First Year Mathematics B (Integration, Series, Discrete Maths & Modelling) (9 Viewers)

RenegadeMx · Aug 31, 2016

Re: MATH1231/1241/1251 SOS Thread

Flop21 said:
I thought no leading columns on a row meant NO solution.

What's going on here? And has it got something to do with the way you put your polynomals into a matrix?

only if u get 0=something
if u have 0=0 its inf many sols

leehuan · Aug 31, 2016

Re: MATH1231/1241/1251 SOS Thread

Oh right didn't register the query fully.

Yeah refer to what they all said Flop. Note that your right hand column is not leading and that's what distinguishes between no solutions and infinite solutions

When the right hand column is leading, in row 3 you have 0=5 (or any thing not zero), which has no solution
When the right hand column is not, in this scenario in row 3 you have 0=0, which is always true

Flop21 · Aug 31, 2016

Re: MATH1231/1241/1251 SOS Thread

thx yeah lol I actually knew this really well, not sure why I confused myself and forgot? Too much info in my brain right now.

leehuan · Sep 1, 2016

Re: MATH1231/1241/1251 SOS Thread

Use the vector space axioms to prove that we do not need to use brackets when writing down the linear combination

$\sum_{k=1}^n\lambda_k\textbf{v}_k=\lambda_1\textbf{v}_1+\lambda_2\textbf{v}_2+\dots+\lambda_n\textbf{v}_n$

That is, prove that the result of the operations in independent of the order in which the additions are performed.

Some tips please before I ask for a full solution

InteGrand · Sep 1, 2016

Re: MATH1231/1241/1251 SOS Thread

leehuan said:
Use the vector space axioms to prove that we do not need to use brackets when writing down the linear combination

$\sum_{k=1}^n\lambda_k\textbf{v}_k=\lambda_1\textbf{v}_1+\lambda_2\textbf{v}_2+\dots+\lambda_n\textbf{v}_n$

That is, prove that the result of the operations in independent of the order in which the additions are performed.

Some tips please before I ask for a full solution

First note we don't need to worry about the lambdas, suffices to just prove it for v1 + v2 + … + vn (can you see why?).

So given that we know (from axioms) that + is an associative and commutative operation, we need to show that any bracketing will lead to the same answer. Try to do this using induction on n. The basic idea is that we can move brackets around by associativity and vectors around inside brackets by commutativity. So basically show by doing this, we can obtain any arrangement of the v's.

leehuan · Sep 1, 2016

Re: MATH1231/1241/1251 SOS Thread

InteGrand said:
First note we don't need to worry about the lambdas, suffices to just prove it for v1 + v2 + … + vn (can you see why?).

So given that we know (from axioms) that + is an associative and commutative operation, we need to show that any bracketing will lead to the same answer. Try to do this using induction on n. The basic idea is that we can move brackets around by associativity and vectors around inside brackets by commutativity. So basically show by doing this, we can obtain any arrangement of the v's.

Is that first bit because scaling has no impact on addition here and we can just let a_n = lambda_nj_n by closure?

And, oh dear I don't know how to write my inductive assumption

InteGrand · Sep 1, 2016

Re: MATH1231/1241/1251 SOS Thread

Yeah.

The inductive hypothesis is basically the given statement (ignoring the lambdas) for a particular integer n.

leehuan · Sep 1, 2016

Re: MATH1231/1241/1251 SOS Thread

Oh so just take

$\sum_{k=1}^m \textbf{a}_k=\textbf{a}_1+\textbf{a}_2+\dots+ \textbf{a}_m$

And then say

$\begin{align*}\sum_{k=1}^{m+1}\textbf{a}_k&=\sum_{k=1}^m \textbf{a}_k+\textbf{a}_{m+1}\\ &\stackrel{*}{=}\left(\textbf{a}_1+\textbf{a}_2+ \dots+ \textbf{a}_m\right)+\textbf{a}_{m+1}\end{align*}$

...wait I'm lost. I can see how to use associativity (sort of) but have I done something I shouldn't have? Because I can't see where commutativity comes in

InteGrand · Sep 1, 2016

Re: MATH1231/1241/1251 SOS Thread

Let's see what it's like for three elements (say a, b, c).

