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Uni Stuff => General University Discussion and Queries => Topic started by: Shadowxo on September 10, 2017, 11:04:09 pm

Title: First Year University Mathematics Questions
Post by: Shadowxo on September 10, 2017, 11:04:09 pm

Here's a place to ask all your first year uni mathematics-related questions!
I couldn't find any others so I thought it would be a good idea to make one :)

Title: Re: First Year University Mathematics Questions
Post by: Shadowxo on September 10, 2017, 11:06:41 pm

I'll start it off with a question of my own:
Do the vectors in the basis of the row space of a matrix + the vectors in the basis of the nullspace / solution space of a matrix make up a basis of Rⁿ, where n is the number of columns in the matrix?

Title: Re: First Year University Mathematics Questions
Post by: Sine on September 10, 2017, 11:12:26 pm

I'm first year uni but doing a 2nd year maths unit LOL, i'll probably be using this thread soon. Great idea Shadowxo

Title: Re: First Year University Mathematics Questions
Post by: RuiAce on September 10, 2017, 11:36:19 pm

Quote from: Shadowxo on September 10, 2017, 11:06:41 pm

I'll start it off with a question of my own:
Do the vectors in the basis of the row space of a matrix + the vectors in the basis of the nullspace / solution space of a matrix make up a basis of Rⁿ, where n is the number of columns in the matrix?

$\text{This would impose the condition that the matrix in question, say, }A\\ \text{is a square matrix.}\\ \text{In general, a matrix need not map vectors from }\mathbb{R}^n\text{ to }\mathbb{R}^n.\text{ We may have }\mathbb{R}^m \to \mathbb{R}^n.$
This can be achieved by simply changing the dimensions of the matrix.
$\text{In addition, quite often the vectors of the row space are written down}\\ \text{as row vectors. Hence, if we desire a basis,}\\ \text{we should exploit the transpose.}$
$\text{Therefore, let }A = \begin{pmatrix}\textbf{v}_1^T \\ \vdots \\ \textbf{v}_n^T\end{pmatrix}\\ \text{where }\textbf{v}_k\in \mathbb{R}^n\qquad \forall k=1,\dots,n\text{ are column vectors}$
________________________________________
$\text{The null-space, however, is the set of vectors }\textbf{x}\text{ such that}\\ A\textbf{x} = \begin{pmatrix}\textbf{v}_1\cdot \textbf{x}\\ \vdots \\ \textbf{v}_n \cdot \textbf{x}\end{pmatrix} = \textbf{0}\\ \text{where the dot product formula can be proven by expanding }\textbf{x}\text{ as the }n\text{-tuple vector, and same for }\textbf{v}_k^T$
$\text{And at the same time, the row-space is the set of linear combinations of}\\ S = \{ \textbf{v}_1, \dots, \textbf{v}_n \}.$
$\text{Because }\textbf{v}_k \cdot \textbf{x} = \textbf{0}\\ \text{for some arbitrary vector in the null-space}\\ \text{and any row vector in the matrix}\\ \text{it follows that any vector in the null-space is orthogonal to that in the row-space.}$
$\text{But by extension, because we know that the dot product is a linear operator}\\ \text{this affirms that EVERY linear combination of the vectors in }S\\ \text{i.e. EVERY vector in the row-space is orthogonal to those in the null-space}$
$\text{Hence, it follows that the row space and the null space}\\ \text{are orthogonal complements of each other.}$
________________________________________
$\text{But a very standard linear algebra theorem says that}\\ \text{for any subspace }S, \, S \oplus S^\perp = V\\ \text{where }S^\perp\text{ is the orthogonal complement}\\ \text{and }V\text{ is the original vector space.}$
$\text{From properties of direct sums}\\ \text{this immediately implies that the union of the bases for}\\ \text{the row space and the column space}\\ \text{form a basis for }\mathbb{R}^n.$
Handwavy - Some results are assumed trivial and left as an exercise. Also potentially poorly explained with my 11:36PM dead brain.

Title: Re: First Year University Mathematics Questions
Post by: AngelWings on December 07, 2017, 09:40:42 pm

Started elsewhere:

Spoiler

Quote from: AngelWings on December 06, 2017, 08:05:06 pm

Still need help with Taylor series/ approximations here. Learnt it back in first year maths (MTH1030) and need to revise this. Forgotten most of it. Mostly I just need a proof and how it works again. I've also forgotten mostly about limits, so yeah... that'd be great if you could help. :)

Quote from: RuiAce on December 06, 2017, 08:27:12 pm

Which parts were you expected to prove? The existence of the $k$-th order Taylor expansion, or that if remainder -> 0 then f is represented by the Taylor series?

Quote from: AngelWings on December 07, 2017, 06:59:40 pm

To be honest, I’ve forgotten some of the basics. The most common one in these books, after a bit of dissecting, incorporates Taylor series on e^x, giving approximately 1 + x + x^2 + x^3 +... Not so sure how we got from Point A to B and would just like to see how to do it again, how we can prove this and so forth.

Quote from: RuiAce on December 07, 2017, 07:08:52 pm

$\text{But that's the series for }\frac{1}{1-x}.\\ \text{The Taylor series for }e^x\text{ is }1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\dots$
Did you mix some of them up?

That is entirely possible considering the only line of working I have been given is: e^x is approximately 1 + x where any term of order 2+ is ignored because it’d be ridiculously small and thus negligible. (X is meant to be tiny e.g. 10^- 8, hence why it’d be negligible. At least in the context where I got these from - a genetics book. See below.)

