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

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Miscellaneous Problems
Question 1.
If $f(x)=e^{a x}$ then show that $f(0), \Delta f(0), \Delta^2 f(0)$ are in G.P
Solution:
$
\begin{aligned}
& \text { Given } \quad f(x)=e^{a x} \\
& \Delta f(x)=\Delta e^{a x} \\
& =e^{a(x+h)}-e^{a x} \\
& =e^{a x}\left[e^{a h}-1\right] \\
& \Delta^2 f(x)=\Delta e^{a x}\left[e^{a h}-1\right] \\
& =\left[e^{a h}-1\right] \Delta e^{a x} \\
& =\left[e^{a h}-1\right]^2 \\
& \text { Now } \quad f(0)=e^{a(0)}-1 \\
& \Delta f(0)=e^{a(0)}\left[e^{a h}-1\right]=e^{a h}-1 \\
& \Delta^2 f(0)=\left[e^{a h}-1\right]^2 \\
& \text { Now } \quad \frac{\Delta f(0)}{f(0)}=\frac{e^{a h}-1}{1}=e^{a h}-1 \\
& \text { and } \frac{\Delta^2 f(0)}{\Delta f(0)}=\frac{\left(e^{a h}-1\right)^2}{e^{a h}-1}=e^{a h}-1 \\
&
\end{aligned}
$
Hence $f(0), \Delta f(0), \Delta^2 f(0)$ are in G.P. with common ratio $e^{a h}-1$
Question 2.
Prove that
(i) $(1+\Delta)(1-\nabla)=1$
(ii) $\Delta \nabla=\Delta-\nabla$
(iii) $\mathrm{E} \nabla=\Delta-\nabla E$
Solution:

(i) To show $(1+\Delta)(1-\nabla)=1$
$
\begin{aligned}
\operatorname{LHS}(1+\Delta)(1-\nabla) & =(\mathrm{E})\left(1-\frac{\mathrm{E}-1}{\mathrm{E}}\right) \\
& =(\mathrm{E})\left(\frac{\mathrm{E}-\mathrm{E}+1}{\mathrm{E}}\right) \\
& =1=\text { RHS }
\end{aligned}
$
(ii) $\Delta \nabla=\Delta-\nabla$
$
\begin{aligned}
\operatorname{LHS} \Delta \nabla & =(E-1)\left(\frac{E-1}{E}\right)=(E-1)\left(1-\frac{1}{E}\right) \\
& =E-1-1+\frac{1}{E} \\
& =(E-1)-\left(1-\frac{1}{E}\right) \\
& =(E-1)-\left(\frac{E-1}{E}\right) \\
& =\Delta-\nabla=\text { RHS }
\end{aligned}
$
(iii) $\mathrm{E} \nabla=\Delta=\nabla \mathrm{E}$
$
\begin{aligned}
& \mathrm{E} \nabla=\mathrm{E}\left(\frac{\mathrm{E}-1}{\mathrm{E}}\right)=\mathrm{E}-1=\Delta \\
& \nabla \mathrm{E}=\left(\frac{\mathrm{E}-1}{\mathrm{E}}\right) \mathrm{E}=\mathrm{E}-1=\Delta
\end{aligned}
$
Hence proved.
Question 3.
A second degree polynomial passes through the point $(1,-1)(2,-1)(3,1)(4,5)$. Find the polynomial. Solution:
Given values can be tabulated as follows

We have to find a second-degree polynomial.
We use Newton's forward interpolation 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
$
Now $x_0+n h=x, x_0=1, h=1$
$
\begin{aligned}
1+n(1) & =x \\
n & =x-1
\end{aligned}
$
The difference table is given below

$
\begin{aligned}
& y=-1+\frac{(x-1)}{1 !}(0)+\frac{(x-1)(x-2)}{2 !}(2) \\
& y=-1+(x-1)(x-2) \\
& y=-1+x^2-3 x+2
\end{aligned}
$
(or) $y=x^2-3 x+1$ is the required second-degree polynomial which passes through the given points
Question 4.
Find the missing figures in the following table

Solution:
Since only four values of $y$ are given, the polynomial which fits the data is of degree three. Hence fourth differences are zero.

