How to Use This Page
These are complete exercise solutions for Samacheer Kalvi 12th Business Maths — Numerical Methods (Chapter 5). This chapter is entirely about calculation skills — constructing difference tables and applying interpolation formulas. Practice building difference tables quickly and accurately, as examiners check the table values.
Key Concepts (Quick Recap)
y(x₀ + uh) = y₀ + uΔy₀ + u(u−1)/2! · Δ²y₀ + u(u−1)(u−2)/3! · Δ³y₀ + ... where u = (x − x₀)/h
y(xₙ + uh) = yₙ + u∇yₙ + u(u+1)/2! · ∇²yₙ + u(u+1)(u+2)/3! · ∇³yₙ + ... where u = (x − xₙ)/h
y(x) = Σᵢ [yᵢ · Πⱼ≠ᵢ (x − xⱼ)/(xᵢ − xⱼ)] Works for unequally spaced data.
- Finite differences: For equally spaced data with step h, the forward differences are defined as:
- First forward difference: Δyᵢ = yᵢ₊₁ − yᵢ
- Second forward difference: Δ²yᵢ = Δyᵢ₊₁ − Δyᵢ
- Third forward difference: Δ³yᵢ = Δ²yᵢ₊₁ − Δ²yᵢ
- n-th forward difference: Δⁿyᵢ = Δ^(n−1)yᵢ₊₁ − Δ^(n−1)yᵢ
- Backward differences:
- First backward difference: ∇yᵢ = yᵢ − yᵢ₋₁
- Second backward difference: ∇²yᵢ = ∇yᵢ − ∇yᵢ₋₁
- ∇yₙ = Δyₙ₋₁, ∇²yₙ = Δ²yₙ₋₂, etc.
- Newton's forward interpolation formula:
- Newton's backward interpolation formula:
- Lagrange's interpolation formula:
- Choosing the method:
- Equally spaced data, interpolate near start → Newton's forward
- Equally spaced data, interpolate near end → Newton's backward
- Unequally spaced data → Lagrange's interpolation
- Difference table construction: Build columns systematically: x, y, Δy, Δ²y, Δ³y, ... Always check that n-th differences are constant (or nearly constant) for a polynomial of degree n.
I. Choose the Best Answer (MCQs)
Q1. The first forward difference Δy₀ is: (a) y₁ − y₀ ✓ (b) y₀ − y₁ (c) y₂ − y₁ (d) y₀ − y₋₁
Answer(a) ✓ — By definition, Δy₀ = y₁ − y₀.
Q2. In Newton's forward interpolation formula, u = (x − x₀)/h. If x₀ = 5, h = 1, and x = 7, then u equals: (a) 1 (b) 2 ✓ (c) 7 (d) 5
Answer(b) ✓ — u = (7 − 5)/1 = 2/1 = 2.
Q3. For a polynomial of degree n, the (n+1)-th forward difference is: (a) n (b) 1 (c) 0 ✓ (d) n!
Answer(c) ✓ — If y is a polynomial of degree n, then Δⁿy is constant and Δ^(n+1)y = 0. This is a fundamental property used to verify difference tables.
Q4. Lagrange's interpolation is used when data points are: (a) Equally spaced only (b) Unequally spaced ✓ (c) Only 2 points (d) Only 3 points
Answer(b) ✓ — Lagrange's formula works for any spacing, but it is specifically useful when data points are unequally spaced (Newton's formulas require equal spacing).
Q5. If Δy₀ = 3 and Δy₁ = 5, then Δ²y₀ is: (a) 8 (b) 2 ✓ (c) −2 (d) 15
Answer(b) ✓ — Δ²y₀ = Δy₁ − Δy₀ = 5 − 3 = 2.
Q6. Newton's backward interpolation formula uses which type of differences? (a) Forward differences (b) Backward differences ✓ (c) Central differences (d) Divided differences
Answer(b) ✓ — Newton's backward formula uses backward differences (∇yₙ, ∇²yₙ, etc.).
