Joe Herbert

Formula | Vectors | Matrices | Complex Numbers |

Proof | Further Algebra | Further Functions | Numerical Methods |

Hyperbolic Functions | Polar Coordinates | Further Calculus |

Discrete |

Statistics |

The magnitude of 3D vector $\underset{\sim}{a} = \begin{pmatrix}x\\y\\z\end{pmatrix}$ is $|\underset{\sim}{a}| = \sqrt{x^2 + y^2 + z^2}$

Multiplying a vector by a scalar alters its magnitude but not its direction.

A unit vector is a vector whose magnitude is 1.

e.g. $\begin{pmatrix}3\\4\end{pmatrix}$ in unit vectors is $\frac{1}{5}\begin{pmatrix}3\\4\end{pmatrix}$

$\underset{\sim}{a}$ in unit vectors is $\frac{1}{|\underset{\sim}{a}|}\underset{\sim}{a}$

j = unit vector in y-axis = $\begin{pmatrix}0\\1\\0\end{pmatrix}$

k = unit vector in z-axis = $\begin{pmatrix}0\\0\\1\end{pmatrix}$

xi + yj + zk = $\begin{pmatrix}x\\y\\z\end{pmatrix}$ e.g. $\begin{pmatrix}-2\\0\\3\end{pmatrix} = -2i + 3k$

In 3D, vector $\underset{\sim}{a} = \begin{pmatrix}a_1\\a_2\\a_3\end{pmatrix}$, vector $\underset{\sim}{b} = \begin{pmatrix}b_1\\b_2\\b_3\end{pmatrix}$, $\underset{\sim}{a} \cdot \underset{\sim}{b} = a_1b_1 + a_2b_2 + a_3b_3$

If two vectors are perpendicular their dot product is 0.

$\begin{pmatrix}x\\y\\z\end{pmatrix} = \begin{pmatrix}3\\2\\1\end{pmatrix} + t \begin{pmatrix}-3\\4\\-1\end{pmatrix}$

$x = 3 - 3t$

$y = 2 + 4t$

$z = 1 - t$

This is the parametric equation of the line.

$t = \frac{3-x}{3}$

$t = \frac{y-2}{4}$

$t = 1 - z$

$\frac{3-x}{3} = \frac{y-2}{4} = 1 - z$ is the equation of this line.

To convert between vector and cartesian equations you must convert to parametric in between.

To find the intersection of lines, convert to parametric equations, solve for x and y simultaneously to work out the scalars for the crossing points. Then work out x and y. Then plug in to find z.

e.g. Find the intersection of the lines $r = i + j + k + \lambda (2i + j + k)$ and $r = -i - 3j + \mu (2i + 2j + k)$

Parametric Equations for x values: $x = 1 + 2 \lambda = -1 + 2 \mu$

Parametric Equations for y values: $y = 1 + \lambda = -3 + 2 \mu$

$\therefore \lambda = 2$ and $\mu = 3$

$z = 1 + \lambda = 3$ and $z = \mu = 3$. Equations have the same z values which shows they're not skew lines and they do intersect.

$\therefore $ the lines intersect at (5, 3, 3).

To find this:

- Find the parametrics of the lines.
- Find the general vector by subtracting the position vector of the point from the parametrics.
- This general vector must be perpendicular to the original line, so their dot product is 0.
- Solve this to find the scalar of the direction vector.
- Plug this back into the general vector to get the vector between the line and the point.
- Then find the magnitude of this vector to get the distance.

To find the cross product, solve a 3x3 matrix:

$$\begin{bmatrix}i & j & k\\a_1 & a_2 & a_3\\b_1 & b_2 & b_3\end{bmatrix}$$ $$i \begin{vmatrix}a_2 & a_3\\b_2 & b_3\end{vmatrix} - j \begin{vmatrix}a_1 & a_3\\b_1 & b_3\end{vmatrix} + k \begin{vmatrix}a_1 & a_2\\b_1 & b_2\end{vmatrix}$$ For vectors $\underset{\sim}{a}$ and $\underset{\sim}{b}$, and angle $\theta$ between them:

$|\underset{\sim}{a} \times \underset{\sim}{b}| = |\underset{\sim}{a}||\underset{\sim}{b}|\sin \theta$

$A = \frac{1}{2}|\underset{\sim}{b} \times \underset{\sim}{a}|$

$B = |\underset{\sim}{b} \times \underset{\sim}{a}|$

For any vector $\underset{\sim}{r}$ on the line, $\underset{\sim}{r}-\underset{\sim}{a}$ is the same direction as vector $\underset{\sim}{b}$:

$(\underset{\sim}{r} - \underset{\sim}{a}) \times \underset{\sim}{b} = 0$

- Find the general vector between the two lines.
- Both directions must be perpendicular to the general vector so their dot products must be 0.
- Solve these to find the scalars.
- Plug the scalars back into the general vector and the shortest distance is the magnitude of this vector.

The shortest vector must be parallel to the cross product.

- Find the cross product of the two directions.
- Find the general vector and set this equal to (k x the cross product), as they are parallel.
- Solve this to find k.
- Find the magnitude of (k x the cross product) to get the distance.

Consider $\underset{\sim}{r_1} = p_1 + \lambda d_1$ and $\underset{\sim}{r_2} = p_2 + \lambda d_2$

Distance between skew lines: $${(p_1 - p_2) \cdot (d_1 \times d_2)} \overwithdelims || {|d_1 \times d_2|}$$

Let $n$ be the normal vector to the plane at $a$.

Let $r$ be the position vector of any point in the plane.

$\overrightarrow{ra} = r - a$

As $n$ is normal, and $r - a$ is a vector in the plane, the angle between them is 90°.

