Should Q2 also say "leading up to and including"? The way it's written would make the answer just be \(a+b\).
(Similarly for Q3.)
Also, is it meant to be assumed that Q1 begins counting from 1?
Is this thread intended to go beyond VCE maths and into university level stuff?
Is this thread intended to go beyond VCE maths and into university level stuff?
Preferably, it'd be great if we could make it solvable at a high school level (even though it may require some level of advanced understanding), however university level is fine as well - though it just limits the demographic of those who can solve the problem.
It would be awesome if everyone who posts could delineate their target audiences from now on :)
Hey everyone. (It was me that asked to post a problem lol).Not saying this is the best approach and I needed some external help for one tiny bit in the middle but here goes. Parts a)-c):
Perhaps if we can get some regulars here, we could rotate posting problems? (Just a thought).
Anyway, here is my problem :D
(https://i.imgur.com/beCfG4r.png)
require(expm)
identity <- diag(4)
kk <- matrix(rep(1,16), nrow=4, ncol=4)
for(A in 0:1) {
for (B in 0:1) {
for (C in 0:1) {
for (D in 0:1) {
for (E in 0:1) {
for (FF in 0:1) {
for (G in 0:1) {
for (H in 0:1) {
for (I in 0:1) {
for (J in 0:1) {
for (K in 0:1) {
for (L in 0:1) {
for (M in 0:1) {
for (N in 0:1) {
for (O in 0:1) {
for (P in 0:1) {
found <- TRUE
mm <- matrix(c(A,B,C,D,E,FF,G,H,I,J,K,L,M,N,O,P), nrow=4, ncol=4)
mm4 <- (mm %^% 4) %% 2
mm3 <- (mm %^% 3) %% 2
for (i in 1:4) {
for (j in 1:4) {
if (mm4[i,j] != identity[i,j]) found <- FALSE
if (mm[i,j] + mm3[i,j] != 1) found <- FALSE
}
}
if (found) stop("stopped!")
}
}
}
}
}
}
}
}
}
}
}
}
}
}
}
}
...
Not saying this is the best approach and I needed some external help for one tiny bit in the middle but here goes.
...
Haven't had a proper go at part d) yet - trying to stick to 2x2 matrices didn't work out easily enough for me. Can probably take advantage of these constructions to produce an example of a 4x4 matrix though. (Otherwise there's a 2x2 that I just missed the first time round.)
...
I did do some math before attempting hard-coding though. Basically I saw that if \(n\) were even, we would have \(M^2 = M^{-2}\). This would give \(M^{-1} = M^3\) and also \(M^4 = I\). So I wrote a program to check each 4x4 matrix over \( \mathbb{Z}_2\), and told it to stop if it found a matrix satisfying \( \boxed{M^4= I} \) and \( \boxed{M + M^3 = K} \) simultaneously.
...
If you think this one is too easy, tryShouldn't the upper boundary be \( \frac\pi2\) here if you're trying to allude to a classic?
Shouldn't the upper boundary be \( \frac\pi2\) here if you're trying to allude to a classic?I mean you probably could still define this integral perfectly the way I originally wrote it, but you're right; I really shouldn't make this more complicated than it already is. Fixed.
(In any case it can't be \(\pi\) because "\(\sec\) is negative in the second quadrant", so \(\ln (\sec x) \) would be non-real)
May be worth mentioning that the first one can be generalised. It turns out that \( \int_0^{\pi/2} \frac{\sin^A x}{\cos^A x + \sin^A x}dx = \int_0^{\pi/2} \frac{\cos^A x}{\cos^A x + \sin^A x} dx = \frac\pi4 \) for all \( A \in \mathbb{R} \). Basically that working out can be replicated for any power, and that power of \(\pi\) was just some arbitrary choice for \(A\).
But yeah my question was basically teasing at stochastic matrices and also eigenvalues :P
NEXT QUESTION: Got given this one ages ago and it was just ugly. (Merry Christmas.)
\[ \int \frac{(x-1)\sqrt{x^4+2x^3 +4x^2+2x+1}}{x^2(x+1)}dx \]
For lzxnl's first question, the derivative would be 0 (numerator would have e^ix(cos(x) +i sin(x)) - (i cos(x) - sin(x)) which when expanded becomes 0 ). However, e^ix is defined as cos(x)+isin(x). You would think that anything besides 0 divided by itself would be 1 but apparently not for this case!The point is more to show that this definition makes sense to a high school student, not to prove the definition. Then, it's about making an appropriate comment regarding zero derivative. I think you're overlooking something in your answer there.
snipHaha, this was a rare integral where I found that using a trigonometric substitution was better than a hyperbolic substitution. That route made it a bit harder ;)
For lzxnl's first question, the derivative would be 0 (numerator would have e^ix(cos(x) +i sin(x)) - (i cos(x) - sin(x)) which when expanded becomes 0 ). However, e^ix is defined as cos(x)+isin(x). You would think that anything besides 0 divided by itself would be 1 but apparently not for this case!You're right about the final answer but I think the intended approach was to use the quotient rule to bash that the derivative was equal to 0, before justifying it by Euler's formula.
