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

InteGrand · Feb 17, 2016

Paradoxica said:
The locus can only be a full circle if the altitudes are allowed to project from the triangle extended outside the circle when the triangle is obtuse. If this is not the case, then you only get a circular arc.

I assumed they were allowed to, as I mentioned for one of the cases.

seanieg89 · Feb 17, 2016

Drsoccerball said:
The taylor series can be derived using the mean value theorem.

$\frac{f(a)-f(b)}{(a-b)} \approx f'(a)$

$f(x) \approx f(b) +f'(x)(x-b)$

$How do we prove the general taylor series using only the mean theorem?$

A good chance to show what some integration by parts can achieve!

$Suppose $f\in\mathcal{C}^{n+1}$, that is $f$ has continuous derivatives up to order $n+1$. Then:\\ \\ $f(x)=f(a)+\int_a^x f'(t)\,dt\\ \\ =f(a)+\int_a^x (-1)^n\frac{d^n}{dt^n}\left(\frac{(x-t)^n}{n!}\right)f'(t)\,dt\\ \\ = \sum_{k=0}^n \frac{(x-a)^kf^{(k)}(a)}{k!}+\frac{1}{n!}\int_a^x (x-t)^nf^{(n+1)}(t)\, dt.$\\ \\This establishes an integral form for the remainder. An application of the mean value theorem in integral form lets us rewrite the remainder in the so-called Cauchy form\\ \\ $R_{n}(x)=\frac{(x-a)(x-c)^nf^{(n+1)}(c)}{n!}.$\\ The Lagrange form for the remainder is found slightly differently but they are similarly useful.\\ \\ Note also that this result lets us construct taylor polynomials in higher dimensions as well. \\$

Paradoxica · Feb 17, 2016

InteGrand said:
I assumed they were allowed to, as I mentioned for one of the cases.

You got the cases wrong.

Construct the cyclic rectangle AA'B'B. The midpoint of CD can only exist when P lies on the minor arc A'B'. Only when P lies on the major arc B'BAA', the interior altitude fails to exist.

InteGrand · Feb 18, 2016

Paradoxica said:
You got the cases wrong.

Construct the cyclic rectangle AA'B'B. The midpoint of CD can only exist when P lies on the minor arc A'B'. Only when P lies on the major arc B'BAA', the interior altitude fails to exist.

$\noindent I didn't mention the case where precisely one of the altitudes is outside, but this isn't too important. I mentioned the case where both were outside (i.e., $P$ on minor arc) because it was slightly different. Even in the case where exactly one altitude is outside (which is covered by the case of $P$ being on the major arc in the given diagram), we still get $\cot y = \frac{PC}{CA}$ directly from right-angled triangle $PCA$ as I mentioned for the case where $P$ is on the major arc. This is because we still have $\angle APC= y$ resulting straight from our definition of $y$ being the angle in the major segment. When we take $P$ to be in the minor arc, we don't have angle $APC$ be $y$ straightaway from the definition of $y$, which is why this was an important case. Thus, it's not important about mentioning the exactly one altitude outside thing, because this gets covered anyway by the $P$ in major arc case.$

Nailgun · Feb 18, 2016

#mathswars

Carrotsticks · Feb 18, 2016

InteGrand said:
$\noindent If so, which year's HSC 4U paper is this from?$

$\\$1990$.$

leehuan · Feb 18, 2016

Carrotsticks said:
$\\$1990$.$

Oh crap, seriously?

That probably also contributed to why I could not figure it out.

leehuan · Feb 18, 2016

Paradoxica said:
The locus can only be a full circle if the altitudes are allowed to project from the triangle extended outside the circle when the triangle is obtuse. If this is not the case, then you only get a circular arc.

Now that Carrotsticks pointed out which paper it was, I found the question on Pg14 of this file.

It appears that these solutions did not care about if the full circle traced out, or if only the arc was
http://www.boredofstudies.org/courses/maths/4u/1202629887_1990_Mathematics_Extension_2_HSC.pdf

Drsoccerball · Feb 18, 2016

Also it was in a more recent CSSA paper.

Green Yoda · Feb 18, 2016

Question:
Solve for x
2(x^3/2)-5x+2(x^1/2)=0

InteGrand · Feb 18, 2016

Rathin said:
Question:
Solve for x
2(x^3/2)-5x+2(x^1/2)=0

$\noindent The equation is $2x^{\frac{3}{2}}-5x + 2\sqrt{x} = 0$, so we'll assume $x$ needs to be non-negative. If we substitute $u=\sqrt{x}$, the equation becomes $2u^3 -5u^2 + 2u=0 \Longleftrightarrow u\left(2u^2 - 5u +2\right)=0$. Factorising the quadratic term: $2u^2 - 5u +2 =\frac{\left(2u-4\right)\left(2u-1\right)}{2}=(u-2)(2u-1)$. So the original equation becomes $u(u-2)(2u-1)=0$. Hence the solutions are $u=0,2,\frac{1}{2}$. Since $x=u^2$, the solutions for $x$ are $x=0,4,\frac{1}{4}$.$

drsabz101 · Feb 19, 2016

Can someone show the steps of how they solved the equation by using cases ( include diagrams please ) :
For |x+1| - |x-2| = 2

InteGrand · Feb 19, 2016

drsabz101 said:
Can someone show the steps of how they solved the equation by using cases ( include diagrams please ) :
For |x+1| - |x-2| = 2

$\noindent Split it up into three intervals (cases): $x<-1$, $-1\leq x \leq 2$, and $x\geq 2$. Find how to simplify the absolute value signs in each interval using the definition of the absolute value sign as $\left | \xi \right |=\begin{cases}\xi & \text{ if } \xi \geq 0 \\ -\xi & \text{ if } \xi <0 \end{cases}$. Then just solve the equation for each case.$

drsabz101 · Feb 19, 2016

How do you solve for each case though, can u go through the case for x<-1?

InteGrand · Feb 19, 2016

drsabz101 said:
How do you solve for each case though, can u go through the case for x<-1?

$\noindent When $x<-1$, we have $x+1<0 \Rightarrow |x+1| = -(x+1) = -x-1$ and $x-2<0 \Rightarrow |x-2| = -(x-2) \Rightarrow -|x-2| = x-2$. So the equation for this case becomes: $-x-1 + x-2 =2 \Longleftrightarrow -3=2$. This is a contradiction. Hence the given equation is never satisfied if $x<-1$ (so we get no solutions from this case).$

drsabz101 · Feb 19, 2016

How did u get that l x-2 l = -(x-2) -> -lx-2l -> x-2?

InteGrand · Feb 19, 2016

drsabz101 said:
How did u get that l x-2 l = -(x-2) -> -lx-2l -> x-2?

Move the negative to the other side (in other words, multiply both sides by -1).

drsabz101 · Feb 19, 2016

For the second case, would you say that lx+1l= x+1? I am not sure though about the lx-2l part though

drsabz101 · Feb 19, 2016

I am still quite confused on what is going on in the cases

InteGrand · Feb 19, 2016

drsabz101 said:
For the second case, would you say that lx+1l= x+1? I am not sure though about the lx-2l part though

$\noindent For the second case, we have $-1 \leq x \leq 2$, so in this interval, $x+1 \geq 0\Rightarrow |x+1| = x+1$. Also, in this interval, $x-2\leq 0$, so $|x-2| = 2-x$ again.$

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