Quote from: AngelWings on December 07, 2017, 12:29:23 am

Intending on a theoretical genetics project for Honours, which involves some first year math, parts of which my memory stalls on. After previous experience, my intended supervisor advised that during this break, I should go through two genetics books. Both of them indirectly expect you to use Taylor approximations, which I can't remember how it works or how to do them. Hence the revision.

I got that answer because my notes are old and written. I must've been haphazardly copying them down quickly during the lectures. Must've missed the factorials.
Still not quite so sure how we would get the one your wrote for e^x above though, but maybe it's because I've forgotten large chunks of content.

Title: Re: First Year University Mathematics Questions
Post by: RuiAce on December 07, 2017, 10:00:25 pm

Quote from: AngelWings on December 07, 2017, 09:40:42 pm

Started elsewhere:
Spoiler
That is entirely possible considering the only line of working I have been given is: e^x is approximately 1 + x where any term of order 2+ is ignored because it’d be ridiculously small and thus negligible. (X is meant to be tiny e.g. 10^- 8, hence why it’d be negligible. At least in the context where I got these from - a genetics book. See below.)
I got that answer because my notes are old and written. I must've been haphazardly copying them down quickly during the lectures. Must've missed the factorials.
Still not quite so sure how we would get the one your wrote for e^x above though, but maybe it's because I've forgotten large chunks of content.

$\text{Well ok, when it comes to approximations}\\ \text{you just approximate using whatever's reasonable enough.}$
$\text{As for the series itself}\\ \text{perhaps the series of the following four functions should be memorised:}\\ \begin{align*}\frac{1}{1-x}&=1+x+x^2+x^3+\dots\\ e^x&=1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\dots\\ \sin x &= x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\dots\\ \cos x &= 1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\dots\end{align*}$
Optionally you may also memorise the one for $\ln(1-x)$, but that can be obtained by integrating $ \frac1{1-x}$.

Remark on first order approximations

This means, that for small $x$, the following are reasonable:
\begin{align*}e^x&\approx 1+x\\ \sin x &\approx x\\ \cos x&\approx 1\end{align*}

These four are generally regarded as the most useful. The first one really isn't anything fancy as it's literally just the geometric series, but exponentials and sinusoids appear quite common naturally. All the other stuff either stem out of these (e.g. $ \sinh$), are inverses of these (e.g. $\tan^{-1}$) or are just random shit mathematicians use for convenience.
$\text{With that being said, the actual computation of a Taylor series}\\ \text{is just done by quoting and using the formula.}$
$\text{In general, a Taylor polynomial about a point }x=a\text{ takes the form}\\ P_n(x)= \sum_{k=0}^n \frac{f^{(k)}(a)}{k!}(x-a)^k.\\ \text{Assuming the limit exists as }n\to \infty,\text{ a Taylor series is what happens when we do exactly that.}\\ \text{i.e. }P_{\infty}(x)=\sum_{k=0}^\infty\frac{f^{(k)}(a)}{k!}(x-a)^k$
We omit justification as to why that is the case for the sake of mere computations.
$\text{If the series about }x=0\text{ exists, i.e. taking }a=0,\\ \text{then }f\text{ is }\textit{exactly}\text{ represented by this series.}\\ \text{That is to say, }\boxed{f(x) = \sum_{k=0}^\infty\frac{f^{(k)}(0)}{k!}x^k}$
So all we really need to compute is $ f(0), f^\prime(0), f^{\prime\prime}(0), f^{\prime\prime\prime}(0), f^{(4)}(0) $ and pretty much, all of the derivatives of $f$, evaluated at 0.
The actual computation starts here.
$\text{But if }f(x) = e^x\text{ then this is easy.}\\ \text{We know that the derivative of }e^x\text{ is still }e^x.\\ \text{Hence, the first, second and so on derivatives must STILL be }e^x,\\ \text{i.e. }\boxed{f^{(k)}(x)=e^x}$
$\text{Which consequently implies }f^{(k)}(0)=e^0=1\\ \text{Hence, }\boxed{e^x = \sum_{k=0}^\infty \frac{1}{k!}x^k}$
Expanding this sum out gives $1 + \frac{x}{1!}+\frac{x^2}{2!}+\dots$ as required.

Small note - The factorials basically appear because they're actually a part of the Taylor series formula.

Title: Re: First Year University Mathematics Questions
Post by: legorgo18 on March 10, 2018, 02:15:10 pm

Hello rui, im really trying my best but these questions just arent clicking for me, ie 32 b,c (not even gonna try d or e with those threatening stars), 33 b, 35 for now!

Title: Re: First Year University Mathematics Questions
Post by: RuiAce on March 10, 2018, 02:32:10 pm

Quote from: legorgo18 on March 10, 2018, 02:15:10 pm

Hello rui, im really trying my best but these questions just arent clicking for me, ie 32 b,c (not even gonna try d or e with those threatening stars), 33 b, 35 for now!