// Finds and prints the first matrix that works
// If none exists, mention that none exists
// Designed specifically for matrices over Z_2
#include <stdio.h>
#include <stdlib.h>
#include <stdbool.h>
// Size of matrix - can be changed
#define SIZE 4
typedef struct matrixRep *Matrix;
typedef struct matrixRep {
int mm[SIZE][SIZE];
} _matrix;
Matrix newMatrix();
void destroyMatrix(Matrix m);
unsigned long long maxNumIterations(unsigned int size);
void matrixInit(Matrix m, int i);
void idInit(Matrix id);
void kkInit(Matrix k);
bool matrixCmp(Matrix a, Matrix b);
Matrix matrixAdd(Matrix a, Matrix b);
Matrix matrixMult(Matrix a, Matrix b);
Matrix matrixCube(Matrix m);
Matrix matrixFourth(Matrix m);
void printMatrix(Matrix m);
int main() {
Matrix m, mcubed, mfourth, msum;
Matrix id, kk;
unsigned long long n = maxNumIterations(SIZE);
bool found = false;
id = newMatrix();
idInit(id);
kk = newMatrix();
kkInit(kk);
m = newMatrix();
for (unsigned long long i = 0; i < n; i++) {
if (found) break;
matrixInit(m, i);
mcubed = matrixCube(m);
msum = matrixAdd(mcubed, m);
mfourth = matrixFourth(m);
if (matrixCmp(msum, kk) && matrixCmp(mfourth, id)) found = true;
destroyMatrix(mcubed);
destroyMatrix(msum);
destroyMatrix(mfourth);
}
if (found) {
printf("Found matrix:\n");
printMatrix(m);
printf("\n");
} else {
printf("Not found.\n");
}
destroyMatrix(id);
destroyMatrix(kk);
destroyMatrix(m);
return 0;
}
// newMatrix: allocates memory for a new matrix
Matrix newMatrix() {
Matrix new = calloc(1, sizeof(struct matrixRep));
return new;
}
// destroyMatrix: frees up memory associated with a matrix
void destroyMatrix(Matrix m) {
free(m);
}
// maxNumIterations: computes 2^(size^2) - the max amount of times
// the loop in the main function can run for
unsigned long long maxNumIterations(unsigned int size) {
unsigned long long n = 1;
unsigned long int sizeSq = size*size;
for (unsigned long int i = 0; i < sizeSq; i++) {
n <<= 1;
}
return n;
}
// matrixInit: sets up m for the current value of i in the main function's loop
void matrixInit(Matrix m, int n) {
int mask = 1;
for (int i = 0; i < SIZE; i++) {
for (int j = 0; j < SIZE; j++) {
m->mm[i][j] = n & mask;
mask <<= 1;
m->mm[i][j] = (m->mm[i][j] == 0) ? 0 : 1;
}
}
}
// idInit: sets up the identity matrix
void idInit(Matrix id) {
int posOfOne = 0;
for (int i = 0; i < SIZE; i++) {
for (int j = 0; j < SIZE; j++) {
if (j == posOfOne) id->mm[i][j] = 1;
else id->mm[i][j] = 0;
}
posOfOne++;
}
}
// kkInit: sets up the matrix of 1's
void kkInit(Matrix k) {
for (int i = 0; i < SIZE; i++)
for (int j = 0; j < SIZE; j++)
k->mm[i][j] = 1;
}
// matrixCmp: checks if two
bool matrixCmp(Matrix a, Matrix b) {
bool same = true;
for (int i = 0; i < SIZE; i++) {
if (!same) break;
for (int j = 0; j < SIZE; j++) {
if (a->mm[i][j] != b->mm[i][j]) {
same = false;
break;
}
}
}
return same;
}
// matrixAdd: computes the sum of two matrices
Matrix matrixAdd(Matrix a, Matrix b) {
Matrix sum = newMatrix();
for (int i = 0; i < SIZE; i++)
for (int j = 0; j < SIZE; j++)
sum->mm[i][j] = a->mm[i][j] ^ b->mm[i][j];
return sum;
}
// matrixMult: computes the product of two matrices
Matrix matrixMult(Matrix a, Matrix b) {
Matrix prod = newMatrix();
for (int i = 0; i < SIZE; i++) {
for (int j = 0; j < SIZE; j++) {
for (int k = 0; k < SIZE; k++)
prod->mm[i][j] += a->mm[i][k] * b->mm[k][j];
prod->mm[i][j] &= 1;
}
}
return prod;
}
// matrixCube: computes the cube of a matrix
Matrix matrixCube(Matrix m) {
Matrix msq = matrixMult(m, m);
Matrix mcb = matrixMult(msq, m);
destroyMatrix(msq);
return mcb;
}
// matrixFourth: computs the fourth power of a matrix
Matrix matrixFourth(Matrix m) {
Matrix msq = matrixMult(m, m);
Matrix mfr = matrixMult(msq, msq);
destroyMatrix(msq);
return mfr;
}
// printMatrix: displays a matrix to stdout
void printMatrix(Matrix m) {
for (int i = 0; i < SIZE; i++) {
for (int j = 0; j < SIZE; j++)
printf("%d ", m->mm[i][j]);
printf("\n");
}
}
...but then soon enough I realised how shit this code was because it's an \( \mathcal{O}(2^{n^2}) \) algorithm (slow when \(n=5\)) :'( wonder if improvements can be made to it. I know that right now it's checking matrices which aren't of full rank which is redundant. Anyone got suggestions on some easy math that I can probably convert into code to make it faster?
Haha, this was a rare integral where I found that using a trigonometric substitution was better than a hyperbolic substitution. That route made it a bit harder ;)I'm not convinced your method is easier. You needed to recognise that compound angle formula, and even at the end, you have to expand both resulting compound angle trig functions back, and that's a bit of algebra too. Initially, I did a trig sub, saw the sec cubed, and thought it was needlessly complicated because of the high power :P
\[ u+1 = \tan \theta \implies du = \sec^2\theta \,d\theta. \]
\begin{align*} \int \frac{\sqrt{(u+1)^2 + 1}}{u+2}du&= \int \frac{\sqrt{\tan^2\theta +1}}{\tan\theta+1}\sec^2\theta \, d\theta\\ &= \int \frac{\sec^3\theta}{\tan\theta+1}d\theta\\&= \int \frac{\sec^2\theta}{1+\tan\theta}\sec\theta\,d\theta\\ &= \int \frac{\tan^2\theta+1}{\tan\theta+1}\sec\theta \\ &= \int \tan\theta\sec\theta-\sec\theta - \frac{2\sec\theta}{\tan \theta+1}\,d\theta\\ &= \sec\theta - \ln |\sec\theta+\tan\theta| - 2\int \frac{d\theta}{\sin\theta+\cos\theta}\end{align*}
\[ \text{of which that last integral can be treated as}\\ \int \frac{d\theta}{\sin\theta+\cos\theta} = \int \frac{d\theta}{\sqrt{2} \sin\left (\theta + \frac\pi4 \right)}\\ \text{and }\int \csc x \,dx = -\ln |\csc x + \cot x|+C \]You're right about the final answer but I think the intended approach was to use the quotient rule to bash that the derivative was equal to 0, before justifying it by Euler's formula.
