leehuan's All-Levels-Of-Maths SOS thread (2 Viewers)

seanieg89 · Mar 8, 2016

Paradoxica said:
Visual confirmation.

But do you know whether or not there is a known cartesian equation for this thing or are you just exploring?

seanieg89 · Mar 8, 2016

leehuan said:
MX2 difficulty:

This question got asked whilst I was busy, and was also solved. But I'm asking it here because I'm wondering if anyone knows what the most elegant method of proof is:

$\\ $Given $ a\ge b\ge c$ \\ Show that:$ \\ {a}^2b+{b}^{2}c+{c}^{2}a \ge a{b}^{2}+b{c}^{2}+c{a}^{2}$

$a-b,a-c,b-c \geq 0 \\ \\ \Rightarrow (a-b)(a-c)(b-c)\geq 0 \\ \\ \Rightarrow \textrm{the inequality you want.}$

Paradoxica · Mar 8, 2016

seanieg89 said:
But do you know whether or not there is a known cartesian equation for this thing or are you just exploring?

I have a cartesian equation. Whether it's elegant or not is another matter entirely.

Paradoxica · Mar 8, 2016

Okay, as it stands, I do not have the cartesian equation, even though I derived it from the parametric equation. This is strange.

seanieg89 · Mar 8, 2016

Okay, had another look at it after my meeting today. My parametric form looks correct, and leads to a cartesian equation pretty straightforwardly.

$As the angle $\theta$ varies along the director circle, let $(x,y)$ be the point we are trying to find the locus of.\\ \\ $x=\frac{a^2b^2-a^4\sin^2\phi-b^4\cos^2\phi}{a^2b^2+a^4\sin^2\phi+b^4\cos^2\phi}\cdot \sqrt{a^2+b^2}\cdot \cos{\phi}$\\ \\ from our previous working.\\ \\ We now exploit the fact that $y/x=\tan\phi$ (again from our previous work, this is just a statement of the fact that $N$ lies on the line segment between $O$ and $X$). \\ \\ Since $y/x=\tan{\phi}$, we have $\sin^2\phi=y^2/(x^2+y^2)$ and $\cos^2\phi=x^2/(x^2+y^2)$. \\ \\Substituting this into our earlier equation gives our final locus of \\ \\ $\frac{x^2+y^2}{a^2+b^2}=\left(\frac{a^2b^2(x^2+y^2)-a^4y^2-b^4x^2}{a^2b^2(x^2+y^2)+a^4y^2+b^4x^2}\right)^2$$

I am sure the analogous computations for the hyperbola are quite similar.

leehuan · Mar 9, 2016

This is messing with my head thoroughly (+ I am tired of course).

$\\ $Q: Find 10 vectors in $\mathbb{R}^{10}$, each of which is $5\sqrt{2}$ apart. Can you now find an 11th vector?$$

Given answer (where I'm pretty sure the ^T is just transpose, because they love column vectors)

$\\ {\left(5,{0}^{9}\right)}^{T}, \, {\left(0,5,{0}^{8}\right)}^{T}, \, \dots \, {\left({0}^{9}, 5\right)}^{T} \\ $Yes, ${\left(\alpha, \alpha, \dots , \alpha\right)}^{T} $ where ${\left(5-\alpha\right)}^{2}+9{\alpha}^{2}=50$

InteGrand · Mar 9, 2016

leehuan said:
This is messing with my head thoroughly (+ I am tired of course).

