Racquel Sanderson

December 10, 2024 · 6:00 am

Puzzle Page 2

If $x^2-3x+1=0$ , then find $x^5+\frac{1}{x^5}$ .

My first thought was to solve for $x$ , but it doesn’t factorise easily, and I didn’t want to find the fifth power of an expression involving surds $(x=\frac{3\pm \sqrt{5}}{2})$ , there must be an easier way.

Because $x\neq0$ , we can divide by $x$

$\begin{equation*}x-3+\frac{1}{x}=0\end{equation}$

Hence

(1) $\begin{equation*}x+\frac{1}{x}=3\end{equation*}$

What is the expansion of $(x+\frac{1}{x})^5$ ?

Using the binomial expansion theorem

$\begin{equation*}(x+\frac{1}{x})^5=x^5+5x^4(\frac{1}{x})+10x^3(\frac{1}{x^2})+10x^2(\frac{1}{x^3})+5x(\frac{1}{x^4})+\frac{1}{x^5}\end{equation}$

$\begin{equation*}(x+\frac{1}{x})^5=x^5+\frac{1}{x^5}+5(x^3+\frac{1}{x^3})+10(x+\frac{1}{x})\end{equation}$

Therefore

(2) $\begin{equation*}x^5+\frac{1}{x^5}=(x+\frac{1}{x})^5-5(x^3+\frac{1}{x^3})-10(x+\frac{1}{x})\end{equation*}$

Let’s do it again for $x^3+\frac{1}{x^3}$

$\begin{equation*}(x+\frac{1}{x})^3=x^3+3x^2(\frac{1}{x})+3x(\frac{1}{x^2})+\frac{1}{x^3}\end{equation}$

(3) $\begin{equation*}x^3+\frac{1}{x^3}=(x+\frac{1}{x})^3-3(x+\frac{1}{x})\end{equation*}$

Substitute $3$ into $2$

$\begin{equation*}x^5+\frac{1}{x^5}=(x+\frac{1}{x})^5-5((x+\frac{1}{x})^3-3(x+\frac{1}{x}))-10(x+\frac{1}{x})\end{equation}$

Remember $x+\frac{1}{x}=3$

Therefore

$\begin{equation*}x^5+\frac{1}{x^5}=3^5-5(3^3)+15\times3-10\times3\end{equation}$

$\begin{equation*}x^5+\frac{1}{x^5}=243-135+45-30=123\end{equation}$

This would be a good extension question for students learning the binomial expansion theorem. We also use this technique for trigonometric identities using complex numbers.

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Filed under Algebra, Binomial Expansion Theorem, Puzzles

December 3, 2024 · 6:54 am

Puzzle Page 1

If $\frac{a+b+2c}{a+b-c}=\frac{31}{15}$ , what does $\frac{a+b}{c}$ equal?

$\begin{equation*}15(a+n+2c)=31(a+b-c)\end{equation}$

$\begin{equation*}15a+15b+30c=31a+31b-31c)\end{equation}$

$\begin{equation*}61c=16a+16b)\end{equation}$

$\begin{equation*}\frac{61}{16}=\frac{a+b}{c}\end{equation}$

Two positive numbers are such that their difference, their sum, and their product are in the ratio $2:5:21$ . What is the smaller of the two numbers?

Let $x$ and $y$ be the two numbers. Then

(1) $\begin{equation*}x-y=2k\end{equation*}$

(2) $\begin{equation*}x+y=5k\end{equation*}$

(3) $\begin{equation*}xy=21k\end{equation*}$

Add equation $1$ and $2$ together to eliminate the $y$

$\begin{equation*}2x=7k\end{equation}$

(4) $\begin{equation*}x=\frac{7k}{2}\end{equation*}$

From $2$ $=5k-x$ , substitute for $y$ into equation $3$ .

(5) $\begin{equation*}x(5k-x)=21k\end{equation*}$

Substitute $x=\frac{7k}{2}$ into equation $5$ .

$\begin{equation*}\frac{7k}{2}(5k-\frac{7k}{2})=21k\end{equation}$

$\begin{equation*}\frac{35k^2}{2}-\frac{49k^2}{4}=21k\end{equation}$

$\begin{equation*}\frac{70k^2}{4}-\frac{49k^2}{4}=\frac{84k}{4}\end{equation}$

$\begin{equation*}21k^2-84k=0\end{equation}$

$\begin{equation*}21k(k-4)=0\end{equation}$

Hence, $k=0$ or $k=4$ .