$
\begin{array}{r}
\text { i.e }\left(\Delta^4 y_k=0 \Rightarrow(\mathrm{E}-1)^4 y_k=0\right. \\
\mathrm{E}^4 y_k-4 \mathrm{E}^3 y_k+6 \mathrm{E}^2 y_k-4 \mathrm{E} y_k+y_k=0
\end{array}
$
Now given $y_0=7, y_1=11, y_3=18, y_5=32$
Let us take the missing values as $y_2$ and $y_4$
Put $k=0$ in (1), we get
$
\begin{aligned}
\mathrm{E}^4 y_0-4 \mathrm{E}^3 y_0+6 \mathrm{E}^2 y_0-4 \mathrm{E} y_0+y_0 & =0 \\
y_4-4 y_3+6 y_2-4 y_1+y_0 & =0 \\
y_4-4(18)+6 y_2-4(11)+7 & =0 \\
y_4+6 y_2 & =72+44-7 \\
y_4+6 y_2 & =109 \quad \ldots \ldots .(2)
\end{aligned}
$
Next Put $k=1$ in (1), we get
$
\begin{aligned}
\mathrm{E}^4 y_1-4 \mathrm{E}^3 y_1+6 \mathrm{E}^2 y_1-4 \mathrm{E} y_1+y_1 & =0 \\
y_5-4 y_4+6 y_3-4 y_2+y_1 & =0 \\
32-4 y_4+6(18)-4 y_2+11 & =0 \\
-4 y_4-4 y_2 & =-32-108-11 \\
-4 y_4-4 y_2 & =-151 \\
\Rightarrow \quad 4 y_4+4 y_2 & =151
\end{aligned}
$
We solve (2) and (3) to get value of $y_4$ and $y_2$
(2) $\times 4$ gives
$
\begin{aligned}
4 y_4+24 y_2 & =436 \\
4 y_4+4 y_2 & =151
\end{aligned}
$
Subtracting,
$
\begin{aligned}
20 y_2 & =285 \\
y_2 & =14.25 \\
y_4 & =109-6 y_2 \\
& =109-6(14.25) \\
& =109-85.5 \\
& =23.5
\end{aligned}
$
From (2),
Thus the missing figures are $14.25$ and $23.5$

Question 5.
Find $f(0.5)$ if $f(-1)=202, f(0)=175, f(1)=82$ and $f(2)=55$
Solution:
The given data can be written as

We have to find $y$ when $x=0.5$.
We use Newton's forward interpolation 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+\ldots
$
Now $x=0.5, x_0=-1, h=1$
So
$
\begin{aligned}
0.5 & =-1+n(1) \\
\Rightarrow \quad n & =1.5
\end{aligned}
$
The difference table is

Using these values we get,

$
\begin{aligned}
& y=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) \\
& y=202-40.5-24.75-8.25 \\
& y=128.5
\end{aligned}
$
Thus the value of $f(0.5)=128.5$
Question 6.
From the following data find $y$ at $x=43$ and $x=84$

Solution:
We have to find the value of $y$ at (a) $x=43$ and (b) $x=84$
(a) $x=43$.
The value of $y$ is required at the beginning of the table.
So we use Newton's forward interpolation 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+\ldots
$
Here $\mathrm{x}=43, \mathrm{x}_0=40, \mathrm{~h}=10$
So $43=40+10 n$
$
\mathrm{n}=0.3
$
The difference table is given below

