Racquel Sanderson

October 7, 2024 · 7:43 am

Arithmetic Sequence

I did this question with on of my year 11 students. I think the algebra and the subscripts can be a bit tricky.

If $T_m=n$ and $T_n=m$ , then prove that $T_{m+n}=0$ . Here where $T_n$ and $T_m$ are terms of an arithmetic sequence.
Mathematics Methods Units 1&2 – Exercise 15B Question 19

\begin{equation*}n=a+(m-1)d\end{equation*}

As you can see from equation $(6), T_{m+n}=0$

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Filed under Algebra, Arithmetic, Sequences, Year 11 Mathematical Methods

Tagged as Arthimetic, sequences, year 11 mathematical methods

October 1, 2024 · 7:31 am

Using De Moivre’s Theorem for Trigonometric Identities

We are going to use De Moivre’s theorem to prove trigonometric identities.

Remember, De Moivre’s Theorem

If $z=r(cos(\theta)+isin(\theta))$ , then $z^n=r^n(cos(n\theta)+isin(n\theta))$

Or a shorter version $z=rcis(\theta)$ , then $z^n=r^ncis(n\theta)$

Now, let $z=cos(\theta)+isin(\theta)$ , find $z+\frac{1}{z}$

$z+z^{-1}=cos(\theta)+isin(\theta)+cos(-\theta)+isin(-\theta)$

Remember $cos(\theta)=cos(\theta)$ and $sin(-\theta)=-sin(\theta)$

$z+\frac{1}{z}=cos(\theta)+isin(\theta)+cos(\theta)-isin(\theta)$

$z+\frac{1}{z}=2cos(\theta)$

It is the same for $z^n+\frac{1}{z^n}$

$z^n+z^{-n}=cos(n\theta)+isin(n\theta)+cos(-n\theta)+isin(-n\theta)$

$z^n+\frac{1}{z^n}=2cos(n\theta)$

We can do something similar with sine.

$z-\frac{1}{z}=cos(\theta)+isin(\theta)-(cos(-\theta)+isin(-\theta))$

$z-\frac{1}{z}=cos(\theta)+isin(\theta)-(cos(\theta)-isin(\theta))$

$z-\frac{1}{z}=cos(\theta)+isin(\theta)-cos(\theta)+isin(\theta)$

$z-\frac{1}{z}=2isin(\theta)$

Hence $z^n-\frac{1}{z^n}=2isin(n\theta)$

Let’s find an identity for $cos(3\theta)$

$cos(3\theta)=\frac{1}{2}(z^3+\frac{1}{z^3})$

$=\frac{1}{2}(z^3+\frac{1}{z^3}+3z^2\times\frac{1}{z}+3z\times\frac{1}{z^2}-3z^2\times\frac{1}{z}-3z\times\frac{1}{z^2})$

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

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

$=\frac{1}{2}(2cos(\theta))^3-3(2cos(\theta)))$

$=\frac{1}{2}(8cos^3(\theta)-6cos(\theta))$

$=4cos^3(\theta)-3cos(\theta)$

$\therefore cos(3\theta)=4cos^3(\theta)-3cos(\theta)$

And $sin(3\theta)$ ?

$sin(3\theta)=\frac{1}{2i}(z^3-\frac{1}{z^3})$

$=\frac{1}{2i}(z^3-\frac{1}{z^3}-3z^2\times\frac{1}{z}+3z\times\frac{1}{z^2}+3z^2\times\frac{1}{z}-3z\times\frac{1}{z^2}$

$=\frac{1}{2i}((z-\frac{1}{z})^3+3z-\frac{3}{z})$

$=\frac{1}{2i}(2isin(\theta))^3+3(z-\frac{1}{z}))$

$=\frac{1}{2i}(-8isin^3(\theta)+6isin(\theta))$

$=-4sin^3(\theta)+3sin(\theta)$

$\therefore sin(3\theta)=3sin(\theta)-4sin^3(\theta)$

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Filed under Complex Numbers, Identities, Trig Identities, Trigonometry

Tagged as complex numbers, de moivre, trig identities

September 29, 2024 · 10:18 am

Intersecting Circles

Two circles of radius $L$ and $2L$ intersect as shown. What is the area of the shaded region?

From Professor Povey’s Perplexing Problems

My plan is to find the sum of the area of the two segments (see below).

Construct triangles

The diagonal of the square (the pink line above) has length

$=\sqrt{(2L)^2+(2L)^2}=\sqrt{8L^2}=2\sqrt{2}L$

From the pink triangle in the above diagram, I am going to find the angles using the cosine rule.

