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

leehuan · Mar 15, 2016

seanieg89 said:
Where did you find this sum in the first place? That should provide a good clue as to the cleanest known way to compute it.

Anyway, if you find a satisfactory explanation for why it works out the way it does, please post it here.

I will probably have another think about it this evening to see if I can find a nicer way.

From my calculus lecturer. That's the thing, I forgot what exactly he said.

InteGrand · Mar 15, 2016

leehuan said:
From my calculus lecturer. That's the thing, I forgot what exactly he said.

Did he provide a proof or proof sketch of it? If so, can you recall any of the details of it (these may provide some clues)?

leehuan · Mar 15, 2016

InteGrand said:
Did he provide a proof or proof sketch of it? If so, can you recall any of the details of it (these may provide some clues)?

Nope.

Gave it blindly to us and said have a go at it and he'll post the answers up on moodle (UNSW help forum for each course)

leehuan · Mar 15, 2016

The question:

$The following sets of point represent simple geometric figures in a plane. ${\lambda}_{1}, {\lambda}_{2} \in \mathbb R. \\ $For each problem, identify the geometric shape.$

One of the parts that I figured out (spoiler tagged for colour coding)

$S=\left\{\textbf{x} \in \mathbb{R}^{3}: \textbf{x}= { \lambda }_{ 1 }\left[ \begin{matrix} 2 \\ 1 \\ -2 \end{matrix} \right] +{ \lambda }_{ 2 }\left[ \begin{matrix} 4 \\ -2 \\ 3 \end{matrix} \right] , \, {\lambda}_{1}\in\left[0,6\right],{\lambda}_{2}\in\left[0,{\lambda}_{1}\right] \right\} \\ $A: Triangle through $(0,0,0), \, (12,6,-12), \, (36,-6,16)$

The part I am stuck on
$S=\left\{\textbf{x} \in \mathbb{R}^{4}: \textbf{x}= { \lambda }_{ 1 }\left[ \begin{matrix} 2 \\ 1 \\ -2 \\ 2 \end{matrix} \right] +{ \lambda }_{ 2 }\left[ \begin{matrix} 4 \\ -2 \\ 3 \\ -1 \end{matrix} \right] , \, {\lambda}_{1}\in\left[0,6\right],{\lambda}_{2} \ge {\lambda}_{1}\right\}$

All I know: Passes through (0,0,0,0), is an unbounded region. What else?

InteGrand · Mar 15, 2016

leehuan said:
The question:

$The following sets of point represent simple geometric figures in a plane. ${\lambda}_{1}, {\lambda}_{2} \in \mathbb R. \\ $For each problem, identify the geometric shape.$

One of the parts that I figured out (spoiler tagged for colour coding)

$S=\left\{\textbf{x} \in \mathbb{R}^{3}: \textbf{x}= { \lambda }_{ 1 }\left[ \begin{matrix} 2 \\ 1 \\ -2 \end{matrix} \right] +{ \lambda }_{ 2 }\left[ \begin{matrix} 4 \\ -2 \\ 3 \end{matrix} \right] , \, {\lambda}_{1}\in\left[0,6\right],{\lambda}_{2}\in\left[0,{\lambda}_{1}\right] \right\} \\ $A: Triangle through $(0,0,0), \, (12,6,-12), \, (36,-6,16)$

The part I am stuck on
$S=\left\{\textbf{x} \in \mathbb{R}^{4}: \textbf{x}= { \lambda }_{ 1 }\left[ \begin{matrix} 2 \\ 1 \\ -2 \\ 2 \end{matrix} \right] +{ \lambda }_{ 2 }\left[ \begin{matrix} 4 \\ -2 \\ 3 \\ -1 \end{matrix} \right] , \, {\lambda}_{1}\in\left[0,6\right],{\lambda}_{2} \ge {\lambda}_{1}\right\}$

All I know: Passes through (0,0,0,0), is an unbounded region. What else?

$\noindent Call the first vector there $\bold{v}_1$ and the other one $\bold{v}_2$. Consider the line segment $\ell$ joining the origin to the point with position vector $6 \left(\bold{v}_1+\bold{v}_2\right)$ $\big{(}$this line segment can be written as $ \left\{\lambda \left(\bold{v}_1+\bold{v}_2\right) : 0 \leq \lambda \leq 6 \right\}$.$\big{)}$ This line segment represents when $\lambda _2$ \textit{equals} $\lambda_1$, for $0\leq \lambda_1 \leq 6$. Since we can have any $\lambda_2 \geq \lambda_1$ (i.e. ``even more'' of $\bold{v}_2$), the desired region is the region that lies on the same side of this line segment as the vector $\bold{v}_2$, as well as being bounded by the rays through the endpoints of $\ell$ parallel to $\bold{v}_2$ and on the same side of $\ell$ as $\bold{v}_2$. So it's kind of like an unlimited parallelogram. Consider the picture below:$

http://www.mathplanet.com/Oldsite/media/44000/parallelogram_499x300.jpg .

