February 19, 2025 · 6:15 am

Finding a Recursive Rule for a Sequence (Year 12 Mathematics Applications)

How do we go about finding the rule for a first order linear recurrence relation?

Something like

    \begin{equation*}5, 7, 11, 19, ...\end{equation}

There isn’t a common difference (arithmetic sequence) or a common ratio (geometric sequence). Sometimes you can just see the rule, but an algorithm will be handy.

Let’s say our relationship is

(1)   \begin{equation*}T_{n+1}=bT_n+c, T_1=a\end{equation*}

Referring back to our sequence 5, 7, 11, 19, ..., we know

(2)   \begin{equation*}7=b\times5+c\end{equation*}

and

(3)   \begin{equation*}11=b\times7+c\end{equation*}

We can solve equation 2 and 3 simultaneously

equation 2- equation 3

    \begin{equation*}-4=-2b\end{equation}

Hence b=2

Substitute b=2 into equation 2

    \begin{equation*}7=2\times 5+c\end{equation}

    \begin{equation*}7=10+c\end{equation}

Hence c=-3

    \begin{equation*}T_{n+1}=2T_n-3, T_1=5\end{equation}

Let’s try to generalise

If T_{n+1}=bT_n+c, T_1=a, then

(4)   \begin{equation*}T_2=bT_1+c\end{equation*}

and

(5)   \begin{equation*}T_3=bT_2+c\end{equation*}

Equation 5 - equation 4

    \begin{equation*}T_3-T_2=(T_2-T_1)b\end{equation}

    \begin{equation*} b=\frac{T_3-T_2}{T_2-T_1}\end{equation}

Hence, b=\frac{T_{n+2}-T_{n+1}}{T_{n+1}-T_n}

Once you know b, substitute into either equation to find C.

Example

Find the recursive rule for the following

-8, -12, -20, -36, ...

    \begin{equation*}b=\frac{T_{n+2}-T_{n+1}}{T_{n+1}-T_n}=\frac{-20-(-12)}{-12-(-8)}=\frac{-8}{-4}=2\end{equation}

    \begin{equation*}-12=2\times-8+c\end{equation}

    \begin{equation*}-12=-16+c\end{equation}

Hence c=4 and T_{n+1}=2T_n+4, T_1=-8

It is also possible to find the rule using a Classpad (if it’s in the calculator section} by using an e-activity.

Leave a Comment

Filed under Sequences, Sequences, Year 12 Mathematics Applications

Tagged as ,

February 15, 2025 · 8:35 am

Counting Techniques

Four teachers decide to swap desks at work. How many ways can this be done if no teacher sits at their previous desk?
Mathematics Specialist Units 1&2 Cambridge

I like this question as it seems easy until you start thinking about it. I think the best approach is a tree diagram.

If we think of the four teachers as A, B, C and D. Then A can no longer sit in A, so the options are B, C and D for the first desk.

For the second desk, If B is in the first desk, then A, C or D could be in the second. If C is in the first desk, then A or D could be in the second (B can’t be in the same desk). If D is in the first desk, then A or C can be in the second desk.

And so on, leaving 9 possibilities

BADC
BCDA
BDAC
CADB
CDAB
CDBA
DABC
DCAB
DCBA

1 Comment

Filed under Counting Techniques, Tree Diagram, Year 11 Specialist Mathematics

Tagged as ,

February 4, 2025 · 9:17 am

Differentiating the Tangent Function

Remember tan(x)=\frac{sin(x)}{cos(x)}.

I use the quotient rule to differentiate f(x)=tan(x).

(1)   \begin{equation*}\frac{d}{dx}(\frac{f(x)}{g(x)})=\frac{f'(x)g(x)-g'(x)f(x)}{[g(x)]^2}\end{equation*}

If h(x)=tan(x)=\frac{sin(x)}{cos(x)} then from equation 1

(2)   \begin{equation*}h'(x)=\frac{cos(x)cos(x)-(-sin(x)sin(x))}{[cos(x)]^2}\end{equation*}

(3)   \begin{equation*}h'(x)=\frac{cos^2(x)+sin^2(x)}{cos^2(x)}\end{equation*}

Remember the Pythagorean identity

(4)   \begin{equation*}sin^2(x)+cos^2(x)=1\end{equation*}

Hence

    \begin{equation*}h'(x)=\frac{1}{cos^2(x)}=sec^2(x)\end{equation}

(5)   \begin{equation*}\frac{d}{dx}tan(x)=sec^2(x)\end{equation*}

Leave a Comment

Filed under Calculus, Differentiation, Differentiation, Identities, Quotient Rule, Trigonometry, Year 12 Mathematical Methods

