Exercise 7.4-Additional Problems - Chapter 7 - Applications of Differential Calculus - 12th Maths Guide Samacheer Kalvi Solutions - Tamil Medium

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Additional Problems
Question 1.
The Taylor's series expansion of $f(x)=\sin x$ about $x=\frac{\pi}{2}$ is obtained by the following way. Solution:
$
\begin{aligned}
& f(x)=\sin x \quad ; \quad f\left(\frac{\pi}{2}\right)=\sin \frac{\pi}{2}=1 \\
& f^{\prime}(x)=\cos x \quad ; \quad f^{\prime}\left(\frac{\pi}{2}\right)=\cos \frac{\pi}{2}=0 \\
& f^{\prime \prime}(x)=-\sin x \quad ; \quad f^{\prime \prime}\left(\frac{\pi}{2}\right)=-1 \\
& f^{\prime \prime \prime}(x)=-\cos x \quad ; \quad f^{\prime \prime}\left(\frac{\pi}{2}\right)=0 \\
& \therefore f(x)=\sin x=f\left(\frac{\pi}{2}\right)+\frac{f^{\prime}\left(\frac{\pi}{2}\right)}{1 !}\left(x-\frac{\pi}{2}\right)+\frac{f^{\prime \prime}\left(\frac{\pi}{2}\right)}{2 !}\left(x-\frac{\pi}{2}\right)^2+\ldots . \\
& =1+0\left(x-\frac{\pi}{2}\right)+\frac{(-1)}{2 !}\left(x-\frac{\pi}{2}\right)^2+\ldots \\
& \sin x=1-\frac{1}{2 !}\left(x-\frac{\pi}{2}\right)^2+\frac{1}{4 !}\left(x-\frac{\pi}{2}\right)^4-\ldots \\
&
\end{aligned}
$
Question 2.
Obtain the Maclaurin's series expansion for the following functions.
(i) $\mathrm{e}^{\mathrm{x}}$
(ii) $\sin ^2 \mathrm{x}$
(iii) $\frac{1}{1+x}$
Solution:
(i) $f(0)=e^0=1 ; \quad f^{\prime}(x)=2 e^{2 x} \Rightarrow f^{\prime}(0)=2$
$
f^{\prime \prime}(x)=4 e^{2 x} ; f^{\prime \prime}(0)=4
$
The Maclaurin's expansion of $f(x)$ is
$
\begin{aligned}
& f(x)=f(0)+\frac{x}{\lfloor 1} f^{\prime}(0)+\frac{x^2}{\lfloor 2} f^{\prime \prime}(0)+\ldots \\
& \text { i.e., } e^{2 x}=1+\frac{x}{\lfloor 1}(2)+\frac{x^2}{\lfloor 2}(4)+\ldots . \\
& =1+\frac{2 x}{\lfloor 1}+\frac{4 x^2}{\lfloor 2}+\ldots=1+\frac{2 x}{\lfloor 1}+\frac{(2 x)^2}{\lfloor 2}+\ldots . \\
&
\end{aligned}
$

(ii)
$
\begin{aligned}
& f(x)=\sin ^2 x \\
& f(x)=\sin ^2 x \quad \Rightarrow f(0)=0 \\
& f^{\prime}(x)=2 \sin x \cos x \Rightarrow f^{\prime}(0)=0 \\
& =\sin 2 x \\
& f^{\prime \prime}(x)=2 \cos 2 x \quad \Rightarrow f^{\prime \prime}(0)=2 \\
& f^3(x)=-4 \sin 2 x \quad \Rightarrow f^3(0)=0 \\
& f^4(x)=-8 \cos 2 x \quad \Rightarrow f^4(0)=-8 \\
&
\end{aligned}
$
The Maclaurin's expansion of $f(x)$ is
$
\begin{aligned}
f(x) & =f(0)+\frac{x}{\lfloor 1} f^{\prime}(0)+\frac{x^2}{\lfloor 2} f^{\prime \prime}(0)+\ldots \\
\text { i.e., } \sin ^2 x & =\frac{x^2}{\lfloor 2}(2)+\frac{x^3}{\lfloor 3}(-8)+\ldots \\
\text { (i.e.) } \sin ^2 x & =x^2-\frac{x^4}{\lfloor 3} \ldots
\end{aligned}
$

(iii)
$
\begin{aligned}
f(x) & =\frac{1}{1+x} ; \quad f(0)=\frac{1}{1+0}=1 \\
f^{\prime}(x) & =\frac{-1}{(1+x)^2} ; \quad f^{\prime}(0)=\frac{-1}{(1)^2}=-1 \\
f^{\prime \prime}(x) & =-\left[\frac{-2}{(1+x)^3}\right]=\frac{2}{(1+x)^3} ; f^{\prime \prime}(0)=2
\end{aligned}
$
The Maclaurin's expansion of $f(x)$ is
$
f(x)=f(0)+\frac{x}{\lfloor 1} f^{\prime}(0)+\frac{x^2}{\lfloor 2} f^{\prime \prime}(0)+\ldots .
$
So, $\frac{1}{1+x}=1+\frac{x}{\lfloor 1}(-1)+\frac{x^2}{\lfloor 2}(2)+\ldots$. .
i.e., $\frac{1}{1+x}=1-x+\frac{2 x^2}{2}-\ldots=1-x+x^2 \ldots$