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\title[]{MA140-Engineering Calculus}
\author[ Lecture 3 ]{\large{ Lecture 3} 
\and %\\ \small{Supervisor: Dr. John Burns}
\\ 
 \includegraphics[scale=0.0]{Figures/lec14.jpg}}
%\date[June 2-4, 2016]{June 2-4, 2016}
%\institute[Uppsala University]{ }


\begin{document}
\begin{frame}
\titlepage
\end{frame}


\section{MA140}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{frame}

{\textbf {\textcolor{red}{1. Linear factors to the power of $1.$}}} 
\begin{example}
Express $\displaystyle{\frac{3x}{(x-1)(x+2)}}$ in partial fractions.
\end{example}

{\textcolor{blue}{$\hspace{2.7cm}\displaystyle{\frac{3x}{(x-1)(x+2)}}=\displaystyle{\frac{A}{x-1}}+\displaystyle{\frac{B}{x+2}}$ $$\hspace{2.6cm}=\displaystyle{\frac{A(x+2)+B(x-1)}{(x-1)(x+2)}}$$\\$\hspace{5.4cm}=\displaystyle{\frac{x[A+B]+[2A-B]}{(x-1)(x+2)}}$}}
\end{frame}
\begin{frame}

{\textcolor{blue}{Compare $x[3]+[0]$ with $x[A+B]+[2A-B]$, so $$\left\{
  \begin{array}{rcr}
    A+B & = & 3 \\
    2A-B & = & 0 \\
  \end{array}
\right.$$
So $3=3A \enspace \Rightarrow A=1$\\ From first equation $3=1+B \enspace \Rightarrow B=2$\\ so $$\displaystyle{\frac{3x}{(x-1)(x+2)}}=\displaystyle{\frac{1}{x-1}}+\displaystyle{\frac{2}{x+2}}$$ }}
\textcolor{red}{Exercise:}\\
Find the constants $A, B$ and $C$, so that:
$$\frac{2x+1}{(x-2)(x+1)(x-3)}=\frac{A}{x-2}+\frac{B}{x+1}+\frac{C}{x-3}$$
\end{frame}
\begin{frame}

{\textbf {\textcolor{red}{2. Linear Factors to Powers Greater than $1$ (i.e. repeated linear factors):}}}\\
If $(x-\alpha)^{k}$ appears in the denominator, it will give rise to the following terms:$$\frac{A_{1}}{x-\alpha}+\frac{A_{2}}{(x-\alpha)^{2}}+\cdots+\frac{A_{k}}{(x-\alpha)^{k}}$$
\textcolor{red}{Exercise:}\\
Show from the start $$\frac{3x+1}{(x-1)^{2}(x+2)}=\frac{5/9}{x-1}+\frac{4/3}{(x-1)^{2}}+\frac{-5/9}{x+2}$$ 
\end{frame}
\begin{frame}

{\textbf {\textcolor{red}{3. Irreducible quadratic factors:}}}\\
Irreducible quadratic factors can not be factorised using real numbers e.g. $x^{2}+x+1$\\
An irreducible quadratic $ax^{2}+bx+c$ gives rise to partial fractions of the form $$\frac{Ax+B}{ax^{2}+bx+c}$$
\begin{example}
Express the following in partial fractions:$$\frac{5x}{(x^{2}+x+1)(x-2)}$$
\end{example}
\end{frame}
\begin{frame}

{\textcolor{blue}{$\displaystyle{\frac{5x}{(x^{2}+x+1)(x-2)}}=\displaystyle{\frac{Ax+B}{x^{2}+x+1}}+\displaystyle{\frac{C}{x-2}}$  $$\hspace{1cm}=\frac{(Ax+B)(x-2)+C(x^{2}+x+1)}{(x^{2}+x+1)(x-2)}$$ Therefore: $5x=(Ax+B)(x-2)+C(x^{2}+x+1)$ \\ Now let $x=2$ then $$5(2)=0+C(2^{2}+2+1)$$ $\hspace{3.37cm} \Rightarrow \quad 10=C(7)$ \\ $\hspace{3.37cm} \Rightarrow \quad C=10/7$ \\
constant term on the RHS=$0$ so $$0=-2B+C$$ $\hspace{4.2cm} \Rightarrow 2B=C$ \\ $\hspace{4.2cm} \Rightarrow B=C/2=5/7$} }

\end{frame}

\begin{frame}

{\textcolor{blue}{The coefficient of $x^{2}=0$ so $0=A+C \quad \Rightarrow \quad A=-C=-10/7$, so $$\frac{5x}{(x^{2}+x+1)(x-2)}=\frac{\frac{-10}{7}x+\frac{5}{7}}{x^{2}+x+1}+\frac{\frac{10}{7}}{x-2}$$}}
\end{frame}
\begin{frame}

{\textbf {\textcolor{red}{4. Irreducible quadratic factors to powers greater than $1$:}}}\\
If $ax^{2}+bx+c$ is an irreducible quadratic factor, then $(ax^{2}+bx+c)^{k}$ in the denominator gives rise to $$\frac{A_{1}x+B_{1}}{ax^{2}+bx+c}+\frac{A_{2}x+B_{2}}{(ax^{2}+bx+c)^{2}}+\cdots+\frac{A_{k}x+B_{k}}{(ax^{2}+bx+c)^{k}}$$ 
\end{frame}



















\end{document}