Given values are

x	1	2	3	4
y	-1	-1	1	5

The difference table is

x₀ + nh = x ⇒ 1 + n = x ⇒ n = x - 1
∴ By Newton's forward interpolation formula,
y= y₀\(\frac { n }{ 1! } { \triangle y }_{ 0 }+\frac { n(n+1) }{ 2! } { \triangle }^{ 2 }{ y }_{ 0 }+\frac { n(n+1)(n+2) }{ 3! } { \triangle }^{ 3 }{ y }_{ 0 }\)

[∵ Δ³y₀ = 0]
⇒ y = -1 + 0 + x² - 3x + 2
⇒ y = x² - 3x + 1
Hence, the required second degree polynomial is  y = x² - 3x + 1

Scale:
In x-axis 1 cm = 100 units
In y-axis 1 cm = 1000 units
Plot the points (200, 2500), (300, 2800) (400, 3100) (640; 3820), (540,3220) and (580, 3640).

At x = 350, draw a vertical line and from the intersecting point on the curve, draw a horizontal line.
From the graph, we find that when x = 350, y = 2900.
Hence, the expense at a level of 350 units is Rs. 2900

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.

Since only five values of f(x) are given, the polynomial which fits the data is of degree 4.
Hence fifth differences are zeros.
∴ Δ⁵y₀ = 0
⇒ (E - 1)⁵y₀ = 0
⇒ (E⁵ - 5E⁴ + 10E³ - 10E² + 5E - 1) y₀ = 0
⇒ y₅ - 5y₄ + 10y₃ - 10y₂ + 5y₁ - y₀ = 0
⇒ 501 - 5(467) + 10y₃ - 10 (421) + 5 (391) - 363 =  0
⇒ 501 - 2335 + 10y₃ - 4210 + 1955 - 363 = 0
⇒ 10y₃ - 4452 = 0
⇒ 10y₃ = 4452
⇒ \({ y }_{ 3 }=\frac{4352}{10}\) = 445.2

Since only four values of f(x) are given, the polynomial which fits the data is of degree 3.
Hence fourth differences are zero.
∴ Δ⁴ (y₀) = 0
⇒ (E - 1)⁴ (y₀) = 0
⇒ (E⁴ - 4E³ + 6E² - 4E + 1) y₀ = 0
⇒ y₄ - 4y₃ + 6y₂ - 4y₁ +y₀ = 0
⇒ 81 - 4(y₃) + 6(9) - 4(3) + 1 = 0
⇒ 81 - 4y₃ + 54 - 12 + 1 = 0
⇒ 81 + 54 - 11 = 4y₃
⇒ 124 = 4y₃ ⇒ \(\frac{124}{4}\) = 31
⇒ y₃ = 31

Given f(x) = x²+ 3h
LHS = Δ f(x)
= f(x + h) - f(x)
= [(x + h)² + 3 (x + h)] - [x² + 3x]

= h² + 2xh + 3h
when h = 1,
LHS = 1² + 2x(1) + 3(1)
= 1+2x+3
= 2x+4= RHS
Hence proved.

Given h = 1
LHS = (E⁻¹Δ) x³
= Δ(E^-1(x³))
= Δ(x - h)³  [∵ E^-1f(x) = f(x - nh)]
= Δ(1 - h)³ [∵ h = 1]
= (x - 1+ 1)³ - (x - 1)³ [∵ Δf(x) =f(x + h) - f(x)]
= x³ - (x - 1)³
= x³ - (x³ - 3x² + 3x - 1)
[∵ (a - b)³ = a³ - 3a²b + 3ab² - b³]
= x³ - x³ + 3x² - 3x + 1
= 3x² - 3x + 1
= RHS
Hence proved

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 \(\left( 1+\frac { h }{ ax } \right) \)

Since only four values of f(x) are given, the polynomial which fits the data is of degree three. Hence fourth differences are zeros.
(ie) Δ⁴y₀ = 0, ∴ (E−1)⁴y = 0
(E⁴ - 4E³ + 6E² - 4E +1)y₀ = 0
E⁴y₀ - 4E³y₀+ 6E²y₀ - 4Ey₀ +y₀  = 0
y₄ − 4y₃ + 6y₂ − 4y₁ + y₀ = 0
67.4 − 4y₃+ 6(54.1) − 4(4.2) + 45 = 0
240.2 = 4y₃ ∴ y₃ = 60.05

Given y₃ = 2, y₄ = −6, y₅ = 8, y₆ = 9 and y₇ = 17
Δ⁴y₃ = (E−1)⁴y₃
= (E⁴ − 4E³ + 6E² − 4E+1)y₃
= E⁴y₃ − 4E³y₃ + 6E²y₃− 4Ey₃ + y₃
= y₇ − 4y₆ + 6y₅ −4y₄+ y₃
= 17 – 4(9) + 6(8) –4(–6) + 2
= 17 – 36 + 48 + 24 + 2 = 55

Δ²U₀ = (E-1)⁴U₀
= (E⁴ − 4E³+ 6E²− 4E+1)U₀
= E⁴U₀ - 4E³U₀ + 6E²U₀− 4EU₀+ U₀
= U₄ − 4U₃ + 6U₂ − 4U₁ + U₀
= 29 − 4(28) + 6(21) − 4(11) + 1.
= 156 – 156 = 0

We know that f(4) − f(3) =  Δf(3)
f(4) − f(3) = Δf(3)
= Δ[f(2) + Δf(2)] ∵[f(3) - f(2) = Δf(2)]
= Δf(2) +Δ²f(2)
= Δf(2) + Δ² [f(1) + Δf(1)]
∴ f(4) = f(4) + Δf(2) +  Δ² f(1) +  Δ³f(1)

Let k be the sixth term of the series in the difference table.
First we find the forward differences

x	y	∆	∆2
1	8
		4
2	12		3
		7
3	19		3
		10
4	29		3
		13
5	42		k-55
		k-42
6	k

Given that the second differences are constant
∴ k – 55 = 3
k = 58
∴ the sixth term of the series is 58

y = f(x) = x³+2x+1 for x = 1,2,3,4,5

x	y	Δy	Δ2y	Δ3y	Δ4y
1	4
		9
2	13		12
		21		6
3	34		18		0
		39		6
4	73		24
		63
5	136

The Forward difference table is given below

x	y	Δy	Δ2y	Δ3y
0	0
		0.174
10	0.174		-0.001
		0.173		-0.001
20	0.347		-0.002
		0.171
30	0.518

(i) Take a suitable scale for the values of x and y, and plot the various points on the graph paper for given values of x and y.
(ii) Draw a suitable curve passing through the plotted points.
(iii) Find the point corresponding to the value x = 38 on the curve and then read the corresponding value of y on the y- axis, which will be the required interpolated value.
From the graph in Figure we find that for x = 38, the value of y is equal to 35