Q7. The second backward difference ∇²y₃ equals: (a) y₃ − 2y₂ + y₁ ✓ (b) y₃ − y₂ + y₁ (c) y₃ + 2y₂ − y₁ (d) y₃ − y₂
Answer(a) ✓ — ∇²y₃ = ∇y₃ − ∇y₂ = (y₃ − y₂) − (y₂ − y₁) = y₃ − 2y₂ + y₁.
Q8. For data points (0,1), (1,3), (2,7), the first forward differences Δy₀ and Δy₁ are: (a) 2, 4 ✓ (b) 4, 2 (c) 2, 2 (d) 3, 7
Answer(a) ✓ — Δy₀ = y₁ − y₀ = 3 − 1 = 2. Δy₁ = y₂ − y₁ = 7 − 3 = 4.
II. Fill in the Blanks
1. The first forward difference Δyᵢ = ______. yᵢ₊₁ − yᵢ
2. The first backward difference ∇yᵢ = ______. yᵢ − yᵢ₋₁
3. For a cubic polynomial, the ______ forward difference is constant. third (3rd)
4. In Newton's forward interpolation, u = (x − x₀)/______. h
5. Lagrange's interpolation does not require ______ data points. equally spaced
6. Δ²y₀ = Δy₁ − ______. Δy₀
7. If y = x², then Δy₀ = ______ (with h = 1, x₀ = 0). 1 (y₀ = 0, y₁ = 1, Δy₀ = 1 − 0 = 1)
8. The relation between forward and backward differences: ∇yₙ = Δyₙ₋₁, ∇²yₙ = ______. Δ²yₙ₋₂
III. True or False
1. Newton's forward interpolation formula can be used for equally spaced data points. → True. Newton's forward formula requires equal spacing (h constant).
2. If the third differences are constant, the data represents a quadratic polynomial. → False. Constant third differences indicate a cubic (degree 3) polynomial. Constant second differences indicate a quadratic.
3. Lagrange's interpolation formula requires equal spacing of x-values. → False. Lagrange's formula works for any spacing — this is its main advantage over Newton's formulas.
4. ∇yₙ = Δyₙ₋₁ → True. Both expressions equal yₙ − yₙ₋₁. This is the fundamental relationship between forward and backward differences.
5. The sum of Lagrange basis polynomials Lᵢ(x) equals 1 for any x. → True. Σᵢ Lᵢ(x) = 1 for all x. This is a property used in verification.
6. Interpolation should not be used to estimate values outside the range of data points. → True. Estimating outside the range is called extrapolation, not interpolation. Extrapolation is less reliable.
IV. Match the Following
| Left | Right |
|---|
| 1. Forward difference Δyᵢ | d. yᵢ₊₁ − yᵢ |
| 2. Backward difference ∇yᵢ | a. yᵢ − yᵢ₋₁ |
| 3. Second forward difference Δ²y₀ | c. Δy₁ − Δy₀ |
| 4. Newton's forward formula | e. Uses Δy₀, Δ²y₀, ... |
| 5. Lagrange's formula | b. Works for unequal spacing |
Answer1→d, 2→a, 3→c, 4→e, 5→b
V. Answer Briefly
Q1. Construct the forward difference table for: x: 1, 2, 3, 4 and y: 1, 4, 9, 16.
Answer
| x | y | Δy | Δ²y | Δ³y | |---|---|-----|------|------| | 1 | 1 | 3 | 2 | 0 | | 2 | 4 | 5 | 2 | | | 3 | 9 | 7 | | | | 4 | 16| | | | Δy₀ = 4−1 = 3, Δy₁ = 9−4 = 5, Δy₂ = 16−9 = 7 Δ²y₀ = 5−3 = 2, Δ²y₁ = 7−5 = 2 Δ³y₀ = 2−2 = 0 Since Δ²y is constant (= 2) and Δ³y = 0, the data represents a quadratic polynomial.
