leehuan's All-Levels-Of-Maths SOS thread (1 Viewer)

InteGrand · Mar 6, 2016

leehuan said:
$\\ $RTP: $\arcsin{x}+\arccos{x}$ is a constant. \\ All I want to know if there's a more elegant approach to taking derivatives and then reversing the differentiation process using integration?$$

Followup: Obviously arctan(x)+arccot(x) becomes a piecewise function. How do I justify that this isn't the case for arcsin(x)+arccos(x)? Does it basically just have to do with the fact arccot(x) is undefined for x=0 but continuous everywhere else?

$\noindent The function $\sin^{-1} x + \cos^{-1} x$ is continuous because it is the sum of two continuous functions.$

Green Yoda · Mar 6, 2016

Question: solve |x-2|>1/x

Green Yoda · Mar 6, 2016

Another Question: x^2-|x|>0

InteGrand · Mar 6, 2016

Rathin said:
Question: solve |x-2|>1/x

$\noindent Note that $x\neq 0$. Also note that if $x$ is negative, the inequation holds always since the LHS is positive whilst the RHS is negative. Now, let's focus on $x>0$.$

$\noindent For $0<x<2$, we have $|x-2| = 2-x$, so the inequation becomes $2-x>\frac{1}{x}$. Since $x>0$, we can multiply through by $x$ to obtain$

$\begin{align*} 2x-x^2 &> 1 \\ \Longleftrightarrow -x^2 + 2x -1 &> 0 \\ \Longleftrightarrow -\left(x^2 -2x+1\right) &> 0 \\ \Longleftrightarrow -\left(x-1\right)^2 &> 0.\end{align*}$

$\noindent This last inequality never holds for real $x$, so there are no solutions to the inequation in the interval $0<x<2$. Now let's consider $x\geq 2$, so $|x-2| = x-2$. So the inequation becomes$

$\begin{align*} x-2 &> \frac{1}{x} \\ \Longleftrightarrow x^2 - 2x &> 1 \quad (\text{multiplying through by }x>0) \\ \Longleftrightarrow x^2 -2x -1&> 0.\end{align*}$

$\noindent Now the roots of the quadratic on the left-hand side are $1\pm \sqrt{2}$ (this can be shown using the quadratic formula or completing the square). So by sketching the parabola, we see that $LHS >0$ precisely when $x>1+\sqrt{2}$ or $x<1-\sqrt{2}$. Since we were only considering $x>2$, the solution is $x>1+\sqrt{2}$. So the solution to the original inequation is $x<0$ or $x>1+\sqrt{2}$.$

InteGrand · Mar 6, 2016

Rathin said:
Another Question: x^2-|x|>0

$\noindent The inequation is $x^2 > |x|$. By sketching the graphs of $y=x^2$ and $y=|x|$ (note that they intersect at and only at $\left( \pm 1,1\right)$), we can easily see that the solution is $x<-1$ or $x>1$.$

Green Yoda · Mar 6, 2016

InteGrand said:
$\noindent Note that $x\neq 0$. Also note that if $x$ is negative, the inequation holds always since the LHS is positive whilst the RHS is negative. Now, let's focus on $x>0$.$

$\noindent For $0<x<2$, we have $|x-2| = 2-x$, so the inequation becomes $2-x>\frac{1}{x}$. Since $x>0$, we can multiply through by $x$ to obtain$

$\begin{align*} 2x-x^2 &> 1 \\ \Longleftrightarrow -x^2 + 2x -1 &> 0 \\ \Longleftrightarrow -\left(x^2 -2x+1\right) &> 0 \\ \Longleftrightarrow -\left(x-1\right)^2 &> 0.\end{align*}$

$\noindent This last inequality never holds for real $x$, so there are no solutions to the inequation in the interval $0<x<2$. Now let's consider $x\geq 2$, so $|x-2| = x-2$. So the inequation becomes$

$\begin{align*} x-2 &> \frac{1}{x} \\ \Longleftrightarrow x^2 - 2x &> 1 \quad (\text{multiplying through by }x>0) \\ \Longleftrightarrow x^2 -2x -1&> 0.\end{align*}$

$\noindent Now the roots of the quadratic on the left-hand side are $1\pm \sqrt{2}$ (this can be shown using the quadratic formula or completing the square). So by sketching the parabola, we see that $LHS >0$ precisely when $x>1+\sqrt{2}$ or $x<1-\sqrt{2}$. Since we were only considering $x>2$, the solution is $x>1+\sqrt{2}$. So the solution to the original inequation is $x<0$ or $x>1+\sqrt{2}$.$

ah i see.. If there is 1/x in an inequality don't you have to multiply with x^2? You just moved x to the other side

Paradoxica · Mar 6, 2016

InteGrand said:
$\noindent The inequation is $x^2 > |x|$. By sketching the graphs of $y=x^2$ and $y=|x|$ (note that they intersect at and only at $\left( \pm 1,1\right)$), we can easily see that the solution is $x<-1$ or $x>1$.$

alternatively, square both sides and go from there.

InteGrand · Mar 6, 2016

Rathin said:
ah i see.. If there is 1/x in an inequality don't you have to multiply with x^2? You just moved x to the other side

$\noindent In general, since we normally deal with arbitrary $x$, we need to multiply by $x^2$ instead. The whole point of this is so that we don't need to worry about flipping inequality signs due to potential multiplication by a negative number. But we didn't need to in the above working since we were dealing with $x>0$ there (so we knew the sign of the number we were multiplying by).$

Green Yoda · Mar 6, 2016

InteGrand said:
$\noindent In general, since we normally deal with arbitrary $x$, we need to multiply by $x^2$ instead. The whole point of this is so that we don't need to worry about flipping inequality signs due to potential multiplication by a negative number. But we didn't need to in the above working since we were dealing with $x>0$ there (so we knew the sign of the number we were multiplying by).$

How do we know x>0?