$
\begin{aligned}
& \quad y=184+\frac{0.3}{1 !}(20)+\frac{(0.3)(0.3-1)}{2 !}(2)+0 \\
& y=184+6-0.21 \\
& y=189.79
\end{aligned}
$
(b) $\mathrm{x}=84$.
The value of $y$ is required at the end of the table.
So we use Newton's backward interpolation formula
$
\begin{aligned}
y_{\left(x=x_n+n h\right)}=y_n+\frac{n}{1 !} \nabla_{y_n}+ & \frac{n(n+1)}{2 !} \nabla^2 y_n+\frac{n(n+1)(n+2)}{3 !} \nabla^3 y_n+\ldots \\
x=84, x_n=90, h=10 & \\
84= & 90+10 n \\
n & =\frac{84-90}{10}=-0.6
\end{aligned}
$
We use the back difference values from the table
$
\begin{aligned}
& y_{(x=84)}=304+\frac{(-0.6)}{1 !}(28)+\frac{(-0.6)(-0.6+1)}{2 !}(2)+0 \\
y= & 304-16.8-0.24 \\
y= & 286.96
\end{aligned}
$
Hence the value of $y$ at $x=43$ is $189.75$ and the value of $y$ at $x=84$ is $286.96$
Question 7.
The area $\mathrm{A}$ of a circle of diameter ' $\mathrm{d}$ ' is given for the following values

Find the approximate values for the areas of circles of diameter 82 and 91 respectively.
Solution:
Let diameter be $\mathrm{x}$ and area be $\mathrm{y}$
We have to find value of $y$ when (a) $x=82$ and (b) $x=91$
We first find the difference as given below

We use backward difference formula here
$
\begin{aligned}
91 & =100+5 n \\
n & =\frac{91-100}{5}=-1.8
\end{aligned}
$
Using the backward differences from the table. We get,
$
\begin{aligned}
y(91) & =7854+\frac{(-1.8)}{1 !}(766)+\frac{(-1.8)(-1.8+1)}{2 !}(40)+\frac{(-1.8)(-1.8+1)(-1.8+2)}{3 !}(2) \\
& +\frac{(-1.8)(-1.8+1)(-1.8+2)(-1.8+3)}{4 !}(4) \\
& =7854-1378.8+\frac{(-1.8)(-0.8)(40)}{2}+\frac{(-1.8)(-0.8)(0.2)(2)}{6}+\frac{(-1.8)(-0.8)(0.2)(1.2)(4)}{24} \\
& =7854-1378.8+28.8+0.096+0.0576 \\
& =6504.15
\end{aligned}
$
Hence the area of a circle when the diameter is 82 is 5281 area of a circle when the diameter is 91 is 6504 .
Question 8.
If $\mathrm{u}_0=560, \mathrm{u}_1=556, \mathrm{u}_2=520, \mathrm{u}_4=385$, show that $\mathrm{u}_3=465$
Solution:
Given $\mathrm{u}_0=560, \mathrm{u}_1=556, \mathrm{u}_2=520, \mathrm{u}_4=385$

Since only four values are given,
(ie)
$
\begin{aligned}
\left(\mathrm{E}^4-4 \mathrm{E}^3+6 \mathrm{E}^2-4 \mathrm{E}+1\right) u_o & =0 \\
\mathrm{E}^4 u_o-4 \mathrm{E}^3 u_o+6 \mathrm{E}^2 u_o-4 \mathrm{E} u_o+u_o & =0 \\
u_4-4 u_3+6 u_2-4 u_1+u_o & =0 \\
385-4 u_3+6(520)-4(556)+560 & =0 \\
385-4 u_3+3120-2224+560 & =0 \\
4 u_3 & =1841 \\
u_3 & =460.25
\end{aligned}
$
$
\Delta^4 u_o=0 \Rightarrow(\mathrm{E}-1)^4 u_o=0
$
Question 9.
From the following table obtain a polynomial of degree $y$ in $\mathrm{x}$

Solution:
To find a polynomial $y=f(x)$
Here $\mathrm{x}_0=1, \mathrm{~h}=1$
$
\begin{aligned}
& \mathrm{x}=\mathrm{x}_0+\mathrm{nh} \\
& \mathrm{x}=1+\mathrm{n}(1) \\
& \mathrm{n}=\mathrm{x}-1
\end{aligned}
$
We find the forward differences as below

