HSC 2016 MX2 Integration Marathon (archive) (2 Viewers)

porcupinetree · Feb 5, 2016

Re: MX2 2016 Integration Marathon

InteGrand said:
$\nondent \textbf{A NEW QUESTION}$

$\noindent Let $x\geq 0$. Let $n$ be a positive integer.$

$\noindent (i) Show that$

$$-x^n \leq \frac{1}{1+x} -\left(1 - x + x^2 -x^3 +\cdots +(-1)^{n-1} x^{n-1} \right) \leq x^n$.$

$\noindent (ii) Hence, use integration to show that$

$$ -\frac{1}{n+1} \leq \ln 2 -\left(1-\frac{1}{2} +\frac{1}{3} -\frac{1}{4} +\cdots + (-1)^{n-1} \cdot \frac{1}{n} \right) \leq \frac{1}{n+1}$.$

$\noindent (iii) Justifying your answer, deduce the value of the infinite sum $1-\frac{1}{2} + \frac{1}{3} -\frac{1}{4} +\cdots$.$

Part (i) reminds me of question 16b (I think?) in the 2014 HSC

Anyhow, I bags answering this Q; will post up my sol'n soon

InteGrand · Feb 5, 2016

Re: MX2 2016 Integration Marathon

porcupinetree said:
Part (i) reminds me of question 16b (I think?) in the 2014 HSC

Anyhow, I bags answering this Q; will post up my sol'n soon

Yeah, this question is similar to that Q., I just changed it slightly to yield the infinite sum for a different number.

It's not too hard a Q. since it's a guided one, but I put this in mainly to give a HSC-style Q. since we haven't had one of those for a while (so we probably scared off some 2016'er lurkers

) as well as for the infinite series.

leehuan · Feb 5, 2016

Re: MX2 2016 Integration Marathon

RealiseNothing said:
Just found it, question 7)a) from 2007.

http://www.boardofstudies.nsw.edu.au/hsc_exams/exam-papers-2007/index5.html

They give barely any guidance and it's EXACTLY what seanie posted.

By considering the first two derivatives is plenty of guidance in my eyes. Literally when I was doing that question the moment they said derivatives I was like pfft, even though I already had a fair idea on where to start with part (i).

And also what InteGrand said - those inequalities were given. True, this is a marathon, but it doesn't mean everyone can figure out everything themselves.

Furthermore, I would not have been able to infer an integral from THAT expression without being shown it as a mere MX2 student.

leehuan · Feb 5, 2016

Re: MX2 2016 Integration Marathon

InteGrand said:
Yeah, this question is similar to that Q., I just changed it slightly to yield the infinite sum for a different number. It's not too hard a Q. since it's a guided one, but I put this in mainly to give a HSC-style Q. since we haven't had one of those for a while (so we probably scared off some 2016'er lurkers ) as well as for the infinite series.

It rang a huge bell in my head.

seanieg89 · Feb 5, 2016

Re: MX2 2016 Integration Marathon

A followup from Integrand's question along the same theme:

$Find an expression for $\frac{1}{1+x^2}-\sum_{k=0}^n (-1)^k x^{2k}$.$

$Hence show that for $0\leq x \leq 1$ we have $\arctan(x)=\sum_{k=0}^\infty (-1)^k \frac{x^{2k+1}}{2k+1}$ .$

$Use the above to find an infinite series for $\pi$.$

BONUS:

$If you actually try to use this series to approximate $\pi$, you will notice that we need to go incredibly far into the series to obtain a decent degree of accuracy. Can you explain why this is the case? Can you think of a way to use what has been proven in this question to obtain a faster converging series for $\pi$ ?$

porcupinetree · Feb 5, 2016

Re: MX2 2016 Integration Marathon

InteGrand said:
$\nondent \textbf{A NEW QUESTION}$

$\noindent Let $x\geq 0$. Let $n$ be a positive integer.$

$\noindent (i) Show that$

$$-x^n \leq \frac{1}{1+x} -\left(1 - x + x^2 -x^3 +\cdots +(-1)^{n-1} x^{n-1} \right) \leq x^n$.$

$\noindent (ii) Hence, use integration to show that$

$$ -\frac{1}{n+1} \leq \ln 2 -\left(1-\frac{1}{2} +\frac{1}{3} -\frac{1}{4} +\cdots + (-1)^{n-1} \cdot \frac{1}{n} \right) \leq \frac{1}{n+1}$.$

$\noindent (iii) Justifying your answer, deduce the value of the infinite sum $1-\frac{1}{2} + \frac{1}{3} -\frac{1}{4} +\cdots$.$

hopefully the image works properly this time

Paradoxica · Feb 5, 2016

Re: MX2 2016 Integration Marathon

seanieg89 said:
A followup from Integrand's question along the same theme:

$Find an expression for $\frac{1}{1+x^2}-\sum_{k=0}^n (-1)^k x^{2k}$.$

$Hence show that for $0\leq x \leq 1$ we have $\arctan(x)=\sum_{k=0}^\infty (-1)^k \frac{x^{2k+1}}{2k+1}$ .$

$Use the above to find an infinite series for $\pi$.$

BONUS:

$If you actually try to use this series to approximate $\pi$, you will notice that we need to go incredibly far into the series to obtain a decent degree of accuracy. Can you explain why this is the case? Can you think of a way to use what has been proven in this question to obtain a faster converging series for $\pi$ ?$

$\noindent $1$ lies at the edge of convergence for the Maclaurin series for $\tan^{-1}{x}$, and diverges anywhere outside that radius of convergence. In order to accelerate the convergence towards $\pi$, set $x$ equal to a number whose magnitude is relatively close to zero, but spit out a rational multiple of $\pi$. For example, $\sqrt{2}-1$ has an inverse of $\frac{\pi}{12}$ and $\sqrt{2}-1$ is fairly close to $\frac{1}{7}$. By picking values of $x$ that get closer and closer to the origin, we greatly improve the convergence towards $\pi$. The interested reader may try as an exercise to obtain values of $x$ that converge to $\pi$ at ever increasing rates until it becomes impractical.$

$$\noindent Another way to obtain rapidly converging values is to use complex numbers and find gaussian integers that multiply to give something whose argument is a rational multiple of $\pi$. This suggests the use of the Machin-like formulae for $\pi$, which have infinitely many solutions over the rational numbers, and are very popular in calculating the digits of $\pi.$

Paradoxica · Feb 6, 2016

Re: MX2 2016 Integration Marathon

$\noindent The integral $U_n$ is defined as $\int_0^\pi \frac{\cos{n x}}{5-4\cos{x}} $d$x $ for all non-negative $n. \\$a) Show $U_{n+1}+U_{n-1}-\frac{5}{2}U_n = 0 \\$b) Determine integers $A$ and $B$ such that $AU_0 -BU_1 = \pi\\$c) Evaluate $U_0$, and hence, $U_1.\\$ d) Using the reductive recurrence relation from a), determine the first few values of $U_n .\\$e) Conjecture the closed form of $U_n$. Prove your conjecture, by mathematical induction, or any other means you see fit.$

KingOfActing · Feb 6, 2016

Re: MX2 2016 Integration Marathon

Paradoxica said:
$\noindent The integral $U_n$ is defined as $\int_0^\pi \frac{\cos{n x}}{5-4\cos{x}} $d$x $ for all non-negative $n. \\$a) Show $U_{n+1}+U_{n-1}-\frac{5}{2}U_n = 0 \\$b) Determine integers $A$ and $B$ such that $AU_0 -BU_1 = \pi\\$c) Evaluate $U_0$, and hence, $U_1.\\$ d) Using the reductive recurrence relation from a), determine the first few values of $U_n .\\$e) Conjecture the closed form of $U_n$. Prove your conjecture, by mathematical induction, or any other means you see fit.$

I've done all of these, let me type them up in Latex now.

KingOfActing · Feb 6, 2016

Re: MX2 2016 Integration Marathon

$\\ \begin{align*}\text{a)} \int_0^\pi \frac{\cos((n+1)x)+\cos((n-1)x)-\frac52\cos(nx)}{5-4\cos(x)}\,dx &= \int_0^\pi \frac{2\cos(nx)\cos(x)-\frac52\cos(nx)}{5-4\cos(x)}\,dx \\&= \int_0^\pi -2\cos(nx)\,dx = 0\end{align*}$

$\begin{align*} &\text{b)}\int_0^\pi \frac{A-B\cos{x}}{5-4\cos{x}}\,dx = \pi\\&\text{Taking A = 5 and B = 4 gives } \int_0^\pi\,dx = \pi \end{align*}$

$\begin{align*} &\text{c)}\\&U_0 = \int_0^\pi \frac{1}{5-4\cos{x}} \\&\text{Let }t = \tan{\frac{x}{2}}\\U_0 = \int_0^\infty \frac{2}{1+9t^2}\,dt = \frac{2}{9}\int_0^\infty \frac{1}{t^2+\frac{1}{9}} = \frac{2}{3}\left (\lim_{t \to \infty}\left( \arctan(3t) \right )-\arctan(0)\right ) = \frac{2}{3}\left(\frac{\pi}{2} - 0 \right ) = \frac{\pi}{3}\end{align*}$