When $k=4$ , $x=\frac{7\times 4}{2}=14$ and $y=5\times 4-14=6$

Therefore the smaller number is $6$ .

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Filed under Algebra, Puzzles, Ratio, Solving Equations

Tagged as problem solving, ratio, Solving Equations

November 29, 2024 · 8:11 am

Interesting Equation

I think this one is doing the rounds, I first saw it here.

$\begin{equation*}2^x3^{x^2}=6\end{equation}$

$x=1$ is the obvious answer, $2^1\times 3^1=6$ , but are there more answers?

This was my approach

$\begin{equation*}ln(2^x3^{x^2})=ln(6)\end{equation}$

$\begin{equation*}ln(2^x)+ln(3^{x^2})=ln(6)\end{equation}$

$\begin{equation*}xln(2)+x^2ln(3)-ln(6)=0\end{equation}$

$\begin{equation*}ln(3)x^2+ln(2)x-ln(6)=0\end{equation}$

A quadratic equation.

Hence,

$\begin{equation*}x=\frac{-ln(2)\pm\sqrt{(ln(2))^2-4(ln(3))(ln(6))}}{2ln(3)}\end{equation}$

I then used my calculator

Hence $x=1$ 0r $x=-1.631$

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Filed under Algebra, Index Laws, Interesting Mathematics, Quadratics, Solving

Tagged as exponentials, index laws, Interesting maths, quadratic solving

November 26, 2024 · 7:07 am

Infinite Product Expansion of cos (x)

Remember

(1) $\begin{equation*}cos(x)=1-\frac{1}{2!}x^2+\frac{1}{4!}x^4-\frac{1}{6!}x^6+...\end{equation*}$

We know that $cos(x)=0$ for odd integer multiples of $\frac{\pi}{2}$ , i.e. $\frac{\pi}{2}, \frac{3\pi}{2}, ...$ , which is $\frac{(2n-1)\pi}{2}$ for $n\neq 0$

Hence,

$\begin{equation*}0=1-\frac{1}{2!}x^2+\frac{1}{4!}x^4-\frac{1}{6!}x^6+...\end{equation}$

for $x=\frac{(2n-1)\pi}{2}, n>0$

We can factorise our $cos(x)$ expansion

$\begin{equation*}(1-\frac{x^2}{r_1})(1-\frac{x^2}{r_2})...\end{equation}$

We know $r_1=\frac{\pi}{2}, r_2=\frac{3\pi}{2}, ...$

$\begin{equation*}cos(x)=(1-\frac{x^2}{(\frac{\pi}{2})^2})(1-\frac{x^2}{(\frac{3\pi}{2})^2})...(1-\frac{x^2}{(\frac{(2n-1)\pi}{2})^2})\end{equation}$

$\begin{equation*}cos(x)=\Pi_{n=1}^{\infty}(1-\frac{x^2}{(\frac{(2n-1)\pi}{2})^2})\end{equation}$

$\begin{equation*}cos(x)=\Pi_{n=1}^{\infty}(1-\frac{4x^2}{(2n-1)^2\pi^2})\end{equation}$

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Filed under Factorising, Identities, Infinite Product Expansion, Interesting Mathematics, Polynomials, Trigonometry

Tagged as infinite product expansion of cos(x), trigonometry

November 20, 2024 · 6:57 am

More Integration

I went down a rabbit hole while reading An Imaginary Tale by Paul J Nahin and I decided I wanted to do this…

$\begin{equation*}\int_0^1{x^x dx}\end{equation}$

$\begin{equation*}x^x=e^{ln(x^x)}\end{equation}$

$\begin{equation*}x^x=e^{xln(x)}\end{equation}$

The power series expansion of $e^x$ is

$\begin{equation*}e^x=1+x+\frac{1}{2!}x^2+\frac{1}{3!}x^3+\frac{1}{4!}x^4+...\end{equation}$

$\begin{equation*}\therefore e^{xln(x)}=1+xln(x)+\frac{1}{2!}(xln(x))^2+\frac{1}{3!}(xln(x))^3+...\end{equation}$

$\begin{equation*}\therefore e^{xln(x)}=\Sigma_{n=0}^{\infty}(\frac{1}{n!}(xln(x))^n)\end{equation}$

Hence $\int_0^1{x^x dx}=\int_0^1(\Sigma_{n=0}^{\infty}(\frac{1}{n!}(xln(x))^n dx)$

$\begin{equation*}=\Sigma_{n=0}^{\infty}(\frac{1}{n!}\int_0^1xln(x))^n dx)\end{equation}$

