Author Topic: First Year University Mathematics Questions (Read 10411 times) Tweet Share

legorgo18 · « **Reply #15 on:** April 05, 2018, 09:57:15 pm »

Question: Find maximal domain and range for f(x)= root(1-2sinx).

I got for range f > or equal 0, then domain x < or equal pi/6 or x> or equal 5pi/6 for 0< or equal x< or equal pi. But when i graph it out using graphing calculator, graph looks really weird, so how are you meant to do this one?

RuiAce · « **Reply #16 on:** April 05, 2018, 10:11:48 pm »

Quote from: legorgo18 on April 05, 2018, 09:57:15 pm

Question: Find maximal domain and range for f(x)= root(1-2sinx).

I got for range f > or equal 0, then domain x < or equal pi/6 or x> or equal 5pi/6 for 0< or equal x< or equal pi. But when i graph it out using graphing calculator, graph looks really weird, so how are you meant to do this one?

$\text{This one's a bit intense because we're not restricted to }0\le x \le 2\pi\\ \text{as we are in high school.}$
$\text{Recall that square root functions are well defined}\\ \text{when anything under the square root is not negative.}\\ \text{So ultimately, we need to solve the equation}\\ \boxed{1 - 2\sin x \ge 0}\quad \forall x\in \mathbb{R}.$
$\text{Rearranging this inequality, we have }\boxed{\sin x \le \frac12}.\\ \text{Observe that the solutions to }\sin x = \frac12\text{ are}\\ x = \dots, -\frac{11\pi}{6}, -\frac{7\pi}{6}, \frac{\pi}{6}, \frac{5\pi}{6}, \frac{13\pi}6, \frac{17\pi}{6}, \dots$
Or in general, the solutions are $ \frac\pi6 + 2k\pi $ and $ \frac{5\pi}{6} + 2k\pi $ where $k \in \mathbb{Z} $.
$\text{But of course, we want when }\sin x \le \frac12.\\ \text{This means that when we plot the graphs }y=\sin x\text{ and }y=\frac12,\\ \text{we want whenever the sinusoid lies }\textbf{below}\text{ the line.}$

$\text{Reading off the graph, it appears that we must have any of ALL of the following.}\\ -\frac{7\pi}6 \le x \le \frac\pi6\\ \frac{5\pi}{6} \le x \le \frac{13\pi}{6}\\ \frac{17\pi}{6} \le x \le \frac{25\pi}{6}\\ \text{..and so on, going backwards as well.}$
$\text{So the domain is actually the union of this infinite number of intervals:}\\ \dots \cup \left[ -\frac{7\pi}{6}, \frac\pi6\right] \cup \left[ \frac{5\pi}{6}, \frac{13\pi}{6}\right]\cup \left[ \frac{17\pi}{6}, \frac{25\pi}{6}\right]\cup \dots$
$\text{One possible way of writing this more compactly is}\\ \bigcup_{k=-\infty}^\infty \left[ -\frac{7\pi}{6} + 2k\pi, \frac\pi6 + 2k\pi \right].\\ \text{But you don't need to convert to this form.}$
_________________________________________________________
$\text{Now for the range, }y\ge 0\text{ is certainly important as you have stated.}$
$\text{But of course the range isn't just }[0,\infty).\\ \text{There's no way we can make an ordinary trig function just explode, unless we flip it.}$
(By ordinary I really just mean $\cos$ and $\sin$.)
$\text{Pretty much, to find where it caps off, we just need to find the}\\ \text{maximum value of }f(x)=\sqrt{1-2\sin x}.\\ \text{Since this is a trig question, we will just use 'common sense' instead of calculus.}$
$\sin x\text{ has maximum value at }1.\\ \text{Therefore }2\sin x\text{ has maximum value at }2.\\ \text{Therefore }-2\sin x\text{ has maximum value at 2 as well. (Why?)}$
$\text{So if }2\text{ is the maximum value of }-2\sin x,\\ 3\text{ is the maximum value of }1-2\sin x.\\ \text{Therefore }\sqrt{3}\text{ must be the maximum value of }\sqrt{1-2\sin x}.$
$\text{Hence, the range is }[0,\sqrt3]$

legorgo18 · « **Reply #17 on:** April 05, 2018, 10:31:21 pm »

Thx rui! For range is it 3 or root 3, also wtf that domain bit, is this type of question still likely to show up for the quizzes today?