Q2. Given x: 0, 1, 2, 3 and y: 1, 3, 7, 13, find Δy₀, Δ²y₀, and Δ³y₀.
Answer
Δy₀ = 3−1 = 2, Δy₁ = 7−3 = 4, Δy₂ = 13−7 = 6 Δ²y₀ = 4−2 = 2, Δ²y₁ = 6−4 = 2 Δ³y₀ = 2−2 = 0 Δy₀ = 2, Δ²y₀ = 2, Δ³y₀ = 0
Q3. For x₀ = 10, h = 5, find u when x = 17.
Answer
u = (x − x₀)/h = (17 − 10)/5 = 7/5 = 1.4
Q4. Write the first three terms of Newton's forward interpolation formula.
Answer
y(x₀ + uh) = y₀ + uΔy₀ + u(u−1)/2! · Δ²y₀ + ... First three terms: y₀ + uΔy₀ + [u(u−1)/2] Δ²y₀
Q5. State when to use Newton's forward vs backward interpolation.
Answer
- Newton's forward interpolation: Use when the interpolating point x is near the beginning of the table (near x₀). - Newton's backward interpolation: Use when the interpolating point x is near the end of the table (near xₙ). The choice affects accuracy and reduces computational error.
Q6. What is extrapolation? Why is it less reliable than interpolation?
Answer
Extrapolation is the process of estimating a value outside the range of known data points (i.e., x < x₀ or x > xₙ). It is less reliable because the polynomial model fitted to known data may not hold outside the observed range. The further from known data, the greater the potential error.
VI. Answer in Detail
Q1. Using Newton's forward interpolation, find y(2.5) from the following data:
Answer
Step 1: Build the forward difference table | x | y | Δy | Δ²y | Δ³y | |---|---|-----|------|------| | 1 | 1 | 7 | 12 | 6 | | 2 | 8 | 19 | 18 | | | 3 | 27| 37 | | | | 4 | 64| | | | Δy₀ = 8−1 = 7, Δy₁ = 27−8 = 19, Δy₂ = 64−27 = 37 Δ²y₀ = 19−7 = 12, Δ²y₁ = 37−19 = 18 Δ³y₀ = 18−12 = 6 Step 2: Compute u x₀ = 1, h = 1, x = 2.5 u = (x − x₀)/h = (2.5 − 1)/1 = 1.5 Step 3: Apply Newton's forward formula y(2.5) = y₀ + uΔy₀ + u(u−1)/2! · Δ²y₀ + u(u−1)(u−2)/3! · Δ³y₀ = 1 + 1.5(7) + 1.5(0.5)/2 · 12 + 1.5(0.5)(−0.5)/6 · 6 = 1 + 10.5 + (0.75/2)(12) + (−0.375/6)(6) = 1 + 10.5 + 0.375 · 12 + (−0.0625)(6) = 1 + 10.5 + 4.5 + (−0.375) = 1 + 10.5 + 4.5 − 0.375 = 15.625 Verification: Since y = x³, y(2.5) = (2.5)³ = 15.625 ✓
Q2. Using Newton's backward interpolation, find y(4.5) from the following data:
Answer
Step 1: Build the backward difference table (using ∇) | x | y | ∇y | ∇²y | ∇³y | |---|---|-----|------|------| | 1 | 1 | | | | | 2 | 8 | 7 | | | | 3 | 27 | 19 | 12 | | | 4 | 64 | 37 | 18 | 6 | ∇y₄ = 64−27 = 37, ∇y₃ = 27−8 = 19, ∇y₂ = 8−1 = 7 ∇²y₄ = 37−19 = 18, ∇²y₃ = 19−7 = 12 ∇³y₄ = 18−12 = 6 Step 2: Compute u xₙ = 4, h = 1, x = 4.5 u = (x − xₙ)/h = (4.5 − 4)/1 = 0.5 Step 3: Apply Newton's backward formula y(xₙ + uh) = yₙ + u∇yₙ + u(u+1)/2! · ∇²yₙ + u(u+1)(u+2)/3! · ∇³yₙ = 64 + 0.5(37) + 0.5(1.5)/2 · 18 + 0.5(1.5)(2.5)/6 · 6 = 64 + 18.5 + (0.75/2)(18) + (1.875/6)(6) = 64 + 18.5 + 6.75 + 1.875 = 91.125 Verification: Since y = x³, y(4.5) = (4.5)³ = 91.125 ✓
Q3. Using Lagrange's interpolation, find y(3) from: (1,1), (2,8), (5,125).
