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

RenegadeMx · Aug 25, 2016

Re: MATH1231/1241/1251 SOS Thread

leehuan said:
^Feelsbad delaying the exercise

Anyway

$$Show that the set $S=\{p\in \mathbb{P}_2: p(0)=1\}$ is not a subspace of $\mathbb{P}_2$

Is this just because the zero polynomial can't be in S?

yes

RenegadeMx · Aug 25, 2016

Re: MATH1231/1241/1251 SOS Thread

to prove its not a v.s in general u either show 0 isnt there (easiest) or that it doesnt satisfy addition/scalar multiplication

only need to show 1 of those conditions as well

leehuan · Aug 25, 2016

Re: MATH1231/1241/1251 SOS Thread

RenegadeMx said:
yes

Lol sweet.
_______________________

Even if there IS a typo in the notation, this question is actually confusing me to the maximum...

This was the answer I got to part (ii)

$z^2-2\left(\cos \frac{2\pi}{5}+\cos \frac{4\pi}{5}\right)z+2\left(\cos \frac{2\pi}{5}+\cos \frac{6\pi}{5}\right)$

leehuan · Aug 25, 2016

Re: MATH1231/1241/1251 SOS Thread

RenegadeMx said:
to prove its not a v.s in general u either show 0 isnt there (easiest) or that it doesnt satisfy addition/scalar multiplication

only need to show 1 of those conditions as well

Yeah sweet, that reaffirms what I know

Paradoxica · Aug 25, 2016

Re: MATH1231/1241/1251 SOS Thread

leehuan said:
Lol sweet.
_______________________

Even if there IS a typo in the notation, this question is actually confusing me to the maximum...

This was the answer I got to part (ii)

$z^2-2\left(\cos \frac{2\pi}{5}+\cos \frac{4\pi}{5}\right)z+2\left(\cos \frac{2\pi}{5}+\cos \frac{6\pi}{5}\right)$

$z^2 -(\alpha + \alpha^4 + \alpha^2 + \alpha^3)z + (\alpha^3 + \alpha^2 + \alpha + \alpha^4)$

You forgot that α⁵=1

leehuan · Aug 25, 2016

Re: MATH1231/1241/1251 SOS Thread

Paradoxica said:
$z^2 -(\alpha + \alpha^4 + \alpha^2 + \alpha^3)z + (\alpha^3 + \alpha^2 + \alpha + \alpha^4)$

You forgot that α⁵=1

Yes... Yes I did...

Far out. Thanks

Flop21 · Aug 26, 2016

Re: MATH1231/1241/1251 SOS Thread

How do I know what values to pick for constructing the triangle? This is in my textbook too, but they still don't explain how to pick the triangle's values.

RenegadeMx · Aug 26, 2016

Re: MATH1231/1241/1251 SOS Thread

Flop21 said:
How do I know what values to pick for constructing the triangle? This is in my textbook too, but they still don't explain how to pick the triangle's values.

its year 9/10 trig. tan = x/2 = opp/adj

Flop21 · Aug 26, 2016

Re: MATH1231/1241/1251 SOS Thread

RenegadeMx said:
its year 9/10 trig. tan = x/2 = opp/adj

oh is this the unit circle?

InteGrand · Aug 26, 2016

Re: MATH1231/1241/1251 SOS Thread

Flop21 said:
oh is this the unit circle?

Pythagoras's theorem. Alternatively, recall the identity tan^2 (u) + 1= sec^2 (u).

Flop21 · Aug 27, 2016

Re: MATH1231/1241/1251 SOS Thread

InteGrand said:
Pythagoras's theorem. Alternatively, recall the identity tan^2 (u) + 1= sec^2 (u).

I'm really confused still. How do you get tan = x/2. The x and the 2, where do they come from.

InteGrand · Aug 27, 2016

Re: MATH1231/1241/1251 SOS Thread

Flop21 said:
I'm really confused still. How do you get tan = x/2. The x and the 2, where do they come from.

$\noindent Recall that we let $x = 2\tan u$ for the substitution. Therefore, $\tan u = \frac{x}{2}$.$

Flop21 · Aug 27, 2016

Re: MATH1231/1241/1251 SOS Thread

InteGrand said:
$\noindent Recall that we let $x = 2\tan u$ for the substitution. Therefore, $\tan u = \frac{x}{2}$.$

OH. Get it now thanks

Flop21 · Aug 27, 2016

Re: MATH1231/1241/1251 SOS Thread

To find if something is a linear combination... do we do the following always?

What if we wants to see if a bunch of polynomials form spanning set for P2? Can we do the same?

Thanks.

InteGrand · Aug 27, 2016

Re: MATH1231/1241/1251 SOS Thread

Flop21 said:
To find if something is a linear combination... do we do the following always?

What if we wants to see if a bunch of polynomials form spanning set for P2? Can we do the same?