RuiAce · « **Reply #18 on:** April 05, 2018, 10:43:16 pm »

Sqrt(3); there was a typo at the end

It's possible, although it'd be a bit harsh. Just know how to solve trigonometric inequalities if it does appear.

swico · « **Reply #19 on:** April 06, 2018, 10:04:41 am »

Hi can someone please help me. How can I find a bijection between [0,1] and [0,1)?

RuiAce · « **Reply #20 on:** April 06, 2018, 10:07:36 am »

Quote from: swico on April 06, 2018, 10:04:41 am

Hi can someone please help me. How can I find a bijection between [0,1] and [0,1)?

Is this a first year question? What subject and uni is it?

swico · « **Reply #21 on:** April 06, 2018, 10:17:02 am »

It is introduction to analysis, actually it is a "second" year unit, but I can take it in first year because I have the preq for it.

TrueTears · « **Reply #22 on:** April 06, 2018, 11:21:31 am »

This may seem like a tricky question at first because there may seem to be not enough "space" to fit $[0,1]$ into $[0,1)$ , but actually there is more "space" than you think. Let $(x_n)_{n=1}^{\infty}$ be a sequence of distinct elements contained in the interval $[0,1]$ where $x_1=1$ and let $M = \{x_n, \ n \in \mathbb{N}\}$ . Define the function $f:[0,1] \rightarrow [0, 1)$ ,
\begin{align*}
f(x) = \begin{cases} x_{n+1} & \text{if \ } x \in M \\ x & \text{if \ } x \in [0,1] \setminus M.\end{cases}
\end{align*}
To show that $f$ is injective, we consider the following three cases:

1) Let $a_1, a_2 \in M$ . Note that $f(a_1)$ and $f(a_2)$ are both elements of $M$ . Since $M$ only consists of distinct elements, then $a_1 \neq a_2$ implies $f(a_1) \neq f(a_2)$ .

2) Let $a_1 \in M$ , $a_2 \in [0,1] \setminus M$ . Note that $f(a_1) \in M$ and $f(a_2) = a_2 \in [0,1] \setminus M$ . Thus, if $a_1 \neq a_2$ then $f(a_1) \neq f(a_2)$ .

3) Let $a_1, a_2 \in [0,1] \setminus M$ . Note that $f(a_1) = a_1$ and $f(a_2) = a_2$ , so if $a_1 \neq a_2$ then $f(a_1) \neq f(a_2)$ .

To show that $f$ is surjective, we need to show that the range of $f$ is equal to $[0,1)$ . The range of the function $f_1(x)=x_{n+1}$ , $x \in M$ , is $\{x_{n+1}, n \in \mathbb{N}\}$ and the range of the function $f_2(x) = x$ , $x \in [0,1] \setminus M$ , is $[0, 1) \setminus \{x_{n+1}, n \in \mathbb{N}\}$ . Since $f$ is a piecewise function of $f_1$ and $f_2$ , its range must be precisely $[0,1)$ and we are done.

