Exercise 5.1 - Chapter 5 Numerical Methods 12th Maths Guide Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated On May 15, 2024

By SaraNextGen

Numerical Methods
$
\text { Ex } 5.1
$
Question 1.
Evaluate (log ax)
Solution:
$
\begin{aligned}
\Delta \log a x & =\log a(x+h)-\log a x \\
& =\log \left[\frac{a(x+h)}{a x}\right] \\
& =\log \left(1+\frac{h}{x}\right)
\end{aligned}
$
Question 2.
If $y=x^3-x^2+x-1$ calculate the values of $y$ for $x=0,1,2,3,4,5$ and form the forward differences table.
Solution:
Given $y=x^3-x^2+x-1$
When
$
\begin{aligned}
& x=0, y=0-0+0-1=-1 \\
& x=1, y=1-1+1-1=0 \\
& x=2, y=8-4+2-1=5 \\
& x=3, y=27-9+3-1=20 \\
& x=4, y=64-16+4-1=51 \\
& x=5, y=125-25+5-1=104
\end{aligned}
$

Question 3.
If $h=1$ then prove that $(E-1 \Delta) x^3=3 x^2-3 x+1$.
Solution:
$
\begin{aligned}
& \mathrm{h}=1 \text { To prove }\left(\mathrm{E}^{-1} \Delta\right) \times 3=3 \times 2-3 \mathrm{x}+1 \\
& \text { L.H.S }=\left(\mathrm{E}^{-1} \Delta\right) \mathrm{x}^3=\mathrm{E}^{-1}\left(\Delta \mathrm{x}^3\right) \\
& =\mathrm{E}^{-1}\left[(\mathrm{x}+\mathrm{h})^3-\mathrm{x}^3\right] \\
& =\mathrm{E}^{-1}(\mathrm{x}+\mathrm{h})^3-\mathrm{E}^{-1}\left(\mathrm{x}^3\right) \\
& =(\mathrm{x}-\mathrm{h}+\mathrm{h})^3-(\mathrm{x}-\mathrm{h})^3 \\
& =\mathrm{x}^3-(\mathrm{x}-\mathrm{h})^3
\end{aligned}
$
But given $\mathrm{h}=1$
$
\begin{aligned}
& \text { So }\left(E^{-1} \Delta\right) x^3=x^3-(x-1)^3 \\
& =x^3-\left[x^3-3 x^2+3 x-1\right] \\
& =3 x^2-3 x+1 \\
& =\text { RHS }
\end{aligned}
$

Question 4.
If $f(x)=x^2+3 x$ then show that $\Delta f(x)=2 x+4$
Solution:
$
\begin{aligned}
& f(x)=x^2+3 x \\
& \Delta f(x)=f(x+h)-f(x) \\
& =(x+h)^2+3(x+h)-x^2-3 x \\
& =x^2+2 x h+h^2+3 x+3 h-x^2-3 x \\
& =2 x h+3 h+h^2 \\
& \text { Put } h=1, \Delta f(x)=2 x+4
\end{aligned}
$
Question 5.
Evaluate $\Delta\left[\frac{1}{(x+1)(x+2)}\right]$ by taking ' 1 ' as the interval of differencing.
Solution:
$
\Delta\left[\frac{1}{(x+1)(x+2)}\right], h=1
$
By Partial fraction,
$
\begin{aligned}
& \frac{1}{(x+1)(x+2)}=\frac{\mathrm{A}}{x+1}+\frac{\mathrm{B}}{x+2} \\
\mathrm{~A}= & 1, \quad \mathrm{~B}=-1 \\
& \text { So } \Delta\left[\frac{1}{(x+1)(x+2)}\right]=\Delta\left[\frac{1}{x+1}-\frac{1}{x+2}\right]
\end{aligned}
$

