First Year Mathematics B (Integration, Series, Discrete Maths & Modelling) (3 Viewers)

RenegadeMx · Sep 6, 2016

Re: MATH1231/1241/1251 SOS Thread

Paradoxica said:
Find the solution to this recurrence relation WITHOUT induction.

U²_n+1 = U_n + 2

U_₀ = 0

this doesnt sound like math12x1

leehuan · Sep 6, 2016

Re: MATH1231/1241/1251 SOS Thread

RenegadeMx said:
this doesnt sound like math12x1

Sounds like discrete maths. Yeah it's out of place.

(Though we're starting topic 4 tomorrow)

RenegadeMx · Sep 6, 2016

Re: MATH1231/1241/1251 SOS Thread

leehuan said:
Sounds like discrete maths. Yeah it's out of place.

(Though we're starting topic 4 tomorrow)

hmm it can also be linked to DE's tho using the idea of a char eqn

leehuan · Sep 7, 2016

Re: MATH1231/1241/1251 SOS Thread

$\text{Why does }\lim_{n\to \infty}\frac{a_{n+1}}{a_n}=L\text{ imply }\lim_{n\to \infty}\sqrt[n]{a_n}=L$

InteGrand · Sep 7, 2016

Re: MATH1231/1241/1251 SOS Thread

leehuan said:
$\text{Why does }\lim_{n\to \infty}\frac{a_{n+1}}{a_n}=L\text{ imply }\lim_{n\to \infty}\sqrt[n]{a_n}=L$

This is famous result (convergence of ratio test implies convergence of root test). A proof may be found here: http://math.stackexchange.com/quest...tio-test-implies-convergence-of-the-root-test .

(The converse of this result is not true – an example showing this is given in the answer here: http://math.stackexchange.com/quest...tio-and-root-tests-converge-to-the-same-limit .)

leehuan · Sep 7, 2016

Re: MATH1231/1241/1251 SOS Thread

InteGrand said:
This is famous result (convergence of ratio test implies convergence of root test). A proof may be found here: http://math.stackexchange.com/quest...tio-test-implies-convergence-of-the-root-test .

Looks a bit hard for a mere 30 minute online quiz...

Maybe we're getting taught it soon. But thanks

InteGrand · Sep 7, 2016

Re: MATH1231/1241/1251 SOS Thread

leehuan said:
Looks a bit hard for a mere 30 minute online quiz...

Maybe we're getting taught it soon. But thanks

You need to write up proofs in online quizzes?

leehuan · Sep 7, 2016

Re: MATH1231/1241/1251 SOS Thread

InteGrand said:
You need to write up proofs in online quizzes?

Oh no, it was a multiple choice question asking "which of the following were true"

leehuan · Sep 9, 2016

Re: MATH1231/1241/1251 SOS Thread

$\text{Show that the following four matrices form a basis for }M_{22}(\mathbb{R})\\ \begin{pmatrix} 1 &0\\0&0\end{pmatrix},\begin{pmatrix}0& 1 \\0&0\end{pmatrix},\begin{pmatrix}0&0\\1 & 0\end{pmatrix},\begin{pmatrix}0&0\\0& 1 \end{pmatrix}$

Essentially the proof of the standard basis. So the proof for linear independence is trivial, but these answers (no prescribed answers - got off second year students) are confusing me.

$\dim(M_{22})=4\text{ and }S\text{ has 4 elements. }\therefore S\text{ spans }M_{22}$

I get that the dimension of M₂₂ is 4 but does S having four elements really suffice? Because if we had two matrices that were scalar products of each other then that'd be problematic right...

InteGrand · Sep 9, 2016

Re: MATH1231/1241/1251 SOS Thread

leehuan said:
$\text{Show that the following four matrices form a basis for }M_{22}(\mathbb{R})\\ \begin{pmatrix} 1 &0\\0&0\end{pmatrix},\begin{pmatrix}0& 1 \\0&0\end{pmatrix},\begin{pmatrix}0&0\\1 & 0\end{pmatrix},\begin{pmatrix}0&0\\0& 1 \end{pmatrix}$

Essentially the proof of the standard basis. So the proof for linear independence is trivial, but these answers (no prescribed answers - got off second year students) are confusing me.

$\dim(M_{22})=4\text{ and }S\text{ has 4 elements. }\therefore S\text{ spans }M_{22}$

I get that the dimension of M₂₂ is 4 but does S having four elements really suffice? Because if we had two matrices that were scalar products of each other then that'd be problematic right...

Yeah in general if you have a vector space of dimension n (n finite), then S being a set of n linearly independent vectors implies it is also spanning, and hence a basis. Another fact is that S having n vectors and being spanning implies it is linearly independent, and hence a basis.