Also, I don't think \(e^{ix}\) is 'defined' as \(\cos x + i \sin x\). It is something that can be proven via Taylor series
________________________________________
Anyway won't steal that integral (although hahaha, \(n\)-th moment of a normal distribution). But I just wanted to put up some C code for problem set 2 because I thought it was a fun programming exercise.Code: (22dec2018question.c) [Select]// Finds and prints the first matrix that works
...but then soon enough I realised how shit this code was because it's an \( \mathcal{O}(2^{n^2}) \) algorithm (slow when \(n=5\)) :'( wonder if improvements can be made to it. I know that right now it's checking matrices which aren't of full rank which is redundant. Anyone got suggestions on some easy math that I can probably convert into code to make it faster?
// If none exists, mention that none exists
// Designed specifically for matrices over Z_2
#include <stdio.h>
#include <stdlib.h>
#include <stdbool.h>
// Size of matrix - can be changed
#define SIZE 4
typedef struct matrixRep *Matrix;
typedef struct matrixRep {
int mm[SIZE][SIZE];
} _matrix;
Matrix newMatrix();
void destroyMatrix(Matrix m);
unsigned long long maxNumIterations(unsigned int size);
void matrixInit(Matrix m, int i);
void idInit(Matrix id);
void kkInit(Matrix k);
bool matrixCmp(Matrix a, Matrix b);
Matrix matrixAdd(Matrix a, Matrix b);
Matrix matrixMult(Matrix a, Matrix b);
Matrix matrixCube(Matrix m);
Matrix matrixFourth(Matrix m);
void printMatrix(Matrix m);
int main() {
Matrix m, mcubed, mfourth, msum;
Matrix id, kk;
unsigned long long n = maxNumIterations(SIZE);
bool found = false;
id = newMatrix();
idInit(id);
kk = newMatrix();
kkInit(kk);
m = newMatrix();
for (unsigned long long i = 0; i < n; i++) {
if (found) break;
matrixInit(m, i);
mcubed = matrixCube(m);
msum = matrixAdd(mcubed, m);
mfourth = matrixFourth(m);
if (matrixCmp(msum, kk) && matrixCmp(mfourth, id)) found = true;
destroyMatrix(mcubed);
destroyMatrix(msum);
destroyMatrix(mfourth);
}
if (found) {
printf("Found matrix:\n");
printMatrix(m);
printf("\n");
} else {
printf("Not found.\n");
}
destroyMatrix(id);
destroyMatrix(kk);
destroyMatrix(m);
return 0;
}
// newMatrix: allocates memory for a new matrix
Matrix newMatrix() {
Matrix new = calloc(1, sizeof(struct matrixRep));
return new;
}
// destroyMatrix: frees up memory associated with a matrix
void destroyMatrix(Matrix m) {
free(m);
}
// maxNumIterations: computes 2^(size^2) - the max amount of times
// the loop in the main function can run for
unsigned long long maxNumIterations(unsigned int size) {
unsigned long long n = 1;
unsigned long int sizeSq = size*size;
for (unsigned long int i = 0; i < sizeSq; i++) {
n <<= 1;
}
return n;
}
// matrixInit: sets up m for the current value of i in the main function's loop
void matrixInit(Matrix m, int n) {
int mask = 1;
for (int i = 0; i < SIZE; i++) {
for (int j = 0; j < SIZE; j++) {
m->mm[i][j] = n & mask;
mask <<= 1;
m->mm[i][j] = (m->mm[i][j] == 0) ? 0 : 1;
}
}
}
// idInit: sets up the identity matrix
void idInit(Matrix id) {
int posOfOne = 0;
for (int i = 0; i < SIZE; i++) {
for (int j = 0; j < SIZE; j++) {
if (j == posOfOne) id->mm[i][j] = 1;
else id->mm[i][j] = 0;
}
posOfOne++;
}
}
// kkInit: sets up the matrix of 1's
void kkInit(Matrix k) {
for (int i = 0; i < SIZE; i++)
for (int j = 0; j < SIZE; j++)
k->mm[i][j] = 1;
}
// matrixCmp: checks if two
bool matrixCmp(Matrix a, Matrix b) {
bool same = true;
for (int i = 0; i < SIZE; i++) {
if (!same) break;
for (int j = 0; j < SIZE; j++) {
if (a->mm[i][j] != b->mm[i][j]) {
same = false;
break;
}
}
}
return same;
}
// matrixAdd: computes the sum of two matrices
Matrix matrixAdd(Matrix a, Matrix b) {
Matrix sum = newMatrix();
for (int i = 0; i < SIZE; i++)
for (int j = 0; j < SIZE; j++)
sum->mm[i][j] = a->mm[i][j] ^ b->mm[i][j];
return sum;
}
// matrixMult: computes the product of two matrices
Matrix matrixMult(Matrix a, Matrix b) {
Matrix prod = newMatrix();
for (int i = 0; i < SIZE; i++) {
for (int j = 0; j < SIZE; j++) {
for (int k = 0; k < SIZE; k++)
prod->mm[i][j] += a->mm[i][k] * b->mm[k][j];
prod->mm[i][j] &= 1;
}
}
return prod;
}
// matrixCube: computes the cube of a matrix
Matrix matrixCube(Matrix m) {
Matrix msq = matrixMult(m, m);
Matrix mcb = matrixMult(msq, m);
destroyMatrix(msq);
return mcb;
}
// matrixFourth: computs the fourth power of a matrix
Matrix matrixFourth(Matrix m) {
Matrix msq = matrixMult(m, m);
Matrix mfr = matrixMult(msq, msq);
destroyMatrix(msq);
return mfr;
}
// printMatrix: displays a matrix to stdout
void printMatrix(Matrix m) {
for (int i = 0; i < SIZE; i++) {
for (int j = 0; j < SIZE; j++)
printf("%d ", m->mm[i][j]);
printf("\n");
}
}
I'm not convinced your method is easier. You needed to recognise that compound angle formula, and even at the end, you have to expand both resulting compound angle trig functions back, and that's a bit of algebra too. Initially, I did a trig sub, saw the sec cubed, and thought it was needlessly complicated because of the high power :PSee, maybe I'm wrong because of the HSC vs VCE context, but the auxiliary transform is actually a part of the HSC MX1 syllabus and basically a one-liner :P it's a buzz-kill for those problems here to those who've integrated for a bit too long in their life
See, maybe I'm wrong because of the HSC vs VCE context, but the auxiliary transform is actually a part of the HSC MX1 syllabus and basically a one-liner :P it's a buzz-kill for those problems here to those who've integrated for a bit too long in their life
\[ \int \frac{(x-1)\sqrt{x^4+2x^3 +4x^2+2x+1}}{x^2(x+1)}dx \]Ewwwwwwwwwwww. That's probably the worst I've seen :P (what a great christmas gift lol)
But I just wanted to put up some C code for problem set 2 because I thought it was a fun programming exercise.Code: (22dec2018question.c) [Select]// Finds and prints the first matrix that works
...but then soon enough I realised how shit this code was because it's an \( \mathcal{O}(2^{n^2}) \) algorithm (slow when \(n=5\)) :'( wonder if improvements can be made to it. I know that right now it's checking matrices which aren't of full rank which is redundant. Anyone got suggestions on some easy math that I can probably convert into code to make it faster?