$\\ $Q: Find 10 vectors in $\mathbb{R}^{10}$, each of which is $5\sqrt{2}$ apart. Can you now find an 11th vector?$$

Given answer (where I'm pretty sure the ^T is just transpose, because they love column vectors)

$\\ {\left(5,{0}^{9}\right)}^{T}, \, {\left(0,5,{0}^{8}\right)}^{T}, \, \dots \, {\left({0}^{9}, 5\right)}^{T} \\ $Yes, ${\left(\alpha, \alpha, \dots , \alpha\right)}^{T} $ where ${\left(5-\alpha\right)}^{2}+9{\alpha}^{2}=50$

I think this was asked before. Check out this thread: http://community.boredofstudies.org/1003/maths/335733/another-hard-vectors-question.html

leehuan · Mar 9, 2016

Convenient.

agha · Mar 10, 2016

Some exercises in formal logic:

$\\ $Prove formally the following:$$

$\\ $1.$ \ p \Rightarrow (q \wedge r) \vdash (p \Rightarrow q) \wedge (p \Rightarrow r)$

$\\ $2.$ \ \sim p \Rightarrow p \vdash p$

$\\ $3.$ \ (p \vee q) \Leftrightarrow p \vdash q \Rightarrow p$

For a list of the symbols used see here

leehuan · Mar 11, 2016

A few questions this time:

$\\ $1. Prove that $\ln{(xy)}=\ln{x}+\ln{y}$ using the definition of the natural logarithm:$ \\ \ln{}: \left(0,\infty\right)\rightarrow\mathbb R, \ln{x}=\int_{1}^{x}{\frac{dt}{t}}$
---> Not too sure how to prove that the integral from 1 to xy is the same as from x to y

$\\ $2. For the set ${S}=\left\{\frac{\sin{\left(k-2\right)}}{k}, k\in\mathbb{Z}^{+}\right\}$, what is Sup ${S}?$

$\\ $3. Prove that ${x}_{n}= {\left(1+\frac{1}{n}\right)}^{n} $ is:$ \\ $a) increasing \\ b) bounded above$$

Drsoccerball · Mar 11, 2016

$1)\ln{xy}=\ln{x}+\ln{y}$

$xy=e^{\ln{x}+\ln{y}}$

$xy= e^{\ln{x}} \cdot e^{\ln{y}}$

$xy=xy$

Drsoccerball · Mar 11, 2016

$2) I= \int_1^x{\frac{dt}{t}}$

$I= [ \ln{t}]_1^x$

$I= \ln{x} -\ln{1}$

$I=\ln{x}$

leehuan · Mar 11, 2016

Drsoccerball said:
$1)\ln{xy}=\ln{x}+\ln{y}$

$xy=e^{\ln{x}+\ln{y}}$

$xy= e^{\ln{x}} \cdot e^{\ln{y}}$

$xy=xy$

You may NOT assume that the logarithm and the exponential functions are inverses AS exp: R -> [0, inf) is defined OUT of the inverse function theorem and is not proven in advance here.

I.e. I require the answer that STRICTLY uses the fundamental definition please

seanieg89 · Mar 11, 2016

seanieg89 · Mar 11, 2016

For 2) it is just sin(1)/3 because k=3 is the first time its positive, and for k>3, the thing is bounded by 1/4 which is smaller than sin(1)/3.

Paradoxica · Mar 11, 2016

seanieg89 said:
Okay, had another look at it after my meeting today. My parametric form looks correct, and leads to a cartesian equation pretty straightforwardly.

$As the angle $\theta$ varies along the director circle, let $(x,y)$ be the point we are trying to find the locus of.\\ \\ $x=\frac{a^2b^2-a^4\sin^2\phi-b^4\cos^2\phi}{a^2b^2+a^4\sin^2\phi+b^4\cos^2\phi}\cdot \sqrt{a^2+b^2}\cdot \cos{\phi}$\\ \\ from our previous working.\\ \\ We now exploit the fact that $y/x=\tan\phi$ (again from our previous work, this is just a statement of the fact that $N$ lies on the line segment between $O$ and $X$). \\ \\ Since $y/x=\tan{\phi}$, we have $\sin^2\phi=y^2/(x^2+y^2)$ and $\cos^2\phi=x^2/(x^2+y^2)$. \\ \\Substituting this into our earlier equation gives our final locus of \\ \\ $\frac{x^2+y^2}{a^2+b^2}=\left(\frac{a^2b^2(x^2+y^2)-a^4y^2-b^4x^2}{a^2b^2(x^2+y^2)+a^4y^2+b^4x^2}\right)^2$$

I am sure the analogous computations for the hyperbola are quite similar.

