W03 - Examples

Partial fractions with repeated factor

Partial fractions with repeated factor

Find the partial fraction decomposition:

$$\frac{3x-9}{x^3+3x^2-4}$$

Solution

(1) Check that denominator degree is lower. $✓$

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(2) Factor denominator:

Rational Roots Theorem: check for roots at $\pm 1$ and $\pm 2$ and $\pm 4$.

Discover that $x=+1$ is a root. Therefore divide by $x-1$:

$$\frac{x^3+3x^2-4}{x-1}\gg \gg x^2+4x+4$$

Factor again:

$$x^2+4x+4\gg \gg (x+2{)}^2$$

Final factored form:

$$x^3+3x^2-4=(x-1)(x+2{)}^2$$

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(3) Write the generic PFD:

$$\frac{3x-9}{(x-1)(x+2{)}^2}=\frac{A}{x-1}+\frac{B}{x+2}+\frac{C}{(x+2{)}^2}$$

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(4) Solve for $A$, $B$, and $C$:

Multiply across by the common denominator:

$$3x-9=A(x+2{)}^2+B(x-1)(x+2)+C(x-1)$$

For $A$, set $x=1$, obtain:

$$\begin{array}{rlrl}& & 3\cdot 1-9& =A(1+2{)}^2+B\cdot 0+C\cdot 0\\ \gg \gg & & -6& =9A\\ \gg \gg & & A& =-2/3\end{array}$$

For $C$, set $x=-2$, obtain:

$$\begin{array}{rlrl}& & 3\cdot (-2)-9& =A\cdot 0+B\cdot 0+C\cdot (-3)\\ \gg \gg & & -15& =-3C\\ \gg \gg & & C& =5\end{array}$$

For $B$, insert prior results and solve.

Plug in $A$ and $C$:

$$3x-9=-\frac{2}{3}(x+2{)}^2+B(x-1)(x+2)+5(x-1)$$

Now plug in another convenient $x$, say $x=3$:

$$\begin{array}{rl}0& =-\frac{2}{3}\cdot 5^2+B\cdot 2\cdot 5+5\cdot 2\\ \frac{50}{3}-10& =10B\gg \gg B=\frac{2}{3}\end{array}$$

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(4) Plug in $A$, $B$, $C$ for the final answer:

$$\frac{3x-9}{x^3+3x^2-4}=\frac{-2/3}{x-1}+\frac{2/3}{x+2}+\frac{5}{(x+2{)}^2}$$ Link to original

Partial fractions - repeated quadratic, linear tops

Partial fractions - repeated quadratic, linear tops

Compute the integral:

$$\int \frac{x^3+1}{(x^2+4{)}^2}dx$$

Solution

(1) Partial fraction decomposition:

• Numerator degree is lower than denominator.
• Factor denominator completely. (No real roots.)

Write generic PFD:

$$\frac{x^3+1}{(x^2+4{)}^2}=\frac{Ax+B}{x^2+4}+\frac{Cx+D}{(x^2+4{)}^2}$$

• Notice: repeated factor: use incrementing powers up to 2.
• Notice: linear over quadratic.

Common denominators and solve:

$$\begin{array}{c}{x}^3+1=(Ax+B)(x^2+4)+Cx+D\\ \\ \gg \gg x^3+1=A{x}^3+B{x}^2+(4A+C)x+4B+D\\ \\ \gg \gg A=1,B=0\\ \\ \gg \gg C=-4,D=1\end{array}$$

Therefore:

$$\frac{x^3+1}{(x^2+4{)}^2}=\frac{x}{x^2+4}+\frac{-4x+1}{(x^2+4{)}^2}$$

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(2) Integrate:

Integrate the first term using substitution $u=x^2+4$:

$$\begin{array}{c}\int \frac{x}{x^2+4}dx\stackrel{u=x^2+4}{\gg \gg }\frac{1}{2}\int \frac{du}{u}\\ \\ \gg \gg \frac{1}{2}\ln |u|+C\gg \gg \frac{1}{2}\ln \left|x^2+4\right|+C\end{array}$$

Break up the second term:

$$\frac{-4x+1}{(x^2+4{)}^2}\gg \gg \frac{-4x}{(x^2+4{)}^2}+\frac{1}{(x^2+4{)}^2}$$

Integrate the first term of RHS:

$$\begin{array}{c}\int \frac{-4x}{(x^2+4{)}^2}dx\gg \gg -2\int \frac{du}{u^2}\\ \\ \gg \gg \frac{2}{u}+C\gg \gg \frac{2}{x^2+4}+C\end{array}$$