Using Newton's forward interpolation formula, 

$
\begin{aligned}
& y=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+\ldots \\
& y=1+(x-1)(-2)+\frac{(x-1)(x-2)}{2}(4)+\frac{(x-1)(x-2)(x-3)}{6}(-8) \\
& +\frac{(x-1)(x-2)(x-3)(x-4)}{24}(16) \\
& y=1-2(x-1)+2(x-1)(x-2)-\frac{4}{3}(x-1)(x-2)(x-3)+\frac{2}{3}(x-1)(x-2)(x-3)(x-4) \\
& y=1-2(x-1)[1-x+2]+\frac{2}{3}(x-1)(x-2)(x-3)[-2+x-4] \\
& y=1-2(x-1)(3-x)+\frac{2}{3}(x-1)(x-2)(x-3)(x-6) \\
& y=1+2(x-1)(x-3)+\frac{2}{3}(x-1)(x-2)(x-3)(x-6) \\
& y=1+2(x-1)(x-3)\left[1+\frac{(x-2)(x-6)}{3}\right] \\
& y=1+2(x-1)(x-3)\left[\frac{3+x^2-8 x+12}{3}\right] \\
& y=1+\frac{2}{3}\left(x^2-4 x+3\right)\left(x^2-8 x+15\right) \\
& y=1+-\left[x^4-8 x^3+15 x^2-4 x^3+32 x^2-60 x+3 x^2-24 x+45\right] \\
& y=\frac{2}{3} x^4-8 x^3+\frac{100}{3} x^2-56 x+31 \text { is the required polynomial } \\
&
\end{aligned}
$
Question 10.
Using Lagrange's interpolation formula find a polynomial which passes through the points $(0,-12),(1,0)$, $(3,6)$ and $(4,12)$.
Solution:
The given values are
$
\begin{aligned}
\mathrm{x}_0 & =0, \mathrm{y}_0=-12 \\
\mathrm{x}_1 & =1, \mathrm{y}_1=0 \\
\mathrm{x}_2 & =3, \mathrm{y}_2=6 \\
\mathrm{x}_3 & =4, \mathrm{y}_3=12
\end{aligned}
$

By Lagrange's interpolaiton 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 \\
& y=\frac{(x-1)(x-3)(x-4)}{(-1)(-3)(-4)}(-12)+0+\frac{x(x-1)(x-4)}{(3)(2)(-1)}(6)+\frac{x(x-1)(x-3)}{(4)(3)(1)}(12) \\
& \Rightarrow \mathrm{y}=(\mathrm{x}-1)(\mathrm{x}-3)(\mathrm{x}-4)-\mathrm{x}(\mathrm{x}-1)(\mathrm{x}-4)+\mathrm{x}(\mathrm{x}-1)(\mathrm{x}-3) \\
& \Rightarrow \mathrm{y}=(\mathrm{x}-1)(\mathrm{x}-4)[(\mathrm{x}-3)-\mathrm{x}]+\mathrm{x}(\mathrm{x}-1)(\mathrm{x}-3) \\
& \Rightarrow \mathrm{y}=(\mathrm{x}-1)(\mathrm{x}-4)(-3)+\mathrm{x}(\mathrm{x}-1)(\mathrm{x}-3) \\
& \Rightarrow \mathrm{y}=(\mathrm{x}-1)\left[-3 \mathrm{x}+12+\mathrm{x}^2-3 \mathrm{x}\right] \\
& \Rightarrow \mathrm{y}=(\mathrm{x}-1)\left(\mathrm{x}^2-6 \mathrm{x}+12\right) \\
& \Rightarrow \mathrm{y}=\mathrm{x}^3-6 \mathrm{x}^2+12 \mathrm{x}-\mathrm{x}^2+6 \mathrm{x}-12 \\
& y=x^3-7 x^2+18 x-12 \text { is the required polynomial which passes through the given points } \\
&
\end{aligned}
$