EDIT: Don't know what's up with AN's latex typeset, it's way different and very non-standard from back in the days I used it lol.

swico · « **Reply #23 on:** April 06, 2018, 11:33:51 am »

Wow thanks TT, I'll have a look over your solution and let you know if I have any questions

RuiAce · « **Reply #24 on:** April 06, 2018, 12:24:30 pm »

$\text{Mostly to build on:}$
(because I was unavailable after I made my first post.)
$\text{One of the more straightforward examples of such a construction}\\ \text{would be the following function:}\\ f(x) = \begin{cases}(x^{-1} + 1)^{-1} &\text{ if }x = \frac1n \text{ for }n = 1,2,3 \dots\\ x&\text{otherwise}.\end{cases}$
$\text{Such a function aims to take all elements of the form }\frac{1}{n}\\ \text{and essentially }\textbf{shift them over to the next one, i.e. }\frac{1}{n+1}.$
It builds on because it's really just taking $ x_n = \frac{1}{n} $.

Demo of how it works

\begin{align*}1 &\mapsto \frac12\\ \frac12 &\mapsto \frac13\\ \frac13 &\mapsto \frac14\end{align*}
$\text{...and so on.}\\ \text{Everything else including 0 is mapped to itself.}$

$\text{The function is bijective because we can construct its inverse:}\\ f^{-1}(y) = \begin{cases}(y^{-1} - 1)^{-1}&\text{if }y = \frac{1}{n}\text{ for }n=2,3,4,\dots \\ y &\text{otherwise}.\end{cases}\\ \text{It basically just shifts everything back.}$
$\text{Having said that, you can also prove its bijectivity from the definition.}\\ \text{The injective part should be easy. But the surjective part will need proper arguing.}$
There might be a typo or two here and there, but the construction follows from the same as above. Anyway, I would like to reserve this thread for questions related to first year units only. Here is a thread for questions similar to this calibre - I will request for it to be moved to somewhere more visible.

swico · « **Reply #25 on:** April 06, 2018, 03:30:01 pm »

Ok, thanks RuiAce, I have a few other questions, I will post in the other thread as you suggested. Thanks again.

WrongWong · « **Reply #26 on:** May 19, 2018, 05:02:42 pm »

Question: If a matrix $M$ has the property that all columns add to one (is a stochastic matrix). Prove by induction that $M^n$ has the same property for all $n \in \mathbb{N}$

Thanks in advance !

RuiAce · « **Reply #27 on:** May 19, 2018, 05:24:49 pm »

$\text{Suppose }M = (M)_{ij}\text{ satisfies the relevant property.}\\ \text{We remark that }M = \underbrace{\begin{pmatrix}m_{11} & \dots & m_{1n}\\ \vdots & \ddots & \vdots\\ m_{n1} & \dots & m_{nn} \end{pmatrix}}_{n\times n\text{ matrix}}$
$\text{The given property tells us that}\\ \begin{align*}\sum_{i=1}^n m_{i1} &= 1\\ \sum_{i=1}^n m_{i2} &= 1\\ &\vdots \\ \sum_{i=1}^n m_{in} &= 1\end{align*}$
Essentially, in general $ \sum_{i=1}^n m_{ik} = 1 $ for all $k = 1, \dots, n$.
___________________________________________________
$\text{Inductive base: }\boxed{n = 2}.\\ \text{Consider }M^2\text{, which by definition of matrix multiplication satisfies}\\ [M^2]_{ij} = m_{i1}m_{1j} + \dots + m_{in}m_{nj}$
$\text{For any column index }j \in \{ 1, \dots, n\}\text{, the sum of entries in each column is:}\\ \begin{align*}\sum_{i=1}^n [M^2]_{ij}&= \sum_{i=1}^n \left(m_{i1}m_{1j} + \dots + m_{in}m_{nj}\right)\\ &= m_{ij} \sum_{i=1}^n m_{i1}+ \dots + m_{nj} \sum_{i=1}^n m_{in} \tag{sums independent of j}\\ &= m_{1j} + \dots + m_{nj} \tag{from above}\\ &= \sum_{i=1}^n m_{ij} \tag{rewriting}\\ &= 1 \tag{again, from above}\end{align*}$
$\text{Therefore }M^2\text{ is stochastic, and we have a base case.}$
I now leave the inductive step as your exercise.