So if you have a vector space of finite dimension n, then having a subset S with n vectors from it that is linearly independent is equivalent to it being spanning, which is equivalent to it being a basis.

leehuan · Sep 9, 2016

Re: MATH1231/1241/1251 SOS Thread

InteGrand said:
Yeah in general if you have a vector space of dimension n (n finite), then S being a set of n linearly independent vectors implies it is also spanning, and hence a basis. Another fact is that S having n vectors and being spanning implies it is linearly independent, and hence a basis.

So if you have a vector space of finite dimension n, then having a subset S with n vectors from it that is linearly independent is equivalent to it being spanning, which is equivalent to it being a basis.

Ahh right thanks, I keep forgetting that.

InteGrand · Sep 9, 2016

Re: MATH1231/1241/1251 SOS Thread

$\noindent For this one though it'd be better to just prove from first principles that the given set is spanning and linearly independent (and hence a basis). This would in fact be an elementary way to \emph{prove} that $M_{22} (\mathbb{R})$ is four-dimensional. If we assume it's four dimensional to prove this standard basis set is indeed a basis, the reasoning is a bit circular.$

$\noindent You say you've already proved the linear independence bit. To prove it's spanning, for simplicity call those four matrices in $S$ $A,B,C,D$ respectively. Let $X = \begin{pmatrix}a & b \\c & d\end{pmatrix}$ be an arbitrary matrix from $M_{22}(\mathbb{R})$. Then $a,b,c,d\in \mathbb{R}$ and it is clear that $X = aA + bB + cC + dD$ (by definition of scalar multiplication of a matrix and addition of matrices). So any matrix $X\in M_{22}(\mathbb{R})$ can be exhibited as a linear combination of the vectors in $S$. Hence $S$ is spanning. So as $S$ is linearly independent (proved by you you said) and spanning, it is by definition a basis for $M_{22}(\mathbb{R})$.$

$\noindent (As a corollary, $M_{22}\left(\mathbb{R}\right)$ has dimension $4$, remembering that by definition the dimension of a vector space with a finite spanning set is the number of vectors in a basis for it, and recalling that this definition is well-defined due to the result in linear algebra that shows that any basis in a vector space with a finite spanning set has the same number of elements as any other basis for it.)$

leehuan · Sep 10, 2016

Re: MATH1231/1241/1251 SOS Thread

Suppose f(x) is expressed as the sum of a Taylor polynomial and a remainder. Then, in Lagrange form:

$R_{n+1}(x)=\frac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1}$

Is c in (0,a) or [0,a]?

Paradoxica · Sep 10, 2016

Re: MATH1231/1241/1251 SOS Thread

leehuan said:
Suppose f(x) is expressed as the sum of a Taylor polynomial and a remainder. Then, in Lagrange form:

$R_{n+1}(x)=\frac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1}$

Is c in (0,a) or [0,a]?

The former.

wait... that's not standard. usually the LHS is down to n, RHS to n+1. What are you doing.

leehuan · Sep 10, 2016

Re: MATH1231/1241/1251 SOS Thread

Paradoxica said:
wait... that's not standard. usually the LHS is down to n, RHS to n+1. What are you doing.

What I was taught in the lecture

Paradoxica · Sep 10, 2016

Re: MATH1231/1241/1251 SOS Thread

leehuan said:
What I was taught in the lecture

Well, whatever. Ir regardless of the inconsistency in notation, the interval strictness should remain the same.

InteGrand · Sep 10, 2016

Re: MATH1231/1241/1251 SOS Thread

leehuan said:
Suppose f(x) is expressed as the sum of a Taylor polynomial and a remainder. Then, in Lagrange form:

$R_{n+1}(x)=\frac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1}$

Is c in (0,a) or [0,a]?

Well c is strictly between a and x, so it's in (a, x) or (x, a) (depending which of a or x is bigger).

leehuan · Sep 10, 2016

Re: MATH1231/1241/1251 SOS Thread

Ah my bad. I got my formula wrong at the time so I put down 0..

But I see now. Overlooked that we can't strictly state which is larger than the other. Guess I have to write it in word form then.

Paradoxica · Sep 11, 2016

Re: MATH1231/1241/1251 SOS Thread

leehuan said:
Ah my bad. I got my formula wrong at the time so I put down 0..

But I see now. Overlooked that we can't strictly state which is larger than the other. Guess I have to write it in word form then.

No you don't. The interval is

(min(a,x), max(a,x))

or, more algorithmically:

$\left( \frac{a+x-|a-x|}{2}, \frac{a+x+|a-x|}{2} \right)$

Flop21 · Sep 13, 2016

Re: MATH1231/1241/1251 SOS Thread

How did this happen. The guy goes "and here's the trick" and writes the result. How can he just look at that LHS and do what he did? What's the trick I'm missing?

First Year Mathematics B (Integration, Series, Discrete Maths & Modelling) (3 Viewers)

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