$\therefore (r-a) \cdot n = 0$

$r \cdot n - a \cdot n = 0$

$r \cdot n = a \cdot n$

e.g. Find the vector equation of a plane with normal $\begin{pmatrix}1\\2\\-1\end{pmatrix}$ through point $\begin{pmatrix}1\\2\\3\end{pmatrix}$:

$r \cdot \begin{pmatrix}1\\2\\-1\end{pmatrix} = \begin{pmatrix}1\\2\\3\end{pmatrix} \cdot \begin{pmatrix}1\\2\\-1\end{pmatrix}$

$r \cdot \begin{pmatrix}1\\2\\-1\end{pmatrix} = 2$. This is the equation of the plane.

e.g. If $r \cdot \begin{pmatrix}1\\2\\-1\end{pmatrix} = 2$ then

$\begin{pmatrix}x\\y\\z\end{pmatrix} \cdot \begin{pmatrix}1\\2\\-1\end{pmatrix} = 2$, giving us $x + 2y - z = 2$

The co-efficients of $x$, $y$ and $z$ give the normal.

e.g. $3x + 2y - 4z \rightarrow \begin{pmatrix}3\\2\\-4\end{pmatrix}$

e.g. $5x + 2z \rightarrow \begin{pmatrix}5\\0\\2\end{pmatrix}$

A plane can be written as $\underset{\sim}{r} = a + \lambda b + \mu c$ where $\lambda, \mu ∈ \mathbb R$ and where $a$ is the position vector of the plane and $b$ and $c$ are direction vectors completely contained within the plane.

You can find the normal by doing $n = |b \times c|$

- Find the normal vector to the plane.
- Find the vector equation with the given point as the position vector and the direction vector as the normal vector.
- Calculate where that vector intersects the plane.
- Find the distance between the points.

A matrix with one column is a vector.

A matrix is square if it has the same number of rows as columns.

A zero matrix is one in which all its elements are 0.

An identity matrix $I$ is a square matrix which has 1s in the leading diagonal (starting top left) and 0 elsewhere: $$\begin{pmatrix}1 & 0 & 0\\0 & 1 & 0\\0 & 0 & 1\end{pmatrix}$$ $AI = IA = A$ for all matrices 1

Then repeat with the first row and second column, then for the next row and first column, and for the second column, etc.

e.g. $$\begin{pmatrix}1 & 0 & 3 & -2\\2 & 8 & 4 & 3\\7 & -1 & 0 & 2\end{pmatrix} \begin{pmatrix}5 & 1\\1 & 7\\0 & 3\\8 & -3\end{pmatrix} = \begin{pmatrix} & \\ & \\ & \end{pmatrix}$$ (1 x 5) + (0 x 1) + (3 x 0) + (-2 x 8) = -11 $$\begin{pmatrix}1 & 0 & 3 & -2\\2 & 8 & 4 & 3\\7 & -1 & 0 & 2\end{pmatrix} \begin{pmatrix}5 & 1\\1 & 7\\0 & 3\\8 & -3\end{pmatrix} = \begin{pmatrix} \color{red}{-11} & \\ & \\ & \end{pmatrix}$$ (1 x 1) + (0 x 7) + (3 x 3) + (-2 x -3) = 16 $$\begin{pmatrix}1 & 0 & 3 & -2\\2 & 8 & 4 & 3\\7 & -1 & 0 & 2\end{pmatrix} \begin{pmatrix}5 & 1\\1 & 7\\0 & 3\\8 & -3\end{pmatrix} = \begin{pmatrix} -11 & \color{red}{16}\\ & \\ & \end{pmatrix}$$ (2 x 5) + (8 x 1) + (4 x 0) + (3 x 8) = 42 $$\begin{pmatrix}1 & 0 & 3 & -2\\2 & 8 & 4 & 3\\7 & -1 & 0 & 2\end{pmatrix} \begin{pmatrix}5 & 1\\1 & 7\\0 & 3\\8 & -3\end{pmatrix} = \begin{pmatrix} -11 & 16\\\color{red}{42} & \\ & \end{pmatrix}$$ (2 x 1) + (8 x 7) + (4 x 3) + (3 x -3) = 61 $$\begin{pmatrix}1 & 0 & 3 & -2\\2 & 8 & 4 & 3\\7 & -1 & 0 & 2\end{pmatrix} \begin{pmatrix}5 & 1\\1 & 7\\0 & 3\\8 & -3\end{pmatrix} = \begin{pmatrix} -11 & 16\\42 & \color{red}{61}\\ & \end{pmatrix}$$ (7 x 5) + (-1 x 1) + (0 x 0) + (2 x 8) = 50 $$\begin{pmatrix}1 & 0 & 3 & -2\\2 & 8 & 4 & 3\\7 & -1 & 0 & 2\end{pmatrix} \begin{pmatrix}5 & 1\\1 & 7\\0 & 3\\8 & -3\end{pmatrix} = \begin{pmatrix} -11 & 16\\42 & 61\\\color{red}{50} & \end{pmatrix}$$ (7 x 1) + (-1 x 7) + (0 x 3) + (2 x -3) = -6 $$\begin{pmatrix}1 & 0 & 3 & -2\\2 & 8 & 4 & 3\\7 & -1 & 0 & 2\end{pmatrix} \begin{pmatrix}5 & 1\\1 & 7\\0 & 3\\8 & -3\end{pmatrix} = \begin{pmatrix} -11 & 16\\42 & 61\\50 & \color{red}{-6}\end{pmatrix}$$ Matrices can be multiplied if there are the same number of columns in A as rows in B. The dimensions of AB will be the rows of A x the columns of B. This means only square matrices can be raised to a power.

