Scale:
In x axis 1 cm = 10 units
In y axis 1 cm = 2 units
Plot the points (40, 6.2), (50, 7.2), (60, 9.1) and (70, 12). At x = 48, draw a vertical line to the graph and from the intersecting point, draw a horizontal line to meet the y-axis
From the graph, we find that when x = 48, the value of Y is equal to 6.8.

when x = 0, y = 0 + 0 + 0 - 1 ⇒ y = -1
when x = 1, y = 1³-1²+ 1-1 ⇒ y=0.
when x = 2, y = 2³ - 2² + 2 - 1 ⇒ y = 8 - 4 + 1 ⇒ y = 5
when x = 3, y = 3³ - 3²+ 3 - 1 ⇒ y = 27 - 9 + 2 ⇒ y = 20
when x = 4, y = 4³ - 4²+ 4 -1 ⇒ y= 64 -16 + 3 ⇒ y = 51
when x = 5, y = 5³ - 5²+ 5 - 1 ⇒ y = 125 - 25 + 4 ⇒ y = 104.
Hence, the forward difference table is
 x  y Δy Δ²y Δ³y Δ⁴y 0 -1             1       1 0   4         5   6   2 5   10   0     15   6   3 20   16   0     31   6   4 51   22         53       5 104        

Δ(log ax) = log (ax + h) -log (ax)
= log\(\left( \frac { ax+h }{ ax } \right) \)
=log \(\left( \frac { ax }{ ax } +\frac { h }{ ax } \right) \)
= log \(\left( 1+\frac { h }{ ax } \right) \)
∴ ∆ (log ax) = log \(\left( 1+\frac { h }{ ax } \right) \)

Given

x	-1	0	1	2
y	202	175	82	55

Since we have to find f(0.5) which is at the beginning of the table, use Newton's forward interpolation formula.
x_n + nh = 0.5 ⇒ -1 + n(1) = 0.5
⇒ n = 0.5 + 1 = 1.5
∴ y_(0.5) = \(\frac { n }{ 1! } { \triangle y }_{ 0 }+\frac { n(n+1) }{ 2! } { \triangle }^{ 2 }{ y }_{ 0 }+\frac { n(n+1)(n+2) }{ 3! } { \triangle }^{ 3 }{ y }_{ 0 }\)
The difference table is

x	y	Δy	Δ2y	Δ3y
-1	202
0	175
1	82	-93	-66
2	55	-27	66	132

∴ y(0.5) = 202+\(\frac { 1.5 }{ 1! } (-27)+\frac { (1.5)(1.5-1) }{ 2! } (-66)+\frac { (1.5)(1.5-1)(1.5-2) }{ 3! } (132)\)
= 202 - 40.5 + (1.5) (.5) (-33) + \(\frac { (1.5)(.5)(-0.5) }{ 6(132) } \) (132)
= 202 - 40.5 - 24.75 - 8.25
= 202 -73.5
= 128.5
Hence f(0.5) = 128.5

Here the intervals are unequal
∴ By Lagrange'sinterpolation formula, we have
x₀ = 15, x₁ = 25, x₂ = 30, x₃ = 35
y₀= 36, y₁ = 40, y₂ = 45, y₃ = 38, and x = 26
∴ y = f(x) = \(\frac { (x-{ x }_{ 1 })(x-{ x }_{ 2 })(x-{ x }_{ 3 }) }{ ({ x }_{ 0 }-{ x }_{ 1 })({ x }_{ 0 }-{ x }_{ 2 })({ x }_{ 0 }-{ x }_{ 3 }) } \times { y }_{ 0 }+\frac { (x-{ x }_{ 0 })(x-{ x }_{ 2 })(x-{ x }_{ 3 }) }{ ({ x }_{ 1 }-{ x }_{ 0 })({ x }_{ 1 }-{ x }_{ 2 })(x_{ 1 }-{ x }_{ 3 }) } \times { y }_{ 1 }+\frac { (x-{ x }_{ 0 })(x-{ x }_{ 1 })(x-{ x }_{ 3 }) }{ ({ x }_{ 2 }-{ x }_{ 0 })({ x }_{ 2 }-{ x }_{ 1 })({ x }_{ 2 }-{ x }_{ 3 }) } \times { y }_{ 2 }+\frac { (x-{ x }_{ 0 })(x-{ x }_{ 1 })(x-{ x }_{ 2 }) }{ ({ x }_{ 3 }-{ x }_{ 0 })({ x }_{ 3 }-{ x }_{ 1 })({ x }_{ 3 }-{ x }_{ 2 }) } \times { y }_{ 3 }\)
= \(\frac { (26-25)(26-30)(26-35) }{ (15-25)(15-30)(15-35) } (36)+\frac { (26-15)(26-1530)(26-35) }{ (25-15)(5-30)(25-35) } (40)+\frac { 26-15)(26-25)(26-35) }{ (30-15)(30-30)(30-35) } (40)+\frac { (26-15)(26-25)(26-30) }{ (35-15)(35-25)(35-30) } (38)\)
= \(\frac { (1)(-4)(-9) }{ (-10)(-15)(-20) } (36)+\frac { (n)(-4)(-9) }{ (10)(-5)(-10) } (40)+\frac { (11)(1)(-9) }{ (15)(5)(-5) } (45)+\frac { (9)(1)(-4) }{ (-20)(10)(5) } (38)\)
= \(-\frac { (36)(36) }{ (150)(20) } +\frac { (44)(9)(40) }{ 500 } +\frac { 99(45) }{ (15)(25) } -\frac { (36)(38) }{ (200)(15) } \)
= \(\frac { 1296 }{ 3000 } +\frac { 15840 }{ 500 } +\frac { 4455 }{ 375 } -\frac { 1368 }{ 1000 } \)
= -0.432 + 31.68 + 11.88 - 1.368
= 41.76.
Hence the number of workers getting income not exceeding Rs. 26 per month is 42.