$cos\theta=\frac{L^2+(2\sqrt{2}L)^2-(2L)^2}{2(L)(2\sqrt{2}L)}$

$cos\theta=\frac{5L^2}{4\sqrt{2}L^2}$

$cos\theta=\frac{5}{4\sqrt{2}}$

$\theta=0.487$

$cos\alpha=\frac{(2L)^2+(2\sqrt{2}L)^2-L^2}{(2\sqrt{2}L)(2L)}$

$cos\alpha=\frac{11L^2}{8\sqrt{2}L^2}$

$cos\alpha=\frac{11}{8\sqrt{2}}$

$\alpha=0.236$

The green quadrilateral is a kite, which means the diagonals are perpendicular.

This means the segment angles are $2\theta$ and $2\alpha$ (because the triangles are isosceles and the diagonal is perpendicular to the base of the triangles).

Area of green segment

$A=\frac{1}{2}r^2(\theta-sin\theta)$

$A=\frac{1}{2}L^2(0.9734-sin(0.9734))=0.0733L^2$

Area of yellow segment

$A=\frac{1}{2}4L^2(0.4721-sin(0.4721))=0.0035$

Total Area = $0.0733L^2+0.0035L^2=0.108L^2$

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Filed under Finding an area, Geometry, Professor Povey

Tagged as finding an unknown shaded area, geometry, professor povey

September 24, 2024 · 3:30 am

Optimisation

An optimisation question from the 2019 ATAR Mathematics Methods exam.

I always like optimisation questions. There is a nice process to follow:

Find the function to optimise (in terms of one variable).
Find the stationary points.
Find the nature of the stationary points.
Find the maximum or minimum.

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Filed under Differentiation, Optimisation, Year 12 Mathematical Methods

Tagged as Calculus, differentiation, Optimisation

September 23, 2024 · 3:11 am

Cats and Dogs

In my town 10% of the dogs think they are cats and 10% of the cats think they are dogs. All the other cats and dogs are perfectly normal. When all the cats and dogs in my town were rounded up and subjected to a rigorous test, 20% of them thought they were cats. What percentage of them really were cats?
Hamilton Olympiad 2003 B4 – The Ultimate Mathematical Challenge

Let $x$ be the number of cats and $y$ be the number of dogs.
Then $0.9x+0.1y$ think they are cats.
But we also know 20% of the total think they are cats.
$0.2(x+y)$
Therefore, $0.9x+0.1y=0.2(x+y)$
$0.9x+0.1y=0.2x+0.2y$
$0.7x=0.1y$
$7x=y$
Percentage of cats is $\frac{x}{x+y}\times100$
Substitute $7x$ for $y$
$\frac{x}{x+7x}\times100=\frac{x}{8x}\times100=\frac{1}{8}\times100=12.5%$
$\therefore 12.5$ % of the animals are cats

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Filed under Algebra, Arithmetic, Percentages, Simplifying fractions, UK Mathematics Challenge

Tagged as Fractions, percent, uk maths challenge

September 19, 2024 · 9:43 am

Solving Cubic Functions

I have been thinking about cubics a bit lately because some of my students are solving and then sketching cubics. Plus I am reading An Imaginary Tale by Paul Nahin, which talks about solving cubics and complex numbers.

Cubics must have at least one real root. If one of the roots is a rational number, then we can use the Factor and Remainder Theorem.

For example,

Solve $2x^3-3x^2-3x+2=0$

But what if it is not factorisable?

For example,

Solve $2x^3+5x^2-2x+4=0$

How many roots does this equation have?

We could find the derivative and find out how many stationary points the function has.

$f'(x)=6x^2+10x-2$

This is a quadratic function. Find the discriminant to determine the number of roots.

$\Delta=b^2-4ac=100-4(6)(-2)=148$

As $\Delta>0$ , there are two stationary points, which means we could have 1, 2 (one root is repeated) or three roots, depending on if the function crosses the $x-$ axis between stationary points. So not much use.

We could try the discriminant of a cubic.

$\Delta=18abcd-4b^3d+b^2c^2-4ac^2-27a^2d^2$

$\Delta=18(2)(5)(-2)(4)-4(5^3)(4)+(5^2)(-2)^2-4(2)(-2)^2-27(2)^2(4)^2=-5100$

The discriminant is negative so there is one real root.

From my reading, we need to turn the cubic into a depressed cubic (cubics of the form $x^3+px+q=0$ ).

We can do this by using a change of variable.