$
\begin{aligned}
& =\Delta\left(\frac{1}{x+1}\right)-\Delta\left(\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] \\
& =\left[\frac{1}{x+2}-\frac{1}{x+1}\right]-\left[\frac{1}{x+3}-\frac{1}{x+2}\right] \\
& =\left[\frac{-1}{(x+2)(x+1)}\right]\left[\frac{-1}{(x+3)(x+2)}\right] \\
& =\frac{-1}{(x+2)}\left[\frac{1}{x+1}-\frac{1}{x+3}\right] \\
& =\frac{-2}{(x+1)(x+2)(x+3)}
\end{aligned}
$
Question 6 .
Find the missing entry 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 zeros.
(ie)
$
\begin{aligned}
\Delta^4 y_0=0, \therefore(\mathrm{E}-1)^4 y_0 & =0 \\
\left(\mathrm{E}^4-4 \mathrm{E}^3+6 \mathrm{E}^2-4 \mathrm{E}+1\right) y_0 & =0 \\
\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 \\
\text { Given } y_0=1, y_1=3, y_2=9, y_4 & =81
\end{aligned}
$
So we get
$
\begin{aligned}
81-4 y_3+6(9)-4(3)+1 & =0 \\
81-4 y_3+54-12+1 & =0 \\
4 y_3 & =124 \Rightarrow y_3=31
\end{aligned}
$
Question 7.
Following are the population of a district

Find the population of the year 1911 ?
Solution:
Since five values are given, the polynomial which fits the data is of degree four.
Hence
$
\Delta^5 y_0=0 \text {, i.e. }(\mathrm{E}-1)^5 y_0=0
$
$
\Rightarrow \begin{array}{r}
\left(\mathrm{E}^5-5 \mathrm{E}^4+10 \mathrm{E}^3-10 \mathrm{E}^2+5 \mathrm{E}-1\right) y_0=0 \\
\mathrm{E}^5 y_0-5 \mathrm{E}^4 y_0+10 \mathrm{E}^3 y_0-10 \mathrm{E}^2 y_0+5 \mathrm{E} y_0-y_0=0 \\
y_5-5 y_4+10 y_3-10 y_2+5 y_1-y_0=0
\end{array}
$
From the given table
$
\begin{aligned}
& y_0=363, y_1=391, y_2=421, y_4=467 \text { and } y_5=501 \\
& 501-5(467)+10 y_3-10(421)+5(391)-363=0 \\
& 501-2335+10 y_3-4210+1955-363=0 \\
& -501+2335+4210-1955+363=10 y_3 \\
& 10 y_3=4452 \\
& y_3=445.2
\end{aligned}
$
Hence the population of the year 1911 is 445 thousand
Question 8.
Find the missing entries from the following.

(ie)
$
\begin{aligned}
\Delta^4 y_0=0, \therefore(\mathrm{E}-1)^4 y_0=0 \\
\left(\mathrm{E}^4-4 \mathrm{E}^3+6 \mathrm{E}^2-4 \mathrm{E}+1\right) y_0=0 \\
\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
\end{aligned}
$
Given
$
\begin{aligned}
y_0=0, y_2=8, y_3=15, y_5 & =35 \\
\Rightarrow \quad y_4-4(15)+6(8)-4 y_1+0 & =0 \\
y_4-4 y_1 & =12
\end{aligned}
$
Now $\Delta^4 y_1=0$
So we have
$
\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 \\
& \text { Again using } y_0=0, y_2=8, y_3=15, y_5=35 \\
& \text { we get } \quad 35-4 y_4+6(15)-4(8)+y_1=0 \\
& -4 y_4+y_1=-35-90+32 \\
& -4 y_4+y_1=-93 \\
& 4 y_4-y_1=93 \\
&
\end{aligned}
$
Solving (1) and (2)
$(1) \times 4$ gives
$
\begin{aligned}
& 4 y_4-16 y_1=48 \\
& 4 y_4-y_1=93 \\
& \text { Subtracting, } \quad-15 y_1=-45 \Rightarrow y_1=3 \\
&
\end{aligned}
$
$
\text { Substituting } y_1=3 \text { in (2) } \quad 4 y_4-3=93
$
$4 y_4=96 \Rightarrow y_4=24$
The missing entries are $y_1=f\left(x_1\right)=3$ and $y_4=f\left(x_4\right)=24$

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