Given

Year	1974	1978	1982	1990
Output in 1000 tones	25	60	80	170

 Here the intervals are unequal.
∴ By Lagranges interpolation formula, we have
x₀ = 1974, x₁ = 1978, x₂ = 1982, x₃ = 1990
y₀ = 25, y₁ = 60, y₂ = 80, y₃ = 170 and x = 1986.
∴ y = f(x) = \(\frac { (x-{ x }_{ 1 })(x-{ x }_{ 2 })(x-{ x }_{ 3 }) }{ ({ x }_{ 0 }-{ x }_{ 1 })({ x }_{ 0 }-{ x }_{ 2 })({ x }_{ 0 }-{ x }_{ 3 }) } \times { y }_{ 0 }+\frac { (x-{ x }_{ 0 })(x-{ x }_{ 2 })(x-{ x }_{ 3 }) }{ ({ x }_{ 1 }-{ x }_{ 0 })({ x }_{ 1 }-{ x }_{ 2 })(x_{ 1 }-{ x }_{ 3 }) } \times { y }_{ 1 }+\frac { (x-{ x }_{ 0 })(x-{ x }_{ 1 })(x-{ x }_{ 3 }) }{ ({ x }_{ 2 }-{ x }_{ 0 })({ x }_{ 2 }-{ x }_{ 1 })({ x }_{ 2 }-{ x }_{ 3 }) } \times { y }_{ 2 }+\frac { (x-{ x }_{ 0 })(x-{ x }_{ 1 })(x-{ x }_{ 2 }) }{ ({ x }_{ 3 }-{ x }_{ 0 })({ x }_{ 3 }-{ x }_{ 1 })({ x }_{ 3 }-{ x }_{ 2 }) } \times { y }_{ 3 }\)
\(\frac { (x-{ x }_{ 0 })(x-{ x }_{ 2 })(x-{ x }_{ 3 }) }{ ({ x }_{ 0 }-{ x }_{ 1 })({ x }_{ 0 }-{ x }_{ 2 })({ x }_{ 0 }-{ x }_{ 3 }) } \times25+\frac { (x-{ x }_{ 0 })(x-{ x }_{ 2 })(x-{ x }_{ 3 }) }{ ({ x }_{ 1 }-{ x }_{ 0 })({ x }_{ 1 }-{ x }_{ 2 })(x_{ 1 }-{ x }_{ 3 }) } \times60+\frac { (x-{ x }_{ 0 })(x-{ x }_{ 1 })(x-{ x }_{ 3 }) }{ ({ x }_{ 2 }-{ x }_{ 0 })({ x }_{ 2 }-{ x }_{ 1 })({ x }_{ 2 }-{ x }_{ 3 }) } \times80+\frac { (x-{ x }_{ 0 })(x-{ x }_{ 1 })(x-{ x }_{ 2 }) }{ ({ x }_{ 3 }-{ x }_{ 0 })({ x }_{ 3 }-{ x }_{ 1 })({ x }_{ 3 }-{ x }_{ 2 }) } \times 70+\frac { (1986-1974)(1986-1982)(1986-1990) }{ (1990-1974)(1990-1978)(1990-1982) } \times 170\)

= 6.25 - 60 + 120 + 42.5
y = 108.75

Let the percentage of lead be x and temperature be y.
Given

P	40	50	60	70	80	90
T	180	204	226	250	276	304

Find y when x = 84
Since the temperature required is at the end of the table we apply Newton's backward interpolation formula.
∴ x_n + nh = 84 ⇒ 90 + n(10) = 84
⇒ 10n = 84 - 90 = -6
[∵ h = 10 & x_n = 90]
⇒ n = \(\frac{-6}{10}\) = -0.6
The difference table is

