Newton forward and backward interpolation

A table of measurements never has the exact point you need — a sensor logs temperature every ten minutes but you want the value at minute four, or a trig table lists $\sin x$ every ten degrees but the exam wants $\sin 4^\circ$ . Interpolation fills that gap using only the numbers already in the table, no new measurement required. Newton's forward and backward difference formulae are the two workhorses for equally spaced data: forward when the unknown point sits near the start of the table, backward when it sits near the end. Both are built from a single finite-difference table and both are truncated polynomials in disguise — keep enough terms and you reproduce a curve of any desired degree through the nearest tabulated points. This article derives the forward formula from the shift operator, applies both formulae to a $\sin x$ table, and flags the sign and reference-point errors that cost the most marks.

The core insight

Newton's forward formula $f(x)=f_0+p\Delta f_0+\binom{p}{2}\Delta^2f_0+\cdots$ (with $p=(x-x_0)/h$ ) fits a polynomial through equally spaced data starting at $x_0$ , and its mirror-image backward formula does the same anchored at the last point $x_n$ , so you always pick whichever keeps $p$ small.

The core idea

Interpolation near equally spaced data reduces to one question: which end of the table is the unknown point closest to? That choice fixes both the formula and the anchor point, and it matters because the accuracy of a truncated series depends entirely on how small $|p|$ stays.

Forward difference formula. For $x_0, x_1, \dots, x_n$ spaced $h$ apart with values $f_0, f_1, \dots, f_n$ , and $x$ near $x_0$ :

f(x) = f_0 + p\,\Delta f_0 + \frac{p(p-1)}{2!}\Delta^2 f_0 + \frac{p(p-1)(p-2)}{3!}\Delta^3 f_0 + \cdots, \qquad p = \frac{x-x_0}{h}

where: $\Delta f_0, \Delta^2 f_0, \dots$ are entries along the top diagonal of the forward difference table, and $p$ measures how many step-sizes $x$ lies past $x_0$ .

Backward difference formula. For $x$ near the last point $x_n$ :

f(x) = f_n + p\,\nabla f_n + \frac{p(p+1)}{2!}\nabla^2 f_n + \frac{p(p+1)(p+2)}{3!}\nabla^3 f_n + \cdots, \qquad p = \frac{x-x_n}{h}

where: $\nabla f_n, \nabla^2 f_n, \dots$ are entries along the bottom diagonal of the same difference table (read upward), and $p$ is now negative when $x$ lies before $x_n$ . Three points tie the two formulae together. First, both are the same underlying idea — the unique degree- $k$ polynomial through $k+1$ consecutive tabulated points — differing only in which point anchors the expansion. Second, the coefficients flip pattern: forward uses the decreasing product $p(p-1)(p-2)\cdots$ , backward uses the increasing product $p(p+1)(p+2)\cdots$ , because backward differences are naturally indexed from the far end. Third, both formulae demand equally spaced $x$ -values; unequally spaced data needs divided differences instead, which lies outside this article's scope.

Figure

A single difference table with the forward diagonal from the top row highlighted and the backward diagonal from the bottom row highlighted, showing both formulae read from the same table

Derivation ANewton's forward formula from the shift operator

The goal is to derive the forward-difference interpolation formula $f(x_0+ph) = f_0 + p\Delta f_0 + \binom{p}{2}\Delta^2 f_0 + \binom{p}{3}\Delta^3 f_0 + \cdots$ purely from the algebra of the forward-difference operator $\Delta f_i = f_{i+1}-f_i$ and the shift operator $Ef_i = f_{i+1}$ , treating $p=(x-x_0)/h$ as an ordinary real variable so that the binomial theorem for a general (non-integer) exponent applies.

Step 1. Relate $E$ and $\Delta$ . By definition $Ef_i = f_{i+1} = f_i + \Delta f_i$ , so $E = 1+\Delta$ as operators. This single identity is the entire engine of the derivation: every forward difference can be traded for a power of $E$ , and vice versa.

Step 2. Write the unknown value as a shift. If $x = x_0 + ph$ for some real number $p$ , then reaching $f(x)$ from $f_0$ means applying the shift operator $p$ times: $f(x_0+ph) = E^p f_0$ . This is exact when $p$ is a positive integer (shifting $p$ whole steps along the table) and is defined to extend consistently to non-integer $p$ , which is exactly the case of interest since the point we want to interpolate almost never lands exactly on a tabulated $x_i$ .

Step 3. Substitute $E = 1+\Delta$ and expand with the binomial series. $E^p f_0 = (1+\Delta)^p f_0 = \left[1 + p\Delta + \frac{p(p-1)}{2!}\Delta^2 + \frac{p(p-1)(p-2)}{3!}\Delta^3 + \cdots\right] f_0$ , using the general binomial expansion $(1+u)^p = \sum_k \binom{p}{k}u^k$ with $u=\Delta$ treated as a formal operator raised to integer powers $\Delta^2, \Delta^3,\dots$ — a legitimate step because $\Delta^k f_0$ is just an ordinary number once applied to the table.