Answer
Step 1: Identify data points x₀ = 1, y₀ = 1; x₁ = 2, y₁ = 8; x₂ = 5, y₂ = 125 We need y(3). Step 2: Compute Lagrange basis polynomials at x = 3 L₀(3) = (3−2)(3−5)/((1−2)(1−5)) = (1)(−2)/((−1)(−4)) = −2/4 = −1/2 L₁(3) = (3−1)(3−5)/((2−1)(2−5)) = (2)(−2)/((1)(−3)) = −4/(−3) = 4/3 L₂(3) = (3−1)(3−2)/((5−1)(5−2)) = (2)(1)/((4)(3)) = 2/12 = 1/6 Step 3: Compute y(3) y(3) = y₀ · L₀(3) + y₁ · L₁(3) + y₂ · L₂(3) = 1 · (−1/2) + 8 · (4/3) + 125 · (1/6) = −0.5 + 32/3 + 125/6 = −1/2 + 32/3 + 125/6 = (−3 + 64 + 125)/6 = 186/6 = 31 Verification: The data is y = x³ (since 1³=1, 2³=8, 5³=125). So y(3) = 27. But wait — our calculation gives 31. Let me recheck. Actually, since we have only 3 points, Lagrange interpolation with 3 points gives a quadratic through them, not a cubic. The data points (1,1), (2,8), (5,125) do not lie on a quadratic — they lie on y = x³ (cubic). With only 3 points, the interpolation polynomial is degree 2, which cannot exactly reproduce a cubic. So the Lagrange interpolation gives y(3) = 31 based on the quadratic through the 3 given points, which differs from the true value y(3) = 27. This discrepancy arises because 3 points can only produce a degree-2 polynomial.
Q4. Construct the forward difference table and find the third differences for:
Answer
Difference Table: | x | y | Δy | Δ²y | Δ³y | Δ⁴y | |---|---|-----|------|------|------| | 0 | 1 | 2 | 2 | 0 | | | 1 | 3 | 4 | 2 | | | | 2 | 7 | 6 | | | | | 3 | 13| 8 | | | | | 4 | 21| | | | | Calculations: Δy₀ = 3−1 = 2, Δy₁ = 7−3 = 4, Δy₂ = 13−7 = 6, Δy₃ = 21−13 = 8 Δ²y₀ = 4−2 = 2, Δ²y₁ = 6−4 = 2, Δ²y₂ = 8−6 = 2 Δ³y₀ = 2−2 = 0, Δ³y₁ = 2−2 = 0 Since Δ²y is constant (= 2) and Δ³y = 0, the data represents a quadratic polynomial. In fact, y = x² + x + 1: 0+0+1=1 ✓, 1+1+1=3 ✓, 4+2+1=7 ✓, 9+3+1=13 ✓, 16+4+1=21 ✓.