We can then use Cardano’s formula

$x=\sqrt[3]{-\frac{q}{2}+\sqrt{(-\frac{q}{2})^2+(\frac{p}{3})^3}}+\sqrt[3]{-\frac{q}{2}-\sqrt{(-\frac{q}{2})^2+(\frac{p}{3})^3}}$

$p=-\frac{37}{12}$ and $q=\frac{431}{108}$
$\frac{p}{3}=\frac{-37}{36}$
$\frac{q}{2}=\frac{431}{216}$
$(\frac{431}{216})^2+(\frac{-37}{36})^3=\frac{139}{48}$
$t=\sqrt[3]{\frac{-431}{216}+\sqrt{\frac{139}{48}}}+\sqrt[3]{\frac{-431}{216}-\sqrt{\frac{139}{48}}}$
$t=-2.21095...$
$\therefore x=-2.21095...-\frac{5}{6}=-3.044$

We can see from the sketch below that there is only one solution and it is about $-3$ .

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Filed under Cubics, Factorising, Polynomials, Solving

Tagged as Cardano, Cubics, Depressed Cubic, Factorising, Polynomials, Solving

September 14, 2024 · 9:45 am

Sum and Product of the Roots of Polynomials

There is a relationship between the sum and product of the roots of a polynomial and the co-efficient of the polynomial.

Let’s start with a quadratic.

The general form for a quadratic (polynomial of degree 2) is

$y=ax^2+bx+c$

Use the quadratic equation formula to find the roots

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

Hence the roots are

$x_1=\frac{-b+\sqrt{b^2-4ac}}{2a}$ and $x_2=\frac{-b-\sqrt{b^2-4ac}}{2a}$

Sum of the roots:

$\frac{-b+\sqrt{b^2-4ac}}{2a}+\frac{-b-\sqrt{b^2-4ac}}{2a}=\frac{-2b}{2a}=\frac{-b}{a}$

Product of the roots:

$\frac{-b+\sqrt{b^2-4ac}}{2a}\times\frac{-b-\sqrt{b^2-4ac}}{2a}$

$\frac{b^2}{4a^2}-\frac{b^2-4ac}{4a^2}$

$\frac{4ac}{4a^2}$

$\frac{a}{c}$

Let’s move to a cubic function.

The general equation is $f(x)=ax^3+bx^2+cx+d$

Let’s say the roots of this cubic are $\alpha, \beta, \gamma$

Then $ax^3+bx^2+cx+d=a(x-\alpha)(x-\beta)(x-\gamma)$

$=a(x^2-\beta x - \alpha x+\alpha\beta)(x-\gamma)$

$=a(x^2-(\alpha+\beta)x+\alpha\beta)(x-\gamma)$

$=a(x^3-\gamma x^2-x^2(\alpha+\beta)+\gamma(\alpha+\beta)x-\alpha\beta\gamma)$

$=a(x^3-(\alpha+\beta+\gamma)x^2+(\alpha\beta+\alpha\gamma+\beta\gamma)x-\alpha\beta\gamma)$

The sum of the roots

$\alpha+\beta+\gamma=\frac{-b}{a}$

The product of the roots

$\alpha\beta\gamma=\frac{-d}{a}$

Also, it can be handy to know

$\alpha\beta+\alpha\gamma+\beta\gamma=\frac{c}{a}$

We can extend the method we used for finding the sum and product of the roots of cubic to polynomials of greater degree.

If the four roots of a quartic are $\alpha, \beta, \gamma$ and $\delta$ , and the general equation is $ax^4+bx^3+cx^2+dx+e$ , then

$\alpha+\beta+\gamma+\delta=\frac{-b}{a}$

$\alpha\beta+\alpha\gamma+\alpha\delta+\beta\gamma+\beta\delta+\gamma\delta=\frac{c}{a}$

$\alpha\beta\gamma+\alpha\beta\delta+\alpha\gamma\delta+\beta\gamma\delta=\frac{-d}{a}$

$\alpha\beta\gamma\delta=\frac{e}{a}$

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Filed under Polynomials, Sum and Product of Roots

Tagged as Polynomials, product, sum, sum and product of polynomials

September 7, 2024 · 9:31 am

Geometry Problem

This problem is from Geometry Snacks by Ed Southall and Vincent Pantaloni – it’s a great book.

Two squares are constructed such that three vertices are collinear as shown. Find the value of the marked angle.

I started by marking in the right angles. And I added the diagonal of the larger square (pink line).

Because there are right angles at $O$ and $P$ , we know there is a circle, which has the diagonal of the square as its diameter (see second image below).