$\text{One way of doing 32b is to interpret the problem like so:}\\ \text{Take }x\text{ to be some real number. If we want }y \in (x, \infty)\\ \text{then we really just want some }y\text{, to be }\textbf{greater than }x.$
$\text{That is, subject to the condition }y^2 - x^2 < 1.\\ \text{Try thinking about what's going on here. We actually have two inequalities now:}\\ \begin{align*}y &> x\\ y &< \sqrt{1+x^2} \end{align*}$
Note that I took the positive square root here I'll take the negative square root for 32c. The intuition behind this is that for 32c, I actually want $ y \in (-\infty, x)$, i.e. $ y < x$, so I'd prefer the negative square root for that one. But here I want $ y > x$, and hence i prefer the positive square root here.
$\text{So now I'm just gonna sketch the regions.}$
(https://i.imgur.com/t1wlUrU.png)
$\text{Note that if I want both of those inequalities to hold, }\textbf{for all}\, x \in \mathbb{R},\\ \text{I just need to check if for every }x\text{, there's some region that overlaps.}$
And if you look closely, there will always be a bit of the red region overlapping with the green region.
$\text{Thus, this choice of }y\text{ should exist.}$
____________________________________________________________________
$\text{Alternatively, I will also present an algebraic proof.}\\ \text{Given my two inequalities above, I can find an }\textbf{explicit choice}\text{ for }y\text{ to make this work.}$
$\text{I have the inequality }y^2 < 1+x^2.\\ \text{I also have the inequality }y > x\text{, which implies }y^2 > x^2.\\ \text{I can combine these two inequalities to give}\\ \boxed{x^2 < y^2 < 1+x^2}$
$\text{I will simply choose }\textbf{any}\text{ value of }y\text{ that works,}\\ \text{for example, }y^2 = \frac12 + x^2.\\ \text{Which then implies that }\boxed{y = \sqrt{\frac12 + x^2}}$
$\text{And pretty much, that's it. This is colloquially called "proof by construction"}\\ \text{where you }\textbf{give a concrete example}\text{ to prove a }\textbf{"there exists"}\text{ statement.}$

I'll let you try doing part c now whilst I look at the other questions.

Title: Re: First Year University Mathematics Questions
Post by: RuiAce on March 10, 2018, 03:03:06 pm

I'll get back to 33. That one is probably the hardest of the bundle you mentioned.
$\text{In 35b, we have}\\ \begin{align*}\sup_{n\in \mathbb{N}} \left( \frac12 + \frac14 + \dots + \frac1{2^n} \right)&= \sup_{n\in\mathbb{N}} \frac12 \left( \frac{1-2^{-n}}{1-2^{-1}}\right)\tag{geometric series formula}\\ &= \sup_{n\in \mathbb{N}} (1-2^{-n})\end{align*}$
$\text{So the set in which we want to find a least upper bound for is the set}\\ \left\{ x \in \mathbb{R} : x = 1-2^{-n} : n \in \mathbb{N} \right\}\\ \text{which is essentially the supremum of the sequence}\\ \left(\frac12, \frac34, \frac78, \frac{15}{16}, \dots \right)$
Now, by inspection this will just be 1. But let's argue this properly.
$\text{From high school maths, you know that the function }f(x) = 2^{-x}\text{ is monotonic increasing.}\\ \text{Hence, the sequence }b_n = 2^{-n}\text{ is also monotonic decreasing.}$
$\text{But therefore }- 2^{-n}\text{ is monotonic increasing, and thus}\\a_n = 1- 2^{-n}\text{ is also monotonic increasing.}\\ \text{Since this holds }\textbf{for all}\, n \in \mathbb{N},\text{ by this logic,}\\ \text{the supremum of this set should be the limit of }a_n,\\ \text{as }n\to \infty.$
$\therefore \sup_{n \in \mathbb{N}} (1-2^{-n}) = \lim_{n\to \infty} 1-2^{-n} = 1 - 0 = 1.$
_________________________________________________________________________
$\text{On the other hand, this sequence is less easy to consider.}\\ \therefore \text{Let }f : (0,\infty ) \to \mathbb{R}\text{ be the function given by}\\f(x) = \frac{\ln x}{x^{1/3}}$
$\text{The quotient rule gives}\\ \begin{align*}f^\prime (x) &= \frac{\frac{x^{1/3}}{x} - \frac13 x^{-2/3} \ln x}{x^{2/3}}\\ &= \frac{3-\ln x}{3x^{4/3}}\end{align*}$
$\text{Setting }f^\prime(x) = 0\text{ reveals the stationary point at }\left( e^3, 3e^{-1} \right).\\ \text{Furthermore, this stationary point can be proven to be a local maximum.}\\ \text{Since this is the }\textbf{only}\text{ stationary point, I can plot just the region }\textbf{around }x = e^3.$
(Note that I don't know what happens as $x \to \infty$, nor do I know what happens as $x \to 0^+$. GeoGebra does, but I don't.)
(https://i.imgur.com/ledWAZC.png)
$\text{Now what does that mean?}\\ \text{Note that the above reasoning should suggest that }\left( e^3, 3e^{-1}\right)\\ \text{should be the }\textbf{absolute maximum}\text{ of }y = f(x).$
$\text{This means, that }f(x) \le 3e^{-1}\textbf{ for all }x\in \mathbb{R}.\\ \text{You can't get any higher than }3e^{-1}\text{ along this curve.}\\ \text{So for the function }f(x)\text{, the supremum, which is also the maximum here, will be }3e^{-1}.$
$\text{Now what does that say for }a_n = \frac{\ln n}{n^{1/3}}?$
(https://i.imgur.com/DtVdZFK.png)
$\text{Because }a_n = f(n)\text{, just evaluated at the integers and not over all of }\mathbb{R},\\ \text{the max of the sequence should be something similar to the max of the function.}$
$\text{However not quite, since }e^3\text{ is not an integer!}\\ \text{Note that from the calculator, }20 < e^3 < 21.\\ \text{Thus, the max of the }\textbf{sequence}\text{, instead of the function, can only be at }n=20\text{ or }n=21.$
i.e. since the max of the function does not occur at an integer, the max of the sequence must occur at one of the two integers, bordering that original number.
$\text{Plugging into the calculator, we see that}\\ a_{20} \approx 1.1036\text{ and }a_{21}\approx 1.1035\\ \text{Thus, the max of the sequence, and consequently the sup of the sequence, is attained}\\ \text{at }n=20\text{, and hence }\sup_n \frac{\ln n}{n^{1/3}} = \frac{\ln 20}{20^{1/3}}$
(Side note: If the max exists, then the max must equal the sup. Notice how in the last problem the max didn't exist, because we asymptotically approached 1. Here, $ f(x) $ didn't just approach $ 3e^{-1}$, but it actually hit it.)