// If none exists, mention that none exists
// Designed specifically for matrices over Z_2
#include <stdio.h>
#include <stdlib.h>
#include <stdbool.h>
// Size of matrix - can be changed
#define SIZE 4
typedef struct matrixRep *Matrix;
typedef struct matrixRep {
int mm[SIZE][SIZE];
} _matrix;
Matrix newMatrix();
void destroyMatrix(Matrix m);
unsigned long long maxNumIterations(unsigned int size);
void matrixInit(Matrix m, int i);
void idInit(Matrix id);
void kkInit(Matrix k);
bool matrixCmp(Matrix a, Matrix b);
Matrix matrixAdd(Matrix a, Matrix b);
Matrix matrixMult(Matrix a, Matrix b);
Matrix matrixCube(Matrix m);
Matrix matrixFourth(Matrix m);
void printMatrix(Matrix m);
int main() {
Matrix m, mcubed, mfourth, msum;
Matrix id, kk;
unsigned long long n = maxNumIterations(SIZE);
bool found = false;
id = newMatrix();
idInit(id);
kk = newMatrix();
kkInit(kk);
m = newMatrix();
for (unsigned long long i = 0; i < n; i++) {
if (found) break;
matrixInit(m, i);
mcubed = matrixCube(m);
msum = matrixAdd(mcubed, m);
mfourth = matrixFourth(m);
if (matrixCmp(msum, kk) && matrixCmp(mfourth, id)) found = true;
destroyMatrix(mcubed);
destroyMatrix(msum);
destroyMatrix(mfourth);
}
if (found) {
printf("Found matrix:\n");
printMatrix(m);
printf("\n");
} else {
printf("Not found.\n");
}
destroyMatrix(id);
destroyMatrix(kk);
destroyMatrix(m);
return 0;
}
// newMatrix: allocates memory for a new matrix
Matrix newMatrix() {
Matrix new = calloc(1, sizeof(struct matrixRep));
return new;
}
// destroyMatrix: frees up memory associated with a matrix
void destroyMatrix(Matrix m) {
free(m);
}
// maxNumIterations: computes 2^(size^2) - the max amount of times
// the loop in the main function can run for
unsigned long long maxNumIterations(unsigned int size) {
unsigned long long n = 1;
unsigned long int sizeSq = size*size;
for (unsigned long int i = 0; i < sizeSq; i++) {
n <<= 1;
}
return n;
}
// matrixInit: sets up m for the current value of i in the main function's loop
void matrixInit(Matrix m, int n) {
int mask = 1;
for (int i = 0; i < SIZE; i++) {
for (int j = 0; j < SIZE; j++) {
m->mm[i][j] = n & mask;
mask <<= 1;
m->mm[i][j] = (m->mm[i][j] == 0) ? 0 : 1;
}
}
}
// idInit: sets up the identity matrix
void idInit(Matrix id) {
int posOfOne = 0;
for (int i = 0; i < SIZE; i++) {
for (int j = 0; j < SIZE; j++) {
if (j == posOfOne) id->mm[i][j] = 1;
else id->mm[i][j] = 0;
}
posOfOne++;
}
}
// kkInit: sets up the matrix of 1's
void kkInit(Matrix k) {
for (int i = 0; i < SIZE; i++)
for (int j = 0; j < SIZE; j++)
k->mm[i][j] = 1;
}
// matrixCmp: checks if two
bool matrixCmp(Matrix a, Matrix b) {
bool same = true;
for (int i = 0; i < SIZE; i++) {
if (!same) break;
for (int j = 0; j < SIZE; j++) {
if (a->mm[i][j] != b->mm[i][j]) {
same = false;
break;
}
}
}
return same;
}
// matrixAdd: computes the sum of two matrices
Matrix matrixAdd(Matrix a, Matrix b) {
Matrix sum = newMatrix();
for (int i = 0; i < SIZE; i++)
for (int j = 0; j < SIZE; j++)
sum->mm[i][j] = a->mm[i][j] ^ b->mm[i][j];
return sum;
}
// matrixMult: computes the product of two matrices
Matrix matrixMult(Matrix a, Matrix b) {
Matrix prod = newMatrix();
for (int i = 0; i < SIZE; i++) {
for (int j = 0; j < SIZE; j++) {
for (int k = 0; k < SIZE; k++)
prod->mm[i][j] += a->mm[i][k] * b->mm[k][j];
prod->mm[i][j] &= 1;
}
}
return prod;
}
// matrixCube: computes the cube of a matrix
Matrix matrixCube(Matrix m) {
Matrix msq = matrixMult(m, m);
Matrix mcb = matrixMult(msq, m);
destroyMatrix(msq);
return mcb;
}
// matrixFourth: computs the fourth power of a matrix
Matrix matrixFourth(Matrix m) {
Matrix msq = matrixMult(m, m);
Matrix mfr = matrixMult(msq, msq);
destroyMatrix(msq);
return mfr;
}
// printMatrix: displays a matrix to stdout
void printMatrix(Matrix m) {
for (int i = 0; i < SIZE; i++) {
for (int j = 0; j < SIZE; j++)
printf("%d ", m->mm[i][j]);
printf("\n");
}
}
Ewwwwwwwwwwww. That's probably the worst I've seen :P (what a great christmas gift lol)Be careful what you wish for ;)
Be careful what you wish for ;)
\[ \int \sqrt[4]{\tan x}\,dx \]
...
