FREE lectures this September + October | HSC Head Start + Exam Revision: book here | QCE Head Start: book here | VCE Exam Revision: book here

September 17, 2019, 08:19:56 pm

### AuthorTopic: "Challenge" Math Qs - Can You Figure It Out?  (Read 3469 times) Tweet Share

0 Members and 1 Guest are viewing this topic.

#### RuiAce

• ATAR Notes Lecturer
• Honorary Moderator
• Great Wonder of ATAR Notes
• Posts: 8493
• "All models are wrong, but some are useful."
• Respect: +2347
##### Re: "Challenge" Math Qs - Can You Figure It Out?
« Reply #45 on: December 28, 2018, 04:03:00 pm »
+3
Quote tags intentionally removed, but for the sake of reference this is basically from the post immediately above.
\begin{align*}\sin{5x}&=\text{Im}[\cos{5x}+i\sin{5x}]\\&=\text{Im}[(\cos{x}+i\sin{x})^5]\\&=\text{Im}[(\cos{x})^5+5(\cos{x})^4i(\sin{x})+10(\cos{x})^3i^2(\sin{x})^2+10(\cos{x})^2i^3(\sin{x})^3+5(\cos{x})i^4(\sin{x})^4+i^5(\sin{x})^5]\\&=5(\cos{x})^4(\sin{x})-10(\cos{x})^2(\sin{x})^3+(\sin{x})^5\\&=5(1-(\sin{x})^2)^2(\sin{x})-10((\sin{x}))(\sin{x})^3+(\sin{x})^5\\&=5(\sin{x})-10(\sin{x})^3+5(\sin{x})^4+10(\sin{x})^5-10(\sin{x})^3+(\sin{x})^5\\&=16(\sin{x})^5-20(\sin{x})^3+5(\sin{x})\end{align*} \text{Subbing in} \ \frac{2\pi}{5} \ \text{into the left hand side, we have the sine of} \ 2\pi \ \text{which is zero, which means we are essentially solving for the roots of the right hand side.} \\ 16\left(\sin{\frac{2\pi}{5}}\right)^5 20\left(\sin{\frac{2\pi}{5}}\right)^3+5\left(\sin{\frac{2\pi}{5}}\right)=0 \\ \left(\sin\left({\frac{2\pi}{5}}\right)\right)\left(16\left(\sin{\frac{2\pi}{5}}\right)^4 20\left(\sin{\frac{2\pi}{5}}\right)^2+5\left(\sin{\frac{2\pi}{5}}\right)\right)=0 \\ \text{Knowing that the sine function only has roots at} \ n\times\pi \ \text{we can rule out} \ \left(\sin{\frac{2\pi}{5}}\right)=0 \ \text{as a solution.} \\ $\left(16\left(\sin{\frac{2\pi}{5}}\right)^4-20\left(\sin{\frac{2\pi {5}}\right)^2+5\left(\sin{\frac{2\pi}{5}}\right)\right)=0$\\ $\text{Let} \ \left(\sin^2{\frac{2\pi}{5}}\right) = a \\ 16a^2-20a+5=0 \\ a=\frac{20\pm\sqrt{400-320}}{32}$\\$a=\frac{20\pm4\sqrt{5}}{32} \left(\sin^2{\frac{2\pi}{5}})= \frac{5\pm\sqrt{5}}{8}$ \\$\left(\sin{\frac{2\pi}{5}}) = \sqrt{\frac{5\pm\sqrt{5}}{8}}$

The derivation derivation is coming, this question just seemed a lot friendlier.

EDIT: working on that right now, but also if you guys have time can you try and find the mistakes in the code? I've been trying for about 45 minutes (15 minutes typing, 45 minutes looking for errors) and I can't find them, super frustrating
end quote

The enter keys in the align* environment damaged most of the code. Once I took all of them out a lot of it fixed itself automatically. Following that I've chopped your code up into bits and pieces (you can use Ctrl+F and search tex) to find where the chopping up occurred. From what I can see, some \right) 's are missing!

Also a set of newline \\ characters are missing after your align* environment I think

#### fun_jirachi

• MOTM: AUG 18
• Forum Obsessive
• Posts: 484
• All doom and Gloom.
• Respect: +260
##### Re: "Challenge" Math Qs - Can You Figure It Out?
« Reply #46 on: December 28, 2018, 04:23:01 pm »
+2
Okay, thanks so much Rui! I'll edit that right now, it looks awful.