Title: Re: First Year University Mathematics Questions
Post by: RuiAce on March 10, 2018, 06:53:06 pm

Quote from: legorgo18 on March 10, 2018, 02:15:10 pm

Hello rui, im really trying my best but these questions just arent clicking for me, ie 32 b,c (not even gonna try d or e with those threatening stars), 33 b, 35 for now!

$\text{Ok, so for Q33, we first recall the definitions of monotonic increasing and decreasing.}\\ \text{A function }f: [a,b]\to \mathbb{R}\text{ is monotonic increasing if for any }x_1, x_2\in [a,b],\\ \text{If }x_1 < x_2\text{, then }f(x_1) < f(x_2).$
$\text{Similarly, }f:[a,b]\to \mathbb{R}\text{ is monotonic decreasing if for any }x_1,x_2\in [a,b],\\ \text{If }x_1 < x_2, \text{ then }f(x_1) > f(x_2).$
This may come as a surprise. In high school, you were taught that $f$ is a monotonic increasing function, if for any $ x \in (a,b) $ it satisfied $ f^\prime (x) \ge 0$, where $ f^\prime (x) $ is the derivative. And similarly, $ \le $ for decreasing. This is actually not a definition, but a theorem. The proof of this theorem requires a neat-ass formula known as the "mean value theorem", which you will learn later on.

But the idea is, we will borrow the theorem to "prove" something is monotonic decreasing, but then relate back to the actual definition.
________________________________________________
$\text{Let }f(t) = \frac{\ln t}{t} \implies f^\prime (t) = \frac{1 - \ln t}{t^2}.\\ \text{Observe that }f^\prime(t) \le 0\text{ whenever }t \le e.\\ \text{Hence }f\text{ is monotonic decreasing for all }t\in [e, \infty).$
$\text{Now, let }x\in [e, \infty).\\ \text{Take }x_1 = x\text{ and }x_2 = x+1.\text{ From the definition of monotonic decreasing,}\\ \text{since we know that }x_1 < x_2\text{, we also know that }f(x_1) >f(x_2).$
$\text{Hence,}\\ \begin{align*}\frac{\ln x}{x} &> \frac{\ln (x+1) }{x+1}\\ \implies (x+1) \ln x &> x \ln (x+1) \end{align*}$
which, coincidentally, means that $x$ is NOT in our set. Because our set is the set of all $p$ such that $ (p+1) \ln x < p \ln (p+1) $. Note that I'm just using $p$ to not mix up three different $x$'s in the same question.
$\text{Therefore, any real number }x\text{, that is greater than or equal to }e,\\ \text{is }\textbf{guaranteed}\text{ to be not in our set.}\\ \text{This means, only elements }\textbf{less}\text{ than }e\text{ can be in our set.}$
$\text{Thus, }e\text{ is an example of an upper bound for our set.}$
Remark: $e$ may not necessarily be the least upper bound, i.e. the supremum. In fact, the supremum is actually approximately at 2.293, which is certainly smaller than 2.718

Also - the lower bound for this set is pretty much 0, because you can't have negative elements in that set to begin with. (Reason - the domain restriction of the log function)

Title: Re: First Year University Mathematics Questions
Post by: Mackenzie2000 on April 05, 2018, 08:07:21 am

When working with matrices, and determining their geometric description, how do we know whether they are parallel, the same, or have the planes intersect in either a line or at a single point?
I am referring to both questions with three equations in three variable, and questions with two equations, but three variables.

Thank you in advance!!

Title: Re: First Year University Mathematics Questions
Post by: RuiAce on April 05, 2018, 08:45:43 am

Quote from: Mackenzie2000 on April 05, 2018, 08:07:21 am

When working with matrices, and determining their geometric description, how do we know whether they are parallel, the same, or have the planes intersect in either a line or at a single point?
I am referring to both questions with three equations in three variable, and questions with two equations, but three variables.

Thank you in advance!!

Essentially, the geometric interpretation of the solutions (and hence the original planes) depends on the nature of the solutions we had found.