InteGrand · Mar 6, 2016

Rathin said:
How do we know x>0?

$\noindent Because we said in the first paragraph that the inequation clearly holds for all $x<0$ (and $x$ can't be 0 since the RHS is undefined then), so we only needed to do the algebra for $x>0$.$

Green Yoda · Mar 6, 2016

InteGrand said:
$\noindent The inequation is $x^2 > |x|$. By sketching the graphs of $y=x^2$ and $y=|x|$ (note that they intersect at and only at $\left( \pm 1,1\right)$), we can easily see that the solution is $x<-1$ or $x>1$.$

how to do it algebraically?

Green Yoda · Mar 6, 2016

can some mod pls remove that pic thanks...

Paradoxica · Mar 6, 2016

Help i'm trapped in the title and can only speak once every 2 years

Rathin said:
how to do it algebraically?

$$\noindent If $a>b>0$ then $a^2>b^2>0$. Replacing $a$ with $x^2$ and $b$ with $|x|$, we have $\\x^4>x^2 $ and this is only true for $|x|>1$

InteGrand · Mar 6, 2016

Rathin said:
how to do it algebraically?

$\noindent Note that we don't have the strict inequality hold at $x=0$, as we have equality instead. For $x>0$, the inequation is $x^2>x$, as $|x| = x$ for $x>0$. Since $x>0$, we can cancel it from either side of the inequality to get $x>1$ as the solution for this case.$

$\noindent The other case to consider is $x<0$. The inequation is then $x^2 > -x$, as $|x| = -x$ for $x<0$. Since $x<0$, we can cancel it from either side of the inequality, \textit{remembering the flip the sign as we do so}, to get $x<-1$ as the solution for this case.$

$$\noindent So overall, the solution is $x>1$ or $x<-1$. A graphical approach is probably easier and faster.$

keepLooking · Mar 6, 2016

Rathin said:
can some mod pls remove that pic thanks...

Just when I'm scrolling through a Maths thread and that pops up. I really saw that coming.. /s

Paradoxica · Mar 7, 2016

Well, I suppose now is a better time than never to crack down this question of mine...

Take an arbitrary ellipse, centred at the origin and aligned with the cartesian axes for ease of computation.
Construct the director circle of the ellipse.
Take a point on the director circle and construct the obviously perpendicular tangents to the ellipse.
At the two points of osculation, construct normals to the tangents.
What is the algebraic equation that defines the locus of all possible points that are the intersection of these normals?

Repeat the above for the director circle of a hyperbola. Obviously |a|>|b| or else the director circle has imaginary radius, and we are not wishing to deal with 4-dimensional algebraic curves.

seanieg89 · Mar 7, 2016

Paradoxica said:
Well, I suppose now is a better time than never to crack down this question of mine...

Take an arbitrary ellipse, centred at the origin and aligned with the cartesian axes for ease of computation.
Construct the director circle of the ellipse.
Take a point on the director circle and construct the obviously perpendicular tangents to the ellipse.
At the two points of osculation, construct normals to the tangents.
What is the algebraic equation that defines the locus of all possible points that are the intersection of these normals?

Do you know for a fact that a nice cartesian form exists? Or will a parametric description suffice?

It is not hard to obtain a parametric formula. (MX2 level conics.)

My method:
1. Show that the director circle has radius sqrt(a^2+b^2) for an ellipse in the standard form.
2. For a point X on the director circle let the two points of contact with the ellipse be A,B and let the normals at A,B meet at N. Then AXBN is a rectangle, and we can obtain N by reflecting X about the midpoint M of AB.
3. We can find the midpoint M of AB in terms of the coordinates x_0,y_0 of X by equating the chord of contact with the cartesian equation of ellipse (ignoring vertical line issues which can be dealt with separately if you like). We do not need to solve the resulting quadratic, we only need to halve the sum of the roots to obtain our midpoint.
4. Our point N then has coordinates given by x_2=x_0+2(x_1-x_0) and similarly for y_2. (Here M=(x_1,y_1).)

My calculation led to

$N=\left(\frac{2a^2b^2}{a^2y_0^2+b^2x_0^2}-1\right)\cdot (x_0,y_0).$

By setting $(x_0,y_0)=\sqrt{a^2+b^2}\cdot (\cos{\xi},\sin{\xi})$ we then obtain a parametric (polar in fact) equation for the locus of N.

With some manipulations, you might be able to find an implicit cartesian equation for N, but I will not spend any time on trying to do this right now.

Out of interest, the parametric plot looks like (for a=sqrt(2),b=1):

If anyone does the same computation, let me know if your parametric plot matches mine, as I was not particularly careful and will not have time to check over this properly till after a meeting tomorrow arvo.

seanieg89 · Mar 7, 2016

Equally useful would be a visual confirmation of the above using Geogebra. I will do this tomorrow if no-one else does.

leehuan · Mar 8, 2016

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}$

Paradoxica · Mar 8, 2016

seanieg89 said:
Equally useful would be a visual confirmation of the above using Geogebra. I will do this tomorrow if no-one else does.

Visual confirmation.

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