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
Page 2
Failing everything, but I'm still Flareon up.

HSC 2018: Modern History [88] | 2U Maths [98]
HSC 2019: Physics | Chemistry | English Advanced | Maths Extension 1 | Maths Extension 2

#### RuiAce

• ATAR Notes Lecturer
• Honorary Moderator
• Great Wonder of ATAR Notes
• Posts: 8493
• "All models are wrong, but some are useful."
• Respect: +2347
##### Re: "Challenge" Math Qs - Can You Figure It Out?
« Reply #47 on: December 28, 2018, 04:41:13 pm »
+3
Okay, thanks so much Rui! I'll edit that right now, it looks awful.

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
Page 2
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.
\text{Claim: Given a sequence defined by a quadratic function, the common difference of }\textit{those}\text{ terms form an A.P.}\\ \text{Proof: Let }T_k = f(k)\text{ where }f(x) = ax^2+bx+c.\text{ Then,}\\ \begin{align*}T_{k+1} - T_k &= f(k+1) - f(k)\\ &= a(k+1)^2 + b(k+1) + c - ak^2 - bk - c\\ &= 2ak + (1+b) \end{align*}\\ \text{hence the differences form an arithmetic progression with common difference }2a.
You can then replicate this proof to make your claim about the common difference with sequences defined by cubic equations. That's enough to suggest that $T_k = \text{some cubic function in terms of }k$ can be used to model the scenario you have right there! And of course, nothing else to comment on with the simultaneous equations.

The usual method is to consider the expression $\sum_{k=1}^n (k+1)^3 + \sum_{k=1}^n (k-1)^3$. On one hand,
\begin{align*} \sum_{k=1}^n (k+1)^3 - \sum_{k=1}^n (k-1)^3 &= \sum_{k=1}^n [(k+1)^3 - (k-1)^3]\\ &\vdots\\ &= 2\sum_{k=1}^n k^2 + 2\sum_{k=1}^n1 \end{align*}
And on the other hand,
\begin{align*}\sum_{k=1}^n (k+1)^3 - \sum_{k=1}^n (k-1)^3 &=[2^3 + \dots + (n-1)^3+ n^3 + (n+1)^3] - [0^3 + 1^3 + 2^3 + \dots + (n-1)^3]\\ &= n^3 + (n+1)^3 - 1^3\end{align*}
You can fill in the details of this proof if you want to or someone else can

#### AlphaZero

• MOTM: DEC 18
• Forum Obsessive
• Posts: 319
• $[\mathcal{L}f](s)=\int_0^\infty\!e^{-st}f(t)dt$
• Respect: +131
##### Re: "Challenge" Math Qs - Can You Figure It Out?
« Reply #48 on: December 28, 2018, 04:52:27 pm »
+3
Fixing code (and condensing it) for @fun_jirachi. Few closing brackets missing and random returns in align environment where there shouldn't be one.
________________________________________________________________
\begin{align*}\sin(5x)&=\text{Im}\Big[\cos(5x)+i\sin(5x)\Big]\\
&=\text{Im}\Big[\big(\cos(x)+i\sin(x)\big)^5\Big]\\
&=\text{Im}\Big[\cos^5(x)+5\cos^4(x)\cdot i\sin(x)+10\cos^3(x)\cdot i^2\sin^2(x)+10\cos^2(x)\cdot i^3\sin^3(x)+5\cos(x)\cdot i^4\sin^4(x)+i^5\sin^5(x)\Big]\\
&\vdots\\
&=16\sin^5(x)-20\sin^3(x)+5\sin(x)\end{align*}
Subbing in $x=\dfrac{2\pi}{5}$ gives $0=16\sin^5\left(\frac{2\pi}{5}\right)-20\sin^3\left(\frac{2\pi}{5}\right)+5\sin\left(\frac{2\pi}{5}\right).$ Since $\sin\left(\dfrac{2\pi}{5}\right)\neq0$, we have $0=16\sin^4\left(\frac{2\pi}{5}\right)-20\sin^2\left(\frac{2\pi}{5}\right)+5.$
\begin{align*}\sin^2\left(\frac{2\pi}{5}\right)&=\frac{20\pm\sqrt{400-320}}{32}\\
&=\frac{5\pm\sqrt{5}}{8}\\
&\vdots\\
&\vdots\ \ (\text{more work required.}) \end{align*}
_______________________________________________________________