Matrix multiplication is not commutative, so AB $\ne$ BA.

Matrix multiplication is associative, so (AB)C = A(BC).

Matrix multiplication distributes over addition, so A(B + C) = AB + AC.

$$\begin{pmatrix}1 & 2 & 3\\4 & 5 & 6\end{pmatrix}^T = \begin{pmatrix}1 & 4\\2 & 5\\3 & 6\end{pmatrix}$$ A matrix is symmetrical if $A^T = A$. All symmetric matrices must be square.

If you transpose a matrix twice, you get back to where you started: $(A^T)^T = A$.

$A^TA \ne AA^T$, both $A^TA$ and $AA^T$ are symmetric matrices. $(AB)^T = B^T A^T$

If det(A) $\ne$ = 0, A is a non-singular matrix and it has an inverse.

- Find det(A)
- Form a matrix of minors, M. The Minor of an element is the determinant of the 2x2 matrix that remains after the row and column containing that element have been crossed out.
- Form a matrix of cofactors, C, by changing all the signs in the pattern $\begin{pmatrix}+ & - & +\\- & + & -\\+ & - & +\end{pmatrix}$
- $A^{-1} = \frac{1}{det(A)} C^T$

A consistent solution is one which has one or more solutions.

To transform a shape using a matrix you multiply the vectors representing the vertices by that matrix and the resultant vectors represent the co-ordinates of the image.

To undo a transformation, transform the shape using the inverse matrix.

The determinant of the matrix is the scale factor of area or volume between the original shape and its image.

Use the standard basis vectors to find the transformation matrix. e.g.

Matrix representing an anticlockwise rotation by $\theta$ about the origin is $\begin{pmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta\end{pmatrix}$ The reflection in the line $y = (\tan \theta) x$ is represented by $\begin{pmatrix} \cos 2\theta & \sin 2\theta \\ \sin 2\theta & -\cos 2\theta\end{pmatrix}$ The reflection in the line x-axis is represented by $\begin{pmatrix} 1 & 0 \\ 0 & -1\end{pmatrix}$ The reflection in the line y-axis is represented by $\begin{pmatrix} -1 & 0 \\ 0 & 1\end{pmatrix}$ An enlargement by scale factor $a$ is represented by $\begin{pmatrix} a & 0 \\ 0 & a\end{pmatrix}$ A stretch by scale factor $a$ parallel to x-axis is represented by $\begin{pmatrix} a & 0 \\ 0 & 1\end{pmatrix}$ A stretch by scale factor $a$ parallel to y-axis is represented by $\begin{pmatrix} 1 & 0 \\ 0 & a\end{pmatrix}$ Doing a transformation $A$ followed by a transformation $B$ is the same as doing the transformation $BA$ (work from right to left).

The matrices representing rotations about the axes in 3D are in the formula booklet.

Reflections in planes normal to the axes in 3D:

$\begin{pmatrix} -1 & 0 & 0 \\ 0 & 1 & 0\\0 & 0 & 1\end{pmatrix}$ for a reflection in $x = 0$

$\begin{pmatrix} 1 & 0 & 0 \\ 0 & -1 & 0\\0 & 0 & 1\end{pmatrix}$ for a reflection in $y = 0$

$\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0\\0 & 0 & -1\end{pmatrix}$ for a reflection in $z = 0$

$\begin{vmatrix} a & a & a \\ 2 & 0 & 4\\0 & 1 & 3\end{vmatrix} = a\begin{vmatrix} 1 & 1 & 1 \\ 2 & 0 & 4\\0 & 1 & 3\end{vmatrix}$

You can add or subtract any amount of any other row to any row without changing the determinant. The same can be done with columns. e.g. applying $C1 \rightarrow C1 - C3$ and $C2 \rightarrow C2 - C3$:

$\begin{vmatrix} 1 & 1 & 1 \\ x & y & 1\\x^2 & y^2 & 1\end{vmatrix} = \begin{vmatrix} 0 & 0 & 1 \\ x-1 & y-1 & 1\\x^2-1 & y^2-1 & 1\end{vmatrix}$

An eigenvalue is how much that vector is stretched. A negative eigenvalue means the vector goes in the opposite direction (still considered the same direction).

A normalised eigenvector is one with magnitude 1. To normalise a vector just divide all the components by the magnitude of that vector.

Equations:

$Av = \lambda v$, where $v$ is an eigenvector for a matrix $A$ and $\lambda$ is the eigenvalue The characteristic equation for a matrix is: $det(A-\lambda I) = 0$ These can be used to find all the eigenvalues and eigenvectors.

To find the Cartesian equation of the invariant lines passing throught the origin, for each eigenvalue, form equations and solve for y from:

$M \begin{pmatrix} x\\y\end{pmatrix} = \lambda \begin{pmatrix} x\\y\end{pmatrix}$

- Find the eigenvalues of M
- Put the eigenvalues in the leading diagonal of a new matrix, and surround these with 0 to get $D$
- Find the corresponding eigenvectors for the eigenvalues and put these as the columns of $U$, in the same order as the eigenvalues were in $D$
- Find the inverse of U

The argument of a complex number $z$ is given by $arg(z) = \theta$, where $\tan(\theta) = \frac{b}{a}$. A point can be written as $(x, y), (r, \theta)$ or $r(\cos\theta + i\sin\theta)$ where $|z| = r$ and $\theta = arg(z)$. If $z = a+bi, z \times z^* = a^2 + b^2 = |z|^2$ and $z+z^* = 2a$ If $z_1 = (r_1,\theta_1)$, and $z_2 = (r_2, \theta_2)$, then $z_1z_2 = (r_1r_2, \theta_1+\theta_2)$ and $\frac{z_1}{z_2} = (\frac{r_1}{r_2}, \theta_1-\theta_2)$. If $a + bi = c + di$, where $a, b, c, d, ∈ \mathbb R$ then $a=c$ and $b=d$.