Thanks.

$\noindent Chuck those polynomials as vectors into a matrix (as the columns of the matrix). Get the matrix into row-echelon form. The vectors (polynomials) will be spanning if and only if the row-echelon form has no zero rows (i.e. every row has a pivot element).$

We can also express a non-leading column's vector in the original matrix as a linear combination of the leading column vectors from the original matrix through back substitution etc. after row-reducing (work through an example to get the hang of this).

Flop21 · Aug 27, 2016

Re: MATH1231/1241/1251 SOS Thread

Why is: a set of 5 vectors in R5 must be a basis for R5, false?

Basically how can I tell how many linear independent vectors can be in Rn. Or how can you tell if they're linearly independent or not just from the number of vectors and Rn?

InteGrand · Aug 27, 2016

Re: MATH1231/1241/1251 SOS Thread

Flop21 said:
Why is: a set of 5 vectors in R5 must be a basis for R5, false?

Basically how can I tell how many linear independent vectors can be in Rn. Or how can you tell if they're linearly independent or not just from the number of vectors and Rn?

$\noindent Any old set of five vectors in $\mathbb{R}^{5}$ need not be a basis. E.g. if the set contains the zero vector, it won't be a basis (because it's automatically now linearly \emph{de}pendent).$

$\noindent The maximum number of linearly independent vectors we can have in a subset from $\mathbb{R}^{n}$ is $n$. So for instance, it's impossible to have a set of four vectors in $\mathbb{R}^{3}$ be linearly independent. If we have a set of \emph{more} than $n$ vectors from $\mathbb{R}^{n}$, this set is guaranteed linearly \emph{de}pendent. If we have $n$ or fewer vectors, we can't conclude whether they're linear independent just from this, we'd need to do further things (like row-reduce a matrix).$

leehuan · Aug 27, 2016

Re: MATH1231/1241/1251 SOS Thread

I feel like my book is trolling me. Is there any convenience in using second order ODE methods to solve these over bringing in an integrating factor?

$y^\prime + 3y=0\\ y^\prime+2y=10e^{2x}$

InteGrand · Aug 27, 2016

Re: MATH1231/1241/1251 SOS Thread

leehuan said:
I feel like my book is trolling me. Is there any convenience in using second order ODE methods to solve these over bringing in an integrating factor?

$y^\prime + 3y=0\\ y^\prime+2y=10e^{2x}$

You can solve these like second-order ODE's with constant coefficients. Any order ODE with constant coefficients can be solved via characteristic equation etc.

$\noindent For any linear ODE, the general solution is a particular solution plus the general solution to the homogeneous equation. Here, since they're first order with constant coefficients, the homogeneous solutions can be inspected. E.g. For the second one, the homogeneous solution is, by inspection, $y_h = Ae^{-2x}$, for an arbitrary constant $A$. (If you want it, the characteristic equation is of course $\lambda +2 = 0$.) Then to find a particular solution, try $y_p = ce^{2x}$, and determine what $c$ must by substituting into the ODE: $2c + 2c = 10 \Rightarrow c = \frac{5}{2}$. So the general solution to the ODE is $y = Ae^{-2x} + \frac{5}{2} e^{2x}$.$

Without any further info, I don't know what context your book asked these in, but maybe it's just getting you to see that ODE's with constant coefficients of other-than-second order can be solved similarly.

$\noindent (The first ODE you posted there is even easier. We can do it entirely by inspection since it's homogeneous too. The solution is $y = Ae^{-3x}$, where $A$ is an arbitrary constant.)$

leehuan · Aug 27, 2016

Re: MATH1231/1241/1251 SOS Thread

InteGrand said:
You can solve these like second-order ODE's with constant coefficients. Any order ODE with constant coefficients can be solved via characteristic equation etc.

$\noindent For any linear ODE, the general solution is a particular solution plus the general solution to the homogeneous equation. Here, since they're first order with constant coefficients, the homogeneous solutions can be inspected. E.g. For the second one, the homogeneous solution is, by inspection, $y_h = Ae^{-2x}$, for an arbitrary constant $A$. (If you want it, the characteristic equation is of course $\lambda +2 = 0$.) Then to find a particular solution, try $y_p = ce^{2x}$, and determine what $c$ must by substituting into the ODE: $2c + 2c = 10 \Rightarrow c = \frac{5}{2}$. So the general solution to the ODE is $y = Ae^{-2x} + \frac{5}{2} e^{2x}$.$

Without any further info, I don't know what context your book asked these in, but maybe it's just getting you to see that ODE's with constant coefficients of other-than-second order can be solved similarly.

$\noindent (The first ODE you posted there is even easier. We can do it entirely by inspection since it's homogeneous too. The solution is $y = Ae^{-3x}$, where $A$ is an arbitrary constant.$

Yeah I guess so. Makes sense

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

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