I'm going to stop it here since your solution isn't complete.
$\sin^2\left(\dfrac{2\pi}{5}\right)$ can't be two values at the same time and you also need to justify taking the positive root both times.
2015$-$2017:  VCE
2018$-$2021:  Bachelor of Biomedicine and Concurrent Diploma in Mathematical Sciences, University of Melbourne

#### fun_jirachi

• MOTM: AUG 18
• Forum Obsessive
• Posts: 484
• All doom and Gloom.
• Respect: +260
##### Re: "Challenge" Math Qs - Can You Figure It Out?
« Reply #49 on: December 28, 2018, 06:05:50 pm »
+3
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.

Sorry for the repost, but this is the full solution (given that Dan kindly pointed out that I'm missing a few things at the end (oops, totally uncharacteristic of me to overlook such obvious things)). Also, it's a lot neater than the pile of trash I posted a few hours ago

\begin{align*}\sin(5x)
&=\text{Im}[\cos(5x)+i\sin(5x)]
\\&=\text{Im}\left[(\cos(x)+i\sin(x))^5\right]
\\&=\text{Im}\left[(\cos(x))^5+5(\cos(x))^4i(\sin(x))+10(\cos(x))^3i^2(\sin(x))^2+10(\cos(x))^2i^3(\sin(x))^3+5(\cos(x))i^4(\sin(x))^4+i^5(\sin(x))^5\right]
\\&=5(\cos(x))^4(\sin(x))-10(\cos(x))^2(\sin(x))^3+(\sin(x))^5
\\&=5\left(1-(\sin(x))^2\right)^2(\sin(x))-10(\sin(x))(\sin(x))^3+(\sin(x))^5
\\&=5(\sin(x))-10(\sin(x))^3+5(\sin(x))^4+10(\sin(x))^5-10(\sin(x))^3+(\sin(x))^5
\\&=16(\sin(x))^5-20(\sin(x))^3+5(\sin(x))
\end{align*}
\\ \text{Subbing in} \ x=\frac{2\pi}{5} \ \text{we have} \ 16\left(\sin\left(\frac{2\pi}{5}\right)\right)^5-20\left(\sin\left(\frac{2\pi}{5}\right)\right)^3+5\left(\sin\left(\frac{2\pi}{5}\right)\right)=0
\\ \left(\sin\left(\frac{2\pi}{5}\right)\right)\left(16\left(\sin\left(\frac{2\pi}{5}\right)\right)^4-20\left(\sin\left(\frac{2\pi}{5}\right)\right)^2+5\right)
\\ \left(\sin\left(\frac{2\pi}{5}\right)\right)=0 \ \text{is not a solution since the sine curve only has solutions at multiples of pi, so then we have}
\\ \left(16\left(\sin\left(\frac{2\pi}{5}\right)\right)^4-20\left(\sin\left(\frac{2\pi}{5}\right)\right)^2+5\right)=0
\\ \text{Let} \ a=\sin^2\left(\frac{2\pi}{5}\right)
\\ 16a^2-20a+5=0
\\ a = \frac{20\pm\sqrt{400-320}}{32}
\\ a = \frac{5\pm\sqrt{5}}{8}
\\ \text{So basically you verify that both solutions are between 0 and 1 (the range of} \ \sin^2x \text{), which they are.}
\\ \text{Noting that} \ \frac{2\pi}{5} < \ \frac{\pi}{2} \ \text{and that} \ \sin^2x \ \text{is monotonic increasing over the same domain basically just pick the bigger one, which is the positive root because} \ \frac{2\pi}{5} \ \text{is closer to} \ \frac{\pi}{2} \ \text{than} \ 0.
\\ \text{(Intuitively this is just true)}
\\ \text{Then} \ \sin^2\left(\frac{2\pi}{5}\right) = \frac{5+\sqrt{5}}{8}
\\ \sin\left(\frac{2\pi}{5}\right) = \pm\sqrt{\frac{5+\sqrt{5}}{8}}
\\ \text{And you take the positive root because} \ \sin\left(\frac{2\pi}{5}\right) \ \text{is a first quadrant angle and must be positive.}
\\ \sin\left(\frac{2\pi}{5}\right)=\sqrt{\frac{5+\sqrt{5}}{8}}

EDIT:
You can fill in the details of this proof if you want to or someone else can

After today's frustration both at my lack of coding skill and intuition, I literally can't be bothered.