$|z-z_1| = |z-z_2|$ represents a straight line, the perpendicular bisector of the line joining points $z_1$ and $z_2$.

$arg(z) = \alpha$ represents the half line through $O$ inclined at an angle $\alpha$ to the positive direction of $x$. $arg(z-z_1) = \alpha$ represents the half line through the point $z_1$ inclined at an angle $\alpha$ to the positive direction of $x$.

$e^{2\pi i}=1$.

$e^{4\pi i}=1$.

$e^{2k\pi i}=1$.

If $z^n=1$ then $z^n = 1 = e^{2k\pi i}$. The equation $z^n=1$ has roots $z=e^\frac{{2k\pi i}}{n}$ where $k ∈ [0, n)$. On an Argand Diagram, the roots all lie on the circle $|z| = 1$ and are equally spaced around the circle at intervals of $\frac{2\pi}{n}$ starting at (1,0).

- Base case: Prove the statement is true for n = 1
- Assumption: Assume the statement is true for n = k
- Inductive step: Show that the statement is true for n = k + 1
- Conclusion: The statement is then true for all positive integers n

Since true for n = 1 and if true for n = k, true for n = k + 1, $\therefore$ true for all $n ∈ \mathbb N$ by induction.

- When n = 1, $\sum_{i=1}^n i^2 = 1$, and $\frac{1}{6}(1+1)(2(1)+1) = \frac{6}{6} = 1$

$1 = 1 \therefore$ statement is true for n = 1 - Assume true for n = k:

$\sum_{i=1}^k i^2 = \frac{k}{6}(k+1)(2k+1)$ - When n = k + 1: $$\sum_{i=0}^{k+1} i^2 = \sum_{i=0}^k i^2 + (k+1)^2$$ $$= \frac{k}{6} (k+1)(2k+1) + (k+1)^2$$ $$= \frac{k+1}{6} (k(2k+1) + 6(k+1))$$ $$= \frac{k+1}{6} (2k^2 + 7k + 3)$$ $$= \frac{k+1}{6} (2k + 3)(k + 2)$$ $$= \frac{k+1}{6} (2(k+1) + 1)((k+1) + 1)$$ This is the same as what we're trying to show, with k + 1 instead of n. Hence true when n = k + 1.
- Since true for n = 1 and if true for n = k, true for n = k + 1, $\therefore$ true for all $n ∈ \mathbb N$ by induction.

- When $n = 1, 3^{2 \times 1} + 11 = 3^2 + 11 = 20$ which is divisible by 4
- Assume that for $n = k, f(k) = 3^{2k} + 11$ is divisible by 4 for all $k ∈ \mathbb Z^+$
- $$f(k+1) = 3^{2(k+1)} + 11$$ $$= 3^{2k} \times 3^2 + 11$$ $$= 9(3^{2k})+11$$ $$= 3^{2k} + 11 + 8(3^{2k})$$ $$= f(k) + 8(3^{2k})$$ $\therefore$ f(n) is divisible by 4 when n = k+1 since f(k) is divisible by 4 and 8 is divisible by 4.
- Since true for n = 1 and if true for n = k, true for n = k + 1, $\therefore$ true for all $n ∈ \mathbb Z^+$ by induction.

- When $n = 1, u_1 = 3^1 - 2 = 1$ as given.
- Assume that for $n = k, u_k = 3^k - 2$ is true for $k ∈ \mathbb Z^+$
- Then $$u_{k+1} = 3u_k + 4$$ $$= 3(3^k - 2) + 4$$ $$= 3^{k+1} - 6 + 4$$ $$= 3^{k+1} - 2$$
- Since true for n = 1 and if true for n = k, true for n = k + 1, $\therefore$ true for all $n ∈ \mathbb Z^+$ by induction.

- When n = 1, $u_1 = 2^2 + 3^2 = 13$ as given

When n = 2, $u_2 = 2^3 + 3^3 = 35$ as given - Assume that for n = k and n = k+1, $u_k = 2^{k+1} + 3^{k+1} \text{ and } u_{k+1} = 2^{k+2} + 3^{k+2}$ are true for $k ∈ \mathbb Z^+$
- Let n = k+1 $$u_{k+2} = 5u_{k+1} - 6u_k$$ $$= 5(2^{k+1} + 3^{k+1}) - 6(2^k + 3^k)$$ $$= 5 \times 2^{k+1} + 5 \times 3^{k+1} - 3 \times 2 \times 2^k - 2 \times 3 \times 3^k$$ $$= 5 \times 2^{k+1} + 5 \times 3^{k+1} - 3 \times 2^{k+1} - 2 \times 3^{k+1}$$ $$= 2 \times 2^{k+1} + 3 \times 3^{k+1} = 2^{k+1+1} + 3^{k+1+1}$$
- Since true for n = 1 and n = 2, and if true for n = k and n = k + 1, true for n = k + 2, $\therefore$ true for all $n ∈ \mathbb Z^+$

