First Year Mathematics B (Integration, Series, Discrete Maths & Modelling) (1 Viewer)

Flop21 · Sep 19, 2016

Re: MATH1231/1241/1251 SOS Thread

InteGrand said:
$\noindent If you want to do it using the characteristic equation rather than by inspection as done above, note that the characteristic equation is $\lambda^2 + 4 = 0$ (because the ODE has no $y'$ term, so there is no $\lambda$ term in the characteristic equation. Another way to think of it is that the ODE is $y'' + 0y' + 4y =0$, so the characteristic equation is $\lambda^2 + 0\lambda + 4=0$, which is just $\lambda^2 + 4 = 0$.).$

$\noindent So the characteristic roots are $\pm 4i$. The general solution can be written down from this.$

ohhhh thanks!

leehuan · Sep 21, 2016

Re: MATH1231/1241/1251 SOS Thread

$\\ \text{Let }A_\phi, A_\theta$ and $A_{\phi+\theta}$ be the matrices for rotations in the plane by angles $\phi$, $\theta$ and $\phi+\theta$ respectively.$$

$\\ \text{Prove that }A_\theta A_\phi=A_{\phi+\theta}$

Wait. I get A_alpha must be an nxn matrix, but must it be a 2x2 matrix? Cause if not then I have no clue how to approach this.

RenegadeMx · Sep 21, 2016

Re: MATH1231/1241/1251 SOS Thread

leehuan said:
$\\ \text{Let }A_\phi, A_\theta$ and $A_{\phi+\theta}$ be the matrices for rotations in the plane by angles $\phi$, $\theta$ and $\phi+\theta$ respectively.$$

$\\ \text{Prove that }A_\theta A_\phi=A_{\phi+\theta}$

Wait. I get A_alpha must be an nxn matrix, but must it be a 2x2 matrix? Cause if not then I have no clue how to approach this.

normally pure rotation matrixces are limited to 2x2, for 3x3 case u revolve the curve around a certain axis, for higher up essentially the det(A) needs to be 1 and the transpose has to be its inverse

InteGrand · Sep 21, 2016

Re: MATH1231/1241/1251 SOS Thread

leehuan said:
$\\ \text{Let }A_\phi, A_\theta$ and $A_{\phi+\theta}$ be the matrices for rotations in the plane by angles $\phi$, $\theta$ and $\phi+\theta$ respectively.$$

$\\ \text{Prove that }A_\theta A_\phi=A_{\phi+\theta}$

Wait. I get A_alpha must be an nxn matrix, but must it be a 2x2 matrix? Cause if not then I have no clue how to approach this.

$\noindent Note $A_{\phi}$ rotates a point by $\phi$ and $A_{\theta}$ rotates by $\theta$. So if we multiply a vector by $A_{\phi}A_{\theta}$, we're rotating it first by $\theta$ and then by $\phi$, so overall all we've done is rotate it by $\phi+\theta$, which means we've had the same effect as multiplying by $A_{\phi+\theta}$. This explains the equality of those matrices in the question. For a similar reason, $A_{\theta}A_{\phi}$ is also equal to $A_{\phi + \theta}$. (So rotation matrices commute.)$

InteGrand · Sep 21, 2016

Re: MATH1231/1241/1251 SOS Thread

$$\noindent To prove it algebraically (noting the matrices are just $2$-by-$2$ since it says rotation in the \emph{plane}, so we're rotating two-dimensional vectors), recall that the rotation matrix $A_{\alpha}$ for rotating counter-clockwise by angle $\alpha$ is $\begin{bmatrix}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{bmatrix}$. From this, you can prove the desired statement using matrix multiplication and the compound angles formulas.$

Further reading about rotation matrices: https://en.wikipedia.org/wiki/Rotation_matrix .

WildestDreams · Sep 22, 2016

Re: MATH1231/1241/1251 SOS Thread

InteGrand said:
$$\noindent To prove it algebraically (noting the matrices are just $2$-by-$2$ since it says rotation in the \emph{plane}, so we're rotating two-dimensional vectors), recall that the rotation matrix $A_{\alpha}$ for rotating counter-clockwise by angle $\alpha$ is $\begin{bmatrix}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{bmatrix}$. From this, you can prove the desired statement using matrix multiplication and the compound angles formulas.$

Further reading about rotation matrices: https://en.wikipedia.org/wiki/Rotation_matrix .

how do you write in this text? with all the mathematic symbols?∷

InteGrand · Sep 22, 2016

Re: MATH1231/1241/1251 SOS Thread

WildestDreams said:
how do you write in this text? with all the mathematic symbols?��∷

Using LaTeX. A Short Guide to using LaTeX on this forum may be found here: http://community.boredofstudies.org/14/mathematics-extension-2/234259/short-guide-latex.html . There's a thread for practising it here: http://community.boredofstudies.org/3/non-school/345040/latex-practice.html .

leehuan · Sep 23, 2016

Re: MATH1231/1241/1251 SOS Thread

This may be a dumb question but what's the difference between calling it a "power series" to a "Taylor series at x=0" (aka Maclaurin series)

seanieg89 · Sep 23, 2016

Re: MATH1231/1241/1251 SOS Thread

Short answer: The Taylor series of a function is a specific power series associated to a given function centred at a given point. A power series is a more general concept and need not be related to a function in the same way that a Taylor series is.