The claim is that any bracketing and ordering of a+b+c results in the same thing (and hence that brackets aren't needed).

I.e. Let A = (a+b)+c. The claim is that any bracketing and ordering results in A still. For example, with the bracketing and ordering (c+b)+a, we have:

(c+b)+a = c+(b+a) (associativity)

= c+(a+b) (commutativity)

= (a+b)+c (commutativity)

= A,

as expected. The claim is that by using commutativity and associativity, we will always (given any initial order and bracketing) be able to get the same result, say the result of the ordering and bracketing (a+b)+c (and also similarly for any general n elements – this is where induction can help).

Flop21 · Sep 2, 2016

Re: MATH1231/1241/1251 SOS Thread

I know how to do these, but I am having trouble finding values to create <x,y>, any hints?

InteGrand · Sep 2, 2016

Re: MATH1231/1241/1251 SOS Thread

Flop21 said:
I know how to do these, but I am having trouble finding values to create <x,y>, any hints?

$\noindent Try finding values of scalars $a$ and $b$ (which will depend on $x$ and $y$) such that given any $x,y \in \mathbb{R}$, we have $\begin{pmatrix}x \\y\end{pmatrix} = a \begin{pmatrix}1 \\ 0\end{pmatrix} + b \begin{pmatrix}1 \\ 1\end{pmatrix}$. Then by linearity we will have $S \begin{pmatrix}x \\ y\end{pmatrix} = a \begin{pmatrix}-4\\ 3\end{pmatrix} + b \begin{pmatrix}1 \\ 2\end{pmatrix}$.$

Flop21 · Sep 2, 2016

Re: MATH1231/1241/1251 SOS Thread

InteGrand said:
$\noindent Try finding values of scalars $a$ and $b$ (which will depend on $x$ and $y$) such that given any $x,y \in \mathbb{R}$, we have $\begin{pmatrix}x \\y\end{pmatrix} = a \begin{pmatrix}1 \\ 0\end{pmatrix} + b \begin{pmatrix}1 \\ 1\end{pmatrix}$. Then by linearity we will have $S \begin{pmatrix}x \\ y\end{pmatrix} = a \begin{pmatrix}-4\\ 3\end{pmatrix} + b \begin{pmatrix}1 \\ 2\end{pmatrix}$.$

Thank you.

And why is this not correct? I did the vector <6,6,3> * the matrix A. What's the correct way?

InteGrand · Sep 2, 2016

Re: MATH1231/1241/1251 SOS Thread

Flop21 said:
Thank you.

And why is this not correct? I did the vector <6,6,3> * the matrix A. What's the correct way?

$\noindent How did you calculate your value of $A \bold{v}$ $\big{(}$where $\bold{v} = (6,6,3)^{T}$$\big{)}$? You should've gotten a different first entry to what you got (just standard matrix muitplication).$

I think you just typoed or made a silly mistake for the first entry. The first entry should be 25*6 + 25*3. What you wrote is 5*6 + 25*6 + 25*3. Probably either a silly mistake (using second row first column of the matrix for that 5), or a typo.

Flop21 · Sep 2, 2016

Re: MATH1231/1241/1251 SOS Thread

InteGrand said:
$\noindent How did you calculate your value of $A \bold{v}$ $\big{(}$where $\bold{v} = (6,6,3)^{T}$$\big{)}$? You should've gotten different numbers to what you got (just standard matrix muitplication).$

No I don't know how to do it. I thought you just went 6<0,5,0> + 6<25,1,5> etc. what's the correct way?