- $$\begin{pmatrix}1 & -1\\0 & 2\end{pmatrix}^1 = \begin{pmatrix}1 & -1\\0 & 2\end{pmatrix}$$ $$\begin{pmatrix}1 & 1-2^1\\0 & 2^1\end{pmatrix} = \begin{pmatrix}1 & -1\\0 & 2\end{pmatrix}$$ $\therefore$ true for n = 1
- Assume true for n = k so $\begin{pmatrix}1 & -1\\0 & 2\end{pmatrix}^k = \begin{pmatrix}1 & 1-2^k\\0 & 2^k\end{pmatrix}$
- When n = k+1, $$\begin{pmatrix}1 & -1\\0 & 2\end{pmatrix}^{k+1} = \begin{pmatrix}1 & -1\\0 & 2\end{pmatrix}^k \begin{pmatrix}1 & -1\\0 & 2\end{pmatrix} = \begin{pmatrix}1^k & 1-2^k\\0^k & 2^k\end{pmatrix}\begin{pmatrix}1 & -1\\0 & 2\end{pmatrix}$$ $$= \begin{pmatrix}1 & 1-2^{k+1}\\0 & 2^{k+1}\end{pmatrix}$$ $\therefore$ true when n = k + 1.
- Since true for n = 1 and if true for n = k, true for n = k + 1 $\therefore$ true for all $n ∈ \mathbb Z^+$

Conditions of Maclaurin series:

- The function f(x) can be expressed in the form $a_0 + a_1x + a_2x^2 + a_3x^3 + ...$
- The function can be derived term by term
- The function and all of its derivatives exist at $x=0$
- The function must converge

$\sin x: \space x - \frac{x^3}{6} + \frac{x^5}{120}$

$\cos x: \space 1 - \frac{x^2}{2} + \frac{x^4}{24} - \frac{x^6}{720}$

$\ln (1+x): \space x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \frac{x^5}{5} - \frac{x^6}{6}$

$(1+x)^n: \space 1 + nx + \frac{1}{2} (n-1)nx^2 + \frac{1}{6}(n-2)(n-1)nx^3 + \frac{1}{24} (n-3)(n-2)(n-1)nx^4 + \frac{1}{120}(n-4)(n-3)(n-2)(n-1)nx^5$

Horizontal asymptotes occur if the graph converges to a particular value of $y$ as $x \to \pm \infty$. When you have a fraction, divide by the largest power of $x$ in the denominator to help determine the behavious as $x \to \pm \infty$. Anything of the form $\frac{\text{constant}}{\text{some power of }x}$ will tend to zero.

Oblique asymptotes occur if the graph almost follows a straight line as $x \to \pm \infty$. They occur when the degree of the numerator is one more than the degree of the denominator. To find them, divide by the largest power in the denominator.

e.g. Find the turning point of $y=\frac{2}{x^2-4x}$ without using calculus.

$$\frac{2}{x^2-4x} = k$$ $$2 = k(x^2 - 4x)$$ $$kx^2 - 4kx - 2 = 0$$ $$(-4k)^2 - 4(k)(-2) = 0$$ $$16k^2 + 8k = 0$$ $$k=0 \text{ or } k = -\frac{1}{2}$$ 0 is the asymptote so the turning point must be when $y = -\frac{1}{2}$

$$-\frac{1}{2} = \frac{2}{x^2 - 4x} \therefore x = 2$$ $$\therefore \text{turning point is at } (2, -\frac{1}{2})$$

When you sketch the graph of $y = \frac{1}{f(x)}$ apply the following rules:

- A zero of $f(x)$ will become a vertical asymptote of $\frac{1}{f(x)}$
- A vertical asymptote of $f(x)$ will become a zero of $\frac{1}{f(x)}$
- The sign of $f(x)$ is the same as the sign of $\frac{1}{f(x)}$
- For all $x$ such that $f(x) = 1, \frac{1}{f(x)} = 1$ as well so the graphs of $y=f(x)$ and $y=\frac{1}{f(x)}$ will intersect here
- A maximum of $f(x)$ will become a minimum of $\frac{1}{f(x)}$ and vice versa

A hyperbola has oblique asymptotes. The oblique asymptotes of a hyperbola which is symmetric about the axes are $y=\pm \frac{b}{a}x$.

A rectangular hyperbola is one whose asymptotes are perpendicular, therefore:

$\frac{b}{a}x \times -\frac{b}{a} = -1$ so $\frac{b^2}{a^2} = 1$ so $b^2 = a^2$.

Therefore, if both $a^2$ and $b^2$ are equal to the same number, $c^2$, the equation of a rectangular hyperbola is: $x^2 - y^2 = c^2$

$y = \frac{k}{x}$ is a rectangular hyperbola but not symmetrical about the axes.

The area is estimated by splitting the area into rectangles, where the height of the rectangle is the y-value of the curve midway between the top sides of the rectangle.

Formula: $$\int^b_a y \space dx \approx h(y_{\frac{1}{2}} + y_{\frac{3}{2}} + ... + y_{n-\frac{3}{2}} + y_{n-\frac{1}{2}})$$ where $h = \frac{b-a}{n}$ $h$ = the width of each rectangle

$b$ = the upper bound

$a$ = the lower bound

$n$ = the number of rectangles

$y_k$ = the $y$ value after k rectangles

e.g. Use the mid-ordinate rule with four strips to find an estimate for $\int^5_1 \ln x \space dx$, giving your answer to 3 sig figs.

$$h=\frac{b-a}{n}=\frac{5-1}{4}=1$$

$x$ values | Mid-$x$ values | Mid-ordinates ($y = \ln x)$ |

$x_0 = 1$ | ||

$x_{\frac{1}{2}} = 1.5$ | $y_{\frac{1}{2}} = \ln 1.5 = 0.40546...$ | |

$x_1 = 2$ | ||

$x_{\frac{3}{2}} = 2.5$ | $y_{\frac{3}{2}} = \ln 2.5 = 0.916291...$ | |

$x_2 = 3$ | ||

$x_{\frac{5}{2}} = 3.5$ | $y_{\frac{5}{2}} = \ln 3.5 = 1.252762...$ | |

$x_3 = 4$ | ||

$x_{\frac{7}{2}} = 4.5$ | $y_{\frac{7}{2}} = \ln 4.5 = 1.504077...$ | |

$x_4 = 5$ |

It is most accurate when there is a constant gradient.