Related info that might clarify the concepts further:

Any formal expression of the form:

$\sum_{n} a_n (z-c)^n$

is said to be a formal power series centred at c. This expression obviously converges at z=c, but need not converge anywhere else. In fact, the root test tells you that this series additionally converges absolutely precisely in the interior of a disk with radius (possibly 0 or inf) that can be computed in terms of the coefficients. Hence such a power series defines a function on some disk iff the coefficients don't increase too fast.

Now let's go the other direction. Say we already have a function, and we would like to obtain a series expression. (This might be useful because truncating the series at a finite number of terms gives you a polynomial. Polynomials are nice, so if we can approximate a function by a polynomial, we often make it easier to prove things about the function.) Suppose our function f is nice enough that we can find a power series (ie find a sequence of coefficients) that actually converges to our function on some small disk centred at say c (we say such functions f are analytic, and this property is quite strong). Then standard facts from analysis say we can differentiate this power series term by term to obtain the derivatives of this function. Doing so and putting z=c, we see that we must in fact have

$a_n=\frac{f^{(n)}(c)}{n!}.$

Ie. if a smooth function is equal to a power series with positive radius of convergence centred at a point, the coefficients must have this precise relationship with the derivatives of f at c.

This motivates us to look at such series (which we call the Taylor series of f) even when f is not analytic. Now in this case, even if the function is smooth (infinitely differentiable) the Taylor series need not converge to the function itself (consider the function f(x)=e^{-1/x} for positive x, f(x)=0 otherwise and try to Taylor expand at 0.) However, if f is k-times differentiable, we can still put asymptotic estimates on the error of the finite Taylor approximations as z->c. This is the content of Taylor's theorem.

leehuan · Sep 25, 2016

Re: MATH1231/1241/1251 SOS Thread

$\text{In Q19, it was proved that }T(\textbf{a})=\text{proj}_{\textbf{b}}\textbf{a}=\frac{\textbf{a}\cdot\textbf{b}}{|\textbf{b}|^2} \textbf{b}\text{ for }\textbf{a}\in \mathbb{R}^n\text{ is a linear map.}$

$\text{Q29: For Q20 (and others not mentioned), find }\ker (A)\text{ and nullity} (A)$

InteGrand · Sep 25, 2016

Re: MATH1231/1241/1251 SOS Thread

leehuan said:
$\text{In Q19, it was proved that }T(\textbf{a})=\text{proj}_{\textbf{b}}\textbf{a}=\frac{\textbf{a}\cdot\textbf{b}}{|\textbf{b}|^2} \textbf{b}\text{ for }\textbf{a}\in \mathbb{R}^n\text{ is a linear map.}$

$\text{Q29: For Q20 (and others not mentioned), find }\ker (A)\text{ and nullity} (A)$

$\noindent I assume you got out the one about showing it's a linear map. Now, the kernel of $T$ is just the set of all vectors in $\mathbb{R}^{n}$ that get sent to $\bold{0}$. Note that $T(\bold{x}) = \bold{0}$ iff $\bold{x}\cdot \bold{b} = 0$ (of course we're assuming that $\bold{b}$, which is a fixed vector, is not the zero vector, since it doesn't really make sense to project onto the zero subspace). Hence the kernel of $T$ can be written as $\left\{\bold{x} \in \mathbb{R}^{n} : \bold{x}\cdot \bold{b} = 0 \right\}$. Geometrically: the kernel is just all the vectors that are orthogonal to $\bold{b}$ (which form a hyperplane in $\mathbb{R}^{n}$).$

$\noindent The image is just all scalar multiples of $\bold{b}$ (edit: realised now they didn't actually ask for image).$

$\noindent It helps to visualise these things in $\mathbb{R}^{3}$ and $\mathbb{R}^{2}$. If $\bold{b}$ is a fixed (non-zero) vector in $\mathbb{R}^{3}$, then the projection map $T$ takes a vector $\mathbf{x}$ from $\mathbb{R}^{3}$ and maps them to their ``orthogonal projection'' in the direction of $\mathbf{b}$. The only way this can be $\mathbf{0}$ is if the vector $\mathbf{x}$ is actually perpendicular to $\mathbf{b}$ (imagine decomposing a vertical vector into horizontal and vertical components -- its horizontal component is $0$, because it's perpendicular to the horizontal direction). And the set of all vectors perpendicular to $\mathbf{b}$ is...just the plane (through the origin) with normal $\mathbf{b}$ (if you visualise this, you should be able to see that if we project vectors from the plane onto the normal vector $\mathbf{b}$, we clearly get $\mathbf{0}$ as the result.)$