Flop21 · Sep 2, 2016

Re: MATH1231/1241/1251 SOS Thread

InteGrand said:
$\noindent How did you calculate your value of $A \bold{v}$ $\big{(}$where $\bold{v} = (6,6,3)^{T}$$\big{)}$? You should've gotten a different first entry to what you got (just standard matrix muitplication).$

I think you just typoed or made a silly mistake for the first entry. The first entry should be 25*6 + 25*3. What you wrote is 5*6 + 25*6 + 25*3. Probably either a silly mistake (using second row first column of the matrix for that 5), or a typo.

oh lol yeah you're right thanks!

leehuan · Sep 6, 2016

Re: MATH1231/1241/1251 SOS Thread

I don't think I was taught how to do this and I can't comprehend the question, so can I please be started off on this one? (No need to finish off the question I reckon)

$Solve the following integral equation(s) by writing them as IVPs$

$$a) $y(x)=2+\int_1^x(y(t))^2dt$

InteGrand · Sep 6, 2016

Re: MATH1231/1241/1251 SOS Thread

leehuan said:
I don't think I was taught how to do this and I can't comprehend the question, so can I please be started off on this one? (No need to finish off the question I reckon)

$Solve the following integral equation(s) by writing them as IVPs$

$$a) $y(x)=2+\int_1^x(y(t))^2dt$

$\noindent Hint: differentiate both sides to obtain an ODE to solve. To get an intial condition (so that we have an IVP), put $x= 1$ in the given integral equation (so that the integral will vanish, leaving us with the intial condition $y(1) = 2$).$

$\noindent (Regarding comprehending the question: The given equation is an \textit{integral equation}, and it's like a differential equation in the sense that we're looking for the functions $y(x)$ that satisfy that equation.)$

leehuan · Sep 6, 2016

Re: MATH1231/1241/1251 SOS Thread

Ok yep figured that one now.

With this one though:

$y(x)=a-\int_b^x(x-t)y(t)dt$

When differentiating this one, since x is in the integrand and in the boundary do I need to be more careful? Or can I use only Leibniz's rule like this

$y^\prime(x)=-\int_b^x \frac{\partial }{\partial x}(x-t)y(t)dt=-\int_b^xy(t)dt$

$\implies y^{\prime \prime}(x)=-y(x)$

InteGrand · Sep 6, 2016

Re: MATH1231/1241/1251 SOS Thread

leehuan said:
Ok yep figured that one now.

With this one though:

$y(x)=a-\int_b^x(x-t)y(t)dt$

When differentiating this one, since x is in the integrand and in the boundary do I need to be more careful? Or can I use only Leibniz's rule like this

$y^\prime(x)=-\int_b^x \frac{\partial }{\partial x}(x-t)y(t)dt=-\int_b^xy(t)dt$

$\implies y^{\prime \prime}(x)=-y(x)$

$\noindent If you're not sure about how to use Leibniz's rule when $x$ is in both the bounds of integration and the integrand, note that in this case we can get around this by writing the integral as $x \int _b ^x y(t)\, \mathrm{d}t - \int _b ^x ty(t)\, \mathrm{d}t$, which we can differentiate using just the FTC (and product rule).$

$\noindent Using Leibniz correctly in this case would indeed give us what you wrote, because it would give $\frac{\mathrm{d}}{\mathrm{d}x} \left(\int _b ^x (x-t)y(t)\, \mathrm{d}t\right) = (x-x)y(x)\cdot \frac{\mathrm{d}x}{\mathrm{d}x} + \int _b ^x \frac{\partial }{\partial x} \left(x-t\right)y(t)\, \mathrm{d}t$.$

$\noindent (Recall that the Leibniz rule when the bounds are variable is: $\frac{\mathrm{d}}{\mathrm{d}x} \left(\int _{a(x)}^{b(x)} f(x,t)\, \mathrm{d}t\right) = -f(x,a(x)) a^{\prime }(x) + f(x,b(x))b^\prime (x) + \int _{a(x)}^{b(x)} \frac{\partial f}{\partial x} (x,t)\, \mathrm{d}t$, assuming the functions involved are sufficiently smooth. See the below link for further reading.)$

https://en.wikipedia.org/wiki/Leibn...form:_Differentiation_under_the_integral_sign

Paradoxica · Sep 6, 2016

Re: MATH1231/1241/1251 SOS Thread

Find the solution to this recurrence relation WITHOUT induction.

U_n+1 = √(U_n + 2)

U₀ = 0

First Year Mathematics B (Integration, Series, Discrete Maths & Modelling) (9 Viewers)

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