Q5. Using Newton's forward interpolation, estimate the population in 1936 from the following census data:
Answer
| Year (x) | 1931 | 1941 | 1951 | 1961 |
|---|
| Population (y) in lakhs | 15 | 20 | 27 | 39 |
Step 1: Build difference table with x₀ = 1931, h = 10 | x | y | Δy | Δ²y | Δ³y | |---|---|-----|------|------| | 1931 | 15 | 5 | 2 | 5 | | 1941 | 20 | 7 | 7 | | | 1951 | 27 | 12 | | | | 1961 | 39 | | | | Δy₀ = 5, Δ²y₀ = 2, Δ³y₀ = 5 Step 2: Compute u u = (1936 − 1931)/10 = 5/10 = 0.5 Step 3: Apply Newton's forward formula y(1936) = y₀ + uΔy₀ + u(u−1)/2! · Δ²y₀ + u(u−1)(u−2)/3! · Δ³y₀ = 15 + 0.5(5) + 0.5(−0.5)/2 · 2 + 0.5(−0.5)(−1.5)/6 · 5 = 15 + 2.5 + (−0.25/2)(2) + (0.375/6)(5) = 15 + 2.5 + (−0.125)(2) + (0.0625)(5) = 15 + 2.5 − 0.25 + 0.3125 = 17.5625 lakhs So the estimated population in 1936 is approximately 17.56 lakhs.
Q6. Using Lagrange's interpolation, find y(2) from the data: (0,1), (1,3), (3,19).
Answer
Step 1: Identify data points x₀ = 0, y₀ = 1; x₁ = 1, y₁ = 3; x₂ = 3, y₂ = 19 We need y(2). Step 2: Compute Lagrange basis polynomials at x = 2 L₀(2) = (2−1)(2−3)/((0−1)(0−3)) = (1)(−1)/((−1)(−3)) = (−1)/(3) = −1/3 L₁(2) = (2−0)(2−3)/((1−0)(1−3)) = (2)(−1)/((1)(−2)) = −2/(−2) = 1 L₂(2) = (2−0)(2−1)/((3−0)(3−1)) = (2)(1)/((3)(2)) = 2/6 = 1/3 Step 3: Compute y(2) y(2) = y₀ · L₀(2) + y₁ · L₁(2) + y₂ · L₂(2) = 1 · (−1/3) + 3 · 1 + 19 · (1/3) = −1/3 + 3 + 19/3 = (−1 + 9 + 19)/3 = 27/3 = 9
Common Mistakes to Avoid
1. Incorrect difference table construction: The most common error is computing differences in the wrong direction. Forward differences: Δyᵢ = yᵢ₊₁ − yᵢ (subtract the PREVIOUS row from the NEXT). Always double-check the first row of each column.
2. Wrong u value in interpolation formulas: u = (x − x₀)/h for forward, u = (x − xₙ)/h for backward. Students often mix up x₀ and xₙ, or forget to divide by h.
3. Using forward formula when backward should be used: If x is near the end of the table, use Newton's backward formula. Using the forward formula near the end leads to large interpolation errors.
4. Sign errors in Newton's backward formula: The backward formula has u(u+1)/2!, u(u+1)(u+2)/3!, etc. — note the plus signs. This differs from the forward formula which has u(u−1)/2!, u(u−1)(u−2)/3!, etc. Mixing these up is a very common error.
5. Calculation errors in Lagrange's basis polynomials: Each Lᵢ(x) = Πⱼ≠ᵢ (x − xⱼ)/(xᵢ − xⱼ). Do not include j = i in the product. Check that you have the correct signs in the denominator.
6. Confusing forward and backward differences in the table: In the difference table, forward differences Δy are listed starting from the top, while backward differences ∇y are listed starting from the bottom. When using Newton's backward formula, read the bottom diagonal of the table.
Exam Checklist
- [ ] I can construct forward and backward difference tables correctly
- [ ] I can identify the degree of the polynomial from the difference table (constant n-th differences = degree n)
- [ ] I can apply Newton's forward interpolation formula with the correct u value
- [ ] I can apply Newton's backward interpolation formula and distinguish it from forward
- [ ] I can apply Lagrange's interpolation formula for unequally spaced data
- [ ] I can choose the appropriate interpolation method based on data spacing and position
- [ ] I can verify interpolation results by checking against known values
- [ ] I understand the difference between interpolation (within range) and extrapolation (outside range)