$\noindent (Also, anything not on the plane would have non-zero component in the direction of $\mathbf{b}$, so is not in the kernel.)$

$\noindent The nullity of $T$ (dimension of the null space (aka kernel)) is $n-1$ (e.g. by rank-nullity theorem, since the image of $T$ is clearly one-dimensional, being the span of $\left\{\mathbf{b}\right\}$). Again this makes sense in the context of $\mathbb{R}^{3}$: in that case as we said, the kernel was a plane, which is $2$-dimensional (and $2 = 3-1$).$

leehuan · Sep 25, 2016

Re: MATH1231/1241/1251 SOS Thread

Should've just broken the question down into basics more confidently lol but yep makes perfect sense.
________________________

$\text{Find a basis for }\mathbb{R}^3\text{ which contains a basis for im}(C)\text{, where}\\ C=\begin{pmatrix}1&2&3&4\\2&-4&6&-2\\-1&2&-3&1\end{pmatrix}$

So the REF of C has columns 1 and 3 leading. Then for the answer they included the vectors corresponding to column 1 <1 2 -1>, column 3 <2 -4 2>, but also e₂. Where did the vector <0 1 0> come from?

InteGrand · Sep 25, 2016

Re: MATH1231/1241/1251 SOS Thread

leehuan said:
Should've just broken the question down into basics more confidently lol but yep makes perfect sense.
________________________

$\text{Find a basis for }\mathbb{R}^3\text{ which contains a basis for im}(C)\text{, where}\\ C=\begin{pmatrix}1&2&3&4\\2&-4&6&-2\\-1&2&-3&1\end{pmatrix}$

So the REF of C has columns 1 and 3 leading. Then for the answer they included the vectors corresponding to column 1 <1 2 -1>, column 3 <2 -4 2>, but also e₂. Where did the vector <0 1 0> come from?

$\noindent We want a basis for $\mathbb{R}^{3}$ (not just for $\mathrm{im}\left(C\right)$) that contains a basis for the image of $C$. We could ensure we obtain this by instead row-reducing the augmented matrix $\left[C \mid I_{3}\right]$ (where $I_{3}$ is the $3\times 3$ identity matrix). Row-reduce this matrix, and you should find that the left-hand part ($C$) has columns $1$ and $3$ leading, and that the right-hand part ($I$) has its second column leading (and everything else in the matrix will be non-leading). This would correspond to taking the basis to be what you wrote.$

RenegadeMx · Sep 25, 2016

Re: MATH1231/1241/1251 SOS Thread

leehuan said:
Should've just broken the question down into basics more confidently lol but yep makes perfect sense.
________________________

$\text{Find a basis for }\mathbb{R}^3\text{ which contains a basis for im}(C)\text{, where}\\ C=\begin{pmatrix}1&2&3&4\\2&-4&6&-2\\-1&2&-3&1\end{pmatrix}$

So the REF of C has columns 1 and 3 leading. Then for the answer they included the vectors corresponding to column 1 <1 2 -1>, column 3 <2 -4 2>, but also e₂. Where did the vector <0 1 0> come from?

Firstly u want basis for r^3, so u need 3 indep vectors, technically u could make the last vector u need anything u want as long as its independent, but why make it complicated when we can go with the easy option of 0,1,0

leehuan · Sep 25, 2016

Re: MATH1231/1241/1251 SOS Thread

My lecturer tried to explain it but I didn't get it at the time. Just wanted to see how any of you would explain it?

Taking R² for convenience here, what is the difference between saying:

"Graph of y=f(x)"
and
"Level curve of F(x,y)=y-f(x)"
____________________________

Context: We were talking about the gradient nabla F(x0,y0)

Struggling to visualise the difference between the "level curve" and the "graph", to understand why the gradient vector is perpendicular to the level surface but tangential to the graph.

(Will provide more context if necessary)

InteGrand · Sep 25, 2016

Re: MATH1231/1241/1251 SOS Thread

leehuan said:
My lecturer tried to explain it but I didn't get it at the time. Just wanted to see how any of you would explain it?

Taking R² for convenience here, what is the difference between saying:

"Graph of y=f(x)"
and
"Level curve of F(x,y)=y-f(x)"
____________________________

Context: We were talking about the gradient nabla F(x0,y0)

Struggling to visualise the difference between the "level curve" and the "graph", to understand why the gradient vector is perpendicular to the level surface but tangential to the graph.

