$. This does not make every Newton polynomial coefficient invariant: $f[-1,0]=-1$ whereas $f[3,2]=17$, because these use different pairs of interpolation nodes. Even at second order, $f[-1,0,2]=1$ and $f[3,2,0]=5$ use different triples. The node order in Newton interpolation determines which subsets occur along the top row.