You will get an underestimate if the curve is concave upwards.

You will get an overestimate if the curve is concave downwards.

e.g. Use Simpson's Rule with 5 ordinates (4 strips) to find an approximation to $\int^3_1 \frac{1}{\sqrt{1+x^3}} dx$ giving your answer to 3 sig figs.

$$h=\frac{b-a}{n}=\frac{3-1}{4}=\frac{1}{2}$$

$x$ values | $y$ values |

$x_0 = 1$ | $y_0 = \frac{1}{\sqrt{1 + 1^3}} = 0.707107$ |

$x_1 = \frac{3}{2}$ | $y_1 = \frac{1}{\sqrt{1 + {{3}\overwithdelims () {2}}^3}} = 0.478091$ |

$x_2 = 2$ | $y_2 = \frac{1}{\sqrt{1 + 2^3}} = 0.33333$ |

$x_3 = \frac{5}{2}$ | $y_3 = \frac{1}{\sqrt{1 + {{5}\overwithdelims () {2}}^3}} = 0.245256$ |

$x_4 = 3$ | $y_4 = \frac{1}{\sqrt{1 + 3^3}} = 0.188982$ |

$y_1 + y_3 = 0.723347$

$$\int^3_1 \frac{1}{\sqrt{1+x^3}} dx \approx \frac{1}{3} h \{(y_0 + y_n) + 4(y_1 + y_3) + 2(y_2)\}$$ $$\approx \frac{1}{3} \frac{1}{2} [(0.896089) + 4(0.723347) + 2(0.33333)]$$ $$\approx 0.742691 \approx 0.743$$

For when $\frac{dy}{dx}$ is just a function of $x$: For $\frac{dy}{dx} = f(x)$ and small $h$, recurrence relations are:$$y_{n+1} = y_n + hf(x_n); x^{n+1} = x_n + h$$ For when $\frac{dy}{dx}$ is a function of $x$ and $y$: For $\frac{dy}{dx} = f(x, y)$: $$y_{r+1} = y_r + hf(x_r, y_r)$$ $h$ = the step size

e.g. Consider a function such that $f(1) = 2$ and $f'(x) = 6x^2 + 2x$. Use Euler's method with step size $\frac{1}{2}$ to find an approximation of $f(3)$:

$x_n$ | $y_n$ | $\frac{dy}{dx} = f(x_n)$ |

1 | 2 | 6 x 1$^2$ + 2 x 1 = 8 |

1.5 | 2 + 0.5 x 8 = 6 | 6 x (1.5)$^2$ + 2 x 1.5 = 16.5 |

2 | 6 + 0.5 x 16.5 = 14.25 | 6 x 2$^2$ + 2 x 2 = 28 |

2.5 | 6 + 0.5 x 28 = 28.25 | 6 x (2.5)$^2$ + 2 x 2.5 = 42.5 |

3 | 28.25 + 0.5 x 42.5 = 49.5 |

There are similar functions $x = \cosh u, y = \sinh u$ which have the same property for the rectangular hyperbola $x^2 - y^2 = 1$. If you pick any $u$ then you will get a point on that rectangular hyperbola.

Hyperbolic sine:$\sinh x = \frac{e^x - e^{-x}}{2}, x ∈ \mathbb R$ Domain: $x ∈ \mathbb R$ Range: $y ∈ \mathbb R$ |
Hyperbolic cosine:$\cosh x = \frac{e^x + e^{-x}}{2}, x ∈ \mathbb R$ Domain: $x ∈ \mathbb R$ Range: $y ∈ [1, \infty)$ |
Hyperbolic tangent:$\tanh x = \frac{\sinh x}{\cosh x} = \frac{e^{2x} - 1}{e^{2x} + 1}, x ∈ \mathbb R$ Domain: $x ∈ \mathbb R$ Range: $y ∈ (-1, 1)$ |

Hyperbolic secant:$\text{sech} \space x = \frac{1}{\cosh x} = \frac{2}{e^{x} + e^{-x}}, x ∈ \mathbb R$ Domain: $x ∈ \mathbb R$ Range: $y ∈ \mathbb R$ |
Hyperbolic cosecant:$\text{cosech} \space x = \frac{1}{\sinh x} = \frac{2}{e^{x} - e^{-x}}, x ∈ \mathbb R, x \ne 0$ Domain: $x ∈ [1, \infty)$ Range: $y ∈ \mathbb R$ |
Hyperbolic cotangent:$\coth x = \frac{1}{\tanh x} = \frac{e^{2x} + 1}{e^{2x} - 1}, x ∈ \mathbb R, x \ne 0$ Domain: $x ∈ (-1, 1)$ Range: $y ∈ \mathbb R$ |

$r$ and $\theta$ are like the modulus and argument in the Argand diagram.

There are multiple ways to write polar coordinates as you can keep adding $2\pi$ to each of the angles, or add $\pi$ to anoy of the angles and negate $r$.

$r^2=x^2+y^2$

$x=r\cos\theta$

$y=r\sin\theta$

e.g. Evaluate $\int^\infty_1 \frac{1}{x^2} \space dx$:

$$\int^\infty_1 \frac{1}{x^2} \space dx = \lim_{c\to \infty} \int^c_1 x^{-2} \space dx$$ $$=\lim_{c\to\infty} [-x^{-1}]^c_1$$ $$=\lim_{c\to\infty} (-\frac{1}{c} + 1)$$ $$= 0 + 1$$ $$= 1$$

If there is no limit, we say the integral does not have a value. There is no limit if it is oscillating or tends to infinity.

