Additional Problems - Chapter 5 - Numerical Methods - 12th Maths Guide Guide Samacheer Kalvi Solutions

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Additional Problems
One Mark Questions
Question 1.
Match the following.

Answer:
(a) - (iii), (b) - (iv), (c) - (ii), (d) - (i)
Question 2.
$E^{-n} f(x)$ is
(a) $f(x+n h)$
(b) $f(x-n h)$
(c) $f(-\mathrm{nh})$
(d) $f(x-n)$
Answer:
(b) $f(x-n h)$
Question 3.
$E$ is a
(a) shifting operator
(b) Displacement operator
(c) $1+\Delta$
(d) all of these
(e) none of these
Answer:
(d) all of these
Question 4.
$\Delta^4 y_3=$
(a) $(E-1)^4 y_3$
(b) $\left(E^3-1\right) y_3$
(c) $(\mathrm{E}-1)^3 \mathrm{y}_0$

(d) $(E-1)^4 y_0$
Answer:
(a) $(E-1)^4 y_3$
Question 5.
Fill in the blanks.
1. The two methods of interpolation are ____________ and ________
2 . If values of $x$ are not equidistant we use ____________ method.
3. $\Delta(\mathrm{f}(\mathrm{x})+\mathrm{g}(\mathrm{x}))=$ ___________
4. $\Delta^{\mathrm{k}} \mathrm{y}_{\mathrm{n}}=$ ________
5. The first three terms in Newton's method will give a ___________ interpolation.
Answer:
1. graphical method, algebraic method
2. Lagrange's method
3. $\Delta \mathrm{f}(\mathrm{x})+\Delta \mathrm{g}(\mathrm{x})$
4. $\Delta^{\mathrm{k}-1} \mathrm{y}_{\mathrm{n}+1}-\Delta^{\mathrm{k}-1} \mathrm{y}_{\mathrm{n}}$
5. Parabolic
Question 6.
Say true or false
1. $\nabla \mathrm{y}_2=\mathrm{y}_1-\mathrm{y}_0$
2. $\nabla^2 y_n=\nabla y_n-\nabla y_{n+1}$
3. When 5 values are given, the polynomial which fits the data is of degree 4
4. $\mathrm{E} \Delta=\Delta \mathrm{E}$
5. $f(2)+\Delta f(2)=f(3)$
Answer:
1. False
2. True
3. True
4. True
5. True

II. 2 Mark Questions
Question 1.
Find the missing term from the following data.

Solution:
Since three values of $y=f(x)$ are given, the polynomial which fits the data is of degree two. Hence third differences are zero.
$
\begin{aligned}
\Delta^3\left(y_0\right)=0, \Rightarrow(\mathrm{E}-1)^3 y_0 & =0 \\
\mathrm{E}^3 y_0-3 \mathrm{E}^2 y_0+3 \mathrm{E} y_0-y_0 & =0 \\
y_3-3 y_2+3 y_1-y_0 & =0 \\
157-3(126)+3 y_1-100 & =0 \\
3 y_1 & =321 \\
y_1 & =107
\end{aligned}
$
Question 2.
From the following data estimate the export for the year 2000

Solution:
Consider a polynomial of degree two.

Hence third differences are zero.
$
\begin{aligned}
\Delta^3\left(y_0\right)=0, \Rightarrow(\mathrm{E}-1)^3 y_0 & =0 \\
\mathrm{E}^3 y_0-3 \mathrm{E}^2 y_0+3 \mathrm{E} y_0-y_0 & =0 \\
y_3-3 y_2+3 y_1-y_0 & =0 \\
397-3(369)+3 y_1-443 & =0 \\
3 y_1 & =1153 \\
y_1 & =384
\end{aligned}
$
Question 3.
For the tabulated values of $y=f(x)$, find $\Delta y_3$ and $\Delta^3 y_2$