What is the area bounded by the entire curve?

InteGrand · Mar 11, 2016

leehuan said:
A few questions this time:

$\\ $1. Prove that $\ln{(xy)}=\ln{x}+\ln{y}$ using the definition of the natural logarithm:$ \\ \ln{}: \left(0,\infty\right)\rightarrow\mathbb R, \ln{x}=\int_{1}^{x}{\frac{dt}{t}}$
---> Not too sure how to prove that the integral from 1 to xy is the same as from x to y

$\\ $2. For the set ${S}=\left\{\frac{\sin{\left(k-2\right)}}{k}, k\in\mathbb{Z}^{+}\right\}$, what is Sup ${S}?$

$\\ $3. Prove that ${x}_{n}= {\left(1+\frac{1}{n}\right)}^{n} $ is:$ \\ $a) increasing \\ b) bounded above$$

$\noindent For the log one, if we can use the fundamental theorem of calculus and chain rule, we can also do it as follows:$

$\noindent Note $L(x) = \int _1 ^ x \frac{1}{t} \text{ d}t \Rightarrow L^\prime (x) =\frac{1}{x}$. Now, fix $a>0$, then we want to show that $L(ax) = L(a) + L(x)$. Differentiate the LHS w.r.t. $x$, and we see that $LHS^\prime (x) = \frac{1}{ax}\cdot a = \frac{1}{x} = RHS^\prime (x)$ (as $L(a)$ is a constant)$.$

$\noindent So $LHS = RHS + c$ for some constant $c$. Now, when $x= 1$, $LHS = L(x)$ and $RHS = L(x) + L(1)$, but $L(1) = \int _1 ^1 \frac{1}{t} \text{ d}t = 0$. So $LHS = RHS$ at $x=1$, so the constant $c$ is 0, so $LHS = RHS$ for all $x>0$, as desired.$

InteGrand · Mar 11, 2016

leehuan said:
A few questions this time:

$\\ $1. Prove that $\ln{(xy)}=\ln{x}+\ln{y}$ using the definition of the natural logarithm:$ \\ \ln{}: \left(0,\infty\right)\rightarrow\mathbb R, \ln{x}=\int_{1}^{x}{\frac{dt}{t}}$
---> Not too sure how to prove that the integral from 1 to xy is the same as from x to y

$\\ $2. For the set ${S}=\left\{\frac{\sin{\left(k-2\right)}}{k}, k\in\mathbb{Z}^{+}\right\}$, what is Sup ${S}?$

$\\ $3. Prove that ${x}_{n}= {\left(1+\frac{1}{n}\right)}^{n} $ is:$ \\ $a) increasing \\ b) bounded above$$

$\noindent 3. This can be proved in a variety of ways. We will show a generalisation. Let $x$ be a fixed real number, and consider the sequence $\{a_n\}$, where $a_n = \left(1+\frac{x}{n}\right)^{n}$. We will show that this is increasing for $1+\frac{x}{n} \geq 0 \Longleftrightarrow n \geq -x$, i.e. that $a_n \geq a_{n-1}$ for all integers $n\geq -x$ (and $n-1 \geq 1$) (so in particular, for your specific question, it is increasing for all natural numbers $n$), and it is bounded above.$

$\noindent (a) Let $n \geq -x$ and $n \geq 2$. We want to show that $a_n \geq a_{n-1}$. Let $b_1 = 1$ and $b_2 = b_3 = \cdots = b_n = 1+\frac{x}{n-1}$ (note $b_i \geq 0$ for all $i$, as $n \geq -x$, and these are all well-defined quantities as $n-1\neq 0$, as $n\geq 2$). Then denoting the arithmetic mean of the $b_i$'s by $A$ and the geometric mean of the $b_i$'s by $G$, we have$