$\text{d) Using the recurrance relation, the first few values are: }\\U_0 = \frac{\pi}{3}, U_1 = \frac{\pi}{6}, U_2 = \frac{\pi}{12}, U_3 = \frac{\pi}{24}$

$\begin{align*}&\text{e) Conjecture: }U_n = \frac{\pi}{3\times2^n}\\&\text{Proof by induction: The base cases }n = 0, 1 \text{ are clearly true. Assume this holds for } n = k \text{ and } n = k + 1 \\&U_{k+2} = \frac{5}{2}U_{k+1}-U_k = \frac{5}{2}\left (\frac{\pi}{3\times2^{n+1}}\right ) - \frac{\pi}{3\times2^n}\\&= \frac{5\pi}{3\times2^{n+2}}-\frac{\pi}{3\times2^n} = \frac{\pi}{3\times2^{n+2}}\\&\text{Hence by the principle of mathematical induction, the conjecture is true for all natural numbers }n \geq 0\end{align*}$

KingOfActing · Feb 6, 2016

Re: MX2 2016 Integration Marathon

Oh whoops, I forgot to show the working to get U_1. It's not hard, just solve for it in AU_0 + BU_1 = pi

InteGrand · Feb 6, 2016

Re: MX2 2016 Integration Marathon

$\noindent \textbf{NEW FOR-FUN QUESTION (NOT HSC STYLE)}$

$\noindent Find $\int _{\frac{1}{4}} ^{\frac{1}{2}} \left \lfloor \ln \left \lfloor \frac{1}{x}\right \rfloor \right \rfloor \text{ d}x$.$

$\noindent \textbf{Note.} The symbol $\lfloor \cdot \rfloor$ refers to the \textit{floor function}, which basically means you round \textsl{down} its argument to the nearest integer. Further reading:$

https://en.wikipedia.org/wiki/Floor_and_ceiling_functions

Paradoxica · Feb 6, 2016

Re: MX2 2016 Integration Marathon

InteGrand said:
$\noindent \textbf{NEW FOR-FUN QUESTION (NOT HSC STYLE)}$

$\noindent Find $\int _{\frac{1}{4}} ^{\frac{1}{2}} \left \lfloor \ln \left \lfloor \frac{1}{x}\right \rfloor \right \rfloor \text{ d}x$.$

$\noindent \textbf{Note.} The symbol $\lfloor \cdot \rfloor$ refers to the \textit{floor function}, which basically means you round \textsl{down} its argument to the nearest integer. Further reading:$

https://en.wikipedia.org/wiki/Floor_and_ceiling_functions

$$\noindent From $x = \frac{1}{4}$ to $x = \frac{1}{3} $, we have $\left \lfloor \frac{1}{x} \right \rfloor = 3$, and from $x = \frac{1}{3}$ to $x = \frac{1}{2}$, we have $\left \lfloor \frac{1}{x} \right \rfloor = 2$. Thus, for $\left \lfloor \frac{1}{x} \right \rfloor = 3$, the integrand is equal to 1 and for $\left \lfloor \frac{1}{x} \right \rfloor = 2$, the integrand is equal to 0. Thus, the integral is simply a rectangle with height 1 and base $\frac{1}{12}$, which gives us the final answer of $\int _{\frac{1}{4}} ^{\frac{1}{2}} \left \lfloor \ln \left \lfloor \frac{1}{x} \right \rfloor \right \rfloor $d$x = \frac{1}{12}$

InteGrand · Feb 6, 2016

Re: MX2 2016 Integration Marathon

Paradoxica said:
$$\noindent From $x = \frac{1}{4}$ to $x = \frac{1}{3} $, we have $\left \lfloor \frac{1}{x} \right \rfloor = 3$, and from $x = \frac{1}{3}$ to $x = \frac{1}{2}$, we have $\left \lfloor \frac{1}{x} \right \rfloor = 2$. Thus, for $\left \lfloor \frac{1}{x} \right \rfloor = 3$, the integrand is equal to 1 and for $\left \lfloor \frac{1}{x} \right \rfloor = 2$, the integrand is equal to 0. Thus, the integral is simply a rectangle with height 1 and base $\frac{1}{12}$, which gives us the final answer of $\int _{\frac{1}{4}} ^{\frac{1}{2}} \left \lfloor \ln \left \lfloor \frac{1}{x} \right \rfloor \right \rfloor $d$x = \frac{1}{12}$

Correct!