In the 3 equation, 3 variable case, we are considering 3 planes in $ \mathbb{R}^3$. So we have the following:
- There is no solution. When this is the case, we're saying that the 3 planes have no common point of intersection. This could be because they appear to form a triangular prism between them, or because at least 2 out of the 3 planes are parallel to one another.
- There is one unique solution. When this is the case, the planes are essentially intersecting at a common point. An easy example of this is just the x-y, y-z and x-z planes, which only intersect at (0,0,0).
- There are infinite solutions. Infinite solutions are characterised by the number of parameters there are in our solution.
- If there is only one parameter, say $t$, then the solutions are of the form $ \textbf{x} = a + t\textbf{v} $. This represents a line in $\mathbb{R}^3$, passing through the point $A$ and in the direction of the vector $ \textbf{v} $. In general, the intersection forms a lie when the planes are just rotations of one another, but all 3 of them aren't the same plane as in the case below.
- If there are two parameters, say $\lambda$ and $\mu$, then the solutions are of the form $ \textbf{x} = \textbf{a}+ \lambda \textbf{u} + \mu \textbf{v} $. This now represents a plane in $\mathbb{R}^3$, which passes through $A$ and is spanned by $ \textbf{u} $ and $ \textbf{v} $. However, this is only possible if the three planes at the start actually all coincided with one another.

I find that figure 1.1.2 of this article gives a good visual representation of them.

On the other hand, when we have 2 equations, we're simply saying we now have only 2 planes in $ \mathbb{R}^3$. Because there's only 2 planes, our possibilities are severely narrowed
- The only way there can be no solutions is if the planes are parallel to one another.
- There can never be a unique solution. (Why?) If two planes in $ \mathbb{R}^3$ are not parallel then they must have infinite solutions.
- If there are two parameters, then once again the planes coincided.
- If there is only one parameter, again the intersection is a line. This is essentially every case when the planes aren't parallel with one another.

Title: Re: First Year University Mathematics Questions
Post by: sophieren on April 05, 2018, 11:19:33 am

Hi,
I was wondering if you are able to solve these questions for me? It would be greatly appreciated!

1) In the triangle ABC, AB = sqrt(2), AC = 1/sqrt(2) and the angle at A is 60◦. Find the length of BC and the size of the angle at C.
2) If sin A = 3/5 where 90◦ < A < 180◦ and cos B = 5/13 where 0◦ < B < 90◦, find sin(A − B)
3) In the triangle ABC, AB = 2, BC = 1 + sqrt(3) and the angle at B is 30◦, Find the length of AC and the size of the angle at C.
4) If sin A = 5/13 where 90◦ < A < 180◦ and cos B = 3/5 where 0◦ < B < 90◦, find sin(A − B).

Thanks!

Title: Re: First Year University Mathematics Questions
Post by: TrueTears on April 05, 2018, 02:38:50 pm

1) is just a straightforward application of the cosine rule. 2) and 4) are applications of the sine difference rule. Have you had a go at applying those?

Title: Re: First Year University Mathematics Questions
Post by: RuiAce on April 05, 2018, 02:47:37 pm

Quote from: sophieren on April 05, 2018, 11:19:33 am

Hi,
I was wondering if you are able to solve these questions for me? It would be greatly appreciated!

1) In the triangle ABC, AB = sqrt(2), AC = 1/sqrt(2) and the angle at A is 60◦. Find the length of BC and the size of the angle at C.
2) If sin A = 3/5 where 90◦ < A < 180◦ and cos B = 5/13 where 0◦ < B < 90◦, find sin(A − B)
3) In the triangle ABC, AB = 2, BC = 1 + sqrt(3) and the angle at B is 30◦, Find the length of AC and the size of the angle at C.
4) If sin A = 5/13 where 90◦ < A < 180◦ and cos B = 3/5 where 0◦ < B < 90◦, find sin(A − B).

Thanks!

Essentially Q1 and Q3 are just high school trigonometry - the main thing you require is the cosine rule (and possibly the sine rule.) Since Q4 follows the exact same as Q2, I will only do Q2.
$\text{Note that }A\text{ is a second quadrant angle.}\\ \text{For this reason, }\sin A > 0\text{ and }\cos A < 0.$
$\text{Draw a triangle with hypotenuse 5 and opp side 3.}\\ \text{Then the adj side has length 4,}\\ \text{so }\cos A = -\frac{4}{5}.$
$\text{On the other hand }B\text{ is a first quadrant angle, so both }\sin B > 0\text{ and }\cos B > 0.\\ \text{Drawing another triangle, we see that }\sin B = \frac{12}{13}.$
$\text{From here, simply expand the compound angle and substitute in.}\\ \begin{align*}\sin (A-B) &= \sin A \cos B - \cos A \sin B\\ &= \frac35 \times \frac5{13} +\frac45\times\frac{12}{13}\\ &= \frac{63}{65}\end{align*}$

Title: Re: First Year University Mathematics Questions
Post by: 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?