∴ The Newton's backward interpolation formula is
y_(y=xn+nh) = \(\frac { n }{ 1! } \Delta { y }_{ n }+\frac { n(n+1) }{ 2! } { \triangledown }^{ 2 }{ y }_{ n }+\frac { n(n+1)(n+2) }{ 3! } { \triangledown }^{ 3 }{ y }_{ n }+\frac { n(n+1)(n+2)(n+3) }{ 4! } { \triangledown }^{ 4 }{ y }_{ n }+\frac { n(n+1)(n+2)(n+3)(n+4) }{ 5! } { \triangledown }^{ 5 }{ y }_{ n }\)
∴ y(84) = 304 + (-0.6) (28) + \(\frac { (-0.6)(-0.6+1) }{ 2 } (2)+0+0+\frac { (-0.6)(-0.6+1)(-0.6+3)(-0.6+4) }{ 5! } (4)\) [∵ ∇³ & ∇⁴ are zero]
⇒ y(84) = 304 - 16.8+ (-0.6) (0.4) + \(\frac { (-0.6)(0.4)(1.4)(2.4)(3.4) }{ 120 } (4)\)
⇒ y(84) = 304 -16.8 - 0.24 - 0.09139
⇒ y(84) = 286.86
Hence, the melting point of the alloy containing 84 percent lead is 286.86°C.

Given

Marks	0-19	20-39	40-59	60-79	80-99
No.of.candidates	41	62	65	50	17

 This can be rewritten as

Marks:	Below 19	Below 39	Below 59	Below 79	Below 99
No. of.  candidates:	41	(41 + 62) =103	(41 + 62 + 65) = 168	(41 + 62 + 65 + 50) =218	(41 + 62 + 65 + 50 + 17) = 235

Since we have to find below 70, use Newton's backward interpolation formula
∴ x_n + nh = 70 ⇒ 99 + n(20) = 70
⇒ 20n = 70 - 99 = -29
⇒ n = \(\frac{-29}{20}\) = -1.45
∴ y₇₀ = y_n + \(\frac { n }{ 1! } { \triangledown y }_{ n }+\frac { n(n+1) }{ 2! } { \triangledown }^{ 2 }{ y }_{ n }+\frac { n(n+1)(n+2) }{ 3! } { \triangledown }^{ 3 }{ y }_{ n }\)
\(=235-1.45(17)+\frac { (-1.45)(-1.45+1) }{ 2 } (-33)+\frac { (-1.45)(-1.45+2)(-18) }{ 6 } \)
\(=235-24.65+\frac { (-1.45)(-.45) }{ 2 } (-33)+\) (-1.45)(-0.45)(0.55)(-3)
= 235 - 24.65 - 12.375 - 1.125
= 196
Hence, the number of students who have scored below 70 are 196 (app).

The forward interpolation formula is
\({ y }_{ ({ x=x }_{ 0 }+nh) }={ y }_{ 0 }+\frac { n }{ 1! } =\triangle { y }_{ 0 }+\frac { n(n-1) }{ 2! } { \triangle }^{ 2 }{ y }_{ 0 }+\frac { n(n-1)(n-2) }{ 3! } { \triangle }^{ 3 }{ y }_{ 0 }+...\)
Here x₀ + nh = x ⇒ x₀ = 0, h = 1
∴ 0 + n = x ⇒ n = x.
The difference table is

x	y = f(x)	Δy	Δ2y	Δ3y
0	1
		1
1	2		-2
		-1		12
2	1		10
		9
3	10

\({ y }_{ (n=x) }={ y }_{ 0 }+\frac { n }{ 1! } =\triangle { y }_{ 0 }+\frac { n(n-1) }{ 2! } { \triangle }^{ 2 }{ y }_{ 0 }+\frac { n(n-1)(n-2) }{ 3! } { \triangle }^{ 3 }{ y }_{ 0 }+...\)
\(y=1+\frac { n }{ 1! } (1)+\frac { n(n-1) }{ 2! } (-2)+\frac { n(n-1)(n-2) }{ 6 } (12)\)
⇒ y = 1 + x + (x² - x)(-1) + x(x² - 3x + 2) (2)
⇒ y = 1 + x - x² + x + 2x³ - 6x² + 4x
⇒ y = 1+ 6x - 7x² + 2x³
Hence, the cubic polynomial is 2x³ - 7x²+ 6x + 1.