(https://i.imgur.com/ejXFluB.png)(Yeah. Told you to be careful :D)
;D lol, there's no way I'm attempting that.
As for the probability question, I'm not too sure how to argue it succinctly. I'm a jaffy, so I haven't taken (and probably won't be able to take) any probability or statistics subjects in Uni. Nonetheless, I'll give the question a go. Probably gonna need some correcting though. I'm just doing a combinatorics argument.
_______________________________________________________________
My dodgy proof
\(S=X_1+\dots+X_n\) counts the number of Bernoulli variables that are equal to \(1\), so \(S\in\left\{0,\dots,n\right\}\). We start by noticing that if we wish to calculate \(\Pr(S=k)\), where \(k\in\left\{0,\dots,n\right\}\), there are \(\binom{n}{k}\) possible cases to consider. Let \(A\) be one of those specific cases. Since \(X_1,\dots,X_n\) are independent, \[\Pr(A)=p^k(1-p)^{n-k}\] because there are \(k\) Bernoulli variables equal to \(1\) with probability \(p\) and \(n-k\) variables equal to \(0\) with probability \(1-p\). Hence, \[\Pr(S=k)=\binom{n}{k}p^k(1-p)^{n-k},\] which is the probability mass function of the binomial distribution. Thus, \(S\sim\text{Bi}(n,p)\).
_______________________________________________________________
I'm having trouble clearly describing what I mean by the event \(A\) being a "specific case". Help?
(Yeah. Told you to be careful :D)
(Background context - basically \( \int \sqrt{\tan x}\,dx\) is a very popular painful integral. It's so popular that people have went on to discovering more efficient ways of solving it, but they lack a lot less intuition than the original approach. But I didn't want to put that one up because that one is either barely easier or 'on par' with the painful one above. So I just decided to revamp up the power a bit instead 8))
I mean, that proof is fair enough at the first year level I think :P it's basically using enumerative combinatorics here instead. To make \(A\) rigourous, you could comment on something along the lines of for a fixed \(k\), let \(A\) be a particular configuration (or sequence) of the \(n\) Ber(p)'s, and then provide an example - e.g. if \(n=2\) and \(k=1\), a particular configuration could be \(X_1 = 0\) and \(X_2 = 1\)). But if you wanna do a bit of reading, look up "sums of independent random variables" and "convolutions" :)
Haha, indeed, even I've tried (and failed) finding more efficient methods for \(\int\sqrt{\tan(x)}\,dx\). (It really is a pain in the a**). I just wanted to see how messy the answer for the fourth root case is :PIdk maybe I've just done too much combinatorics over my life but I think giving examples are a great thing.to help illustrate your point. Some things are just hard to explain by nature (especially in combinatorics anyway).
For rigorously defining \(A\), I was trying really hard not to give an example lol, though I guess it doesn't hurt to clarify what I mean. I'll be sure to read up on the stuff you mentioned :)
NEXT QUESTION: Ok I'm bored of integrals for now. For the high schoolers:
\[ \text{Find the implied domain of}\\ f(x) = \ln (x^2-4x) \arcsin (x^2-4x-3).\]
\[ \text{Find the implied domain of}\\ f(x) = \ln (x^2-4x) \arcsin (x^2-4x-3).\]
Be careful what you wish for ;)
\[ \int \sqrt[4]{\tan x}\,dx \]
(That's not the next question and I ain't touching it either, but you can if you really want to.)
But yeah I give up on that code. After thinking about it again, if I still need to "check" the obvious before computing the matrix powers (e.g. no 0 in each row and column) it's still gonna be an inefficient algorithm.
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NEXT QUESTION: Ok I'm bored of integrals for now. For the high schoolers:
\[ \text{Find the implied domain of}\\ f(x) = \ln (x^2-4x) \arcsin (x^2-4x-3).\]
For the university students:
\[ \text{Let }X_1, \dots, X_n\text{ be an i.i.d. sequence of }\operatorname{Ber}(p)\text{ random variables.}\\ \text{Prove that }S := \sum_{i=1}^n X_i \sim \operatorname{Bin}(n,p). \]
Also, define 'binomial distribution' and 'Bernoulli distribution'. I would argue that a binomial distribution is defined as a sum of iid Bernoulli trials, thus your result is true trivially. If you define them in terms of pmfs, generating functions are handy too. This calculation, however, is trivial compared to that integral. Fark. Thank goodness for Mathematica to check my working along the way.Yep PMFs. MGFs would make the computations doable in just a few lines so that would work. Didn't think about it at the time but I wasn't trying to make that one hard.