Title: Re: First Year University Mathematics Questions
Post by: RuiAce 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} $.
$\text{But of course, we want when }\sin x \le \frac12.\\ \text{This means that when we plot the graphs }y=\sin x\text{ and }y=\frac12,\\ \text{we want whenever the sinusoid lies }\textbf{below}\text{ the line.}$
(https://i.imgur.com/eBBLQaR.png)
$\text{Reading off the graph, it appears that we must have any of ALL of the following.}\\ -\frac{7\pi}6 \le x \le \frac\pi6\\ \frac{5\pi}{6} \le x \le \frac{13\pi}{6}\\ \frac{17\pi}{6} \le x \le \frac{25\pi}{6}\\ \text{..and so on, going backwards as well.}$
$\text{So the domain is actually the union of this infinite number of intervals:}\\ \dots \cup \left[ -\frac{7\pi}{6}, \frac\pi6\right] \cup \left[ \frac{5\pi}{6}, \frac{13\pi}{6}\right]\cup \left[ \frac{17\pi}{6}, \frac{25\pi}{6}\right]\cup \dots$
$\text{One possible way of writing this more compactly is}\\ \bigcup_{k=-\infty}^\infty \left[ -\frac{7\pi}{6} + 2k\pi, \frac\pi6 + 2k\pi \right].\\ \text{But you don't need to convert to this form.}$
_________________________________________________________
$\text{Now for the range, }y\ge 0\text{ is certainly important as you have stated.}$
$\text{But of course the range isn't just }[0,\infty).\\ \text{There's no way we can make an ordinary trig function just explode, unless we flip it.}$
(By ordinary I really just mean $\cos$ and $\sin$.)
$\text{Pretty much, to find where it caps off, we just need to find the}\\ \text{maximum value of }f(x)=\sqrt{1-2\sin x}.\\ \text{Since this is a trig question, we will just use 'common sense' instead of calculus.}$
$\sin x\text{ has maximum value at }1.\\ \text{Therefore }2\sin x\text{ has maximum value at }2.\\ \text{Therefore }-2\sin x\text{ has maximum value at 2 as well. (Why?)}$
$\text{So if }2\text{ is the maximum value of }-2\sin x,\\ 3\text{ is the maximum value of }1-2\sin x.\\ \text{Therefore }\sqrt{3}\text{ must be the maximum value of }\sqrt{1-2\sin x}.$
$\text{Hence, the range is }[0,\sqrt3]$

Title: Re: First Year University Mathematics Questions
Post by: legorgo18 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?

Title: Re: First Year University Mathematics Questions
Post by: RuiAce 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.

Title: Re: First Year University Mathematics Questions
Post by: 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)?

Title: Re: First Year University Mathematics Questions
Post by: RuiAce 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?

Title: Re: First Year University Mathematics Questions
Post by: swico 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.

Title: Re: First Year University Mathematics Questions
Post by: TrueTears 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.

Title: Re: First Year University Mathematics Questions
Post by: swico 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 :)

Title: Re: First Year University Mathematics Questions
Post by: RuiAce 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.

Title: Re: First Year University Mathematics Questions
Post by: swico 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.

Title: Re: First Year University Mathematics Questions
Post by: WrongWong 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 !

Title: Re: First Year University Mathematics Questions
Post by: RuiAce 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

Title: Re: First Year University Mathematics Questions
Post by: K98100 on May 20, 2018, 09:43:08 pm

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

Title: Re: First Year University Mathematics Questions
Post by: RuiAce 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.

Title: Re: First Year University Mathematics Questions
Post by: goodluck on February 18, 2019, 09:09:13 pm

Hey,

It took me a while to find this question page but I just have this question and was wondering how you do it via vectors. I didn't really understand any of the online working out!

Prove using vector methods that the midpoints of the sides of a convex quadrilateral form a parallelogram.

Title: Re: First Year University Mathematics Questions
Post by: RuiAce on February 18, 2019, 09:35:17 pm

Quote from: goodluck on February 18, 2019, 09:09:13 pm

Hey,

It took me a while to find this question page but I just have this question and was wondering how you do it via vectors. I didn't really understand any of the online working out!

Prove using vector methods that the midpoints of the sides of a convex quadrilateral form a parallelogram.

The one thing that you might not fully understand due to limitations of the HSC is what a "position vector" is yet. For the sake of these proofs we can assume that the space we're working in is some Euclidean space, and to simplify things further we can assume it's just $ \mathbb{R}^2$ or $\mathbb{R}^3$. (You can interpret these as the usual 2-dimensional Cartesian plane, and the 3-dimensional Cartesian space. The 3D version basically also has a $z$-axis.)

A position vector is essentially the vector drawn from the origin to the point you're interested in. (i.e. the vector $ \overrightarrow{OP}$.) Usually, if we have a point $P$, its position vector is denoted by $\mathbf{p}$. You're welcome to think of it as an arrow drawn from the origin to $P$.

However, if any point is somehow related to $P$, its position vector is related to $\mathbf{p}$. For example, if $P$ was the point $ (2,3)$, and you wanted the position vector of $(4,6)$, it would actually be $2\mathbf{p}$ because you doubled both the $x$ and $y$ components. If you look on a diagram, the vector from $(0,0)$ to $(4,6)$ should also look exactly twice as long as the one to $(2,3)$, and also pointing in the same direction.