Step 4. Read off the formula. Distributing $f_0$ through the bracket gives $f(x_0+ph) = f_0 + p\,\Delta f_0 + \frac{p(p-1)}{2!}\Delta^2 f_0 + \frac{p(p-1)(p-2)}{3!}\Delta^3 f_0 + \cdots$ , which is exactly Newton's forward-difference formula. In practice the series is truncated after the highest difference the table supports (an $n+1$ -point table only determines differences up to $\Delta^n f_0$ reliably), which is precisely why using a longer series than the data supports is meaningless — there is no more information in the table beyond that order. The backward formula follows by the identical argument applied to the operator $\nabla f_i = f_i - f_{i-1}$ and $E^{-1} = 1-\nabla$ , anchored at $f_n$ instead of $f_0$ .

Worked example

Use the table (values of $\sin x$ , $x$ in degrees, $h=10$ ):

$x$	0	10	20	30
$f(x)$	0	0.1736	0.3420	0.5000

Estimate $f(4)$ with the forward formula and $f(27)$ with the backward formula.

Step 1. Build the difference table. The first forward differences are $\Delta f_0 = 0.1736$ , $\Delta f_1 = 0.1684$ , $\Delta f_2 = 0.1580$ ; the second differences are $\Delta^2 f_0 = -0.0052$ , $\Delta^2 f_1 = -0.0104$ ; the third difference is $\Delta^3 f_0 = -0.0012$ . Reading the same table from the bottom gives the backward differences at $x_3=30$ : $\nabla f_3 = 0.1580$ , $\nabla^2 f_3 = -0.0104$ , $\nabla^3 f_3 = -0.0052$ .

Step 2. For $f(4)$ , since $4$ is near $x_0=0$ , use forward with $p = (4-0)/10 = 0.4$ :

f(4) \approx 0 + (0.4)(0.1736) + \frac{(0.4)(-0.6)}{2}(-0.0052) + \frac{(0.4)(-0.6)(-1.6)}{6}(-0.0012)

= 0.06944 + 0.000624 - 0.0000768 \approx 0.06999

This is close to the true value $\sin 4^\circ \approx 0.0698$ .

Step 3. For $f(27)$ , since $27$ is near $x_3=30$ , use backward with $p = (27-30)/10 = -0.3$ :

f(27) \approx 0.5000 + (-0.3)(0.1580) + \frac{(-0.3)(0.7)}{2}(-0.0104) + \frac{(-0.3)(0.7)(1.7)}{6}(-0.0052)

= 0.5000 - 0.0474 + 0.00109 + 0.00031 \approx 0.4540

This compares well with $\sin 27^\circ \approx 0.4540$ .

Final answer: $f(4) \approx 0.0699$ using forward differences and $f(27) \approx 0.4540$ using backward differences — both formulae read the same underlying difference table, just from opposite ends.

Common traps

Using the forward formula when $x$ is near the end of the table (or the reverse) — the series does converge in principle but needs far more terms, so an early truncation gives a poor answer.
Miscomputing $p$ : forward measures from the first point, $p=(x-x_0)/h$ ; backward measures from the last, $p=(x-x_n)/h$ ; swapping the reference point silently corrupts every term after the first.
Sign confusion in the backward coefficients — they grow as $p(p+1)$ , $p(p+1)(p+2)$ , an increasing product, unlike the forward formula's decreasing $p(p-1)$ , $p(p-1)(p-2)$ .
Forgetting the equal-spacing requirement — both formulae assume constant $h$ ; unequally spaced tables need divided-difference interpolation instead.

Ask the tutor: if you are unsure which formula fits a given table position, or want a difference table double-checked term by term, ask the AI tutor to walk through it with you.

How do I decide whether to use the forward or the backward formula?

Look at where the point you want lies relative to the table. If it is close to the first entry $x_0$ , use forward and compute $p=(x-x_0)/h$ ; if it is close to the last entry $x_n$ , use backward and compute $p=(x-x_n)/h$ . The rule of thumb is to keep $|p|$ as small as possible, because the higher-order terms in the truncated series shrink fastest when $p$ is small — that keeps a short series accurate.

Why does the backward formula use p(p+1) instead of p(p-1)?

It comes straight from the derivation: the backward operator satisfies $E^{-1} = 1-\nabla$ , so expanding $f_n$ shifted by $-p$ steps as $(1-\nabla)^{-p}f_n$ (equivalently reindexing the same binomial series anchored at the last point) produces coefficients built from the increasing product $p(p+1)(p+2)\cdots$ rather than the decreasing one. It is not an arbitrary convention — it falls out of which end of the table you anchor the expansion at.

Can I use the same difference table for both formulae?

Yes — that is the whole point of building one difference table. The forward formula reads the top diagonal starting at $f_0$ ( $\Delta f_0, \Delta^2 f_0,\dots$ ), and the backward formula reads the bottom diagonal starting at $f_n$ ( $\nabla f_n, \nabla^2 f_n,\dots$ ), but for a single table these are literally the same numbers viewed from opposite corners — no separate table needs to be constructed for backward interpolation.

Newton forward and backward interpolation

The core idea

Worked example

Common traps

Before this

Where this leads