Paradoxica · Feb 7, 2016

Re: MX2 2016 Integration Marathon

$\noindent By considering the substitution $u = x^{\frac{2}{3}}$ and subsequently, $u = \frac{1}{v}$, evaluate $\int_0^\infty \frac{\sqrt[3]{x}}{1+x^2}$d$x$ WITHOUT partial fractions.$

porcupinetree · Feb 7, 2016

Re: MX2 2016 Integration Marathon

Paradoxica said:
$\noindent By considering the substitution $u = x^{\frac{2}{3}}$ and subsequently, $u = \frac{1}{v}$, evaluate $\int_0^\infty \frac{\sqrt[3]{x}}{1+x^2}$d$x$ WITHOUT partial fractions.$

porcupinetree · Feb 7, 2016

Re: MX2 2016 Integration Marathon

Next question, one of my favourites:

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

leehuan · Feb 7, 2016

Re: MX2 2016 Integration Marathon

porcupinetree said:
Next question, one of my favourites:

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

This has probably been asked before

InteGrand · Feb 8, 2016

Re: MX2 2016 Integration Marathon

porcupinetree said:
Next question, one of my favourites:

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

$\noindent Let $I=\int \frac{x^{2}}{1+x^{4}}\text{ d}x$, so$

$\begin{align*} 2I &= \int \frac{x^2+1}{1+x^{4}}\text{ d}x + \int \frac{x^2-1}{1+x^{4}}\text{ d}x \\ &= \int \frac{1+\frac{1}{x^2}}{\frac{1}{x^2}+x^2}\text{ d}x + \int \frac{1-\frac{1}{x^2}}{\frac{1}{x^2}+x^{2}}\text{ d}x\\ &= \int \frac{1+\frac{1}{x^2}}{\left ( x-\frac{1}{x} \right )^2 + 2}\text{ d}x + \int \frac{1-\frac{1}{x^2}}{\left ( x+\frac{1}{x} \right )^2 -2}\text{ d}x \\ \Rightarrow I&= \frac{1}{2}\int \frac{\frac{\mathrm{d} }{\mathrm{d} x}\left ( x-\frac{1}{x} \right )}{\left ( x-\frac{1}{x} \right )^2 + 2}\text{ d}x + \frac{1}{2} \int \frac{\frac{\mathrm{d} }{\mathrm{d} x}\left ( x+\frac{1}{x} \right )}{\left ( x+\frac{1}{x} \right )^2 -2}\text{ d}x \\ \Rightarrow I &= \frac{1}{2\sqrt{2}}\tan^{-1} \left ( \frac{x-\frac{1}{x}}{\sqrt{2}} \right ) +\frac{1}{4\sqrt{2}}\ln \left | \frac{x+\frac{1}{x}-\sqrt{2}}{x+\frac{1}{x}+\sqrt{2}} \right | +\mathcal{C}. \end{align*}$

leehuan · Feb 8, 2016

Re: MX2 2016 Integration Marathon

InteGrand said:
$\noindent Let $I=\int \frac{x^{2}}{1+x^{4}}\text{ d}x$, so$

$\begin{align*} 2I &= \int \frac{x^2+1}{1+x^{4}}\text{ d}x + \int \frac{x^2-1}{1+x^{4}}\text{ d}x \\ &= \int \frac{1+\frac{1}{x^2}}{\frac{1}{x^2}+x^2}\text{ d}x + \int \frac{1-\frac{1}{x^2}}{\frac{1}{x^2}+x^{2}}\text{ d}x\\ &= \int \frac{1+\frac{1}{x^2}}{\left ( x-\frac{1}{x} \right )^2 + 2}\text{ d}x + \int \frac{1-\frac{1}{x^2}}{\left ( x+\frac{1}{x} \right )^2 -2}\text{ d}x \\ \Rightarrow I&= \frac{1}{2}\int \frac{\frac{\mathrm{d} }{\mathrm{d} x}\left ( x-\frac{1}{x} \right )}{\left ( x-\frac{1}{x} \right )^2 + 2}\text{ d}x + \frac{1}{2} \int \frac{\frac{\mathrm{d} }{\mathrm{d} x}\left ( x+\frac{1}{x} \right )}{\left ( x+\frac{1}{x} \right )^2 -2}\text{ d}x \\ \Rightarrow I &= \frac{1}{2\sqrt{2}}\tan^{-1} \left ( \frac{x-\frac{1}{x}}{\sqrt{2}} \right ) +\frac{1}{4\sqrt{2}}\ln \left | \frac{x+\frac{1}{x}-\sqrt{2}}{x+\frac{1}{x}+\sqrt{2}} \right | +\mathcal{C}. \end{align*}$

Now that I think about it.

The original function has no discontinuity at x=0 yet the primitive does. Does this matter?

HSC 2016 MX2 Integration Marathon (archive) (2 Viewers)

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