Hope it's ok if answer this question, even though I just finished high school. Don't plan on studying maths at uni so don't think I have too much of an advantage! However, I'm missing my high school maths subjects already, so I thought I could give it a go...It looked right after you fixed it :D
Also, hopefully its correct!Spoiler
\[ \text{Let} \: f(x)=g(x) \times h(x) \\ \begin{align*} \therefore g(x) &= ln(x^2-4x) \\ h(x) &= arcsin(x^2-4x-3) \end{align*} \\ \implies \text{Domain of } \: f: \text{Dom} \: g \cap \text{Dom} \: h \]
\[ \text{Inside of a log function must be greater than zero.} \\ \therefore x^2-4x>0 \\ \text{By considering the graph, we obtain:} \\ \text{Dom} \: g: (-\infty,0) \cup (4,\infty) \]
\[ \text{Inside of a arcsin function must be between or equal to, -1 and 1.} \\ \therefore -1 \leq x^2-4x-3 \leq 1 \\ 2 \leq x^2-4x \leq 4 \]
\[ \text{Comparing to above, we identify that this domain is a subset of the domain of} \: g \: \\ \therefore \text{The domain of} \: f \: \text{is the same as the domain of} \: h.\\ \text{Let the domain of} \: h \: \text{be}\: [a,b] \cup [c,d] \\ \text{Solve for the endpoints:} \]
\[x^2-4x=2 \\ \therefore b=2-\sqrt{6} \quad \& \quad c=2+\sqrt{6}\]
\[x^2-4x=4 \\ \therefore a=2-2\sqrt{2}\quad \& \quad d=2+2\sqrt{2}\]
\[\therefore \text{Dom} \: h: [2-2\sqrt{2},2-\sqrt{6}] \cup [2+\sqrt{6},2+2\sqrt{2}]\]
\[\implies \text{Dom} \: f: [2-2\sqrt{2},2-\sqrt{6}] \cup [2+\sqrt{6},2+2\sqrt{2}]\]
EDITS: Many edits to correct poor LaTeX, still probability mistakes lol.
Yep PMFs. MGFs would make the computations doable in just a few lines so that would work. Didn't think about it at the time but I wasn't trying to make that one hard.It looked right after you fixed it :DI'm going to leave this one for someone else. The simplest method for doing this would be related to thinking about why this is true geometrically (I don't think RuiAce wants someone to say 'because \(\cos(\theta) \in [-1,1]\)' btw; this should motivate a proof, however, which is certainly doable at VCE level).
Next question: Prove the Cauchy-Schwarz inequality for vectors in \(\mathbb{R}^3\) (however you like, but if there's a VCE method then go with that)
\[ |\underset{\sim}{a} \cdot \underset{\sim}{b}|^2 \leq | \underset{\sim}{a} | |\underset{\sim}{b}| \]
(Someone else take over writing questions after this one pls.)
\[ |\underset{\sim}{a} \cdot \underset{\sim}{b}|^2 \leq | \underset{\sim}{a} | |\underset{\sim}{b}| \]Shouldn't the RHS be \(\left|\underset{\sim}{\text{a}}\right|^2\left|\underset{\sim}{\text{b}}\right|^2\)?
Here's a fun question for you guys. Can \( \int_0^\infty f(x)\,dx\) exist if \(f(x)\) does not decay to zero as \(x\rightarrow\infty\)?
Shouldn't the RHS be \(\left|\underset{\sim}{\text{a}}\right|^2\left|\underset{\sim}{\text{b}}\right|^2\)?Maybe I should've put Minkowski's instead.Very cheap solution (sorry)It's clear that the inequality holds if \(\underset{\sim}{\text{a}}=\underset{\sim}{0}\) or \(\underset{\sim}{\text{b}}=\underset{\sim}{0}\), so we will consider \(\underset{\sim}{\text{a}}\neq \underset{\sim}{0}\) and \(\underset{\sim}{\text{b}}\neq \underset{\sim}{0}\).
\begin{align*}\left|\underset{\sim}{\text{a}}\cdot\underset{\sim}{\text{b}}\right|&=\left|\left|\underset{\sim}{\text{a}}\right|\left|\underset{\sim}{\text{b}}\right|\cos(\theta)\right|\\
&=\left|\underset{\sim}{\text{a}}\right|\left|\underset{\sim}{\text{b}}\right|\left|\cos(\theta)\right|\\
&\leq \left|\underset{\sim}{\text{a}}\right|\left|\underset{\sim}{\text{b}}\right|\quad \Big(0\leq\left|\cos(\theta)\right|\leq 1\Big)\end{align*}Fun example I came across some time ago\[\text{Let }f(x)=\begin{cases}1, & x\in[n,\ n+2^{-n}]\\ 0, & \text{elsewhere}\end{cases},\ \ n\in\mathbb{N}\cup\{0\}.\] \[\int_0^\infty f(x)\;dx=\sum_{n=0}^\infty 2^{-n}=2\ \text{ and }\ f(x)\nrightarrow 0\ \text{as}\ x\to\infty.\] So, the answer is yes, it is possible.
Next Questions
For highschool students:
Find the magnitude of the acute angle formed by the two lines in \(\mathbb{R}^2\) given by
\[5\sqrt{3}x+9y=9\ \ \text{and}\ -\!\sqrt{3}x+6y=6.\]
For uni students:
Prove Lagrange's identity
\[\|\mathbf{u}\times\mathbf{v}\|^2=\|\mathbf{u}\|^2\|\mathbf{v}\|^2-(\mathbf{u}\cdot\mathbf{v})^2.\]
Or, if you're not a fan of the algebra required in the above question:
Solve for \(y(t)\) the initial value problem \[\frac{dy}{dt}=\sin(y),\quad y(0)=\alpha\in\mathbb{R}\setminus\{k\pi\mid k\in\mathbb{Z}\}.\] Then, show that \[\lim_{t\to\infty}y(t)=(2m+1)\pi,\quad \text{for some }m\in\mathbb{Z}.\]
Shouldn't the RHS be \(\left|\underset{\sim}{\text{a}}\right|^2\left|\underset{\sim}{\text{b}}\right|^2\)?Very cheap solution (sorry)It's clear that the inequality holds if \(\underset{\sim}{\text{a}}=\underset{\sim}{0}\) or \(\underset{\sim}{\text{b}}=\underset{\sim}{0}\), so we will consider \(\underset{\sim}{\text{a}}\neq \underset{\sim}{0}\) and \(\underset{\sim}{\text{b}}\neq \underset{\sim}{0}\).
\begin{align*}\left|\underset{\sim}{\text{a}}\cdot\underset{\sim}{\text{b}}\right|&=\left|\left|\underset{\sim}{\text{a}}\right|\left|\underset{\sim}{\text{b}}\right|\cos(\theta)\right|\\
&=\left|\underset{\sim}{\text{a}}\right|\left|\underset{\sim}{\text{b}}\right|\left|\cos(\theta)\right|\\
&\leq \left|\underset{\sim}{\text{a}}\right|\left|\underset{\sim}{\text{b}}\right|\quad \Big(0\leq\left|\cos(\theta)\right|\leq 1\Big)\end{align*}Fun example I came across some time ago\[\text{Let }f(x)=\begin{cases}1, & x\in[n,\ n+2^{-n}]\\ 0, & \text{elsewhere}\end{cases},\ \ n\in\mathbb{N}\cup\{0\}.\] \[\int_0^\infty f(x)\;dx=\sum_{n=0}^\infty 2^{-n}=2\ \text{ and }\ f(x)\nrightarrow 0\ \text{as}\ x\to\infty.\] So, the answer is yes, it is possible.