$\begin{align*}A&= \frac{\sum \limits _{i=1}^{n} b_i}{n} \quad (\text{by definition of arithmetic mean}) \\ &= \frac{1+ \left(n-1\right) \left(1+\frac{x}{n-1}\right)}{n} \quad (\text{by what we defined the }b_i\text{'s as})\\ &= \frac{1+ n-1 + x}{n} \\ &= \frac{n+x}{n} \end{align*}$

$\noindent and$

$\begin{align*}G &= \left(\prod \limits _{i=1} ^{n} b_i \right)^{\frac{1}{n}} \quad (\text{by definition of geometric mean}) \\ &= \left[ 1\cdot \left( 1+\frac{x}{n-1}\right)^{n-1}\right]^{\frac{1}{n}}(\text{by what we defined the }b_i\text{'s as}) \\ &= \left(1+\frac{x}{n-1}\right)^{\frac{n-1}{n}}. \end{align*}$

$\noindent By the AM-GM inequality, we have$

$\begin{align*}A \geq G &\Longleftrightarrow \frac{n+x}{n}\geq \left(1+\frac{x}{n-1}\right)^{\frac{n-1}{n}} \quad (\text{equality iff all }b_i\text{'s are equal, which happens iff }x=0, \text{when the sequence becomes a trivial sequence of constant value }1) \\ &\Longleftrightarrow 1+\frac{x}{n} \geq \left(1+\frac{x}{n-1}\right)^{\frac{n-1}{n}} \\ &\Longleftrightarrow \left(1+\frac{x}{n}\right)^n \geq \left(1+\frac{x}{n-1}\right)^{n-1} \\ &\Longleftrightarrow a_n \geq a_{n-1}. \end{align*}$

$\noindent This proves what we wanted for part (a).$

$\noindent (b) Observe that$

$\begin{align*}\left(1+\frac{x}{n}\right)^n &= \sum_{k=0}^n \binom{n}{k}\frac{x^k}{n^k}\\ &\leq \sum_{k=0}^n \binom{n}{k}\frac{|x|^k}{n^k} \\ &= \sum_{k=0}^n \frac{n(n-1)(n-2)\ldots (n-(k-1))}{k!}\cdot \frac{1}{n^k} \cdot |x|^k \\ &= \sum_{k=0}^n 1 \cdot \left(1-\frac{1}{n}\right)\cdot \left(1-\frac{2}{n}\right)\cdot \cdots \cdot \left(1-\frac{k-1}{n}\right)\cdot \frac{|x|^k}{k!} \\ &\leq \sum_{k=0} ^{n} \frac{|x|^k}{k!} \quad (\text{as each factor is between }0 \text{ and }1) \\ &\leq \sum_{k=0}^{\infty} \frac{|x|^k}{k!}\equiv S_x, \end{align*}$

$\noindent where that last sum converges for any $x \in \mathbb{R}$ (to say $S_x$) by the ratio test $\Big{(}$another way to show that this sum converges is to use a comparison test by replacing the $k!$ in the denominator by something that is eventually smaller (so that the new sum is bigger) and yields a finite sum, such as $M^k$, where $M$ is a real number satisfying $M>|x|$. This new sum is finite because it is a geometric series with positive common ratio $\frac{|x|}{M}<1$.$\Big{)}$. Hence $\left(1+\frac{x}{n}\right)^n$ is bounded above (by $S_x$).$

$\noindent Aside: The fact that the sequence above is monotonically increasing and bounded above shows that the sequence has a limit (by the Monotone Convergence Theorem). It can be shown that this limit is in fact $\mathrm{e}^x$.$

leehuan · Mar 11, 2016

Can someone just send me the link if they're not bothered to type it cause apparently it's a very famous result...

$\\ $Show that :$ \\ \sum_{m=1}^{99}{\frac{\sin{\left(\frac{17 m \pi}{100}\right)} \sin{\left(\frac{39 m \pi}{100}\right)}}{1+\cos{\left( \frac{m\pi}{100} \right) }}}=1037$

Paradoxica · Mar 12, 2016

If a number is the root of a polynomial with algebraic co-efficients, is that number itself necessarily algebraic?

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