Hint

Your inductive hypothesis is just that $M^k$ is stochastic, and you wish to prove that $M^{k+1} $ is stochastic. But you can just write
\[ M^k = \begin{pmatrix} p_{11} & \dots & p_{1n}\\ \vdots & \ddots & \vdots\\ p_{n1} & \dots & p_{nn}\end{pmatrix} \]
because you still know that the product of a bunch of $n\times n$ matrices is still $n\times n$! And then rinse and repeat pretty much everything we did just now.

Edit: Apologies, mixed my letters here and there. I should've made the matrix $ k\times k$ or something

K98100 · « **Reply #28 on:** May 20, 2018, 09:43:08 pm »

Hi,
Could I please get some help on this question?
Thanks

RuiAce · « **Reply #29 on:** May 21, 2018, 07:06:24 am »

Quote from: K98100 on May 20, 2018, 09:43:08 pm

Hi,
Could I please get some help on this question?
Thanks

$T\text{ is basically the projection map }T\left( \begin{bmatrix}x\\y\\z\end{bmatrix} \right) = \text{proj}_{\begin{bmatrix}1\\0\\1\end{bmatrix}} \begin{bmatrix}x\\y\\z\end{bmatrix}$
$\text{So for an explicit formula, we just apply the formula for the projection (and the dot product).}\\ \begin{align*}T\left(\begin{bmatrix}x\\y\\z\end{bmatrix} \right) &= \frac{\begin{bmatrix}x\\y\\z\end{bmatrix} \cdot \begin{bmatrix}1\\0\\1\end{bmatrix} }{\begin{bmatrix}1\\0\\1\end{bmatrix} \cdot \begin{bmatrix}1\\0\\1\end{bmatrix}}\begin{bmatrix}1\\0\\1\end{bmatrix}\\ &= \frac{x+z}{2} \begin{bmatrix}1\\0\\1\end{bmatrix}\end{align*}$
You can of course, put $ \frac{x+y}{2} $ into the vector if you want.
_______________________________________
$\text{With no assumptions given on the bases}\\ \text{I will assume the standard bases for the domain and the codomain.}$
$\text{We compute:}\\ \begin{align*}T\left( \begin{bmatrix}1\\0\\0\end{bmatrix} \right) &= \begin{bmatrix}1/2\\0\\1/2\end{bmatrix}\\ T\left( \begin{bmatrix}0\\1\\0\end{bmatrix}\right)&= \begin{bmatrix}0\\0\\0\end{bmatrix}\\ T\left( \begin{bmatrix}0\\0\\1\end{bmatrix}\right) &= \begin{bmatrix}1/2\\0\\1/2\end{bmatrix}\end{align*}\\ \text{so our matrix is}\\ A = \begin{pmatrix}1/2 & 0 & 1/2\\ 0 & 0 & 0\\ 1/2 & 0 & 1/2 \end{pmatrix}$
$\text{Reading off our matrix, we see that}\\ \text{row}(A) =\text{span} \left\{ [\frac12,0,\frac12] \right\}.\\ \text{If you just use your knowledge of scalar multiples of vectors, you see that}\\ \text{row}(A) = \text{span} \left\{[1,0,1]\right\}\\ \text{which is, in fact, all points on the line }\ell\text{ we had at the start.}$
________________________________________
$\text{For the null space, we are solving }T(\textbf{x}) = A\textbf{x} = \textbf{0}.\\ \text{This system of equations is easy to solve}\\ \text{and has solutions of the form }\textbf{x}= s \begin{bmatrix}-1\\0\\1\end{bmatrix} + t\begin{bmatrix}0\\1\\0 \end{bmatrix}\quad (s,t\in \mathbb{R}).$
This is clearly a plane.

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Author Topic: First Year University Mathematics Questions (Read 10411 times) Tweet Share

legorgo18

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