Tagged as ,

February 3, 2025 · 9:11 am

Differentiating Trigonometric Functions

In the last post we looked at two trig limits:

(1)   \begin{equation*}\lim_{x \to 0}\frac{sin(x)}{x}=1\end{equation*}

(2)   \begin{equation*}\lim_{x \to 0}\frac{1-cos(x)}{x}=0\end{equation*}

We are going to use these two limits to differentiate sine and cosine functions from first principals.

    \begin{equation*}f(x)=sin(x)\end{equation}

    \begin{equation*}f'(x)=\lim\limits_{h \to 0}\frac{sin(x+h)-sin(x)}{h}\end{equation}

Use the trig identity

    \begin{equation*}sin(A+B)=sinAcosB+sinBcosA\end{equation}

    \begin{equation*}f'(x)=\lim\limits_{h \to 0}\frac{sin(x)cos(h)+sin(h)cos(x)-sin(x)}{h}\end{equation}

    \begin{equation*}f'(x)=\lim\limits_{h \to 0}(\frac{sin(x)(cos(h)-1)}{h}+\frac{sin(h)cos(x)}{h})\end{equation}

    \begin{equation*}f'(x)=sin(x)\lim\limits_{h \to 0}(\frac{(cos(h)-1)}{h}+cos(x)\lim\limits_{h \to 0}\frac{sin(h)}{h}\end{equation}

    \begin{equation*}f'(x)=sin(x)\lim\limits_{h \to 0}(\frac{-(-cos(h)+1)}{h}+cos(x)\lim\limits_{h \to 0}\frac{sin(h)}{h}\end{equation}

Evaluate the limits

    \begin{equation*}f'(x)=sin(x)\times 0+cos(x)\times (1)=cos(x)\end{equation}

Hence, \frac{d}{dx}sin(x)=cos(x).

Now we are going to do the same for f(x)=cos(x).

    \begin{equation*}f'(x)=\lim\limits_{h \to 0}\frac{cos(x+h)-cos(x)}{h}\end{equation}

Use the trigonometric identity

    \begin{equation*}cos(A+B)=cosAcosB-sinAsinB\end{equation}

    \begin{equation*}f'(x)=\lim\limits_{h \to 0}\frac{cos(x)cos(h)-sin(x)sin(h)-cos(x)}{h}\end{equation}

    \begin{equation*}f'(x)=\lim\limits_{h \to 0}\frac{cos(x)(cos(h)-1)-sin(x)sin(h)}{h}\end{equation}

    \begin{equation*}f'(x)=cos(x)\lim\limits_{h \to 0}\frac{-(1-cos(h))}{h}-sin(x)\lim\limits_{h \to 0}\frac{sin(h)}{h}\end{equation}

Evaluate the limits

    \begin{equation*}f'(x)=cos(x)\times(0)-sin(x)\times (1)=-sin(x)\end{equation}

Hence \frac{d}{dx} cos(x)=-sin(x)

1 Comment

Filed under Calculus, Differentiation, Identities, Trigonometry, Year 12 Mathematical Methods

Tagged as , , , , , ,

February 2, 2025 · 7:46 am

Trigonometric Limits

\lim\limits_{x \to 0}\frac{sin(x)}{x}=?

Unit Circle

Remember cos(x)=\frac{OA}{OB}=\frac{OA}{1}, hence OA=cos(x) and the co-ordinate of A is (cos(x), 0).

sin(x)=\frac{AB}{OB}=\frac{AB}{1}, hence AB=sin(x) and the co-ordinate of B is (cos(x), sin(x))

And from the definition of tan(x) we know D is the point (1, tan(x))

Consider the areas of triangle OAB, sector OBC, and triangle OCD.

We know from inspection of the above diagram that

Area OAB< Area OCB<Area OCD

Which means,

\frac{1}{2}b_1 h_1<\frac{1}{2}r^2x<\frac{1}{2}b_2 h_2

We can ignore all of the halves.

cos(x)sin(x)<x<(1)tan(x)

Remember tan(x)=\frac{sin(x)}{cos(x)}

cos(x)sin(x)<x<\frac{sin(x)}{cos(x)}

Divide everything by sin(x) (as we are in the first quadrant we know sin(x)>0, so we don’t need to worry about the inequality)

cos(x)<\frac{x}{sin(x)}<\frac{1}{cos(x)}

Invert everything and change the direction of the inequalities)

\frac{1}{cos(x)}>\frac{sin(x)}{x}>cos(x)

I am going to rewrite it as follows

cos(x)<\frac{sin(x)}{x}<\frac{1}{cos(x)}

because I like to use less thans rather than greater thans.