If there is a vertical asymptote then the integral is vertically unbounded and improper.

When an asymptote is in the middle of a region then split it into two integrals.

The volume of revolution formed by rotating a portion of a curve between $x = a$ and $x = b$ around the x-axis is given by: $$V = \pi \int^b_a y^2 \space dx$$ To resolve the y-axis instead of the x-axis, integrate $x^2$ with respect to $dy$.

Break the interval [a,b] into $n$ equal subdivisions: $$\Delta x = \frac{b-a}{n}$$ Choose $x_i$, i ∈ [1,n]. Then the average of the function can be given by: $$f_{avg} = \frac{f(x_1) + f(x_2) + ... + f(x_n)}{n}$$ But $n = \frac{b-a}{\Delta x}$ so: $$f_{avg} = \sum_{i=1}^n \frac{f(x_i)\Delta x}{b-a}$$ $$=\frac{1}{b-a}\sum_{i=1}^n f(x_i)\Delta x$$ E.g. Find the average value of $R(z) = \sin (2z) e^{1 - \cos (2z)}$ on the interval $[-\pi, \pi]$.

$$\frac{1}{\pi + \pi} \int^{-\pi}_\pi \sin (2z) e^{1- \cos (2z)} \space dz$$ $$\text{let } u = 1- \cos (2z)$$ $$dz = \frac{du}{2 \sin(2z)}$$ $$\frac{1}{2\pi} \int^{-\pi}_\pi \sin (2z) e^u \frac{du}{2 \sin (2z)}$$ $$\frac{1}{4\pi} \int^0_0 e^u du = 0$$

E.g. If the average value of a function $f(x) = x$ on [3,c] is 10, then find the value of c.

$$10 = \frac{1}{c-3} \int^c_3 x \space dx$$ $$10 = \frac{1}{c-3}{{x^2}\overwithdelims [] {2}} ^c _3$$ $$10 = \frac{1}{c-3}\biggr(\frac{c^2}{2} - \frac{3^2}{2}\biggr)$$ $$10 = \frac{c^2 - 9}{2(c-3)}$$ $$20c - 60 = c^2 - 9$$ $$c^2 - 20c + 51 = 0$$ $$(c-17)(c-3) = 0$$ $$\enclose{updiagonalstrike}{c = 3} \text{ or } c = 17$$

$$\frac{dy}{dx} = f(x)g(y)$$ $$\frac{1}{g(y)}\frac{dy}{dx} = f(x)$$ $$\int \frac{1}{g(y)} \space dy = \int f(x) \space dx$$

General solutions are when we are just given a differential equation so there is a family of equations which could be the final answer.

e.g. General solution to $\frac{dy}{dx} = 2$ is $y = 2x + C$

When you are given values for $y$ and $x$ you can find the particular solution from the general solution.

You can only reverse the product rule if the coefficient of y is the derivative of the coefficient of $\frac{dy}{dx}$.

Whatever term ends up in front of the $\frac{dy}{dx}$ will be on the front of the $y$ in the integral.

e.g. $$x^3\frac{dy}{dx} + 3x^2 y = \sin x$$ $$\frac{d}{dx}(x^3 y) = \sin x$$ $$x^3 y = \int \sin x dx = - \cos x + C$$ $$y = \frac{C - \cos x}{x^3}$$

To solve $\frac{dy}{dx} + Py = Q$, you can multiply through by the integrating factor, which produces an equation which you can then use with the reverse product rule trick.

Integrating Factor = $$e^{\int P \space dx}$$ This rule only works if the original coefficient of $\frac{dy}{dx}$ is 1. If it isn't, you must divide through by the coefficient of $\frac{dy}{dx}$.

We know the solution of $a\frac{dy}{dx} + by = 0$ is $y = Ae^{-\frac{b}{a}x}$. Assuming the solution of $a\frac{d^2y}{dx^2} + b\frac{dy}{dx} + cy = 0$ is similar, and of the form $Ae^{mx}$:

Let $y = Ae^{mx}$

Then $\frac{dy}{dx} = Ame^{mx}$ and $\frac{d^2y}{dx^2} = Am^2e^{mx}$

Thus $a\frac{d^2y}{dx^2} + b\frac{dy}{dx} + cy = aAm^2e^{mx} + bAme^{mx} + cA = 0$

$\therefore Ae^{mx} (am^2 + bm + c) = 0$

Since $Ae^{mx} \gt 0$, $am^2 + bm + c = 0$

The equation $am^2 + bm + c = 0$ is called the auxiliary equation, and if $m$ is a root of the auxiliary equation then $y = Ae^{mx}$ is a solution of the differential equation $a\frac{d^2y}{dx^2} + b\frac{dy}{dx} + cy = 0$ When the auxiliary equation has two real distinct roots $\alpha$ and $\beta$, the general solution of the differential equation is then $y = Ae^{\alpha x} + Be^{\beta x}$, where $A$ and $B$ are arbitrary constants. When the auxiliary equation has two equal roots $\alpha$, the general solution is $y = (A + Bx)e^{ax}$ If the auxiliary equation has two pure imaginary roots $\pm i\omega$, the general solution is $y = A\cos (\omega x) + B\sin (\omega x)$ where $A$ and $B$ are arbitrary complex constants. If the auxiliary equation has two complex roots $p \pm iq$, the general solution is $y = e^{px}(A \cos (qx) + B \sin (qx))$ where $A$ and $B$ are arbitrary complex constants.