By partial fraction method
\(\frac { 1 }{ (x+1)(x+2) } =\frac { A }{ x+1 } +\frac { B }{ x+2 } \)

⇒ 1 = A (x + 2) + B (x + 1)
when x = -1, 1 =  A [-1 + 2] ⇒ 1 = A
when x = -2, 1 = B [-2+ 1] ⇒1 = - B
⇒ B = -1.
∴ \(\frac { 1 }{ (x+1)(x+2) } =\frac { 1 }{ x+1 } +\frac { 1 }{ x+2 } \)
\(\therefore \triangle \left[ \frac { 1 }{ (x+1)(x+2) } \right] =\triangle \left[ \frac { 1 }{ x+1 } -\frac { 1 }{ x+2 } \right] \)
∴ \(\left( \frac { 1 }{ x+1+1 } -\frac { 1 }{ x+1 } \right) -\left( \frac { 1 }{ x+1+2 } -\frac { 1 }{ x+2 } \right) \) [∵ ∆ f(x) = f(x+1) - f(x)]
= \(\left( \frac { 1 }{ x+2 } -\frac { 1 }{ x+1 } \right) -\left( \frac { 1 }{ x+3 } -\frac { 1 }{ x+2 } \right) \) where h = 1
= \(\frac { 1 }{ x+2 } -\frac { 1 }{ x+1 } -\frac { 1 }{ x+3 } +\frac { 1 }{ x+2 } \)
= \(\frac { 2 }{ x+2 } -\frac { 1 }{ x+1 } -\frac { 1 }{ 1+3 } \)
= \(\frac { 2(x+1)(x+3)-1(x+2)(x+3)-1(x+1)(x+2) }{ (x+1)(x+2)(x+3) } \)
= \(\frac { 2({ x }^{ 2 }+4x+3)-({ x }^{ 2 }+5x+6)-({ x }^{ 2 }+3x-2) }{ (x+1)(x+2)(x+3) } \)
= \(\frac { -2 }{ (x+1)(x+2)(x+3) } \)
\(\therefore\ =\triangle \left[ \frac { 1 }{ (x+1)(x+2) } \right] =\frac { -2 }{ (x+1)(x+2)(x+3) } \)

We know  f(x), ∇f(x + h), ∇f(x + 2h),... are the first order differences of f(x).
= f(x) - 2f(x - h) + f(x - 2h)
Consider ∇²f(x) = ∇[∇f(x))]
= ∇[f(x) - f(x - h)]
= ∇ f(X) - ∇ f(x - h)
= [f(x) - f(x - h)] - [f(x - h) - f(x - 2h)]
= f(x) - f(x - h) - f(x - h) - f(x - 2h)
∴ ∇² f(x), ∇² f(x + h), ∇² f(x + 2h) ... are the second order differences of f(x).

We know Δ (f(x)) = f(x + h) -f(x)
Since h = 1, ∆ (f)) = f(x + 1) - f(x)
[∵ (x + 1)³ = x³ + 3x² + 3x + 1]
⇒ Δ (x³) = (x + 1)³ - x³

⇒ Δ (x³) = 3x² + 3x + 1

By data we have,
x₀ = 0, x₁ = 1, x₂ = 3
y₀ = 5, y₁ = 6, y₂ = 50 and x = 2
Using Lagrange's formula, we get
\(y=\frac { (x-{ x }_{ 1 })(x-{ x }_{ 2 }) }{ ({ x }_{ 0 }-{ x }_{ 1 })({ x }_{ 0 }-{ x }_{ 2 }) } { y }_{ 0 }+\frac { (x-{ x }_{ 1 })(x-{ x }_{ 2 }) }{ ({ x }_{ 0 }-{ x }_{ 1 })({ x }_{ 0 }-{ x }_{ 2 }) } { y }_{ 1 }+\frac { (x-{ x }_{ 0 })(x-{ x }_{ 1 }) }{ ({ x }_{ 2 }-{ x }_{ 0 })({ x }_{ 0 }-{ x }_{ 1 }) } { y }_{ 2 }\)
= \(5\times \frac { (2-0)(2-3) }{ (0-1)(0-3) } +6\times \frac { (2-0)(2-3) }{ (1-0)(1-3) } +50\times \frac { (2-0)(2-1) }{ (3-0)(3-1) } \)

= \(\frac { 10 }{ 3 } -6+\frac { 50 }{ 3 } =\frac { 60 }{ 3 } -6=20-6=14\)
∴ f(2) = 14

LHS = Δ∇ = (E-1) \(\left( \frac { E-1 }{ E } \right) \)

= Δ - ∇ [∵ Δ = E -1 & ∇ = \(\frac { E-1 }{ E } \) ]
= RHS
Hence Proved.

LHS = (1 + Δ) (1 - ∇)

[∵ Δ = E - 1 & ∇ = \(\frac { E-1 }{ E } \)]

= (1 + Δ) (1 - ∇)
= (1 + E - 1) \((1-\frac { E-1 }{ E } )\)
= E\((1-\frac { E-1 }{ E } )\)
= E - E\((\frac { E-1 }{ E } )\)
= E - (E - 1)
= E - E + 1 = 1
= RHS.