$\noindent The region would be like deleting edge $BC$ extending the sides $AB$ and $CD$ indefinitely, and shading in everything in between these infinitely extended sides. (If $D$ was the origin, the point $A$ could refer to the other endpoint of the line $\ell$, and $\bold{v}_2$ would lie along where $DC$ is, in the direction from $D$ to $C$.)$

Green Yoda · Mar 15, 2016

question:
shade 0<=x-y<3

InteGrand · Mar 15, 2016

Rathin said:
question:
shade 0<=x-y<3

$\noindent We have $0\leq x-y<3$. Firstly this means $x-y \geq 0$, so $y\leq x$. So the entire line $y=x$ is included. Also, any remaining parts must be below this line. For the second part of the inequality, $x-y<3 \Rightarrow y>x-3$. So draw the line $y=x-3$ (as a dotted line, since we won't include it; this line is just the line $y=x$ you should've drawn already, shifted down by 3). Then, shade in the part above this line and below the line $y=x$. It will be like an infinite parallelogram band.$

Green Yoda · Mar 15, 2016

InteGrand said:
$\noindent We have $0\leq x-y<3$. Firstly this means $x-y \geq 0$, so $y\leq x$. So the entire line $y=x$ is included. Also, any remaining parts must be below this line. For the second part of the inequality, $x-y<3 \Rightarrow y>x-3$. So draw the line $y=x-3$ (as a dotted line, since we won't include it; this line is just the line $y=x$ you should've drawn already, shifted down by 3). Then, shade in the part above this line and below the line $y=x$. It will be like an infinite parallelogram band.$

thank you

p.s soz for my noob questions leehuan on ur thread haha

Green Yoda · Mar 15, 2016

InteGrand said:
$\noindent We have $0\leq x-y<3$. Firstly this means $x-y \geq 0$, so $y\leq x$. So the entire line $y=x$ is included. Also, any remaining parts must be below this line. For the second part of the inequality, $x-y<3 \Rightarrow y>x-3$. So draw the line $y=x-3$ (as a dotted line, since we won't include it; this line is just the line $y=x$ you should've drawn already, shifted down by 3). Then, shade in the part above this line and below the line $y=x$. It will be like an infinite parallelogram band.$

thank you

p.s soz for my noob questions leehuan on ur thread haha

Paradoxica · Mar 16, 2016

seanieg89 said:
Where did you find this sum in the first place? That should provide a good clue as to the cleanest known way to compute it.

Anyway, if you find a satisfactory explanation for why it works out the way it does, please post it here.

I will probably have another think about it this evening to see if I can find a nicer way.

I found two things on SE remotely related to this identity form.

Trigonometric identity involving sum of “Dirichlet kernel like” fractions

Elegant Trigonometric Sums

seanieg89 · Mar 16, 2016

seanieg89 said:
I will probably have another think about it this evening to see if I can find a nicer way.

So I did have another think about it and couldn't come up with a much nicer way.

I wrote out my proof though, so perhaps that will be of some help to people who want to understand it more.

Remark: You might expect with an answer as simple as $b(n-a)(-1)^{a+b}$ , there must be a much quicker way of computing the sum. Whilst there could very well be a faster way of computing the sum, note that that expression is only valid under the assumption that a >= b. (If not, then think about the fact that each summand is symmetric in a and b, but the resulting sum is not!) If we wanted to relax the assumption that a >= b, our answer would actually be

$\min(a,b)\cdot(n-\max(a,b))\cdot (-1)^{a+b}$

which is not quite as nice.

Also note that we had similar asymmetric dependence on the larger/smaller parameters in a trigonometric integral I posted in the MX2 integration marathon a while back, this behaviour crops up often when multiplying two oscillatory expressions that oscillate at different rates and then summing/integrating.

seanieg89 · Mar 16, 2016

For anyone that doesn't remember, the integral referred to in my above post was

$\int_{-\infty}^\infty \frac{\sin(ax)\sin(bx)}{x^2}\, dx$

where $a,b\in\mathbb{R}$ .

turntaker · Mar 16, 2016

lee help me with 1+1 - 1 = ?

Paradoxica · Mar 17, 2016

What is a continued product? I have looked up the definitions, but I'm getting different definitions.

For reference, I'm looking at mathematical literature from the early 1900's.

seanieg89 · Mar 17, 2016

Paradoxica said:
What is a continued product? I have looked up the definitions, but I'm getting different definitions.

For reference, I'm looking at mathematical literature from the early 1900's.

It's just a name. Lots of old terminology like that is ambiguous and different mathematicians mean different things by it. I don't hear that particular name much in modern mathematics so I doubt you would get universal agreement on what objects to use that name for.

Drsoccerball · Mar 18, 2016

How can you tell the difference between a line and a plane when its in parametric form vs when its in Cartesian form?

leehuan · Mar 18, 2016

Drsoccerball said:
How can you tell the difference between a line and a plane when its in parametric form vs when its in Cartesian form?

Lines in R2:
ax+by=c

Planes in R3:
ax+by+cz=d

(Hyperplanes in R4: ax_1+bx_2+cx_3+dx_4=e)

Then parametrics aren't too bad either

Lines in R^n
x=a+lambdab

Planes in R^n
x=a+lambdab+muc

Drsoccerball · Mar 18, 2016

leehuan said:
Lines in R2:
ax+by=c

Planes in R3:
ax+by+cz=d

(Hyperplanes in R4: ax_1+bx_2+cx_3+dx_4=e)

Then parametrics aren't too bad either

Lines in R^n
x=a+lambdab

Planes in R^n
x=a+lambdab+muc

Aren't lines just any vectors except when its in the form of a plane?

$$ Plane would be $ \overrightarrow{x}= \lambda (...) + \gamma (...)$ With no starting point?

InteGrand · Mar 18, 2016

Drsoccerball said:
$$ Plane would be $ \overrightarrow{x}= \lambda (...) + \gamma (...)$ With no starting point?

$\noindent That would be a plane through the origin (assuming the two vectors are non-parallel and non-zero of course).$

leehuan · Mar 18, 2016

Drsoccerball said:
Aren't lines just any vectors except when its in the form of a plane?

$$ Plane would be $ \overrightarrow{x}= \lambda (...) + \gamma (...)$ With no starting point?

What? What do you mean lines are any vectors except when it's in the form of a plane?

And the plane you gave is essentially any plane... that is parallel through your vectors (...) passing through the origin
Assuming that yes, lambda and gamma are non-zero and the vectors aren't parallel

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

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