« Last Edit: December 28, 2018, 06:08:28 pm by fun_jirachi »
Failing everything, but I'm still Flareon up.

HSC 2018: Modern History [88] | 2U Maths [98]
HSC 2019: Physics | Chemistry | English Advanced | Maths Extension 1 | Maths Extension 2

#### RuiAce

• ATAR Notes Lecturer
• Honorary Moderator
• Great Wonder of ATAR Notes
• Posts: 8493
• "All models are wrong, but some are useful."
• Respect: +2347
##### Re: "Challenge" Math Qs - Can You Figure It Out?
« Reply #50 on: December 28, 2018, 06:14:28 pm »
+3
snip
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!

#### fun_jirachi

• MOTM: AUG 18
• Forum Obsessive
• Posts: 484
• All doom and Gloom.
• Respect: +260
##### Re: "Challenge" Math Qs - Can You Figure It Out?
« Reply #51 on: December 28, 2018, 06:28:09 pm »
+1
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 $\sin^2 \frac{2\pi}{5} = \frac{5+\sqrt5}{8}$

Huh, I really don't know why that didn't occur to me. Every time I read what I wrote, I feel more and more stupid, but I'll keep inequalities in mind when doing stuff like this in the future.

Man, you have so much time on your hands with these questions! You must be done with all two of your TuteSmart assignments already!

Nope, just a fat cbbs
Failing everything, but I'm still Flareon up.

HSC 2018: Modern History [88] | 2U Maths [98]
HSC 2019: Physics | Chemistry | English Advanced | Maths Extension 1 | Maths Extension 2

#### RuiAce

• ATAR Notes Lecturer
• Honorary Moderator
• Great Wonder of ATAR Notes
• Posts: 8493
• "All models are wrong, but some are useful."
• Respect: +2347
##### Re: "Challenge" Math Qs - Can You Figure It Out?
« Reply #52 on: December 30, 2018, 04:04:49 pm »
+2
And yeah non-uniform convergence works. I would've just used this guy:
$f(x,y) = \frac{x^2}{x^2+y^2}\text{ with limits as }x\to 0, \, y\to 0$
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.
1. Prove the cosine rule in trigonometry using properties of the vector dot product
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:
$\text{Show that if }\lambda\text{ is an eigenvalue of }A\in \mathbb{R}^{n\times n}\\ \text{then it is also an eigenvalue of }A^T$

#### lzxnl

• Victorian
• ATAR Notes Legend
• Posts: 3430
• Respect: +207
##### Re: "Challenge" Math Qs - Can You Figure It Out?
« Reply #53 on: December 30, 2018, 06:43:56 pm »
+1
And yeah non-uniform convergence works. I would've just used this guy:
$f(x,y) = \frac{x^2}{x^2+y^2}\text{ with limits as }x\to 0, \, y\to 0$
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:
$\text{Show that if }\lambda\text{ is an eigenvalue of }A\in \mathbb{R}^{n\times n}\\ \text{then it is also an eigenvalue of }A^T$
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.

I love writing diagonal matrices using Kronecker deltas and using Einstein summation; makes all the working so pleasant to read.
2012
Mathematical Methods (50) Chinese SL (45~52)

2013
English Language (50) Chemistry (50) Specialist Mathematics (49~54.9) Physics (49) UMEP Physics (96%) ATAR 99.95

2014-2016: University of Melbourne, Bachelor of Science, Diploma in Mathematical Sciences (Applied Maths)

2017-2018: Master of Science (Applied Mathematics)

2019-:

Accepting students for  VCE tutoring in Maths Methods, Specialist Maths and Physics! PM for more details

#### redpanda83

• Trailblazer
• Posts: 48
• Respect: +2
##### Re: "Challenge" Math Qs - Can You Figure It Out?
« Reply #54 on: December 31, 2018, 05:09:21 pm »
0
Hi
can someone help me out  with these two questions.
What is required to do them? I would like to learn.
I got them from here 2018 melb uni maths comp senior division
Thanks!

