The Gregory Newton Formula is a powerful interpolation technique used to estimate unknown values between tabulated data points by employing the forward difference operator. Also known as the Newton Forward Difference Interpolation Formula, it expresses any function value \( f(x) \) as a polynomial in terms of the forward differences \( \Delta^n f_0 \). This formula is covered in numerical methods topics relevant to Class 11 and Class 12 Mathematics, and it appears in engineering entrance contexts including JEE Advanced. This article covers the definition, derivation, complete formula sheet, three solved examples, CBSE exam tips, common mistakes, and FAQs on the Gregory Newton Formula.
Key Gregory Newton Formula Expressions at a Glance
Quick reference for the most important Gregory Newton interpolation formulas.
- Gregory Newton Forward Interpolation: \( f(x_0 + ph) = f_0 + p\Delta f_0 + rac{p(p-1)}{2!}\Delta^2 f_0 + rac{p(p-1)(p-2)}{3!}\Delta^3 f_0 + \cdots \)
- Forward Difference Operator: \( \Delta f_n = f_{n+1} – f_n \)
- Second Forward Difference: \( \Delta^2 f_n = \Delta f_{n+1} – \Delta f_n \)
- Step size: \( h = x_{n+1} – x_n \)
- Interpolating variable: \( p = rac{x – x_0}{h} \)
- General term (r-th): \( rac{p(p-1)(p-2)\cdots(p-r+1)}{r!}\Delta^r f_0 \)
- Binomial coefficient form: \( inom{p}{r}\Delta^r f_0 \)
What is the Gregory Newton Formula?
The Gregory Newton Formula is a finite-difference interpolation method. It calculates the approximate value of a function at any point within a tabulated range. The formula uses the first tabulated value \( f_0 \) and successive forward differences to build a polynomial approximation.
James Gregory and Isaac Newton independently developed this technique in the seventeenth century. It is therefore named after both mathematicians. The method works best when the argument \( x \) lies near the beginning of the data table. This is why it is called the “forward” difference formula.
In the NCERT curriculum, interpolation concepts appear in Class 12 Mathematics (Chapter: Relations and Functions) and in the Class 11 context of sequences and differences. The Gregory Newton Formula is also a standard topic in B.Sc. Mathematics, B.Tech. numerical methods courses, and competitive exams that test analytical and approximation skills.
The formula requires equally spaced data points. The spacing between consecutive x-values must be constant and is denoted by \( h \). Given a table of \( n+1 \) values, the Gregory Newton Formula constructs a polynomial of degree at most \( n \) that passes through all the given points.
Gregory Newton Formula — Expression and Variables
The Gregory Newton Forward Difference Interpolation Formula is:
\[ f(x_0 + ph) = f_0 + p\Delta f_0 + \frac{p(p-1)}{2!}\Delta^2 f_0 + \frac{p(p-1)(p-2)}{3!}\Delta^3 f_0 + \cdots + \frac{p(p-1)(p-2)\cdots(p-n+1)}{n!}\Delta^n f_0 \]
In compact summation notation, the Gregory Newton Formula is written as:
\[ f(x_0 + ph) = \sum_{r=0}^{n} \binom{p}{r} \Delta^r f_0 \]
where \( \binom{p}{r} = \frac{p(p-1)(p-2)\cdots(p-r+1)}{r!} \) is the generalised binomial coefficient.
| Symbol | Quantity | Remarks |
|---|---|---|
| \( f_0 \) | First tabulated function value | Value of \( f(x) \) at \( x = x_0 \) |
| \( h \) | Step size (spacing) | \( h = x_{n+1} – x_n \); must be constant |
| \( p \) | Interpolating variable | \( p = (x – x_0)/h \); dimensionless |
| \( \Delta f_0 \) | First forward difference at \( x_0 \) | \( \Delta f_0 = f_1 – f_0 \) |
| \( \Delta^2 f_0 \) | Second forward difference at \( x_0 \) | \( \Delta^2 f_0 = \Delta f_1 – \Delta f_0 \) |
| \( \Delta^r f_0 \) | r-th order forward difference | Obtained from the forward difference table |
| \( n \) | Degree of interpolating polynomial | One less than number of data points |
| \( f(x_0 + ph) \) | Interpolated function value | Estimated value at the desired point \( x \) |
Derivation of the Gregory Newton Formula
We begin by assuming that \( f(x) \) can be expressed as a polynomial of degree \( n \) in \( p \), where \( p = (x – x_0)/h \).
