leehuan's All-Levels-Of-Maths SOS thread (3 Viewers)

leehuan · Mar 23, 2016

InteGrand said:
What do you mean? Plug. in lambda = 2 to the matrix you posted. Then interpret the rows as equations as usual.

I mean, how would I know that the specific value was lambda = 2 and not something else like 5

Green Yoda · Mar 23, 2016

InteGrand said:
Because then the denominator of the RHS would be 0, and we can't divide by 0.

Edit: oh you said why *can*. Well, it can't be -1 in fact, as you had.

so my answer is correct right?

InteGrand · Mar 23, 2016

leehuan said:
I mean, how would I know that the specific value was lambda = 2 and not something else like 5

Always try the ones that make diagonal elements 0. If a diagonal element is 0, it means the matrix is either not yet in row echelon form (e.g. Plugging lambda = 2 makes the matrix require further operations to get into row echelon form, since then the third column needs to be cleared up), or we get a zero row. In either case, you'll see eventually we have a zero row (when lambda = 2, we clear the third column and get a zero row in the third row), and when (and only when) that happens, we know there's potential for a lack of a solution.

If we just sub. in something random like 5, we see that the matrix ends up being in row echelon form with all rows leading, so there is a solution then. So in summary, try and only try values that'll make diagonal elements become 0, so that there will be a zero row, leading to a possibility of no solutions.

leehuan · Mar 23, 2016

Ah so to summarise, force out the diagonal matrix of coefficients to see what's going on?

InteGrand · Mar 23, 2016

leehuan said:
Ah so to summarise, force out the diagonal matrix of coefficients to see what's going on?

Yeah, if you're trying to find for which values of lambda the equations have no solution, just try and make diagonal entries 0, and investigate what happens (it's still possible that there solutions, because the right hand column may also be 0 where there's a zero row, depending on the situation with the lambdas, so this is why you need to investigate to see whether there are solutions or not).

(Basically try and only try values of lambda that'll immediately force a zero row, or will cause the matrix to NOT be in row echelon form.)

RealiseNothing · Mar 23, 2016

Rathin said:
quick question:
for 1/|x-1| > 1/|x+1|
I got the answer as x>0 and x≠1 or x≠-1
but the answer is just x>0 and x≠1
why can x=-1?

The textbook's answer is better though. You don't need to say x≠-1 since you already said x>0.

Green Yoda · Mar 23, 2016

RealiseNothing said:
The textbook's answer is better though. You don't need to say x≠-1 since you already said x>0.

hmm i see thanks

Drsoccerball · Mar 27, 2016

How do you find the shortest distance in vectors.

1. How to derive it for a point to a line(in Cartesian form vs parametric form)
2. From line to line (Cartesian vs parametric)

leehuan · Mar 27, 2016

Calling InteGrand and co.

I haven't done vectorial geometry like I said, to know how to do his question

InteGrand · Mar 27, 2016

Drsoccerball said:
How do you find the shortest distance in vectors.

1. How to derive it for a point to a line(in Cartesian form vs parametric form)
2. From line to line (Cartesian vs parametric)

1) The method is described here in the 'vector formulation' section:

https://en.wikipedia.org/wiki/Distance_from_a_point_to_a_line#A_vector_projection_proof

2) The method for finding the distance between two skew lines (non-parallel yet non-intersecting lines, which can exist only in dimension 3 or higher) in R^3 is described here:

https://en.wikipedia.org/wiki/Skew_lines#Distance .

If the lines are parallel, use the method from part 1). You can use any point on Line 1 as the external point to Line 2 and find its distance to Line 2, and this will be the answer.

leehuan · Mar 28, 2016

Need a demo please.

$\\ $Use the $\varepsilon - \delta$ definition of a limit to show that$ \\ \lim_{x\rightarrow0}{\frac{1-\cos{x}}{x^2}}=\frac{1}{2}$

Paradoxica · Mar 28, 2016

leehuan said:
Need a demo please.

$\\ $Use the $\varepsilon - \delta$ definition of a limit to show that$ \\ \lim_{x\rightarrow0}{\frac{1-\cos{x}}{x^2}}=\frac{1}{2}$

Double Angle Expansion seems to be the way to go.

Paradoxica · Mar 28, 2016

leehuan said:
Need a demo please.

$\\ $Use the $\varepsilon - \delta$ definition of a limit to show that$ \\ \lim_{x\rightarrow0}{\frac{1-\cos{x}}{x^2}}=\frac{1}{2}$

Upon further investigation, I have the following result:

$\frac{1-\cos{x}}{x^2} \geq \frac{x-\sin{x}}{x^3} \geq 0$

Which is true for

$x \in \left[ 0, \frac{\pi}{2} \right]$

It can be derived by assuming the following result:

$\cos{x}\leq \frac{\sin{x}}{x}\leq 1$

which can be proven geometrically.

parad0xica · Mar 28, 2016

leehuan said:
Need a demo please.