Now what happens as x tends to 0?

cos(0)=1

1<\frac{sin(x)}{x}<\frac{1}{1}

Hence by the squeeze theorem \lim\limits_{x \to 0}\frac{sin(x)}{x}=1

Now we know this limit, we are going to use it to find \lim\limits_{x \to 0}\frac{1-cos(x)}{x}

Multiply by \frac{1+cos(x)}{1+cos(x)}

\lim\limits_{x \to 0}\frac{1-cos(x)}{x}\times \frac{1+cos(x)}{1+cos(x)}

\lim\limits_{x \to 0}\frac{1-cos^2(x)}{x(cos(x)+1)}

\lim\limits_{x \to 0}\frac{sin^2(x)}{x(cos(x)+1)}

\lim\limits_{x \to 0}\frac{sin(x)}{x}\times sin(x)(cos(x)+1)}

\lim\limits_{x \to 0}\frac{sin(x)}{x}\times \lim\limits_{x \to 0}sin(x)(cos(x)+1)

If we evaluate the limits,

(1)(sin(0)(cos(0)+1)=1\times 0 \times 2=0

Hence, \lim\limits_{x \to 0}\frac{1-cos(x)}{x}=0

In the next post we are going to use these limits to differentiate sine and cosine functions.

1 Comment

Filed under Area, Area of Triangles (Sine), Calculus, Identities, Trigonometry, Year 12 Mathematical Methods

Tagged as ,

January 16, 2025 · 8:08 am

HSC Advanced 2024 Question 30

HSC Advanced 2024

Two circles have the same centre O. The smaller circle has a radius of 1 cm, while the larger has a radius of (x+1) cm. The circles enclose a region QRST, which is subtended by angle of {\theta} at O, as shaded.

The area of QRST is A cm2, where A is a constant and A>0

Let P cm be the perimeter of QRST

(a) By finding expressions for the area and perimeter of QRST show that P(x)=2x+\frac{2A}{x}

(b) Show that if the perimeter is minimised, then {\theta} must be less than 2.

(a) A=\frac{1}{2}\theta((x+1)^2-1^2)
A=\frac{1}{2}\theta(x^2+2x)
2A=\theta x^2 +2x \theta
\frac{2A}{x}=\theta x +2\theta

P=\theta(1)+\theta(1+x)+2x
P=2\theta +\theta x +2x
P=\frac{2A}{x}+2x

I like it when the first part requires the student to show something and the second part has them use it (that way they can still do the second part even if they couldn’t do the first part).

(b) \frac{dP}{dx}=2-\frac{2A}{x^2}
0=2-\frac{2A}{x^2}
x=\sqrt{A}

\frac{d^2P}{dx^2}=\frac{4A}{x^3}
Both x and A are greater than zero, therefore \frac{d^2P}{dx^2}>0 and x=\sqrt{A} is a minimum.

Substitute x=\sqrt{A} into the Area formula
2A=A\theta+2\sqrt{A}\theta
\theta=\frac{2A}{A+2\sqrt{A}}
\theta=\frac{2A}{\sqrt{A}(\sqrt{A}+2)}
\theta=\frac{2\sqrt{A}}{\sqrt{A}+2}

Now \frac{2\sqrt{A}}{\sqrt{A}+2}<\frac{2\sqrt{A}}{\sqrt{A}}
Hence \theta<\frac{2\sqrt{A}}{\sqrt{A}}
and \theta<2

Leave a Comment

Filed under Differentiation, Optimisation, Uncategorized

Tagged as , ,

January 14, 2025 · 7:52 am

Deriving the Quotient Rule for Differentiation

Like we did for the product rule, we are going to derive the differentiating rule for functions in the form y=\frac{f(x)}{g(x)}.