Solution:
$
\begin{aligned}
\Delta y_3 & =y_4-y_3=0.49-0.35=0.14 \\
\Delta^3 y_2 & =\Delta\left(\Delta^2 y_2\right)=\Delta\left(y_4-2 y_3+y_2\right) \\
& =\left(y_5-y_4\right)-2\left(y_4-y_3\right)+\left(y_3-y_2\right) \\
& =y_5-3 y_4+3 y_3-y_2 \\
& =0.67-3(0.49)+3(0.35)-0.26 \\
& =-0.01
\end{aligned}
$
Question 4.
If $f(x)=x^2+a x+b$, find $\Delta^r f(x)$
Solution:
$
\begin{aligned}
& \Delta f(x)=f(x+h)-f(x) \\
& =\left[(x+h)^2+a(x+h)+b\right]-\left[x^2+a x+b\right] \\
& =2 x h+h^2+a h \\
& \Delta^2 \mathrm{f}(\mathrm{x})=\left[2(\mathrm{x}+\mathrm{h}) \mathrm{h}+\mathrm{h}^2+\mathrm{ah}\right]-\left[2 \mathrm{xh}+\mathrm{h}^2+\mathrm{ah}\right]=2 \mathrm{~h}^2 \\
& \Delta^3 \mathrm{f}(\mathrm{x})=0
\end{aligned}
$
Thus $\Delta^{\mathrm{r}} \mathrm{f}(\mathrm{x})=0$ for all $r \geq 3$
Question 5.
Show that $\Delta^3 y_4=\nabla^3 y_7$
Solution:
$
\begin{aligned}
\Delta^3 y_4 & =(\mathrm{E}-1)^3 y_4=\mathrm{E}^3 y_4-3 \mathrm{E}^2 y_4+3 \mathrm{E}_4-y_4 \\
& =y_7-3 y_6+3 y_5-y_4 \\
\Delta^3 y_7 & =\left(1-\mathrm{E}^{-1}\right)^3 y_7 \\
& =\left(1-3 \mathrm{E}^{-1}+3 \mathrm{E}^{-2}-\mathrm{E}^{-1}\right) y_7 \\
& =y_7-3 \mathrm{E}^{-1} y_7+3 \mathrm{E}^{-2} y_7-\mathrm{E}^{-3} y_7 \\
& =y_7-3 y_6+3 y_5-y_4
\end{aligned}
$
Hence proved
III. 3 and 5 Marks Questions
Question 1.
If $f(0)=5, f(1)=6, f(3)=50, f(4)=105$, find $f(2)$ by using Lagrange's formula.

Solution:

By Lagrange's interpolation formula,
$
\begin{aligned}
y & =\frac{\left(x-x_1\right)\left(x-x_2\right)\left(x-x_3\right)}{\left(x_0-x_1\right)\left(x_0-x_2\right)\left(x_0-x_3\right)} \times y_0+\frac{\left(x-x_0\right)\left(x-x_2\right)\left(x-x_3\right)}{\left(x_1-x_0\right)\left(x_1-x_2\right)\left(x_1-x_3\right)} y_1 \\
& +\frac{\left(x-x_0\right)\left(x-x_1\right)\left(x-x_3\right)}{\left(x_2-x_0\right)\left(x_2-x_1\right)\left(x_2-x_3\right)} \times y_2+\frac{\left(x-x_0\right)\left(x-x_1\right)\left(x-x_2\right)}{\left(x_3-x_0\right)\left(x_3-x_1\right)\left(x_3-x_2\right)} y_3
\end{aligned}
$
We have to find the value of $y$ when $x=2$

$
\begin{aligned}
& y=\frac{(1)(-1)(-2)}{(-1)(-3)(-4)}(5)+\frac{(2)(-1)(-2)}{(1)(-2)(-3)}(6)+\frac{(2)(1)(-2)}{(2)(3)(-1)}(50)+\frac{(2)(1)(-1)}{(4)(3)(1)}(105) \\
& y=-0.833+4+33.333-17.5 \\
& y=19
\end{aligned}
$
Question 2.
Find $\mathrm{y}$ when $\mathrm{x}=0.2$ given that

Solution:
Since the required value of $y$ is near the beginning of the table, we use Newton's forward difference formula
$
y_{\left(x=x_0+n h\right)}=y_0+\frac{n}{1 !} \Delta y_0+\frac{n(n-1)}{2 !} \Delta^2 y_0+\frac{n(n-1)(n-2)}{3 !} \Delta^3 y_0+\frac{n(n-1)(n-2)(n-3)}{4 !} \Delta^4 y_0+\ldots
$
The difference table is given below