Everything else, however, you kinda need to learn. Anyway here's the proof.
\[ \text{Let }ABCD\text{ be a convex quadrilateral and suppose that}\\ A,\, B,\, C\text{ and }D\text{ have position vectors}\\ \mathbf{a}\, \mathbf{b},\, \mathbf{c}\text{ and } \mathbf{d}\text{ respectively.} \]
\[ \text{Then the midpoints have position vectors}\\ M_{AB} \text{ being }\frac12 (\mathbf{a}+\mathbf{b})\\ M_{BC}\text{ being }\frac12 (\mathbf{b}+\mathbf{c})\\ M_{CD}\text{ being }\frac{1}{2}(\mathbf{c}+\mathbf{d})\\ M_{CD}\text{ being }\frac12(\mathbf{d}+\mathbf{a})\]
This is not a coincidence. This is actually just the midpoint formula you learn in 2U in disguise.
\[ \text{Now, recall that given position vectors}\\ \text{the vector }\overrightarrow{AB}\text{ is the same as the vector }\mathbf{b}-\mathbf{a}.\]
This is a consequence of vector addition. Using tip-to-tail vector addition, we can rewrite $\overrightarrow{AB} = \overrightarrow{AO} + \overrightarrow{OB} =- \overrightarrow{OA} + \overrightarrow{OB} = -\mathbf{a} + \mathbf{b}$. Note that flipping a vector is essentially the same as taking negatives of vectors.
\[ \text{So in our case, we have}\\ \begin{align*} \overrightarrow{M_{AB}M_{BC}} &= \frac12 (\mathbf{b}+\mathbf{c}) - \frac12 (\mathbf{a}+\mathbf{b})\\ &= \frac12 (\mathbf{c} - \mathbf{a})\end{align*}\]
\[ \text{Similarly,}\\ \begin{align*}\overrightarrow{M_{CD}M_{DA}}&= \frac12 (\mathbf{d}+\mathbf{a}) - \frac12 (\mathbf{c} +\mathbf{d}) \\ &= \frac12 (\mathbf{a}-\mathbf{c})\end{align*}\]
\[ \text{But consequently }\overrightarrow{M_{AB}M_{BC}} = -\overrightarrow{M_{CD}M_{DA}}. \]
At this point we are essentially done. Note that the line segments $ M_{AB}M_{BC}$ and $M_{CD}M_{DA}$ are opposite sides of the quadrilateral, that we're hoping to prove is a parallelogram. But if the vectors are negatives of each other, that means that they must have the same magnitude, and only point in opposite directions.

The vectors pointing in opposite directions doesn't matter though. They're still parallel as well.

So consequently, we essentially have a pair of equal and opposite sides. This is one of the many tests to prove that a quadrilateral is a parallelogram.

Note: This is definitely not the only approach. It's just my preferred approach.

Title: Re: First Year University Mathematics Questions
Post by: justwannawish on February 24, 2019, 07:51:13 pm

Hey, I just have two questions, one of which had been posted on the forum previously. I'm a bit confused about the difference between the two solutions of 32b and 32c if done algebraically. I read and understood the solution for 32b but I'm not sure what the difference the new domain would create

for c) x will have to be smaller than y, right, so if we compute it out, wouldn't it still give up the same algebraic solution as b) did? or would the signs be flipped as we are considering it from the negative side?
I also tried to graphically solve it and have attached it. Could you see if I'm on the right track?

My second question is also attached. I'm just a bit unsure of how it works though I get why theta=0 is a solution ofc.

Title: Re: First Year University Mathematics Questions
Post by: RuiAce on February 24, 2019, 08:09:25 pm

Quote from: justwannawish on February 24, 2019, 07:51:13 pm

Hey, I just have two questions, one of which had been posted on the forum previously. I'm a bit confused about the difference between the two solutions of 32b and 32c if done algebraically. I read and understood the solution for 32b but I'm not sure what the difference the new domain would create

for c) x will have to be smaller than y, right, so if we compute it out, wouldn't it still give up the same algebraic solution as b) did? or would the signs be flipped as we are considering it from the negative side?
I also tried to graphically solve it and have attached it. Could you see if I'm on the right track?

My second question is also attached. I'm just a bit unsure of how it works though I get why theta=0 is a solution ofc.

Firstly note that your rearranged condition is off. You're right about that $y^2 < x^2 + 1$, but this quadratic inequality solves to give $ \boxed{-\sqrt{x^2-1} < y < \sqrt{x^2+1}} $. On Desmos, you can just type the original condition $x^2-y^2 < 1$ and the correct plot will still show.

Anyway. The order quantifiers mean everything here. Whilst it turns out both statements are true, the positioning of the quantifiers mean that they mean different things.

The first says that for all real numbers $x$ there exists some $y$ where $ y > x$, such that $y^2-x^2 < 1$.
This means that $x$ has to be any arbitrary real number. And if the statement is true, we should be able to pick some $y$ (very likely in terms of $x$) that will satisfy the condition $y^2 - x^2 < 1$.

The second says that there exists a real number $y$ (not necessarily unique), such that for all $x$ where $x < y$, such that $y^2-x^2 <1$.
This means that this time, we need to choose some real number $y$ (and it can't be in terms of anything). That $y$, then has to guarantee us that regardless of what $x$ is, the condition $y^2-x^2<1$ is satisfied.

So for the first one, you have to start by assuming $x$ is whatever real number you want it to be. Then, you need to choose some $y\in (x,\infty)$ that will work. One example that will work is $ y = \sqrt{x^2 + \frac12} $ - not sure what the answers they give is but that's an example of one of them.
(On Desmos, you can try plotting the required condition $y>x$, the condition that we want i.e. $ y ^2<x^2+1$, and the curve $y = \sqrt{x^2+\frac12}$. You'll notice that the curve lies completely between where the two regions intersect, thus affirming this answer is valid.)