Let’s consider the integral

(1) $\begin{equation*}\int_0^1 x^n(ln(x))^n dx\end{equation*}$

Let $u=ln(x)$ then $\frac{du}{dx}=\frac{1}{x}$ and $dx=x du$ where $x=e^u$

When $x=0, u=-\infty$ and when $x=1, u=0$

(2) $\begin{equation*}\int_0^1 x^n(ln(x))^n dx=\int_{-\infty}^0 e^{nu}u^ne^u du\end{equation*}$

(3) $\begin{equation*}\int_0^1 x^n(ln(x))^n dx=\int_{-\infty}^0 e^{(n+1)u}u^n du\end{equation*}$

Integrate by parts using the tabular method.

Sign	Differentiate	Integrate
+	$u^n$	$e^{(n+1)u}$
–	$nu^{n-1}$	$\frac{e^{(n+1)u}}{n+1}$
+	$n(n-1)u^{n-2}$	$\frac{e^{(n+1)u}}{(n+1)^2}$
–	$n(n-1)(n-2)u^{n-3}$	$\frac{e^{(n+1)u}}{(n+1)^3}$
+	$\frac{n!}{(n-4)!}u^{n-4}$	$\frac{e^{(n+1)u}}{(n+1)^4}$
	$\vdots$	$\vdots$
	$\frac{n!}{(n-n)!}u^{n-n}$	$\frac{e^{(n+1)u}}{(n+1)^{n}}$
$(-1)^n$	$0$	$\frac{e^{(n+1)u}}{(n+1)^{n+1}}$

When we substitute $u=-\infty$ or $u=0$ the differentiation column is zero except for $\frac{n!}{(n-n)!}u^{n-n}$ , which is $n!$ ,

Thus $\int_0^1 x^n(ln(x))^n dx=\frac{e^{(n+1)u}}{(n+1)^{n+1}}}]_{-\infty}^0$

$\begin{equation*}=n!\times\frac{e^0}{(n+1)^{n+1}}-0\end{equation}$

$\begin{equation*}=\frac{n!}{(n+1)^{n+1}}\end{equation}$

Now we just need to think about the sign.

$\begin{equation*}=(-1)^n\frac{n!}{(n+1)^{n+1}}\end{equation}$

The integral is now

$\int_0^1{x^x dx}=(\Sigma_{n=0}^{\infty}(\frac{1}{n!}( (-1)^n\frac{n!}{(n+1)^{n+1}})$

So $\int_0^1{x^x dx}=\Sigma_{n=0}^{\infty}( (-1)^n\frac{1}{(n+1)^{n+1}}$

Let’s work out some partial sums

$\int_0^1{x^x dx}=0.778343$

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Filed under Definite, Integration, Integration by Parts, Tabular Integration

Tagged as Integration, integration by parts, tabular integration

November 17, 2024 · 5:23 am

Integration by Parts using the Tabular Method

(1) $\begin{equation*}\int_0^1x^2e^xdx\end{equation*}$

I am going to do this integral in two ways; the traditional method and the tabular method.

Traditional Method

Remember $\int{u dv}=u\times v-\int{v du}$

Let $u=x^2$ and $dv=e^x$

Then $du=2x$ and $v=\int{e^x dx}=e^x$

$\begin{equation*}\int_0^1{x^2e^x dx}=x^2 \times e^x ]_0^1-\int_0^1{e^x 2x dx}\end{equation}$

Now we need to do integration by parts on $\int_0^1{e^x 2x dx}$

Let $u=2x$ and $dv=e^x$

Then $du=2$ and $v=e^x$

$\begin{equation*}\int_0^1{x^2e^x dx}=x^2 \times e^x ]_0^1-(2x\times e^x]_0^1-\int_0^1{e^x 2 dx})\end{equation}$

$\begin{equation*}\int_0^1{x^2e^x dx}=x^2 \times e^x ]_0^1-(2x\times e^x]_0^1-(2e^x )]_0^1)\end{equation}$

$\begin{equation*}\int_0^1{x^2e^x dx}=e-(2e-(2e-2))\end{equation}$

$\begin{equation*}\int_0^1{x^2e^x dx}=e+2\end{equation}$

Tabular Integration

Similar to before, select a $u$ and a $dv$ , $u=x^2$ and $dv=e^x$

Sign	D(ifferentiate)	I(ntegrate)
+	$x^2$	$e^x$
–	$2x$	$e^x$
+	$2$	$e^x$
–	$0$	$e^x$

Stop when the differentiating column reaches zero.