Next Questions
For highschool students:
Find the magnitude of the acute angle formed by the two lines in \(\mathbb{R}^2\) given by
\[5\sqrt{3}x+9y=9\ \ \text{and}\ -\!\sqrt{3}x+6y=6.\]
For uni students:
Prove Lagrange's identity
\[\|\mathbf{u}\times\mathbf{v}\|^2=\|\mathbf{u}\|^2\|\mathbf{v}\|^2-(\mathbf{u}\cdot\mathbf{v})^2.\]
Or, if you're not a fan of the algebra required in the above question:
Solve for \(y(t)\) the initial value problem \[\frac{dy}{dt}=\sin(y),\quad y(0)=\alpha\in\mathbb{R}\setminus\{k\pi\mid k\in\mathbb{Z}\}.\] Then, show that \[\lim_{t\to\infty}y(t)=(2m+1)\pi,\quad \text{for some }m\in\mathbb{Z}.\]
For highschool students:
Find the magnitude of the acute angle formed by the two lines in \(\mathbb{R}^2\) given by
\[5\sqrt{3}x+9y=9\ \ \text{and}\ -\!\sqrt{3}x+6y=6.\]
It's harder for the VCE students because they don't have that nice formula that we do. In the HSC your approach would be the intended approach
Was thinking I might get baited because the answer seemed a bit too clean, but here is what I got anyway. :)
Yeah I wanted to keep the question within the bounds of VCE terminology so I tried just doing the R^3 case, but now I see that was a mistake lol.Yeah that's how you would prove the convergence. As for actually calculating the integral, you would consider the real part of \(\int_0^\infty e^{ix^2}\,dx\) and consider a sector of angle \(\frac{\pi}{4}\) in the complex plane; the arc at infinity vanishes, the total integral is zero by Cauchy's integral theorem, and the integral along the ray is a Gaussian integral.
Just gonna prove the convergence because ceebs looking up how to solve that integral right now.
\[ x = \sqrt{u} \implies dx = \frac1{2\sqrt{u}}\,du.\\ u = \frac{1}{s} \implies ds = -\frac{1}{s^2}\,ds.\\ \begin{align*} \int_0^\infty \cos (x^2)\,dx &= \int_0^\infty\frac{1}{2}u^{-1/2}\cos (u)\,du\\ &= \int_0^1 \frac12 u^{-1/2} \cos (u^2)\,du + \int_1^\infty \frac12 u^{-1/2}\cos u\,du \\ &= \int_1^\infty \frac12 s^{1/2} \cos \left( \frac{1}{s^2} \right) s^{-2}\,ds +\int_1^\infty \frac12 u^{-1/2}\cos u\,du \end{align*} \]
\[ \text{For the second integral}\\ \begin{align*} \int_1^\infty u^{-1/2} \cos u\,du &= u^{-1/2} \sin u \big|_1^\infty + \frac12 \int_1^\infty u^{-3/2} \sin u\,du. \end{align*}\\ \text{The former is finite and equals }\sin 1\\ \text{and the latter converges via comparison with the }p\text{-integral }\int_1^\infty u^{-3/2}\,du. \]The first integral converges using the same comparison. So puzzling everything together the whole thing converges.It's harder for the VCE students because they don't have that nice formula that we do. In the HSC your approach would be the intended approachSqueeze theorem limit used\[ 0 = \lim_{u\to \infty} -u^{-1/2} \leq \lim_{u\to \infty} u^{-1/2}\sin u \leq \lim_{u\to \infty} u^{-1/2} = 0\\ \implies \lim_{u\to \infty} u^{-1/2}\sin u = 0 \]
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Continuing,
\begin{align*}e^t &= \frac{\tan \frac{y}{2}}{\tan \frac{\alpha}{2}} \\ y &= 2 \arctan \left( e^t \tan \frac{\alpha}{2} \right) + 2m\pi \end{align*}
where \(m \in \mathbb{Z}\).
\[ \text{As }\alpha\text{ is not an integer multiple of }\pi,\\ \tan \frac{\alpha}{2} \text{ is well defined and not equal to 0.}\\ \text{If }\tan \alpha > 0,\, 2\arctan \left( e^t \tan \frac{\alpha}{2} \right) \to 2\times \frac\pi2 = \pi\text{ and we're done.}\\ \text{Otherwise it just approaches }-\frac\pi2\text{ instead, but then sub }m=n+1\text{ and we're also done.} \]
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Next question:
High school:
\[ \text{1. Prove that }\sum_{k=1}^n k^2 = \frac16 n (n+1)(2n+1)\\ \text{2. Now derive it from scratch.} \]
First year/Second year university:
\[ \text{Show that in general, }\lim_{x\to a} \lim_{y\to b} f(x,y) \neq \lim_{y\to b} \lim_{x\to a} f(x,y)\\ \text{where }a, b \in \mathbb{R} \cup \{\infty\} \]
Try to construct a routine counterexample, and also a creative counterexample.
Beyond:
Label the vertices of \(K_5\) 1, 2, 3, 4 and 5, where \(K_n\) is the complete graph on \(n\) vertices. Suppose that two Euler circuits are the same if their sequences of edges are up to the same rotational symmetry. (For example 1-2-3-4-5-1-3-5-2-4-1 would be the same as 2-3-4-5-1-3-5-2-4-1-2).