Step 1: Write \( f(x) = a_0 + a_1 p + a_2 p(p-1) + a_3 p(p-1)(p-2) + \cdots \)
Step 2: Set \( x = x_0 \), so \( p = 0 \). This gives \( f_0 = a_0 \).
Step 3: Apply the forward difference operator \( \Delta \) to both sides. At \( p = 0 \), \( \Delta f_0 = a_1 \).
Step 4: Apply \( \Delta^2 \) to both sides. At \( p = 0 \), \( \Delta^2 f_0 = 2! \cdot a_2 \), so \( a_2 = \Delta^2 f_0 / 2! \).
Step 5: By induction, \( a_r = \Delta^r f_0 / r! \). Substituting all \( a_r \) back gives the Gregory Newton Formula.
Complete Numerical Methods Formula Sheet
| Formula Name | Expression | Variables | Use Case | Topic |
|---|---|---|---|---|
| Gregory Newton Forward Formula | \( f(x_0+ph) = \sum_{r=0}^{n}\binom{p}{r}\Delta^r f_0 \) | p = (x−x⊂0;)/h, h = step size | Interpolation near start of table | Numerical Methods |
| Newton Backward Interpolation | \( f(x_n+ph) = \sum_{r=0}^{n}(-1)^r\binom{-p}{r}\nabla^r f_n \) | p = (x−x⊂n;)/h, \( \nabla \) = backward diff. | Interpolation near end of table | Numerical Methods |
| Forward Difference Operator | \( \Delta f_n = f_{n+1} – f_n \) | f⊂n; = tabulated value at x⊂n; | Building difference table | Finite Differences |
| Backward Difference Operator | \( \nabla f_n = f_n – f_{n-1} \) | f⊂n; = tabulated value at x⊂n; | Building backward difference table | Finite Differences |
| Central Difference Operator | \( \delta f_n = f_{n+1/2} – f_{n-1/2} \) | Half-integer subscripts | Central interpolation | Finite Differences |
| Lagrange Interpolation | \( f(x) = \sum_{i=0}^{n} f_i \prod_{j \neq i} \frac{x – x_j}{x_i – x_j} \) | x⊂i; = given points, f⊂i; = values | Unequally spaced data | Numerical Methods |
| Stirling’s Formula | \( f(x_0+ph) = f_0 + p\mu\delta f_0 + \frac{p^2}{2!}\delta^2 f_0 + \cdots \) | \( \mu \) = mean, \( \delta \) = central diff. | Interpolation near centre of table | Numerical Methods |
| Bessel’s Formula | \( f(x_0+ph) = \frac{f_0+f_1}{2} + (p-\tfrac{1}{2})\Delta f_0 + \cdots \) | p between 0 and 1 | Interpolation between two central values | Numerical Methods |
| Newton Divided Difference | \( f(x) = f[x_0] + (x-x_0)f[x_0,x_1] + \cdots \) | f[x⊂i;,x⊂j;] = divided differences | Unequally spaced data | Numerical Methods |
| Trapezoidal Rule | \( \int_{x_0}^{x_n} f\,dx \approx \frac{h}{2}[f_0 + 2(f_1+\cdots+f_{n-1}) + f_n] \) | h = step size | Numerical integration | Numerical Methods |
| Simpson’s 1/3 Rule | \( \int_{x_0}^{x_n} f\,dx \approx \frac{h}{3}[f_0 + 4f_1 + 2f_2 + 4f_3 + \cdots + f_n] \) | n must be even | Numerical integration | Numerical Methods |
Gregory Newton Formula — Solved Examples
Example 1 (Class 11-12 Level): Basic Interpolation
Problem: Given the following table of values, use the Gregory Newton Forward Difference Formula to find \( f(1.5) \).
| x | f(x) |
|---|---|
| 1 | 1 |
| 2 | 8 |
| 3 | 27 |
| 4 | 64 |
Given: \( x_0 = 1,\ h = 1,\ x = 1.5 \), so \( p = (1.5 – 1)/1 = 0.5 \).
Step 1: Build the forward difference table.
| x | f(x) | \( \Delta f \) | \( \Delta^2 f \) | \( \Delta^3 f \) |
|---|---|---|---|---|
| 1 | 1 | 7 | 12 | 6 |
| 2 | 8 | 19 | 18 | — |
| 3 | 27 | 37 | — | — |
| 4 | 64 | — | — | — |
Step 2: Read off the first-row differences: \( f_0 = 1,\ \Delta f_0 = 7,\ \Delta^2 f_0 = 12,\ \Delta^3 f_0 = 6 \).