Then we multiply diagonally

$(+x^2)(e^x)+(-2x)(e^x)+(+2e^x)$

$=x^2e^x-2xe^x+2e^x]_0^1$

$=e-2e+2e-(0-0-2)$

$=e+2$

It is only worth using this method if integration by parts is required more than once. Also, the $u$ has to eventually differentiate to $0$ .

Let’s try another one

(2) $\begin{equation*}\int{x^3cos(2x) dx}\end{equation*}$

Let $u=x^3$ and $dv=cos(2x)$

Sign	D	I
+	$x^3$	$cos(2x)$
–	$3x^2$	$\frac{1}{2}sin(2x)$
+	$6x$	$-\frac{1}{4}cos(2x)$
–	$6$	$-\frac{1}{8}sin(2x)$
+	$0$	$\frac{1}{16}cos(2x)$

$\int{x^3cos(2x) dx}=(x^3)(\frac{1}{2}sin(2x))+(-3x^2)(-\frac{1}{4}cos(2x))+(6x)(-\frac{1}{8}sin(2x))+(-6)(\frac{1}{16}cos(2x))+c$

$=\frac{x^3}{2}sin(2x)+\frac{3x^2}{4}cos(2x)-\frac{3x}{4}sin(2x)-\frac{3}{8}cos(2x)+c$

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Filed under Integration, Integration by Parts, Tabular Integration

Tagged as Calculus, integration by parts, tabular integration

November 14, 2024 · 4:19 am

Vectors – Closest Approach (2D)

At 10am, object $M$ travelling with constant velocity $(20, 10)$ km/h is sighted at the point with position vector $(-100, 100)$ km. At 11am object $N$ travelling with constant velocity $v=(x,y)$ km/h is sighted at the point with position vector $(-20,-50)$ km respectively. Use a scalar product method to determine $v$ given that the two objects were closest together at a distance of $40$ km at 4pm.

OT Lee Mathematics Specialist Year 11

At 4pm $M$ is at the point with position vector $(-100+6\times 20, -100+6\times 10)=(20, -40)$

and $N$ is at the point with position vector $(-20+5\times x, -50+5\times y)=(-20+5x, -50+5y)$

We know the distance between $M$ and $N$ at 4pm is $40$ km.

Hence,

$\begin{equation*}40^2=(20-(-20+5x))^2+(-40-(-50+5y))^2\end{equation}$

$\begin{equation*}40^2=(40-5x)^2+(10-5y)^2\end{equation}$

$\begin{equation*}1600=25x^2-400x+1600+25y^2-100y+100\end{equation}$

(1) $\begin{equation*}0=x^2+y^2-16x-4y+4\end{equation*}$

In the diagram below, I have found the position vector of $N$ relative to $M$ $(_Nr_M)$ and the velocity of $N$ relative to $M$ $(N_v_M)$

We know that when $_Nr_M\cdot_Nv_M=0$ $M$ and $N$ are the closest distance apart.

$\begin{equation*}(-40+5x,-10+5y)\cdot(x-20,y-10)=0\end{equation}$

$\begin{equation*}(-40+5x)(x-20)+(-10+5y)(y-10)\end{equation}$

$\begin{equation*}5x^2+5y^2-140x-60y+900=0\end{equation}$

(2) $\begin{equation*}x^2+y^2-28x-12y+180=0\end{equation*}$

Two equations and two unknowns which we can solve simultaneously. Both equations are circles.

Equation $(1)$ becomes

(3) $\begin{equation*}(x-8)^2+(y-2)^2=64\end{equation*}$

and equation $(2)$ becomes

(4) $\begin{equation*}(x-14)^2+(y-6)^2=52\end{equation*}$

From equation $(3)$

$\begin{equation*}x=\sqrt{64-(y-2)^2}+8\end{equation}$

We will worry about the negative version later.