1. Explain why \(K_3\) has only two distinct Euler circuits (easy warm-up)
2. Carefully prove that \(K_5\) has 264 distinct Euler circuits.
High school:
\[ \text{1. Prove that }\sum_{k=1}^n k^2 = \frac16 n (n+1)(2n+1)\\ \text{2. Now derive it from scratch.} \]
Before I start Q2, am I allowed to know the result from Q1 and work towards it? Or do I have to derive derive it? :(It says from scratch. So derive derive it. :P
2. Fully expand \(\sin(5x)\) in terms of powers of \(\sin(x)\). Use this result to find \(\sin\left(\frac{2\pi}{5}\right)\). You may encounter a quintic equation. One of the solutions, which you should be able to show isn't \(\sin\left(\frac{2\pi}{y}\right)\), should be easy to find/guess. Factorise that, and the resulting quartic should be easily solvable.
Okay, thanks so much Rui! I'll edit that right now, it looks awful.Nice solution! To make the pattern you identified in part iii) appear a bit more obvious to the reader, you can argue it like this.
I'm going to drop the solutions for the derivation of the sum of squares. I panicked a little bit for about the first half hour, but once I realised the end result was a polynomial, and the intuition behind plugging a number into a function and getting a number back, it wasn't that much harder than the induction proof.
Hope you guys can read my handwriting, haven't posted a picture in a while.Page 1(https://i.imgur.com/LKRYgT5.jpg)Page 2(https://i.imgur.com/dASansM.jpg)
2. Fully expand \(\sin(5x)\) in terms of powers of \(\sin(x)\). Use this result to find \(\sin\left(\frac{2\pi}{5}\right)\). You may encounter a quintic equation. One of the solutions, which you should be able to show isn't \(\sin\left(\frac{2\pi}{y}\right)\), should be easy to find/guess. Factorise that, and the resulting quartic should be easily solvable.
You can fill in the details of this proof if you want to or someone else can :P
snipTo argue the choice of the positive root you're okay to assume that \(y = \sin^2 x\) is monotonic increasing for all \(0 < x < \frac\pi2\), however you haven't used that to convince me about anything. You're just saying that \(\sin^2 \frac{2\pi}{5} \) is closer to \( \sin^2\frac\pi2 \) than \(\sin^2 0\), but it doesn't tell us any information of "by how much is it closer".
To argue the choice of the positive root you're okay to assume that \(y = \sin^2 x\) is monotonic increasing for all \(0 < x < \frac\pi2\), however you haven't used that to convince me about anything. You're just saying that \(\sin^2 \frac{2\pi}{5} \) is closer to \( \sin^2\frac\pi2 \) than \(\sin^2 0\), but it doesn't tell us any information of "by how much is it closer".
One way to work around this issue is to note furthermore that \( \sin^2 \frac{\pi}{4} = \frac12 \). Note that \( \frac{5-\sqrt5}{8} < \frac12 < \frac{5 + \sqrt5}{8} \), and also because \(\sin^2x \) is monotonic increasing, we also have \( \sin^2 \frac{\pi}{4} < \sin^2 \frac{2\pi}{5} \). Thus by pairing it off we end up with \(\sin^2 \frac{2\pi}{5} = \frac{5+\sqrt5}{8} \)
Man, you have so much time on your hands with these questions! You must be done with all two of your TuteSmart assignments already! :P
1. Prove the cosine rule in trigonometry using properties of the vector dot productOtherwise this is a relatively straightforward matrix question provided you know what induction is.
And yeah non-uniform convergence works. I would've just used this guy:Ah yes, multivariate calc 101. I should know this, I taught a tutorial for that subject last semester. So many of those 0/0 limits that don't exist.
\[ f(x,y) = \frac{x^2}{x^2+y^2}\text{ with limits as }x\to 0, \, y\to 0 \]
Alternatively just use the example in the previous question:
\[ f(x,y) = 2\arctan \left(e^x \tan \frac{y}{2} \right)\text{ with limits as }y\to 0, x\to \infty\]
High school questions: Well firstly there's still this one which I will leave alone.Otherwise this is a relatively straightforward matrix question provided you know what induction is.
\[ \text{Let }D\text{ be a diagonal matrix. Use mathematical induction to prove that for }n\in \mathbb{Z}_+\\ \text{the terms of }D^n\text{ are just the individual components raised to the }n\text{-th power.} \]
First year university: Another straightforward one.
\[ \text{Let }A\text{ be a square matrix with eigenvalue }\lambda \text{ and eigenvector }\mathbf{v}.\\ \text{Prove by induction that }\mathbf{v}\text{ is an eigenvalue of }A^n\text{ where }n \in \mathbb{Z}_+\\ \text{and find the corresponding eigenvalue.}\]
First year university:
\[ \text{Show that if }\lambda\text{ is an eigenvalue of }A\in \mathbb{R}^{n\times n}\\ \text{then it is also an eigenvalue of }A^T\]
Hi
can someone help me out with these two questions.
What is required to do them? I would like to learn. ;D
I got them from here 2018 melb uni maths comp senior division
link to exam - http://www.mathscomp.ms.unimelb.edu.au/archive/2018SP.pdf
Thanks!
Hey redpanda83,Thanks for your suggestion Felix
I am going to be absolutely no help at all explaining the solution. But have you had a look at the solutions provided by Melbourne Uni?
Link to solutions
For anyone interested, all past papers & solutions for each division and year can be found here.
Thanks for your suggestion Felix
I looked at the solutions before but i dont completely understand them. The techniques they use, i dont know what they are. I was able to get other questions done without no issue but question 2 and 6 doesnt work out for me. I thought someone here might be able to explain them. :D
For university studentsIsn't this just a regular path integral? Or are you trying to throw off the people who think the answer is 0
Let \(C\) be the curve given by \(3x^2+y^2=1\) oriented anticlockwise, and let \[\mathbf{F}(x,y)=\left(\frac{-y}{x^2+y^2},\ \frac{x}{x^2+y^2}\right).\] Evaluate \(\displaystyle \int_C \mathbf{F}\cdot d\mathbf{s}\).