Step 3: Apply the Gregory Newton Formula with \( p = 0.5 \):
\[ f(1.5) = 1 + (0.5)(7) + \frac{(0.5)(0.5-1)}{2!}(12) + \frac{(0.5)(0.5-1)(0.5-2)}{3!}(6) \]
Step 4: Calculate each term separately.
Term 1: \( 1 \)
Term 2: \( 0.5 \times 7 = 3.5 \)
Term 3: \( \frac{0.5 \times (-0.5)}{2} \times 12 = \frac{-0.25}{2} \times 12 = -0.125 \times 12 = -1.5 \)
Term 4: \( \frac{0.5 \times (-0.5) \times (-1.5)}{6} \times 6 = \frac{0.375}{6} \times 6 = 0.375 \)
Step 5: Sum all terms: \( f(1.5) = 1 + 3.5 – 1.5 + 0.375 = 3.375 \).
Answer
\( f(1.5) = 3.375 \). This matches the exact value \( (1.5)^3 = 3.375 \), confirming the formula’s accuracy for polynomial data.
Example 2 (Class 12 / B.Sc. Level): Finding a Missing Value
Problem: The following table gives \( \log_{10} x \) for selected values of \( x \). Estimate \( \log_{10}(337.5) \) using the Gregory Newton Forward Difference Formula.
| x | \( \log_{10} x \) |
|---|---|
| 310 | 2.4914 |
| 320 | 2.5051 |
| 330 | 2.5185 |
| 340 | 2.5315 |
| 350 | 2.5441 |
Given: \( x_0 = 310,\ h = 10,\ x = 337.5 \), so \( p = (337.5 – 310)/10 = 2.75 \).
Step 1: Build the forward difference table (values rounded to 4 decimal places).
| x | f(x) | \( \Delta f \) | \( \Delta^2 f \) | \( \Delta^3 f \) | \( \Delta^4 f \) |
|---|---|---|---|---|---|
| 310 | 2.4914 | 0.0137 | −0.0003 | 0.0000 | 0.0001 |
| 320 | 2.5051 | 0.0134 | −0.0003 | 0.0001 | — |
| 330 | 2.5185 | 0.0130 | −0.0000 | — | — |
| 340 | 2.5315 | 0.0126 | — | — | — |
| 350 | 2.5441 | — | — | — | — |
Step 2: Use \( f_0 = 2.4914,\ \Delta f_0 = 0.0137,\ \Delta^2 f_0 = -0.0003 \). Higher differences are negligible.
Step 3: Apply the Gregory Newton Formula with \( p = 2.75 \):
\[ f(337.5) = 2.4914 + (2.75)(0.0137) + \frac{(2.75)(1.75)}{2}(-0.0003) \]
Step 4: Calculate each term.
Term 1: \( 2.4914 \)
Term 2: \( 2.75 \times 0.0137 = 0.037675 \)
Term 3: \( \frac{2.75 \times 1.75}{2} \times (-0.0003) = 2.40625 \times (-0.0003) = -0.000722 \)
Step 5: Sum: \( f(337.5) \approx 2.4914 + 0.037675 – 0.000722 = 2.5284 \).
Answer
\( \log_{10}(337.5) \approx 2.5284 \). The actual value is approximately \( 2.5283 \), confirming the Gregory Newton Formula gives excellent accuracy.
Example 3 (JEE / Advanced Level): Constructing the Interpolating Polynomial
Problem: Using the Gregory Newton Forward Difference Formula, find the polynomial that fits the following data. Then evaluate it at \( x = 2.5 \).
| x | f(x) |
|---|---|
| 0 | 1 |
| 1 | 0 |
| 2 | 1 |
| 3 | 10 |
Given: \( x_0 = 0,\ h = 1 \). For \( x = 2.5 \), \( p = (2.5 – 0)/1 = 2.5 \).
Step 1: Build the forward difference table.
| x | f(x) | \( \Delta f \) | \( \Delta^2 f \) | \( \Delta^3 f \) |
|---|---|---|---|---|
| 0 | 1 | −1 | 2 | 6 |
| 1 | 0 | 1 | 8 | — |
| 2 | 1 | 9 | — | — |
| 3 | 10 | — | — | — |
Step 2: \( f_0 = 1,\ \Delta f_0 = -1,\ \Delta^2 f_0 = 2,\ \Delta^3 f_0 = 6 \).
Step 3: Write the Gregory Newton polynomial in \( p \):
\[ f(p) = 1 + p(-1) + \frac{p(p-1)}{2}(2) + \frac{p(p-1)(p-2)}{6}(6) \]
Step 4: Simplify each term.