Whereas for the second one, you need to commence by choosing some $y$ that is guaranteed to always ensure that if $x < y$, then $y^2-x^2 < 1$. $y=0$ should be an intuitively obvious answer here, because then $y^2 - x^2 = -x^2$, which is always less than 1 regardless of what $x$ is. (The condition that $x\in (-\infty,y)$ coincidentally happens to not be that important, because it actually holds for all $x\in \mathbb{R}$ anyway.)
(This time, on Desmos, to affirm this result we have to plot the line $y=0$. What we want to check is that no matter what $x$ is, the condition $y^2-x^2 < 1$ is satisfied. Converting this graphically, we want to therefore check that the line $y=0$, is always inside the region $y^2-x^2 < 1$ (and doesn't, say, suddenly pop out of it at some value of $x$. You can check that this will indeed be the case.)

Note that the condition is always $x < y$ - this stays the same more or less as a consequence of how the questions were written. But the approaches differ because of how the quantifiers are provided.
______________________________________________

With the other question, you can just note that $y=\sin \theta$ is contained entirely in the 1st and 3rd quadrants in that domain, whilst $y = m\theta$ is contained entirely in the 2nd and 4th quadrants under the condition that $m < 0$. Also note that whilst $\sin \theta = 0$ when $\theta = -\pi, 0, \pi$, on the other hand $m\theta = 0$ only when $\theta = 0$.

Title: Re: First Year University Mathematics Questions
Post by: justwannawish on February 28, 2019, 05:56:20 pm

It took me a while but I actually fully understand that so thank you for your extremely in-depth explanation :)

I was trying to figure out how to do this question, but I'm confused on what we do regarding the exponential/logarithm form.

Express the rational function (27x^3+1)/ (x^(3/2)) in terms of a hyperbolic function, logarithms and other functions as needed, for x>0.

Title: Re: First Year University Mathematics Questions
Post by: RuiAce on March 01, 2019, 01:34:01 pm

Quote from: justwannawish on February 28, 2019, 05:56:20 pm

It took me a while but I actually fully understand that so thank you for your extremely in-depth explanation :)

I was trying to figure out how to do this question on the matlab quiz, but I'm confused on what we do regarding the exponential/logarithm form.

Express the rational function (27x^3+1)/x^3/2 in terms of a hyperbolic function, logarithms and other functions as needed, for x>0.

(P.S. This is a generic thread and not just for UNSW students. Not everyone will know about the MapleTA quizzes.)

Assuming that there's no typo. Also assuming that the 3/2 is in the bottom power - please make this a bit clearer in the future.
\[ \text{Let }\boxed{e^t = \sqrt{27}x^{3/2}}\implies \boxed{t = \ln \left(\sqrt{27}x^{3/2} \right)}.\text{ Then,}\\ \begin{align*}\frac{27x^3+1}{x^{3/2}} &= \tag{clever factorising}\frac{27^{1/2} x^{3/2} \left(27^{1/2} x^{3/2} + 27^{-1/2} x^{-3/2} \right)}{x^{3/2}}\\ &= \sqrt{27} (e^t + e^{-t})\\ &= 2\sqrt{27} \cosh t\\ &= 2\sqrt{27} \cosh \left( \ln \left(\sqrt{27}x^{3/2}\right) \right) \end{align*} \]
Optionally, we may use log laws to re-express what's on the inside as $ \frac32 \ln (3x) $.

Note that the factorisation was the critical step. In general, because $ \cosh t = \frac{e^t+e^{-t}}{2}$, we need to have some $ f(x) + \frac1{f(x)}$ pattern appearing, before we can sub $e^t = f(x)$. Same goes for $\sinh t$ except we have a minus instead.

Title: Re: First Year University Mathematics Questions
Post by: justwannawish on March 01, 2019, 06:45:09 pm

Quote from: RuiAce on March 01, 2019, 01:34:01 pm

(P.S. This is a generic thread and not just for UNSW students. Not everyone will know about the MapleTA quizzes.)

Assuming that there's no typo. Also assuming that the 3/2 is in the bottom power - please make this a bit clearer in the future.
\[ \text{Let }\boxed{e^t = \sqrt{27}x^{3/2}}\implies \boxed{t = \ln \left(\sqrt{27}x^{3/2} \right)}.\text{ Then,}\\ \begin{align*}\frac{27x^3+1}{x^{3/2}} &= \tag{clever factorising}\frac{27^{1/2} x^{3/2} \left(27^{1/2} x^{3/2} + 27^{-1/2} x^{-3/2} \right)}{x^{3/2}}\\ &= \sqrt{27} (e^t + e^{-t})\\ &= 2\sqrt{27} \cosh t\\ &= 2\sqrt{27} \cosh \left( \ln \left(\sqrt{27}x^{3/2}\right) \right) \end{align*} \]
Optionally, we may use log laws to re-express what's on the inside as $ \frac32 \ln (3x) $.

Note that the factorisation was the critical step. In general, because $ \cosh t = \frac{e^t+e^{-t}}{2}$, we need to have some $ f(x) + \frac1{f(x)}$ pattern appearing, before we can sub $e^t = f(x)$. Same goes for $\sinh t$ except we have a minus instead.

Sorry about that! I didn't mean to restrict it to UNSW and I apologise if anyone thought that. I've edited my above post to make it clearer (I hope I'm not bothering you or anyone who checks this thread by sending too many questions, it's just a bit tough atm, but I'll try my best to keep up with the course work!)

ATAR Notes: Forum

Uni Stuff => General University Discussion and Queries => Topic started by: Shadowxo on September 10, 2017, 11:04:09 pm