(Will provide more context if necessary)

$\noindent If $F(x,y) := y - f(x)$, where $f$ is a function of a \emph{single} real variable, then the graph is $y = f(x)$ is one of the level curves of $F$, namely $F(x,y) = 0$. (Remember, the level curves of $F$ are the family of curves $F(x,y) = C$, where $C$ is a constant. So $y=f(x)$ graph would be the level curves corresponding to $C = 0$.) We would get the whole family of level curves by plotting the family of graphs $y = f(x) + C$.$

InteGrand · Sep 25, 2016

Re: MATH1231/1241/1251 SOS Thread

leehuan said:
My lecturer tried to explain it but I didn't get it at the time. Just wanted to see how any of you would explain it?

Taking R² for convenience here, what is the difference between saying:

"Graph of y=f(x)"
and
"Level curve of F(x,y)=y-f(x)"
____________________________

Context: We were talking about the gradient nabla F(x0,y0)

Struggling to visualise the difference between the "level curve" and the "graph", to understand why the gradient vector is perpendicular to the level surface but tangential to the graph.

(Will provide more context if necessary)

The gradient vector isn't tangent to the graph (a level curve of F), it's normal to it.

leehuan · Sep 26, 2016

Re: MATH1231/1241/1251 SOS Thread

InteGrand said:
The gradient vector isn't tangent to the graph (a level curve of F), it's normal to it.

Oh. Right, whoops keep getting my terminology mixed.
_____________________________

Ok so just trying to reunite two different methods:

$\text{Q: In each case, find the equation of the tangent plane to the given surface at the given point:}\\ \text{b) }x^2+y^2+z^2=21\qquad (4,2,1)$

$\textit{Method 1}\\ \begin{align*}2x+2z\frac{\partial z}{\partial x}&=0\\ 2y+2z\frac{\partial z}{\partial y}&=0\end{align*}\\ \text{Hence a normal vector to the surface is}\\ \textbf{n}=\begin{pmatrix}\frac{x}{z}\\ \frac{y}{z}\\ 1\end{pmatrix}$

$\textit{Method 2}\\ \text{Let }F(x,y,z)=x^2+y^2+z^2\text{ so that }x^2+y^2+z^2=21\text{ is the level surface }\\F(x,y,z)=21\\ \text{Then}\\ \nabla F(x,y,z)=\begin{pmatrix}2x\\2y\\2z\end{pmatrix}$

In both cases, substitution of the given point will result in the same final answer of 4x+2y+z=21 due to the point-normal form. However, why is it depending on which approach I take, the vector produced isn't the same?

(Although I did notice that the second is a scalar multiple of the first)

InteGrand · Sep 26, 2016

Re: MATH1231/1241/1251 SOS Thread

leehuan said:
Oh. Right, whoops keep getting my terminology mixed.
_____________________________

Ok so just trying to reunite two different methods:

$\text{Q: In each case, find the equation of the tangent plane to the given surface at the given point:}\\ \text{b) }x^2+y^2+z^2=21\qquad (4,2,1)$

$\textit{Method 1}\\ \begin{align*}2x+2z\frac{\partial z}{\partial x}&=0\\ 2y+2z\frac{\partial z}{\partial y}&=0\end{align*}\\ \text{Hence a normal vector to the surface is}\\ \textbf{n}=\begin{pmatrix}\frac{x}{z}\\ \frac{y}{z}\\ 1\end{pmatrix}$

$\textit{Method 2}\\ \text{Let }F(x,y,z)=x^2+y^2+z^2\text{ so that }x^2+y^2+z^2=21\text{ is the level surface }\\F(x,y,z)=21\\ \text{Then}\\ \nabla F(x,y,z)=\begin{pmatrix}2x\\2y\\2z\end{pmatrix}$

In both cases, substitution of the given point will result in the same final answer of 4x+2y+z=21 due to the point-normal form. However, why is it depending on which approach I take, the vector produced isn't the same?

(Although I did notice that the second is a scalar multiple of the first)

With your first method, it's set up so that it always produces the normal vector that has 1 as its third component (because we're writing 1 in the third component from the outset). The second method isn't set up so that it'll guarantee the third component is 1. The two answers are just scalar multiples of each other, so are as good as each other.

leehuan · Sep 27, 2016

Re: MATH1231/1241/1251 SOS Thread

$\text{Let }S=\{\textbf{u}_1,\textbf{u}_2,\textbf{u}_3, \textbf{u}_4 \}\text{ be an orthonormal basis for }\mathbb{R}^4\\ \text{and form the matrix }U=(\textbf{u}_1\quad \textbf{u}_2\quad\textbf{u}_3\quad \textbf{u}_4)\\ \text{(treating each vector as the columns)}$

$\text{Show that }U^TU=I$

First Year Mathematics B (Integration, Series, Discrete Maths & Modelling) (1 Viewer)

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