Integrate the second term of RHS using $x=2\tan \theta$:

$$\begin{array}{c}\int \frac{dx}{(x^2+4{)}^2}\gg \gg \int \frac{2{\sec }^2\theta d\theta }{16{\sec }^4\theta }\\ \\ \gg \gg \frac{1}{8}\int {\cos }^2\theta d\theta \\ \\ \gg \gg \frac{1}{8}\int \frac{1}{2}(1+\cos (2\theta ))d\theta \\ \\ \gg \gg \frac{1}{16}\theta +\frac{1}{32}\sin (2\theta )+C\\ \\ \gg \gg \frac{1}{16}{\tan }^{-1}\left(\frac{x}{2}\right)+\frac{1}{32}2\sin \theta \cos \theta +C\\ \\ \gg \gg \frac{1}{16}{\tan }^{-1}\left(\frac{x}{2}\right)+\frac{1}{32}2\frac{x}{\sqrt{x^2+2^2}}\frac{2}{\sqrt{x^2+2^2}}+C\\ \\ \gg \gg \frac{1}{16}{\tan }^{-1}\left(\frac{x}{2}\right)+\frac{1}{8}\frac{x}{x^2+2^2}+C\end{array}$$Link to original

“Rationalize a quotient” - convert into PFD

Extra - “Rationalize a quotient” - convert into PFD

Sometimes an integrand may be converted to a rational function using a substitution.

Consider this integral:

$$\int \frac{\sqrt{x+4}}{x}dx$$

Set $u=\sqrt{x+4}$, so $x=u^2-4$ and $dx=2udu$:

$$\gg \gg \int \frac{2udu}{u^2-4}$$

Now this rational function has a partial fraction decomposition:

$$\frac{2u}{u^2-4}\gg \gg \frac{2u}{(u-2)(u+2)}\gg \gg \frac{1}{u-2}+\frac{1}{u+2}$$

It is easy to integrate from there!

Practice exercises:

• To compute $\int \frac{\sqrt{x}}{x-1}dx$, try the substitution $u=\sqrt{x}$.
• To compute $\int \frac{dx}{\sqrt[2]{x}-\sqrt[3]{x}}$, try the substitution $u=\sqrt[6]{x}$.
• To compute $\int \frac{1}{x-\sqrt{x+2}}dx$, try the substitution $u=\sqrt{x+2}$.

Link to original

Simpson’s Rule on the Gaussian distribution

Simpson’s Rule on the Gaussian distribution

The function $e^{x^2}$ is very important for probability and statistics, but it cannot be integrated analytically.

Apply Simpson’s Rule to approximate the integral:

$${\int }_0^1{e}^{x^2}dx$$

with $\Delta x=0.1$ and $n=10$. What error bound is guaranteed for this approximation?

Solution

(1) We need a table of values of $x_i$ and $y_i=f(x_i)$:

$x_i:$	$0.0$	$0.1$	$0.2$	$0.3$	$0.4$	$0.5$	$0.6$	$0.7$	$0.8$	$0.9$	$1.0$
$f(x_i):$	$e^{0.0^2}$	$e^{0.1^2}$	$e^{0.2^2}$	$e^{0.3^2}$	$e^{0.4^2}$	$e^{0.5^2}$	$e^{0.6^2}$	$e^{0.7^2}$	$e^{0.8^2}$	$e^{0.9^2}$	$e^{1.0^2}$
$\approx$	$1.000$	$1.010$	$1.041$	$1.094$	$1.174$	$1.284$	$1.433$	$1.632$	$1.896$	$2.248$	$2.718$

These can be plugged into the Simpson Rule formula to obtain our desired approximation:

$$\begin{array}{c}{S}_{10}=\frac{1}{3}\cdot 0.1\cdot (1.000+4\cdot 1.010+2\cdot 1.041+4\cdot 1.094+\cdots +2\cdot 1.896+4\cdot 2.248+2.718)\\ \\ \approx 1.463\end{array}$$

To find the error bound we need to find the smallest number we can manage for $K_4$.

Take four derivatives and simplify:

$$f^{(4)}(x)=(12+48x^2+16x^4)e^{x^2}$$

On the interval $x\in [0,1]$, this function is maximized at $x=1$. Use that for the optimal $K_4$:

$$f^{(4)}(1.000)=206.589$$

Finally we plug this into the error bound formula:

$$\frac{K_4(b-a{)}^5}{180n^4}=\frac{206.589\cdot 1.000^5}{180\cdot 10^4}\approx 0.0001$$ $$\gg \gg \mathrm{Error}(S_{10})\le 0.0001$$Link to original