$
\begin{aligned}
& \text { Now } x=0.2, x_0=0, h=1 \\
& \Rightarrow 0.2=0+n(1) \\
& \Rightarrow n=0.2 \\
y & =176+\frac{0.2}{1 !}(9)+0+\frac{(0.2)(0.2-1)(0.2-2)}{3 !}(-1)+\frac{(0.2)(0.2-1)(0.2-2)(0.2-3)}{4 !}(4) \\
y & =176+1.8+\frac{(0.2)(-0.8)(-1.8)(-1)}{6}+\frac{(0.2)(-0.8)(-1.8)(-2.8)}{24} 4 \\
y & =176+1.8-0.048-0.1344 \\
y & =177.6176
\end{aligned}
$
Question 3.
Find the number of men getting wages between Rs. 30 and Rs. 35 from the following table:

Solution:
The difference table

For $x=35, h=10, x_0=30$
$
\begin{aligned}
35 & =30+n(10) \\
n & =\frac{35-30}{10} \\
n & =0.5
\end{aligned}
$
By Newton's forward formula
$
\begin{aligned}
y(35) & =9+\frac{0.5}{1 !}(30)+\frac{(0.5)(0.5-1)}{2 !}(5)+\frac{(0.5)(0.5-1)(0.5-2)}{3 !}(2) \\
& =9+15-0.6+0.1 \\
& =24 \text { ( approximately) }
\end{aligned}
$
No. of men getting wages less than 35 is 24 . Therefore the number of men getting wages between Rs. 30 and Rs. 35 is $y(35)-y(30)$
(i.e) $24-9=15$
Question 4.
Using Newton's formula estimate the population of town for the year 1995:

Solution:
1995 lies in (1991, 2001). Hence we use Newton's backward interpolation formula.
Here $\mathrm{x}=1995, \mathrm{x}_{\mathrm{n}}=2001, \mathrm{~h}=10$
$
\begin{aligned}
& 1995=\mathrm{x}_{\mathrm{n}}+\mathrm{nh} \\
& \Rightarrow 1995=2001+10 \mathrm{n} \\
& \Rightarrow \mathrm{n}=\frac{1995-2001}{10} \\
& \Rightarrow \mathrm{n}=-0.6
\end{aligned}
$

The backward difference table is given below

$
\begin{aligned}
& \begin{array}{l}
\therefore y=101+\frac{(-0.6)}{1 !}(8)+\frac{(-0.6)(-0.6+1)}{2 !}(-4)+\frac{(-0.6)(-0.6+1)(-0.6+2)}{3 !}(-1) \\
+\frac{(-0.6)(-0.6+1)(-0.6+2)(-0.6+3)}{4 !}(-3)
\end{array} \\
& \begin{array}{l}
y=101-4.8+0.48+0.056+0.1008 \\
y=96.8368
\end{array}
\end{aligned}
$
Hence the population for the year 1995 is $96.837$ thousands.
Question 5.
Using Lagrange's formula find $y(11)$ from the following table

Solution:
Given
$
\begin{aligned}
& \mathrm{x}_0=6, \mathrm{y}_0=13 \\
& \mathrm{x}_1=7, \mathrm{y}_1=14 \\
& \mathrm{x}_2=10, \mathrm{y}_2=15 \\
& \mathrm{x}_3=12, \mathrm{y}_3=17 \\
& \mathrm{x}=11
\end{aligned}
$

Using Lagrange's formula,
$
\begin{aligned}
y & =\frac{(4)(1)(-1)}{(-1)(-4)(-6)}(13)+\frac{(5)(1)(-1)}{(1)(-3)(-5)}(14)+\frac{(5)(4)(-1)}{(4)(3)(-2)}(15)+\frac{(5)(4)(1)}{(6)(5)(12)}(17) \\
& =2.1666-4.6666+12.5+5.6666 \\
y & =15.6666
\end{aligned}
$