Something like, y=\frac{x^2+3x+2}{x^3-1}

Remember the first principals limit

\lim_{\limits h \to 0}\frac{f(x+h)-f(x)}{h}

If y=\frac{f(x)}{g(x)}, then

y'=\lim_{\limits h \to 0}\frac{\frac{f(x+h)}{g(x+h)}-\frac{f(x)}{g(x)}}{h}

Find a common denominator for the numerator (i.e. g(x+h)g(x))

y'=\lim_{\limits h \to 0}\frac{\frac{f(x+h)g(x)-f(x)g(x+h)}{g(x+h)g(x)}}{h}

To make things a bit easier I am going to multiply by \frac{1}{h} rather than having h as the denominator

y'=\lim_{\limits h \to 0}\frac{f(x+h)g(x)-f(x)g(x+h)}{g(x+h)g(x)} \times \frac{1}{h}

Now I am going to add and subtract f(x)g(x)

y'=\lim_{\limits h \to 0}\frac{f(x+h)g(x)-f(x)(g(x)+f(x)g(x)-f(x)g(x+h)}{g(x+h)g(x)} \times \frac{1}{h}

Factorise

y'=\lim_{\limits h \to 0}\frac{g(x)(f(x+h)-f(x))+f(x)(g(x)-g(x+h))}{g(x+h)g(x)} \times \frac{1}{h}

Change the sign in the middle

y'=\lim_{\limits h \to 0}\frac{g(x)(f(x+h)-f(x))-f(x)(g(x+h)-g(x))}{g(x+h)g(x)} \times \frac{1}{h}

Separate the limits

y'=g(x)\lim_{\limits h \to 0}\frac{\frac{f(x+h)-f(x)}{h}}{g(x+h)g(x)}-f(x)\lim_{\limits h \to 0}\frac{\frac{g(x+h)-g(x)}{h}}{g(x+h)g(x)}

which simplifies to

y'=g(x)\frac{f'(x)}{g(x)g(x)}-f(x)\frac{g'(x)}{g(x)g(x)}

y'=\frac{f'(x)g(x)-g'(x)f(x)}{[g(x)]^2}

In words

The derivative of the top times the bottom take the derivative of the bottom times the top all over the bottom squared

Example

y=\frac{x^2+3x+2}{x^3-1}

y'=\frac{(2x+3)(x^3-1)-3x^2(x^2+3x+2)}{(x^3-1)^2}

y'\frac{2x^4-2x+3x^3-3-3x^4-9x^3-6x^2}{(x^3-1)^2}

y'=\frac{-x^4-6x^3-6x^2-2x-3}{(x^3-1)^2}

Exam questions usually specify no simplifying.

2 Comments

Filed under Calculus, Differentiation, Quotient Rule, Year 12 Mathematical Methods

January 11, 2025 · 9:56 am

Deriving the Product Rule for Differentiation

In my previous post we looked at the Chain Rule for Differentiation, this post is on the Product Rule. Differentiating a function in the form y=f(x)\times g(x).

For example, y=(3x^3+2x-1)(x^4+2x^2)

Remember differentiating from first prinicpals:

f'(x)=\lim_{\limits h \to 0} \frac{f(x+h)-f(x)}{h}

y=f(x)g(x)

\frac{dy}{dx}=\lim_{\limits h\to 0}\frac{f(x+h)g(x+h)-f(x)g(x)}{h}}

\small{ \frac{dy}{dx}=\lim_{\limits h \to 0} \frac{f(x+h)g(x+h)-g(x+h)f(x)+g(x+h)f(x)-f(x)g(x)}{h}}

By subtracting and then adding g(x+h)f(x) we haven’t changed the limit, but it means we can do some factorising.

\frac{dy}{dx}=\lim_{\limits h \to 0}\frac{g(x+h)(f(x+h)-f(x))+f(x)(g(x+h)-g(x))}{h}

\small{\frac{dy}{dx}=\lim_{\limits h \to 0}g(x+h)\lim_{\limits h \to 0}\frac{f(x+h)-f(x)}{h}+\lim_{\limits h \to 0}f(x)\lim_{\limits h \to 0}\frac{g(x+h)-g(x)}{h}}

When we evaluate the limits

\frac{dy}{dx}=g(x)f'(x)+f(x)g'(x)

Example

Find the derivative of y=(3x^3+2x-1)(x^4+2x^2)

I remember the rule in words ‘derivative of the first times the second plus the derivative of the second times the first’.

y'=(9x^2+2)(x^4+2x^2)+(4x^3+4x)(3x^3+2x-1)

y'=9x^6+18x^4+2x^4+4x^2+12x^6+8x^4-4x^3+12x^4+8x^2-4x

y'=21x^6+40x^4-4x^312x^2-4x

Most exam questions have ‘don’t simplify’, so the first line of working above would be enough.

Onto the Quotient Rule.