#### FelixHarvey

• Forum Regular
• Posts: 56
• Respect: +9
##### Re: "Challenge" Math Qs - Can You Figure It Out?
« Reply #55 on: December 31, 2018, 05:43:47 pm »
+1
Hi
can someone help me out  with these two questions.
What is required to do them? I would like to learn.
I got them from here 2018 melb uni maths comp senior division
Thanks!

Hey redpanda83,

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?

For anyone interested, all past papers & solutions for each division and year can be found here.
« Last Edit: December 31, 2018, 05:48:00 pm by FelixHarvey »
2018: Methods 45 | Specialist 41 | English 41 | Physics 44 | Software Development 48

ATAR: 99.40

Tutoring! PM me for details.

#### redpanda83

• Trailblazer
• Posts: 48
• Respect: +2
##### Re: "Challenge" Math Qs - Can You Figure It Out?
« Reply #56 on: December 31, 2018, 05:52:14 pm »
0
Hey redpanda83,

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?

For anyone interested, all past papers & solutions for each division and year can be found here.
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.
« Last Edit: December 31, 2018, 05:59:44 pm by redpanda83 »

#### FelixHarvey

• Forum Regular
• Posts: 56
• Respect: +9
##### Re: "Challenge" Math Qs - Can You Figure It Out?
« Reply #57 on: December 31, 2018, 06:23:16 pm »
+1
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.

Ok, I've had a look at question 6 and I'll have a crack. No guarantee on the accuracy though...

Each number in the sequence has the form:
$101010101...1010101=10^0+10^2+10^4+...+10^{2n} \quad where \: n>1$
$\text{If we divide by 99 we get } \frac{1}{99}(99(10^0+10^2+10^4+...+10^{2n})) \\ =\frac{1}{99}(10^2-1)(10^0+10^2+10^4+...+10^{2n})$
$\text{"Expanding" the brackets we get: } \frac{1}{99}(10^{2+0}+10^{2+2}+10^{4+2}...+10^{2n+2} - (10^0+10^2+10^4+...+10^{2n}))$
$\text{Then cancelling, leaves us with: }\frac{1}{99}(10^{2(n+1)}-1)$
$\text{Using difference of two squares we can obtain: }\frac{1}{99}(10^{n+1}-1)(10^{n+1}+1)$
$\text{As }n>1\text{ both the factors } 10^{n+1}-1 \quad \& \quad 10^{n+1}+1 \text{ are greater than 99. Therefore, all numbers excluding 101 are not prime.}$
2018: Methods 45 | Specialist 41 | English 41 | Physics 44 | Software Development 48

ATAR: 99.40

Tutoring! PM me for details.

#### AlphaZero

• MOTM: DEC 18
• Forum Obsessive
• Posts: 319
• $[\mathcal{L}f](s)=\int_0^\infty\!e^{-st}f(t)dt$
• Respect: +131
##### Re: "Challenge" Math Qs - Can You Figure It Out?
« Reply #58 on: May 25, 2019, 01:02:35 am »
0
I feel like reviving this thread.

For highschool students

a)     Where $a>0$, show that $\displaystyle \int_{0}^a \!\Big(f(x)+f(-x)\Big)dx=\int_{-a}^a f(x)\,dx$.

b)     Hence, find $\displaystyle \int_{-1}^1 \frac{x^n}{e^x+1}dx$, in terms of $n$, where $n>0$.

For university students

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  $\displaystyle \int_C \mathbf{F}\cdot d\mathbf{s}$.
2015$-$2017:  VCE
2018$-$2021:  Bachelor of Biomedicine and Concurrent Diploma in Mathematical Sciences, University of Melbourne

#### RuiAce

• ATAR Notes Lecturer
• Honorary Moderator
• Great Wonder of ATAR Notes
• Posts: 8493
• "All models are wrong, but some are useful."
• Respect: +2347
##### Re: "Challenge" Math Qs - Can You Figure It Out?
« Reply #59 on: May 25, 2019, 09:03:01 am »
0
For university students

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  $\displaystyle \int_C \mathbf{F}\cdot d\mathbf{s}$.
Isn't this just a regular path integral? Or are you trying to throw off the people who think the answer is 0