$\\ $Use the $\varepsilon - \delta$ definition of a limit to show that$ \\ \lim_{x\rightarrow0}{\frac{1-\cos{x}}{x^2}}=\frac{1}{2}$

$\forall \epsilon > 0, \exists \delta > 0 $ such that $ 0 < |x-0| < \delta \Rightarrow \left|\frac{1- \cos x}{x^2} - \frac{1}{2}\right| < \epsilon$

Choose $\delta = \frac{1}{\sqrt{2\epsilon}}$ , then we have $0 < |x| < \frac{1}{\sqrt{2\epsilon}}$ .

Now
$\left|\frac{1- \cos x}{x^2} - \frac{1}{2}\right| \\\stackrel{A}{=} \left|\frac{\sin ^2 \left(\frac{x}{2} \right)}{2x^2} - \frac{1}{2} \right| \\\stackrel{B}{=} \frac{1}{2} \left| \frac{\cos ^2 \left(\frac{x}{2} \right)}{x^2}\right| \\\stackrel{C}{=} \frac{1}{2} \frac{1}{|x|^2} \\\stackrel{D}{<} \epsilon$

as required.

A: double angle
B: s^2 + c^2 = 1
C: cosine attains max at 1
D: hypothesis

parad0xica · Mar 28, 2016

With these types of proofs, you choose $\delta$ at the end ^_^

Suppose the general formula: $\lim_{x \to a} f(x) = L$ .

Then $\forall \epsilon > 0, \exists \delta > 0 $ such that $ 0 < |x-a| < \delta \Rightarrow |f(x) - L| < \epsilon.$

$$You simplify $|f(x) - L| < \epsilon $, because this is what you want, to reach something like $3|x-a| < \epsilon.$ So $ \delta = \frac{\epsilon}{3}$ .

Then you re-write the proof somehow making a magical guess of $\delta$ !

Drsoccerball · Mar 28, 2016

When parad0xica is smarter than the original lel...

Paradoxica · Mar 28, 2016

parad0xica said:
$\forall \epsilon > 0, \exists \delta > 0 $ such that $ 0 < |x-0| < \delta \Rightarrow \left|\frac{1- \cos x}{x^2} - \frac{1}{2}\right| < \epsilon$

Choose $\delta = \frac{1}{\sqrt{2\epsilon}}$ , then we have $0 < |x| < \frac{1}{\sqrt{2\epsilon}}$ .

Now
$\left|\frac{1- \cos x}{x^2} - \frac{1}{2}\right| \\\stackrel{A}{=} \left|\frac{\sin ^2 \left(\frac{x}{2} \right)}{2x^2} - \frac{1}{2} \right| \\\stackrel{B}{=} \frac{1}{2} \left| \frac{\cos ^2 \left(\frac{x}{2} \right)}{x^2}\right| \\\stackrel{C}{=} \frac{1}{2} \frac{1}{|x|^2} \\\stackrel{D}{<} \epsilon$

as required.

A: double angle
B: s^2 + c^2 = 1
C: cosine attains max at 1
D: hypothesis

Note for part A and B, you should not have x/2. The substitution process is clear to me, and I have verified it myself, so it should definitely not be there.

Paradoxica · Mar 28, 2016

Drsoccerball said:
When parad0xica is smarter than the original lel...

When both paradoxicas are smarter then you

Drsoccerball · Mar 28, 2016

Paradoxica said:
Note for part A and B, you should not have x/2. The substitution process is clear to me, and I have verified it myself, so it should definitely not be there.

double angle formula bruh

Nailgun · Mar 28, 2016

parad0xica said:
$\forall \epsilon > 0, \exists \delta > 0 $ such that $ 0 < |x-0| < \delta \Rightarrow \left|\frac{1- \cos x}{x^2} - \frac{1}{2}\right| < \epsilon$

Choose $\delta = \frac{1}{\sqrt{2\epsilon}}$ , then we have $0 < |x| < \frac{1}{\sqrt{2\epsilon}}$ .

Now
$\left|\frac{1- \cos x}{x^2} - \frac{1}{2}\right| \\\stackrel{A}{=} \left|\frac{\sin ^2 \left(\frac{x}{2} \right)}{2x^2} - \frac{1}{2} \right| \\\stackrel{B}{=} \frac{1}{2} \left| \frac{\cos ^2 \left(\frac{x}{2} \right)}{x^2}\right| \\\stackrel{C}{=} \frac{1}{2} \frac{1}{|x|^2} \\\stackrel{D}{<} \epsilon$

as required.

A: double angle
B: s^2 + c^2 = 1
C: cosine attains max at 1
D: hypothesis

wait you actually can do maths

i thought you were a troll account

wat

leehuan's All-Levels-Of-Maths SOS thread (3 Viewers)

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