2 Comments

Filed under Calculus, Differentiation, Product Rule, Year 12 Mathematical Methods

January 9, 2025 · 9:40 am

Deriving the Chain Rule for Differentiation

How to differentiate something in the form y=[f(x)]^n

For example, y=(3x^2-2x+6)^5, we could expand the expression, but the Chain Rule provides a quick and easy method.

Differentiate y=[f(x)]^n

Let u=f(x), then y=u^n

We want to find \frac{dy}{dx}, but \frac{dy}{dx}=\frac{dy}{du}\times \frac{du}{dx}

They’re not fractions, but limits of fractions, but they work like fractions.

\frac{du}{dx}=f'(x) and \frac{dy}{du}=nu^{n-1}

Therefore, \frac{dy}{dx}=f'(x)\times nu^{n-1}

Replace u with f(x)

(1)   \begin{equation*}\frac{dy}{dx}=n[f(x)]^{n-1}f'(x)\end{equation*}

What about a function in the form y=f(g(x))?

We’re going to follow the same process.

Let u=g(x), then y=f(u)

\frac{du}{dx}=g'(x) and \frac{dy}{du}=f'(u)

Therefore \frac{dy}{dx}=f'(u)g'(x)

(2)   \begin{equation*}\frac{dy}{dx}=f'(g(x))g'(x) \end{equation*}

Equations 1 and 2 are versions of the Chain Rule.

Example

Find the derivative of y=(3x^2-2x+6)^5

    \begin{equation*}\frac{dy}{dx}=5(3x^2-2x+6)^4\times (6x-2)\end{equation}

    \begin{equation*}\frac{dy}{dx}=5(6x-2)(3x^2-2x+6)^4\end{equation}

    \begin{equation*}\frac{dy}{dx}=10(3x-1)(3x^2-2x+6)\end{equation}

Next time we are going to look at the Product Rule.

1 Comment

Filed under Calculus, Chain Rule, Differentiation, Year 12 Mathematical Methods

Tagged as , , ,

January 6, 2025 · 3:58 am

An Imaginary Tale – The story of i

An Imaginary Tale – The Story of \sqrt{-1}

I bought a second hand copy of this book from Abe Books (it’s possible to find cheap maths books this way).

Here’s the blurb …

Today complex numbers have such widespread practical use–from electrical engineering to aeronautics–that few people would expect the story behind their derivation to be filled with adventure and enigma. In An Imaginary Tale , Paul Nahin tells the 2000-year-old history of one of mathematics’ most elusive numbers, the square root of minus one, also known as i . He recreates the baffling mathematical problems that conjured it up, and the colorful characters who tried to solve them. In 1878, when two brothers stole a mathematical papyrus from the ancient Egyptian burial site in the Valley of Kings, they led scholars to the earliest known occurrence of the square root of a negative number. The papyrus offered a specific numerical example of how to calculate the volume of a truncated square pyramid, which implied the need for i . In the first century, the mathematician-engineer Heron of Alexandria encountered I in a separate project, but fudged the arithmetic; medieval mathematicians stumbled upon the concept while grappling with the meaning of negative numbers, but dismissed their square roots as nonsense. By the time of Descartes, a theoretical use for these elusive square roots–now called “imaginary numbers”–was suspected, but efforts to solve them led to intense, bitter debates. The notorious i finally won acceptance and was put to use in complex analysis and theoretical physics in Napoleonic times. Addressing readers with both a general and scholarly interest in mathematics, Nahin weaves into this narrative entertaining historical facts and mathematical discussions, including the application of complex numbers and functions to important problems, such as Kepler’s laws of planetary motion and ac electrical circuits. This book can be read as an engaging history, almost a biography, of one of the most evasive and pervasive “numbers” in all of mathematics.

It took me a long time to read this book – according to storygraph I started it in August (12th to be exact). I really enjoyed this book, but I do think you need to understand maths. I enjoyed working through all of the different formulae and examples. I have put some on this blog

I found the historical aspects very interesting.

Minus times minus is plus

The reason for this we need not discuss

W. H. Auden

Maybe I should use the above quote with my Year 8s who are just starting on their negative number journey.

This book covers quite complex (pun intended) ideas – particularly in Chapter 6 Wizard Mathematics and Chapter 7 The Nineteenth Century, Cauchy, and the Beginning of Complex Function Theory.

If you can do algebra and a bit of calculus and complex numbers interest you, then I think this book is for you.

A review

Leave a Comment

Filed under Book Review