Term 1: \( 1 \)
Term 2: \( -p \)
Term 3: \( p(p-1) \)
Term 4: \( p(p-1)(p-2) \)
So \( f(p) = 1 – p + p^2 – p + p^3 – 3p^2 + 2p = p^3 – 2p^2 + 0p + 1 \).
Since \( p = x \) when \( h = 1 \) and \( x_0 = 0 \), the polynomial is \( f(x) = x^3 – 2x^2 + 1 \).
Step 5: Evaluate at \( x = 2.5 \):
\[ f(2.5) = (2.5)^3 – 2(2.5)^2 + 1 = 15.625 – 12.5 + 1 = 4.125 \]
Answer
The interpolating polynomial is \( f(x) = x^3 – 2x^2 + 1 \) and \( f(2.5) = 4.125 \). This demonstrates how the Gregory Newton Formula recovers the exact polynomial from a finite set of data points.
CBSE Exam Tips 2025-26
- Always build the difference table first. Errors in the difference table cascade through the entire calculation. Double-check every subtraction before applying the formula.
- Verify equal spacing. The Gregory Newton Forward Difference Formula is valid only for equally spaced \( x \)-values. If the spacing varies, switch to Newton’s Divided Difference or Lagrange Interpolation.
- Compute \( p \) carefully. The value \( p = (x – x_0)/h \) must be calculated precisely. A small error in \( p \) propagates into every term of the formula.
- Use the formula near the start of the table. The Gregory Newton Forward Formula gives the best accuracy when \( p \) is between 0 and 1. For values near the end of the table, use the backward difference formula instead.
- Truncate wisely. In board exams, you usually need only up to the third or fourth difference. Higher-order differences are often zero or negligible for standard polynomial data.
- We recommend practising at least five difference table problems before your 2025-26 board examination. Speed and accuracy in building the table are the keys to full marks.
Common Mistakes to Avoid
- Mistake 1 — Wrong sign in differences. Students often subtract in the wrong direction. Always compute \( \Delta f_n = f_{n+1} – f_n \), not \( f_n – f_{n+1} \). A sign error here ruins all subsequent differences.
- Mistake 2 — Using the formula for unequal spacing. The Gregory Newton Forward Formula requires constant \( h \). Using it on unevenly spaced data gives incorrect results. Use Lagrange’s formula for unequal spacing.
- Mistake 3 — Forgetting the factorial in the denominator. The general term is \( \frac{p(p-1)\cdots(p-r+1)}{r!}\Delta^r f_0 \). Students frequently omit \( r! \), especially for the third and fourth terms.
- Mistake 4 — Choosing the wrong \( x_0 \). Always take \( x_0 \) as the tabulated value closest to but less than (or equal to) the desired \( x \). Taking \( x_0 \) far from the interpolation point reduces accuracy and increases the number of terms needed.
- Mistake 5 — Rounding intermediate results too early. Rounding \( p \) or the differences to fewer decimal places mid-calculation introduces significant error in the final answer. Carry at least four decimal places throughout.
JEE Application of the Gregory Newton Formula
In our experience, JEE aspirants encounter the Gregory Newton Formula primarily in the context of numerical methods and sequence-based problems. While NEET focuses on biology and physics rather than numerical analysis, JEE Advanced occasionally tests interpolation concepts in its mathematics paper.
Application Pattern 1 — Polynomial Recovery: JEE problems may give a table of values of an unknown polynomial and ask for the polynomial itself. The Gregory Newton Forward Difference Formula systematically recovers the polynomial, as demonstrated in Example 3 above. Once the polynomial is known, any further question (finding roots, maxima, definite integrals) becomes straightforward.
Application Pattern 2 — Error Estimation: Advanced problems ask students to estimate the error in interpolation. The truncation error of the Gregory Newton Formula after \( n \) terms is proportional to \( h^{n+1} f^{(n+1)}(\xi) / (n+1)! \) for some \( \xi \) in the interval. JEE questions may ask which order of formula is sufficient to achieve a given accuracy.
Application Pattern 3 — Numerical Differentiation: The Gregory Newton Formula can be differentiated term by term with respect to \( p \) to obtain numerical derivatives. JEE problems sometimes provide a data table and ask for \( f'(x) \) or \( f”(x) \) at a given point. Differentiating the Gregory Newton polynomial is the standard approach.
In our experience, JEE aspirants who master the forward difference table construction and the binomial coefficient pattern in the Gregory Newton Formula can solve these problems in under three minutes, well within the time constraints of JEE Advanced Paper 2.
FAQs on Gregory Newton Formula
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