Substitute for $x$ into equation $(4)$

$\begin{equation*}(\sqrt{64-(y-2)^2}+8-14)^2+(y-6)^2=52\end{equation}$

$\begin{equation*}(\sqrt{64-(y-2)^2}-6)^2+(y-6)^2=52\end{equation}$

$\begin{equation*}64-(y-2)^2-12\sqrt{64-(y-2)^2}+36+y^2-12y+36=52\end{equation}$

$\begin{equation*}136-y^2+4y-4+y^2-12y-12\sqrt{64-(y-2)^2}=52\end{equation}$

$\begin{equation*}-12\sqrt{64-(y-2)^2}=-80+8y\end{equation}$

$\begin{equation*}\sqrt{64-(y-2)^2}=\frac{20}{3}-\frac{2}{3}y\end{equation}$

Square both sides of the equation

$\begin{equation*}64-(y-2)^2=\frac{400}{9}-\frac{80}{9}y+\frac{4}{9}y^2\end{equation}$

$\begin{equation*}60-y^2+4y=\frac{400}{9}-\frac{80}{9}y+\frac{4}{9}y^2\end{equation}$

$\begin{equation*}0=\frac{13}{9}y^2-\frac{116}{9}y-\frac{140}{9}\end{equation}$

$\begin{equation*}0=13y^2-116y-140\end{equation}$

$\begin{equation*}0=13y^2-130y+14y-140\end{equation}$

$\begin{equation*}0=13y(y-10)+14(y-10)\end{equation}$

$\begin{equation*}0=(y-10)(13y+14)\end{equation}$

(5) $\begin{equation*}\therefore y=10, y=-\frac{14}{13}\end{equation*}$

Substitute $y=10$ into $x=\sqrt{64-(y-2)^2}+8$

$\begin{equation*}x=\sqrt{64-(y-2)^2}+8\end{equation}$

(6) $\begin{equation*}x=8\end{equation*}$

Substitute $y=-\frac{14}{13}$ into $x$

$\begin{equation*}x=\sqrt{64-(-\frac{14}{13}-2)^2}+8\end{equation}$

(7) $\begin{equation*}x=\frac{200}{13}\end{equation*}$

Now we need to consider the negative version of $x$ . If you work through (like I did above) you end with the same equation for $y$ .

Hence our two values for $v$ are $v=(8,10)$ or $v=(\frac{200}{13},-\frac{14}{13})$ .

Would someone be expected to do this in an exam? I hope not, but I think its worth doing.

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Filed under Closest Approach, Vectors, Year 11 Specialist Mathematics

November 7, 2024 · 9:51 am

Mathematics Applications – Counting can be Tricky Sometimes (off by one error)

Sequences are part of the Year 12 Mathematics Applications course and sometimes it’s tricky to work out which terms the question requires.

For example, ATAR 2020 Question 11

Judith monitors the water quality in her garden pond at the same time everyday. She likes to maintain the concentration of algae between 200 and 250 unites per 100 litres (L). Her measurements show that the concentration increases daily according to the recursive rule

$C_{n+1}=1.025C_n$ where $C_1=200$ units per 100 L (the minimum concentration)

When the concentration gets above the 250 units per 100 L limit, she treats the water to bring the concentration back to the minimum 200 units per 100 l.

(a) If Judith treated the water on Sunday 6 December 2020, determine

(i) the concentration on Wednesday, 9 December 2020.
(ii) the day when she next treated the water.

(b) During the first week of January 2021, Judith monitored the water and recorded the following readings

Day	1	2	3	4	5	6	7
Concentration (C)	200	206	212.28	218.55	225.10	231.85	238.81

(i) Determine the revised recursive rule.
(ii) If she treated the water on 10 January and went on holiday until 20 January, when she next treated the water, calculate the concentration of the water on her return. Assuming the recursive rule from (b)(i) is used.

I get my students to count on their fingers to ensure they get the correct term or day.

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Filed under Sequences, Sequences, Uncategorized, Year 12 Mathematics Applications

Tagged as off by one error, Year 12 Mathematics Applications

November 2, 2024 · 11:45 am

Effect of Function Transformations on Integration

My year 12 Mathematical Methods students have questions like this

Given that $f(x)$ is continuous everywhere and that $\int_{4}^{10} f(x) dx=-10$ , find:

(a) $\int_{4}^{10}2x +f(x) dx$

(b) $\int_{5}^{11} f(x-1) dx$

(c) $\int_{1}^{3} f(3x+1) dx$

(d) $\int_{-10}^{-4} -f(-x) dx$

(e) $\int{10}^{22} f(\frac{x-2}{2}) dx$

(f) $\int_{-3}^{-9} f(1-x) dx$

OT Lee Mathematics Methods Textbook Ex 8.3 question 6

For the most part these questions aren’t too difficult, but the horizontal dilations cause issues.

\int_{4}^{10} 2x dx +\int_{2}^{10} f(x) dx

Once you get the hang of it, you can skip the change of variable and multiply the value of the definite integral by the scale factor of the horizontal dilation (only if the integration bounds are also changed).

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Filed under Definite, Integration, Uncategorized, Year 12 Mathematical Methods