- First you have to solve for when it is equal to 0. So for $a\frac{d^2y}{dx^2} + b\frac{dy}{dx} + cy = f(x)$, solve $a\frac{d^2y}{dx^2} + b\frac{dy}{dx} + cy = 0$. This is known as the Complementary Function.
- Then solve $a\frac{d^2y}{dx^2} + b\frac{dy}{dx} + cy = f(x)$ by doing the following.
- For a constant RHS, set $y = \lambda$. For a linear RHS, set $y = \lambda x + \mu$. For a quadratic RHS, set $y = \lambda x^2 + \mu x + \nu$. For an exponential RHS, set $y = Ae^x$. For a trigonometric RHS, e.g. $13 \sin 3x$, set $y = \lambda \sin 3x + \mu \cos 3x$.
- Then differentiate to find $\frac{dy}{dx}$ and $\frac{d^2y}{dx^2}$.
- Plug back in to the differential equation given.
- Then equate coefficients and solve for the necessary values of $\lambda$, $\mu$, $\nu$.
- Plug these values back into $y$. This is known as the Particular Integral.

- To find the general solution, $y = C.F. + P.I.$.

With boundary conditions (when you are given a point), you can find the constants of the C.F. First find the general solution and plug in $y$ and $x$. If you're given $\frac{dy}{dx}$ in the question, differentiate $y$ then plug in again. Then solve the simultaneous equations.

- Vertex: A point on a graph
- Edge: A line between vertices
- Weight: A number associated with an edge, representing something like distance or time
- Network: A graph with weights
- Directed Graph (Digraph): A graph where edges have directions
- Walk: A sequence of edges where the end of one edge is the start of the next (no jumps between vertices)
- Trail: A walk in which no edge is repeated
- Closed Trail: A trail which starts and ends at the same vertex
- Path: A trail where no vertex is repeated
- Cycle: A closed path (a path which starts and ends at the same vertex)
- Connected: A graph is connected if every vertex has an edge connecting it to another (so it has no isolated vertices)
- Multiple Edges: Where there is more than one edge between two vertices
- Loop: Where a vertex has an edge connecting it to itself
- Simple Graph: A graph with no multiple edges or loops
- Tree: A simple connected graph with no cycles
- Subgraph: A graph formed from some of the vertices and edges of another graph. Every edge must have a vertex at both ends but a vertex can be left isolated
- Spanning Tree: A subgraph, H, of a connected graph, G, is said to be a spanning tree of G if H is a tree and contains all the vertices of G. There may be many possible spanning trees for a connected graph
- Degree (Order): The degree of a vertex is the number of edges that join to it
- Semi-Eulerian: A trail which covers every edge
- If a graph has just two odd degree vertices, then it can be shown to be Semi-Eulerian

- Eulerian: A Semi-Eulerian graph which finishes at its starting vertex
- If a graph has no odd degree vertices, then it can be shown to be Eulerian

- Hamiltonian Cycle (A Tour): A cycle where no vertices are repeated
- Hamiltonian Graph: A graph which possesses a Hamiltonian cycle
- Adjacency Matrix (Incidence Matrix): A graph can be displayed in a table for computers to understand.

- A loop counts as 2
- A simple graph will only have ones and zeros in its adjacency matrix
- If you add up a row or column of an adjacency matrix you get the corresponding degree

- Distortion: Where the vertices and edges of a graph are moved around, but no connections are changed. Its adjacency matrix will stay the same
- Planar: A graph which can be distorted so no edges cross each other
- Face: If a graph is both connected and planar, the plane containing the graph can be divided up into regions, known as faces, which are bounded by edges. You also include the infinite face with no boundary (the outside of the graph)
- Euler's Formula is true for all connected planar graphs:F + V = E + 2

Faces + Vertices = Edges + 2 - Complete Graph: A simple graph where every vertex is connected to every other vertex. A complete graph with $n$ vertices is denoted $K_n$
- Complement (Inverse): The complement of a simple graph is obtained by adding the edges necessary to make a complete graph, then removing the original edges. This may leave isolated vertices
- Subdivision: A subdividion of a graph is obtained by inserting a new vertex into an edge one or more times
- Bipartite: A graph where vertices are split into two sets, with edges joining each vertex to at least one vertex from the other set, but none from its own set
- Complete Bipartite: A bipartite graph where every edge from each set is connected exactly once to every edge from the other set
- Isomorphic: Two graphs are isomorphic if one can be distorted to produce the other. They will have identical adjacency matrices, but the vertices may have been relabelled
- Kuratowski's Theorem: A graph is non-planar if and only if it contains a subgraph that is a subdivision of either $K_{3,3}$ or $K_5$
- $K_5$ and $K_{3,3}$ are shown below respectively

- $K_5$ and $K_{3,3}$ are shown below respectively

- Network: A weighted graph
- Node: A vertex on a network
- Arc: An edge on a network

Remove arcs in order of decreasing weight, ensuring the network remains connected. Arcs of the same weight may be arbitrarily chosen.

- Start with any node
- Add the arc leading to the nearest node (the node connected by the arc with the least weight).
- Add the arc leading (from any of the nodes collected so far) to the nearest uncollected node, and repeat.
- Stop once all nodes have been collected.

- Start with the shortest arc.
- Choose the next shortest arc, provided it doesn't create a cycle, and repeat. If two arcs are of equal length, either may be chosen.

A variable is random if the sum of the probabilities is 1.

e.g.

$x$ | 1 | 2 | 3 | 4 | 5 | 6 |

$P(X=x)$ | $\frac{1}{6}$ | $\frac{1}{6}$ | $\frac{1}{6}$ | $\frac{1}{6}$ | $\frac{1}{6}$ | $\frac{1}{6}$ |

If you have an n-sided fair die, then $$E(X) = \frac{n+1}{2}$$ $$Var(X) = \frac{n^2 - 1}{12} = \frac{(n+1)(n-1)}{12}$$